Q2: What is the type of hybridisation of each carbon in the following compounds? (a) CH_3 Cl, (b) left(CH_3right)_2 CO, (c) CH_3 CN, (d) HCONH_2, (e)…

To Find: The hybridization of each carbon atom in the given compounds. Solution: The type of hybridization can be determined by counting the number of sigma ( σ ) bonds and lone pairs around each carbon atom (steric number). 4 σ bonds, 0 lone pairs implies sp^3 hybridization (tetrahedral) 3 σ…

Q8 (i): CH_3CH_2CH(OH)CH_2CH_2CH(CH_3)CH_2CH_3

To Find: The IUPAC name for each of the given chemical structures. Solution: (i) Structure: OH CH₃ ⁸CH₃-⁷CH₂-⁶CH-⁵CH₂-⁴CH₂-³CH-²CH₂-¹CH₃ 1. Identify the principal functional group: The hydroxyl (-OH) group is the principal functional group, so the compound is an alcohol. The suffix will be '-ol'.…

Q8 (ii): CH_3CH_2C(=O)CH_2C(=O)CH_3

To Find: The IUPAC name for each of the given chemical structures. Solution: (i) Structure: OH CH₃ ⁸CH₃-⁷CH₂-⁶CH-⁵CH₂-⁴CH₂-³CH-²CH₂-¹CH₃ 1. Identify the principal functional group: The hydroxyl (-OH) group is the principal functional group, so the compound is an alcohol. The suffix will be '-ol'.…

Q8 (iii): CH_3C(=O)CH_2CH_2CH_2COOH

To Find: The IUPAC name for each of the given chemical structures. Solution: (i) Structure: OH CH₃ ⁸CH₃-⁷CH₂-⁶CH-⁵CH₂-⁴CH₂-³CH-²CH₂-¹CH₃ 1. Identify the principal functional group: The hydroxyl (-OH) group is the principal functional group, so the compound is an alcohol. The suffix will be '-ol'.…

Q8 (iv): CH equiv C-CH=CH-CH=CH_2

To Find: The IUPAC name for each of the given chemical structures. Solution: (i) Structure: OH CH₃ ⁸CH₃-⁷CH₂-⁶CH-⁵CH₂-⁴CH₂-³CH-²CH₂-¹CH₃ 1. Identify the principal functional group: The hydroxyl (-OH) group is the principal functional group, so the compound is an alcohol. The suffix will be '-ol'.…

Q14: Which bond is more polar in the following pairs of molecules: (a) H_3 C-H, H_3 C-Br (b) H_3 C-NH_2, H_3 C-OH (c) H_3 C-OH, H_3 C-SH

To Find: The more polar bond in each pair of molecules. Principle: The polarity of a covalent bond depends on the difference in electronegativity ( Δ EN) between the two bonded atoms. A larger difference in electronegativity leads to a more polar bond. Electronegativity Values (Pauling scale): H:…

Q17: Write resonance structures of CH_2=CH-CHO. Indicate relative stability of the contributing structures.

To Find: The resonance structures for propenal ( CH_2=CH-CHO ) and their relative stability. Principle: Resonance involves the delocalization of π electrons in a conjugated system. The stability of resonance structures is judged by these general rules: 1. The structure with the most covalent bonds…

Organic Chemistry – Some Basic Principles and Techniques Class 11 Chemistry NCERT Solutions

Q1Chapter Problems

How many $\sigma$ and $\pi$ bonds are present in each of the following molecules?

(a)

\mathrm{HC} \equiv \mathrm{CCH}=\mathrm{CHCH}_{3}

(b)

\mathrm{CH}_{2}=\mathrm{C}=\mathrm{CHCH}_{3}

Solution

To Find: The number of sigma (

\sigma

) and pi (

\pi

) bonds in the given molecules.

Solution: To find the number of

\sigma

and

\pi

bonds, we can draw the complete structural formula for each molecule. Remember that:

A single bond contains 1 $\sigma$ bond.
A double bond contains 1 $\sigma$ bond and 1 $\pi$ bond.
A triple bond contains 1 $\sigma$ bond and 2 $\pi$ bonds.

(a) $\mathrm{HC} \equiv \mathrm{CCH}=\mathrm{CHCH}_{3}$ (Pent-3-en-1-yne)

The expanded structure is:

  H   H   H
  |   |   |

H-C

\equiv

C-C=C-C-H | | H H

Sigma ( $\sigma$ ) bonds:
- C-H bonds: There are 6 C-H single bonds.
- C-C bonds: There is 1 C $\equiv$ C bond, 1 C-C single bond, and 1 C=C bond. Each of these contains one $\sigma$ bond. So, there are 4 C-C sigma bonds in total.
- Total $\sigma$ bonds = (Number of C-H bonds) + (Number of C-C bonds) = 6 + 4 = 10.
Pi ( $\pi$ ) bonds:
- The C $\equiv$ C triple bond contains 2 $\pi$ bonds.
- The C=C double bond contains 1 $\pi$ bond.
- Total $\pi$ bonds = 2 + 1 = 3.

(b) $\mathrm{CH}_{2}=\mathrm{C}=\mathrm{CHCH}_{3}$ (Buta-1,2-diene)

The expanded structure is:

H       H
|       |

H-C=C=C-C-H | | | H H H

Sigma ( $\sigma$ ) bonds:
- C-H bonds: There are 6 C-H single bonds.
- C-C bonds: There are two C=C double bonds and one C-C single bond. This gives a total of 3 C-C sigma bonds.
- Total $\sigma$ bonds = (Number of C-H bonds) + (Number of C-C bonds) = 6 + 3 = 9.
Pi ( $\pi$ ) bonds:
- Each of the two C=C double bonds contains 1 $\pi$ bond.
- Total $\pi$ bonds = 1 + 1 = 2.

Final Answer: (a) The molecule

\mathrm{HC} \equiv \mathrm{CCH}=\mathrm{CHCH}_{3}

has 10 $\sigma$ bonds and 3 $\pi$ bonds. (b) The molecule

\mathrm{CH}_{2}=\mathrm{C}=\mathrm{CHCH}_{3}

has 9 $\sigma$ bonds and 2 $\pi$ bonds.

Q2Chapter Problems

What is the type of hybridisation of each carbon in the following compounds?

(a)

\mathrm{CH}_{3} \mathrm{Cl}

,

(b)

\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}

,

(c)

\mathrm{CH}_{3} \mathrm{CN}

,

(d)

\mathrm{HCONH}_{2}

,

(e)

\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCN}

Solution

To Find: The hybridization of each carbon atom in the given compounds.

Solution: The type of hybridization can be determined by counting the number of sigma (

\sigma

) bonds and lone pairs around each carbon atom (steric number).

4 $\sigma$ bonds, 0 lone pairs $\implies sp^3$ hybridization (tetrahedral)
3 $\sigma$ bonds, 0 lone pairs $\implies sp^2$ hybridization (trigonal planar)
2 $\sigma$ bonds, 0 lone pairs $\implies sp$ hybridization (linear)

(a) $\mathrm{CH}_{3} \mathrm{Cl}$ (Chloromethane) The carbon atom is bonded to 3 H atoms and 1 Cl atom with single bonds. It forms 4

\sigma

bonds. Thus, its hybridization is $sp^3$ .

(b) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}$ (Acetone/Propanone) Structure:

\mathrm{CH}_3-\mathrm{C}(=\mathrm{O})-\mathrm{CH}_3

. Let's label the carbons:

\mathrm{C}^1\mathrm{H}_3-\mathrm{C}^2(=\mathrm{O})-\mathrm{C}^3\mathrm{H}_3

.

C $^1$ and C $^3$ (methyl carbons): Each is bonded to 3 H atoms and 1 C atom via single bonds (4 $\sigma$ bonds). Hybridization is $sp^3$ .
C $^2$ (carbonyl carbon): It is bonded to two C atoms via single bonds and one O atom via a double bond (1 $\sigma$ , 1 $\pi$ ). It forms 3 $\sigma$ bonds in total. Hybridization is $sp^2$ .

(c) $\mathrm{CH}_{3} \mathrm{CN}$ (Acetonitrile) Structure:

\mathrm{CH}_3-\mathrm{C}\equiv\mathrm{N}

. Let's label the carbons:

\mathrm{C}^1\mathrm{H}_3-\mathrm{C}^2\equiv\mathrm{N}

.

C $^1$ (methyl carbon): Bonded to 3 H atoms and 1 C atom via single bonds (4 $\sigma$ bonds). Hybridization is $sp^3$ .
C $^2$ (nitrile carbon): Bonded to one C atom via a single bond and one N atom via a triple bond (1 $\sigma$ , 2 $\pi$ ). It forms 2 $\sigma$ bonds in total. Hybridization is $sp$ .

(d) $\mathrm{HCONH}_{2}$ (Formamide) Structure:

\mathrm{H}-\mathrm{C}(=\mathrm{O})-\mathrm{NH}_2

.

Carbon atom: It is bonded to 1 H atom (single bond), 1 O atom (double bond), and 1 N atom (single bond). It forms 3 $\sigma$ bonds. Hybridization is $sp^2$ .

(e) $\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCN}$ Structure:

\mathrm{C}^1\mathrm{H}_3-\mathrm{C}^2\mathrm{H}=\mathrm{C}^3\mathrm{H}-\mathrm{C}^4\equiv\mathrm{N}

.

C $^1$ : Bonded to 3 H atoms and 1 C atom via single bonds (4 $\sigma$ bonds). Hybridization is $sp^3$ .
C $^2$ : Bonded to 1 C atom (single bond), 1 H atom (single bond), and 1 C atom (double bond). It forms 3 $\sigma$ bonds. Hybridization is $sp^2$ .
C $^3$ : Bonded to 1 C atom (double bond), 1 H atom (single bond), and 1 C atom (single bond). It forms 3 $\sigma$ bonds. Hybridization is $sp^2$ .
C $^4$ : Bonded to 1 C atom (single bond) and 1 N atom (triple bond). It forms 2 $\sigma$ bonds. Hybridization is $sp$ .

Final Answer: (a)

sp^3

(b) Methyl carbons:

sp^3

; Carbonyl carbon:

sp^2

(c) Methyl carbon:

sp^3

; Nitrile carbon:

sp

(d)

sp^2

(e) From left to right:

sp^3, sp^2, sp^2, sp

Q3Chapter Problems

Write the state of hybridisation of carbon in the following compounds and shapes of each of the molecules.

(a)

\mathrm{H}_{2} \mathrm{C}=\mathrm{O}

,

(b)

\mathrm{CH}_{3} \mathrm{F}

,

(c)

\mathrm{HC} \equiv \mathrm{N}

.

Solution

To Find: The hybridization of carbon and the shape of each molecule.

Solution:

(a) $\mathrm{H}_{2} \mathrm{C}=\mathrm{O}$ (Formaldehyde)

Hybridization: The carbon atom is bonded to two hydrogen atoms (single bonds) and one oxygen atom (double bond). It forms a total of 3 sigma ( $\sigma$ ) bonds and 1 pi ( $\pi$ ) bond. Since there are 3 $\sigma$ bonds and no lone pairs on the carbon, its hybridization is $sp^2$ .
Shape: $sp^2$ hybridization leads to a trigonal planar geometry around the central carbon atom.

(b) $\mathrm{CH}_{3} \mathrm{F}$ (Fluoromethyl)

Hybridization: The carbon atom is bonded to three hydrogen atoms and one fluorine atom, all via single bonds. It forms a total of 4 sigma ( $\sigma$ ) bonds. With 4 $\sigma$ bonds and no lone pairs, its hybridization is $sp^3$ .
Shape: $sp^3$ hybridization leads to a tetrahedral geometry.

(c) $\mathrm{HC} \equiv \mathrm{N}$ (Hydrogen Cyanide)

Hybridization: The carbon atom is bonded to one hydrogen atom (single bond) and one nitrogen atom (triple bond). It forms a total of 2 sigma ( $\sigma$ ) bonds and 2 pi ( $\pi$ ) bonds. With 2 $\sigma$ bonds and no lone pairs, its hybridization is $sp$ .
Shape: $sp$ hybridization leads to a linear geometry.

Final Answer: (a) Hybridization: $sp^2$ , Shape: Trigonal planar (b) Hybridization: $sp^3$ , Shape: Tetrahedral (c) Hybridization: $sp$ , Shape: Linear

Q4Chapter Problems

Expand each of the following condensed formulas into their complete structural formulas.

(a)

\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}_{2} \mathrm{CH}_{3}

(b)

\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}\left(\mathrm{CH}_{2}\right)_{3} \mathrm{CH}_{3}

Solution

To Find: The complete structural formula for each given condensed formula.

Solution: Complete structural formulas show all the atoms and all the bonds connecting them.

(a) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}_{2} \mathrm{CH}_{3}$ This is a ketone (pentan-3-one). The 'CO' group represents a carbonyl group where carbon is double-bonded to oxygen.

The complete structural formula is:

  H   H       H   H
  |   |       |   |

H — C — C — C — C — C — H | | || | | H H O H H

(b) $\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}\left(\mathrm{CH}_{2}\right)_{3} \mathrm{CH}_{3}$ This is an alkene (oct-2-ene). The (CH2)3 part represents a chain of three methylene groups (-CH2-CH2-CH2-).

The complete structural formula is:

  H       H   H   H   H   H   H
  |       |   |   |   |   |   |

Final Answer: The expanded structures are shown above.

Q5Chapter Problems

For each of the following compounds, write a condensed formula and also their bond-line formula.

(a)

\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CH}_{3}

(b)

\mathrm{N} \equiv \mathrm{CC}(\mathrm{OH}) \mathrm{C} \equiv \mathrm{N}

Solution

To Find: The condensed formula and bond-line formula for each given compound.

Solution:

(a) $\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CH}_{3}$

Condensed Formula: We can group similar repeating units. The three $\mathrm{CH}_2$ groups can be written as $(\mathrm{CH}_2)_3$ . The methyl group attached to the last carbon can be written as $(\mathrm{CH}_3)_2$ . However, the provided structure has two different carbons with methyl groups. A more accurate condensation is to group the initial part. The structure is 5,6-dimethylheptan-1-ol. Condensed formula: $\mathrm{HO}(\mathrm{CH}_2)_3\mathrm{CH}(\mathrm{CH}_3)\mathrm{CH}(\mathrm{CH}_3)_2$ Note: The structure in the question is slightly different from the solution in the source. Let's solve for the question's structure: $\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}(\mathrm{CH}_{3}) \mathrm{CH}(\mathrm{CH}_{3}) \mathrm{CH}_{3}$ . This is 4,5-dimethylhexan-1-ol. Condensed formula for 4,5-dimethylhexan-1-ol: $\mathrm{HO}(\mathrm{CH}_2)_3\mathrm{CH}(\mathrm{CH}_3)\mathrm{CH}(\mathrm{CH}_3)_2$ is incorrect. Let's write it step-by-step: $\mathrm{HO}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}(\mathrm{CH}_3)-\mathrm{CH}(\mathrm{CH}_3)-\mathrm{CH}_3$ A better condensed formula is: $\mathrm{HO(CH_2)_3CH(CH_3)CH(CH_3)CH_3}$
Bond-line Formula: In this notation, carbon and hydrogen atoms are not shown. Each vertex and endpoint represents a carbon atom. Heteroatoms (like O) are shown. The structure has a 6-carbon chain with an -OH at C1, and methyl groups at C4 and C5. The bond-line formula is:
```
  CH₃
  |
```
HO-CH₂-CH₂-CH₂-CH-CH-CH₃ | CH₃

As a pure bond-line structure:
```
/\ 
```
HO-- --/-- | /

(b) $\mathrm{N} \equiv \mathrm{CC}(\mathrm{OH}) \mathrm{C} \equiv \mathrm{N}$ This is 2-hydroxypropanedinitrile. The central carbon is attached to H, OH, and two CN groups. Correction: The structure shown in the source is $\mathrm{N} \equiv \mathrm{C}-\mathrm{CH}(\mathrm{OH})-\mathrm{C} \equiv \mathrm{N}$ . Let's assume this is the intended molecule.

Condensed Formula: The structure is already quite condensed. A common way to write it is $\mathrm{HOCH(CN)_2}$ .
Bond-line Formula: Here, we must show the heteroatoms (N, O) and the triple bonds.
```
OH
|
```
N $\equiv$ C-CH-C $\equiv$ N

In pure bond-line notation, the central CH group is a vertex.

N $\equiv$ C--C--C $\equiv$ N | OH

Final Answer: (a)

Condensed Formula: $\mathrm{HO(CH_2)_3CH(CH_3)CH(CH_3)CH_3}$
Bond-line Formula: /\ HO-- --/-- | /

(b)

Condensed Formula: $\mathrm{HOCH(CN)_2}$
Bond-line Formula: N $\equiv$ C--C--C $\equiv$ N | OH

Q6Chapter Problems

Expand each of the following bond-line formulas to show all the atoms including carbon and hydrogen

(a)

A six-membered ring with an attached isopropyl group.

(b)

A four-membered ring with a double bond.

(c)

A five-carbon chain with a triple bond at one end and an OH group on the third carbon.

(d)

A branched alkane with the structure of a cross.

Solution

To Find: The complete structural formula for each given bond-line formula.

Solution: We need to draw the structure showing all carbon and hydrogen atoms. Each vertex and end of a line in the bond-line formula represents a carbon atom, which is assumed to be bonded to enough hydrogen atoms to satisfy its valency of four.

(a) Isopropylcyclohexane The bond-line formula shows a hexagon (cyclohexane) attached to a 'Y' shape (isopropyl group).

The cyclohexane ring has 6 carbon atoms, each bonded to two other carbons. To satisfy valency, 5 of these carbons are bonded to 2 hydrogens each (CH₂), and the one with the substituent is bonded to 1 hydrogen (CH).
The isopropyl group, -CH(CH₃)₂, has a central carbon (CH) bonded to the ring and to two methyl (CH₃) groups.

The complete structural formula is:

         H   H₂C—CH₂   H
         |  /     \  |
        C--C-------C-C-CH₃
       / \         /  |
     H₂C---CH₂---CH   CH₃

(b) Cyclobutene The bond-line formula is a square with one double bond.

The two carbons in the double bond are each bonded to one other carbon (single bond) and one hydrogen (CH).
The other two carbons are each bonded to two other carbons (single bonds) and two hydrogens (CH₂).

The complete structural formula is:

      H - C = C - H
          |   |
      H₂C - CH₂

(c) Pent-1-yn-3-ol The bond-line formula is a 5-vertex zig-zag line with a triple bond between C1 and C2, and an OH group at C3.

C1: Part of a triple bond, bonded to one hydrogen (CH).
C2: Part of a triple bond, bonded to C1 and C3 (C).
C3: Bonded to C2, C4, one H, and one OH group (CHOH).
C4: Bonded to C3, C5, and two hydrogens (CH₂).
C5: End of the chain, bonded to C4 and three hydrogens (CH₃).

The complete structural formula is:

            H   OH  H   H
            |   |   |   |

H — C ≡ C — C — C — C — H | | | H H H

(d) 2,2-Dimethylpropane (Neopentane) The bond-line formula is a cross shape.

The central carbon is bonded to four other carbon atoms.
Each of the four outer carbons is a methyl group (CH₃).

The complete structural formula is:

          H
          | 
      H - C - H
          | 
  H   H - C - H   H
  |       |       |

H - C ----- C ----- C - H | | | H H - C - H H | H - C - H | H

Final Answer: The complete structural formulas are shown above.

Q7Chapter Problems

Structures and IUPAC names of some hydrocarbons are given below. Explain why the names given in the parentheses are incorrect.

(a)

\mathrm{CH_3-CH(CH_3)-CH_2-CH_2-CH(CH_3)-CH(CH_3)-CH_2-CH_3}

2,5,6- Trimethyloctane [and not 3,4,7-Trimethyloctane]

(b)

\mathrm{CH_3-CH_2-CH(CH_2CH_3)-CH_2-CH(CH_3)-CH_2-CH_3}

3-Ethyl-5-methylheptane [and not 5-Ethyl-3-methylheptane]

Solution

To Explain: Why the names in parentheses are incorrect according to IUPAC nomenclature rules.

Solution:

(a) 2,5,6-Trimethyloctane vs. 3,4,7-Trimethyloctane

Structure:

  CH₃         CH₃ CH₃
  |\           |   |

¹CH₃-²CH-³CH₂-⁴CH₂-⁵CH-⁶CH-⁷CH₂-⁸CH₃

IUPAC Rule: The parent chain should be numbered from the end that gives the lowest set of locants (positions) to the substituents. This is known as the lowest locant rule.

Numbering from left to right: The methyl substituents are at positions 2, 5, and 6. The set of locants is (2, 5, 6).
Numbering from right to left: The methyl substituents would be at positions 3, 4, and 7. The set of locants is (3, 4, 7).
Comparison: We compare the two sets of locants term by term. The first number in (2, 5, 6) is 2, which is lower than the first number in (3, 4, 7), which is 3. Therefore, the set (2, 5, 6) is lower.

Conclusion: The correct numbering is from left to right, giving the name 2,5,6-Trimethyloctane. The name 3,4,7-Trimethyloctane is incorrect because it violates the lowest locant rule.

(b) 3-Ethyl-5-methylheptane vs. 5-Ethyl-3-methylheptane

Structure:

      CH₂CH₃
      |

¹CH₃-²CH₂-³CH-⁴CH₂-⁵CH-⁶CH₂-⁷CH₃ | CH₃

IUPAC Rule: When two different substituents are in equivalent positions from either end of the parent chain, the numbering is chosen to give the lower number to the substituent that comes first in alphabetical order.

Substituents: We have an ethyl group and a methyl group.
Alphabetical Order: 'Ethyl' comes before 'methyl'.
Numbering from left to right: The ethyl group is at position 3, and the methyl group is at position 5. Locants are (3, 5).
Numbering from right to left: The methyl group would be at position 3, and the ethyl group would be at position 5. Locants are (3, 5).
Comparison: The set of locants (3, 5) is the same from both directions. In this situation, we apply the alphabetical order rule. Since 'ethyl' comes before 'methyl' alphabetically, the ethyl group should be assigned the lower number.

Conclusion: The correct numbering is from left to right, giving the ethyl group position 3. The correct name is 3-Ethyl-5-methylheptane. The name 5-Ethyl-3-methylheptane is incorrect because it assigns the higher number to the alphabetically preceding substituent when locants are equivalent.

Final Answer: (a) The name 3,4,7-Trimethyloctane is incorrect because it violates the lowest locant rule; the set of locants (2,5,6) is lower than (3,4,7). (b) The name 5-Ethyl-3-methylheptane is incorrect because when substituent locants are equivalent from both ends, the alphabetically superior group (ethyl) must be given the lower number.

Q8Chapter Problems

Write the IUPAC names of the compounds i-iv from their given structures.

(i)

$\mathrm{CH_3CH_2CH(OH)CH_2CH_2CH(CH_3)CH_2CH_3}$

(ii)

$\mathrm{CH_3CH_2C(=O)CH_2C(=O)CH_3}$

(iii)

$\mathrm{CH_3C(=O)CH_2CH_2CH_2COOH}$

(iv)

$\mathrm{CH \equiv C-CH=CH-CH=CH_2}$

Solution

To Find: The IUPAC name for each of the given chemical structures.

Solution:

(i) Structure:

    OH                  CH₃
    |                   |

⁸CH₃-⁷CH₂-⁶CH-⁵CH₂-⁴CH₂-³CH-²CH₂-¹CH₃

Identify the principal functional group: The hydroxyl (-OH) group is the principal functional group, so the compound is an alcohol. The suffix will be '-ol'.
Find the longest carbon chain containing the functional group: The longest chain has 8 carbon atoms. The parent alkane is octane.
Number the chain: Number from the end that gives the lowest number to the principal functional group (-OH). Numbering from the right gives the -OH group position 6. Numbering from the left gives it position 3. So, we number from the left.
Identify and name substituents: There is a methyl group (-CH₃) at position 6.
Assemble the name: (Substituent position)-(Substituent name)(Parent chain name)-(Functional group position)-ol. The name is 6-Methyloctan-3-ol.

(ii) Structure:

  O           O
  ||          ||

¹CH₃-²C-³CH₂-⁴C-⁵CH₂-⁶CH₃

Identify the principal functional group: The ketone (>C=O) group is present. Since there are two, the suffix will be '-dione'.
Find the longest carbon chain containing both functional groups: The longest chain has 6 carbon atoms. The parent alkane is hexane.
Number the chain: Number from the end that gives the lowest locants to the ketone groups. Numbering from the left gives positions 2 and 4. Numbering from the right gives positions 3 and 5. The set (2, 4) is lower.
Assemble the name: (Parent alkane name)-(Functional group positions)-dione. The name is Hexane-2,4-dione.

(iii) Structure:

  O
  ||

⁶CH₃-⁵C-⁴CH₂-³CH₂-²CH₂-¹COOH

Identify the principal functional group: There is a carboxylic acid (-COOH) and a ketone (>C=O). According to the priority order, carboxylic acid is the principal group. The suffix will be '-oic acid'.
Find the longest carbon chain containing the principal functional group: The chain has 6 carbon atoms. The parent alkane is hexane.
Number the chain: The carbon of the -COOH group is always C-1.
Identify and name other functional groups as substituents: The ketone group at position 5 is named as a prefix 'oxo'.
Assemble the name: (Substituent position)-(Substituent name)(Parent chain name)oic acid. The name is 5-Oxohexanoic acid.

(iv) Structure:

⁶CH ≡ ⁵C - ⁴CH = ³CH - ²CH = ¹CH₂

Identify the functional groups: There are double bonds (alkene) and a triple bond (alkyne).
Find the longest carbon chain containing the multiple bonds: The chain has 6 carbon atoms. The parent name is derived from hexane.
Number the chain: When there is a choice, the double bond is given the lower number. Numbering from the right gives the first multiple bond (a double bond) at position 1. Numbering from the left gives the first multiple bond (a triple bond) at position 1. In case of a tie, the double bond gets preference. So, we number from the right.
Assemble the name: The parent name is hex-. There are two double bonds ('diene') at positions 1 and 3, and one triple bond ('yne') at position 5. The name is constructed as: (Parent name)a-(double bond positions)-diene-(triple bond position)-yne. The 'e' from 'diene' is dropped. The name is Hexa-1,3-dien-5-yne.

Final Answer:

(i)

6-Methyloctan-3-ol

(ii)

Hexane-2,4-dione

(iii)

5-Oxohexanoic acid

(iv)

Hexa-1,3-dien-5-yne

Q9Chapter Problems

Derive the structure of (i) 2-Chlorohexane, (ii) Pent-4-en-2-ol, (iii) 3- Nitrocyclohexene, (iv) Cyclohex-2-en-1-ol, (v) 6-Hydroxyheptanal.

Solution

To Find: The chemical structure for each given IUPAC name.

Solution:

(i) 2-Chlorohexane

Parent chain: 'hexane' means a 6-carbon single-bonded chain.
Substituent: '2-Chloro' means a chlorine atom is attached to the second carbon.
Structure: $\mathrm{CH_3-CHCl-CH_2-CH_2-CH_2-CH_3}$

(ii) Pent-4-en-2-ol

Parent chain: 'pent' means a 5-carbon chain.
Functional groups: '-4-en' means a double bond between C4 and C5. '-2-ol' means a hydroxyl (-OH) group is at C2. The '-ol' has higher priority, so numbering starts from the end closer to it.
Structure: $\mathrm{CH_2=CH-CH_2-CH(OH)-CH_3}$

(iii) 3-Nitrocyclohexene

Parent structure: 'cyclohexene' means a 6-membered carbon ring with one double bond.
Numbering: The carbons of the double bond are numbered 1 and 2.
Substituent: '3-Nitro' means a nitro group (-NO₂) is on C3.
Structure: A hexagon with a double bond between C1 and C2, and an NO₂ group attached to C3.

(iv) Cyclohex-2-en-1-ol

Parent structure: 'cyclohex' means a 6-membered carbon ring.
Functional groups: '-2-en' means a double bond between C2 and C3. '-1-ol' means a hydroxyl (-OH) group is at C1. The '-ol' group has priority and is assigned position 1.
Structure: A hexagon with an OH group at C1 and a double bond between C2 and C3.

(v) 6-Hydroxyheptanal

Parent chain: 'heptan' means a 7-carbon chain.
Principal functional group: '-al' means an aldehyde group (-CHO) is at the end of the chain. Its carbon is C1.
Substituent: '6-Hydroxy' means a hydroxyl (-OH) group is on C6.
Structure: $\mathrm{CH_3-CH(OH)-CH_2-CH_2-CH_2-CH_2-CHO}$

Final Answer:

(i)

2-Chlorohexane:

\mathrm{CH_3CH_2CH_2CH_2CH(Cl)CH_3}

(ii)

Pent-4-en-2-ol:

\mathrm{CH_2=CHCH_2CH(OH)CH_3}

(iii)

3-Nitrocyclohexene: A six-membered ring with a double bond and a nitro group on the carbon adjacent to the double bond.

(iv)

Cyclohex-2-en-1-ol: A six-membered ring with a hydroxyl group and a double bond, where the OH is on C1 and the double bond starts at C2.

(v)

6-Hydroxyheptanal:

\mathrm{CH_3CH(OH)(CH_2)_4CHO}

Q10Chapter Problems

Write the structural formula of:

(a)

o-Ethylanisole, (b) p-Nitroaniline,

(c)

2,3-Dibromo-1 - phenylpentane,

(d)

4-Ethyl-1-fluoro-2-nitrobenzene.

Solution

To Find: The structural formula for each given compound name.

Solution:

(a) o-Ethylanisole

Base compound: Anisole is methoxybenzene ( $\ ext{C}_6\ ext{H}_5\ ext{OCH}_3$ ).
Substituent: An ethyl group ( $-\ ext{CH}_2\ ext{CH}_3$ ).
Position: 'o-' (ortho) means the substituent is at position 2, adjacent to the primary group (methoxy group).
Structure: A benzene ring with a $-\ ext{OCH}_3$ group and an adjacent $-\ ext{CH}_2\ ext{CH}_3$ group.

(b) p-Nitroaniline

Base compound: Aniline is aminobenzene ( $\ ext{C}_6\ ext{H}_5\ ext{NH}_2$ ).
Substituent: A nitro group ( $-\ ext{NO}_2$ ).
Position: 'p-' (para) means the substituent is at position 4, directly opposite the primary group (amino group).
Structure: A benzene ring with an $-\ ext{NH}_2$ group and a $-\ ext{NO}_2$ group opposite to it.

(c) 2,3-Dibromo-1-phenylpentane

Parent chain: 'pentane' means a 5-carbon chain.
Substituents:
- '2,3-Dibromo': Bromine atoms at C2 and C3.
- '1-phenyl': A phenyl group ( $\ ext{C}_6\ ext{H}_5-$ ) at C1.
Structure: $\ ext{C}_6\ ext{H}_5-\ ext{CH}_2-\ ext{CH(Br)}-\ ext{CH(Br)}-\ ext{CH}_2-\ ext{CH}_3$

(d) 4-Ethyl-1-fluoro-2-nitrobenzene

Parent compound: Benzene.
Substituents:
- '4-Ethyl': An ethyl group ( $-\ ext{CH}_2\ ext{CH}_3$ ) at C4.
- '1-fluoro': A fluorine atom (-F) at C1.
- '2-nitro': A nitro group ( $-\ ext{NO}_2$ ) at C2.
Numbering: The substituents are listed alphabetically in the name (Ethyl, fluoro, nitro). Numbering follows the lowest locant rule, which is satisfied by 1,2,4.
Structure: A benzene ring with -F at C1, $-\ ext{NO}_2$ at C2, and $-\ ext{CH}_2\ ext{CH}_3$ at C4.

Final Answer: The structural formulas are as described above.

Q11Chapter Problems

Using curved-arrow notation, show the formation of reactive intermediates when the following covalent bonds undergo heterolytic cleavage.

(a)

\mathrm{CH}_{3}-\mathrm{SCH}_{3}

,

(b)

\mathrm{CH}_{3}-\mathrm{CN}

,

(c)

\mathrm{CH}_{3}-\mathrm{Cu}

Solution

To Show: The formation of reactive intermediates from heterolytic cleavage using curved-arrow notation.

Solution: Heterolytic cleavage is the breaking of a covalent bond where one atom takes both electrons from the bond, forming a cation and an anion. The curved arrow shows the movement of the electron pair.

(a) $\mathrm{CH}_{3}-\mathrm{SCH}_{3}$ (Dimethyl sulfide)

Sulphur is more electronegative than carbon. Therefore, during cleavage of the C-S bond, the sulphur atom will take the electron pair.
Cleavage: The arrow starts from the C-S bond and points to the S atom.

\mathrm{CH}_{3} - \stackrel{\curvearrowleft}{\mathrm{S}} - \mathrm{CH}_{3} \longrightarrow \stackrel{+}{\mathrm{C}}\mathrm{H}_{3} + :\stackrel{-}{\mathrm{S}}\mathrm{CH}_{3}

Intermediates: A methyl carbocation ( $^+\mathrm{CH}_3$ ) and a methanethiolate anion ( $^-\mathrm{SCH}_3$ ).

(b) $\mathrm{CH}_{3}-\mathrm{CN}$ (Acetonitrile)

The cyanide group (-CN) as a whole is more electronegative than the methyl group due to the nitrogen atom. The carbon of the methyl group will lose the electrons.
Cleavage: The arrow starts from the C-C bond and points to the carbon of the cyanide group.

\mathrm{CH}_{3} - \stackrel{\curvearrowleft}{\mathrm{C}}\mathrm{N} \longrightarrow \stackrel{+}{\mathrm{C}}\mathrm{H}_{3} + :\stackrel{-}{\mathrm{C}}\mathrm{N}

Intermediates: A methyl carbocation ( $^+\mathrm{CH}_3$ ) and a cyanide anion ( $^-\mathrm{CN}$ ).

(c) $\mathrm{CH}_{3}-\mathrm{Cu}$ (Methylcopper)

This is an organometallic compound. Carbon is more electronegative than most metals, including copper. Therefore, the carbon atom will take the electron pair.
Cleavage: The arrow starts from the C-Cu bond and points to the C atom.

\mathrm{CH}_{3} \stackrel{\curvearrowright}{-} \mathrm{Cu} \longrightarrow :\stackrel{-}{\mathrm{C}}\mathrm{H}_{3} + \mathrm{Cu}^{+}

Intermediates: A methyl carbanion ( $^-\mathrm{CH}_3$ ) and a copper(I) cation ( $\mathrm{Cu}^+$ ).

Final Answer: The heterolytic cleavages and the resulting intermediates are shown above using curved-arrow notation.

Q12Chapter Problems

Giving justification, categorise the following molecules/ions as nucleophile or electrophile: $\mathrm{HS}^{-}, \mathrm{BF}_{3}, \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}^{-}, \left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}:, \mathrm{Cl}^{+}, \mathrm{CH}_{3}-\stackrel{+}{\mathrm{C}}=\mathrm{O}, \mathrm{H}_{2} \mathrm{N}:^{-}, \stackrel{+}{\mathrm{N}} \mathrm{O}_{2}$

Solution

To Categorize: The given species as nucleophiles or electrophiles with justification.

Definitions:

Nucleophile (nucleus-loving): An electron-rich species that donates an electron pair. They can be negatively charged or neutral with lone pairs of electrons.
Electrophile (electron-loving): An electron-deficient species that accepts an electron pair. They can be positively charged or neutral with an incomplete octet or an atom with a partial positive charge.

Solution:

Nucleophiles:

$\mathrm{HS}^{-}$ (Hydrosulfide ion): It is a negatively charged ion with lone pairs on the sulfur atom. It is electron-rich and can donate an electron pair. Justification: Anion with lone pairs.
$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}^{-}$ (Ethoxide ion): It is a negatively charged ion with lone pairs on the oxygen atom. It is electron-rich and can donate an electron pair. Justification: Anion with lone pairs.
$\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}:$ (Trimethylamine): It is a neutral molecule, but the nitrogen atom has a lone pair of electrons that it can donate. Justification: Neutral molecule with a lone pair.
$\mathrm{H}_{2} \mathrm{N}:^{-}$ (Amide ion): It is a negatively charged ion with lone pairs on the nitrogen atom. It is a very strong nucleophile. Justification: Anion with lone pairs.

Electrophiles:

$\mathrm{BF}_{3}$ (Boron trifluoride): It is a neutral molecule, but the central boron atom has only 6 valence electrons (an incomplete octet). It can accept an electron pair to complete its octet. Justification: Incomplete octet.
$\mathrm{Cl}^{+}$ (Chloronium ion): It is a positively charged ion and is electron-deficient. It readily accepts an electron pair. Justification: Cation, electron-deficient.
$\mathrm{CH}_{3}-\stackrel{+}{\mathrm{C}}=\mathrm{O}$ (Acylium ion): The carbon atom has a positive charge and an incomplete octet (6 valence electrons). It is a strong electrophile. Justification: Cation with an incomplete octet.
$\stackrel{+}{\mathrm{N}} \mathrm{O}_{2}$ (Nitronium ion): The nitrogen atom has a positive charge, making the ion highly electron-deficient and eager to accept an electron pair. Justification: Cation, electron-deficient.

Final Answer:

Nucleophiles: $\mathrm{HS}^{-}, \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}^{-}, \left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}:, \mathrm{H}_{2} \mathrm{N}:^{-}$
Electrophiles: $\mathrm{BF}_{3}, \mathrm{Cl}^{+}, \mathrm{CH}_{3}-\stackrel{+}{\mathrm{C}}=\mathrm{O}, \stackrel{+}{\mathrm{N}} \mathrm{O}_{2}$

Q13Chapter Problems

Identify electrophilic centre in the following: $\mathrm{CH}_{3} \mathrm{CH}=\mathrm{O}, \mathrm{CH}_{3} \mathrm{CN}, \mathrm{CH}_{3} \mathrm{I}$ .

Solution

To Find: The electrophilic center in each of the given molecules.

Definition: An electrophilic center is an atom within a molecule that is electron-deficient and can be attacked by a nucleophile. This deficiency is often due to bonding with a more electronegative atom, creating a polar bond and a partial positive charge (

\delta^+

) on the atom.

Solution:

(a) $\mathrm{CH}_{3} \mathrm{CH}=\mathrm{O}$ (Acetaldehyde)

The molecule contains a carbonyl group (C=O). Oxygen is more electronegative than carbon.
This pulls the electron density of the C=O double bond towards the oxygen atom.
As a result, the oxygen atom gets a partial negative charge ( $\delta^-$ ) and the carbonyl carbon atom gets a partial positive charge ( $\delta^+$ ).
Therefore, the carbonyl carbon is the electrophilic center. $\mathrm{CH}_{3} - \mathrm{C}^{\delta+}\mathrm{H} = \mathrm{O}^{\delta-}$

(b) $\mathrm{CH}_{3} \mathrm{CN}$ (Acetonitrile)

The molecule contains a nitrile group (C $\equiv$ N). Nitrogen is more electronegative than carbon.
The electron density of the C $\equiv$ N triple bond is pulled towards the nitrogen atom.
This gives the nitrogen a partial negative charge ( $\delta^-$ ) and the nitrile carbon atom a partial positive charge ( $\delta^+$ ).
Therefore, the nitrile carbon is the electrophilic center. $\mathrm{CH}_{3} - \mathrm{C}^{\delta+}\equiv\mathrm{N}^{\delta-}$

(c) $\mathrm{CH}_{3} \mathrm{I}$ (Iodomethane)

The molecule contains a C-I bond. Iodine is a halogen and is more electronegative than carbon.
The electron density of the C-I single bond is pulled towards the iodine atom.
This gives the iodine atom a partial negative charge ( $\delta^-$ ) and the methyl carbon atom a partial positive charge ( $\delta^+$ ).
Therefore, the methyl carbon is the electrophilic center. $\mathrm{CH}_{3}^{\delta+} - \mathrm{I}^{\delta-}$

Final Answer:

In $\mathrm{CH}_{3} \mathrm{CH}=\mathrm{O}$ , the carbonyl carbon is the electrophilic center.
In $\mathrm{CH}_{3} \mathrm{CN}$ , the nitrile carbon is the electrophilic center.
In $\mathrm{CH}_{3} \mathrm{I}$ , the methyl carbon is the electrophilic center.

Q14Chapter Problems

Which bond is more polar in the following pairs of molecules: (a) $\mathrm{H}_{3} \mathrm{C}-\mathrm{H}, \mathrm{H}_{3} \mathrm{C}-\mathrm{Br}$

(b)

\mathrm{H}_{3} \mathrm{C}-\mathrm{NH}_{2}, \mathrm{H}_{3} \mathrm{C}-\mathrm{OH}

(c)

\mathrm{H}_{3} \mathrm{C}-\mathrm{OH}, \mathrm{H}_{3} \mathrm{C}-\mathrm{SH}

Solution

To Find: The more polar bond in each pair of molecules.

Principle: The polarity of a covalent bond depends on the difference in electronegativity (

\Delta

EN) between the two bonded atoms. A larger difference in electronegativity leads to a more polar bond.

Electronegativity Values (Pauling scale):

H: 2.20
C: 2.55
N: 3.04
O: 3.44
S: 2.58
Br: 2.96

Solution:

(a) $\mathrm{H}_{3} \mathrm{C}-\mathrm{H}$ vs. $\mathrm{H}_{3} \mathrm{C}-\mathrm{Br}$

C-H bond: $\Delta$ EN = |EN(C) - EN(H)| = |2.55 - 2.20| = 0.35
C-Br bond: $\Delta$ EN = |EN(C) - EN(Br)| = |2.55 - 2.96| = 0.41
Since 0.41 > 0.35, the C-Br bond is more polar.

(b) $\mathrm{H}_{3} \mathrm{C}-\mathrm{NH}_{2}$ vs. $\mathrm{H}_{3} \mathrm{C}-\mathrm{OH}$ We are comparing the C-N bond and the C-O bond.

C-N bond: $\Delta$ EN = |EN(C) - EN(N)| = |2.55 - 3.04| = 0.49
C-O bond: $\Delta$ EN = |EN(C) - EN(O)| = |2.55 - 3.44| = 0.89
Since 0.89 > 0.49, the C-O bond is more polar.

(c) $\mathrm{H}_{3} \mathrm{C}-\mathrm{OH}$ vs. $\mathrm{H}_{3} \mathrm{C}-\mathrm{SH}$ We are comparing the C-O bond and the C-S bond.

C-O bond: $\Delta$ EN = |EN(C) - EN(O)| = |2.55 - 3.44| = 0.89
C-S bond: $\Delta$ EN = |EN(C) - EN(S)| = |2.55 - 2.58| = 0.03
Since 0.89 > 0.03, the C-O bond is significantly more polar.

Final Answer: (a) The C-Br bond in

\mathrm{H}_{3} \mathrm{C}-\mathrm{Br}

is more polar. (b) The C-O bond in

\mathrm{H}_{3} \mathrm{C}-\mathrm{OH}

is more polar. (c) The C-O bond in

\mathrm{H}_{3} \mathrm{C}-\mathrm{OH}

is more polar.

Q15Chapter Problems

In which C-C bond of $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}$ , the inductive effect is expected to be the least?

Solution

Given: The molecule 1-bromopropane,

\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{Br}

.

To Find: The C-C bond where the inductive effect is the least.

Principle: The inductive effect is the transmission of charge through a chain of atoms in a molecule, resulting from a difference in electronegativity. It is a permanent effect that weakens rapidly with distance from the source.

Solution:

Let's label the carbon atoms in the chain: $\mathrm{C}^3\mathrm{H}_{3} - \mathrm{C}^2\mathrm{H}_{2} - \mathrm{C}^1\mathrm{H}_{2} - \mathrm{Br}$
The bromine (Br) atom is highly electronegative compared to carbon. It withdraws electron density from the C¹ atom through the C-Br bond. This is a -I (negative inductive) effect. $\mathrm{C}^1\mathrm{H}_{2}^{\delta+} - \mathrm{Br}^{\delta-}$
The C¹ atom, having become partially positive ( $\delta^+$ ), in turn withdraws some electron density from the adjacent C² atom. This makes C² also partially positive ( $\delta\delta^+$ ), but to a lesser extent than C¹. $\mathrm{C}^2\mathrm{H}_{2}^{\delta\delta+} - \mathrm{C}^1\mathrm{H}_{2}^{\delta+}$
This effect is transmitted further down the chain to the C³ atom, but it becomes even weaker. The positive charge on C³ ( $\delta\delta\delta^+$ ) is negligible. $\mathrm{C}^3\mathrm{H}_{3}^{\delta\delta\delta+} - \mathrm{C}^2\mathrm{H}_{2}^{\delta\delta+}$
The inductive effect of the Br atom is strongest on the C¹-C² bond and weakest on the C²-C³ bond because the C²-C³ bond is the farthest from the electron-withdrawing Br atom.

Final Answer: The inductive effect is expected to be the least in the bond between carbon-2 and carbon-3 (the C²-C³ bond).

Q16Chapter Problems

Write resonance structures of $\mathrm{CH}_{3} \mathrm{COO}^{-}$ and show the movement of electrons by curved arrows.

Solution

To Find: The resonance structures of the acetate ion (

\mathrm{CH}_{3} \mathrm{COO}^{-}

) with electron movement shown by curved arrows.

Principle: Resonance structures are different Lewis structures for the same molecule or ion that differ only in the placement of electrons. The actual structure is a hybrid of these contributing structures. Curved arrows are used to show the delocalization (movement) of electron pairs.

Solution:

Draw the initial Lewis structure of the acetate ion. The central carbon is bonded to the methyl carbon, one oxygen atom (double bond), and another oxygen atom (single bond) which carries the negative charge.
```
  :Ö:
  ||
```
H₃C — C — Ö:⁻
Identify the electrons that can be delocalized. In the acetate ion, a lone pair on the negatively charged oxygen atom is adjacent to the C=O pi ( $\pi$ ) bond. This lone pair can be moved to form a new $\pi$ bond, and the existing C=O $\pi$ bond can be moved onto the other oxygen atom as a lone pair.
Show the electron movement with curved arrows.
- A curved arrow starts from the lone pair on the negatively charged oxygen and points to the space between that oxygen and the carbon to form a new double bond.
- Simultaneously, another curved arrow starts from the C=O double bond and points to the other oxygen atom to move the $\pi$ electrons as a lone pair.

Resonance Structures:

  :Ö:                  :Ö:⁻
  ||                   |

H₃C — C — Ö:⁻

\longleftrightarrow

H₃C — C = Ö:

\curvearrowright

\curvearrowleft

Explanation:

In Structure I, the negative charge is on the bottom oxygen.
In Structure II, the negative charge is on the top oxygen.
These two structures are equivalent in energy and contribute equally to the resonance hybrid.
In the actual resonance hybrid, both C-O bonds are identical, with a bond length intermediate between a single and a double bond, and the negative charge is delocalized equally over both oxygen atoms.

Final Answer: The resonance structures for the acetate ion are shown above, with curved arrows indicating the delocalization of the lone pair and pi electrons.

Q17Chapter Problems

Write resonance structures of $\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CHO}$ . Indicate relative stability of the contributing structures.

Solution

To Find: The resonance structures for propenal (

\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CHO}

) and their relative stability.

Principle: Resonance involves the delocalization of

\pi

electrons in a conjugated system. The stability of resonance structures is judged by these general rules:

The structure with the most covalent bonds is the most stable.
Structures with complete octets for all atoms (except H) are more stable.
Structures with less charge separation are more stable.
If there is charge separation, the structure where the negative charge is on the more electronegative atom is more stable.

Solution:

Draw the primary Lewis structure (Structure I). This is the neutral structure with no formal charges.

$\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}=\mathrm{O}$ (Structure I)
Identify the conjugated system. We have a C=C double bond conjugated with a C=O double bond.
Show electron delocalization. Oxygen is more electronegative than carbon, so it will pull the $\pi$ electrons from the C=O bond. This creates a positive charge on the carbonyl carbon, which is then stabilized by the $\pi$ electrons from the adjacent C=C bond.
- Move the $\pi$ electrons from the C=O bond to the oxygen atom.
- Move the $\pi$ electrons from the C=C bond to form a new C-C double bond.
This results in a structure with charge separation (Structure II).

$\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}=\mathrm{O} \quad \longleftrightarrow \quad \stackrel{+}{\mathrm{C}}\mathrm{H}_{2}-\mathrm{CH}=\mathrm{CH}-\stackrel{-}{\mathrm{O}}$ (Structure I) (Structure II)
Consider other possible (but less stable) structures. We could imagine the electrons moving in the opposite direction, putting a positive charge on oxygen and a negative charge on carbon (Structure III). This is highly unstable.

$\stackrel{-}{\mathrm{C}}\mathrm{H}_{2}-\mathrm{CH}=\mathrm{CH}-\stackrel{+}{\mathrm{O}}$ (Structure III)

Relative Stability:

Structure I: This is the most stable contributor. It has the maximum number of covalent bonds, all atoms have a complete octet (except H), and there is no separation of formal charges.
Structure II: This structure is less stable than I because it involves charge separation. However, it is a significant contributor because the negative charge is placed on the more electronegative atom (oxygen), and the positive charge is on a carbon atom.
Structure III: This structure is the least stable and makes a negligible contribution. It has charge separation, and more importantly, the positive charge is on the highly electronegative oxygen atom, while the negative charge is on the less electronegative carbon atom. This is an extremely unfavorable arrangement.

Order of Stability: I > II > III

Final Answer: The resonance structures are:

\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}=\mathrm{O} \quad \longleftrightarrow \quad \stackrel{+}{\mathrm{C}}\mathrm{H}_{2}-\mathrm{CH}=\mathrm{CH}-\stackrel{-}{\mathrm{O}}

(I) (II)

The order of stability is I > II. Structure III is highly unstable and generally not considered a major contributor.

Q18Chapter Problems

Explain why the following two structures, I and II cannot be the major contributors to the real structure of

\mathrm{CH}_{3} \mathrm{COOCH}_{3}

.

H₃C - C⁺ = O⁻         H₃C - C⁻ - O⁺
      |                  ||
     :O-CH₃              O-CH₃
      (I)                (II)

Solution

Given: Two potential resonance structures (I and II) for methyl acetate,

\mathrm{CH}_{3} \mathrm{COOCH}_{3}

.

To Explain: Why these structures are not major contributors to the resonance hybrid.

Principle: The stability and contribution of a resonance structure are determined by several factors:

Octet Rule: Structures where all second-row elements have a complete octet of valence electrons are more stable.
Formal Charges: Structures with fewer formal charges are more stable. Structures with large charge separation are less stable.
Electronegativity: In charged structures, placing a negative charge on a more electronegative atom and a positive charge on a less electronegative atom is more favorable.

Analysis of the Structures:

First, let's draw the most stable Lewis structure for methyl acetate, which has no formal charges and satisfies the octet rule for all C and O atoms.

:Ö:
||

H₃C — C — Ö — CH₃ (Primary Structure)

Now, let's analyze the given structures I and II.

Structure I:

H₃C - C⁺ = O⁻
      |
     :O-CH₃

Formal Charges: This structure has charge separation, with a +1 charge on the carbonyl carbon and a -1 charge on the carbonyl oxygen. The primary structure has no formal charges, making it more stable.
Octet Rule: The positively charged carbon atom has only 6 valence electrons (one single bond, one double bond), which is an incomplete octet. This makes the structure highly unstable.

Structure II:

H₃C - C⁻ - O⁺
      ||
     O-CH₃

Formal Charges: This structure also has charge separation, with a -1 charge on the carbon and a +1 charge on the oxygen.
Electronegativity: This arrangement is extremely unfavorable. It places a positive formal charge on the highly electronegative oxygen atom and a negative formal charge on the less electronegative carbon atom. This is directly contrary to the principles of chemical stability.

Conclusion: Both structures I and II are very unstable compared to the primary Lewis structure of methyl acetate.

Structure I is a minor contributor because it has an incomplete octet on a carbon atom and involves charge separation.
Structure II is an extremely minor (negligible) contributor because it not only has charge separation but also places the formal charges on atoms contrary to their electronegativity.

Final Answer: Structures I and II cannot be major contributors because:

They both involve separation of formal charges, which is less stable than the neutral primary structure.
Structure I has a carbon atom with an incomplete octet (6 valence electrons), which is a major source of instability.
Structure II places a positive charge on the highly electronegative oxygen atom, which is extremely unfavorable.

Q19Chapter Problems

Explain why $\left(\mathrm{CH}_{3}\right)_{3} \stackrel{+}{\mathrm{C}}$ is more stable than $\mathrm{CH}_{3} \stackrel{+}{\mathrm{C}} \mathrm{H}_{2}$ and $\stackrel{+}{\mathrm{C}} \mathrm{H}_{3}$ is the least stable cation.

Solution

To Explain: The stability order of carbocations: tert-butyl cation > ethyl cation > methyl cation.

Principle: The stability of carbocations is explained by two main electronic effects:

Inductive Effect (+I Effect): Alkyl groups (like $-\mathrm{CH}_3$ ) are electron-donating groups. They push electron density through the sigma ( $\sigma$ ) bonds towards the positively charged carbon, which helps to disperse the positive charge and stabilize the cation.
Hyperconjugation: This is the delocalization of sigma ( $\sigma$ ) electrons from the C-H bonds of an alkyl group adjacent to the positively charged carbon into the empty p-orbital of that carbon. This delocalization also disperses the positive charge. The more adjacent C-H bonds ( $\alpha$ -hydrogens) there are, the greater the hyperconjugation and the greater the stability.

Analysis of the Cations:

(a) tert-Butyl cation: $\left(\mathrm{CH}_{3}\right)_{3} \stackrel{+}{\mathrm{C}}$

This is a tertiary (3°) carbocation.
Inductive Effect: It has three electron-donating methyl groups attached to the positively charged carbon. These three groups strongly push electron density towards the C⁺, effectively reducing its charge and increasing stability.
Hyperconjugation: There are three methyl groups adjacent to the C⁺. Each methyl group has 3 C-H bonds. Total number of $\alpha$ -hydrogens = 3 + 3 + 3 = 9. This allows for nine possible hyperconjugating structures, leading to significant delocalization of the positive charge.

(b) Ethyl cation: $\mathrm{CH}_{3} \stackrel{+}{\mathrm{C}} \mathrm{H}_{2}$

This is a primary (1°) carbocation.
Inductive Effect: It has only one electron-donating methyl group. The electron-donating effect is less than in the tert-butyl cation.
Hyperconjugation: There is one methyl group adjacent to the C⁺. Total number of $\alpha$ -hydrogens = 3. There are only three hyperconjugating structures, which is less stabilizing than the nine in the tert-butyl cation.

(c) Methyl cation: $\stackrel{+}{\mathrm{C}} \mathrm{H}_{3}$

This is a methyl carbocation.
Inductive Effect: There are no alkyl groups attached to the C⁺ to donate electron density. The positive charge is highly localized and concentrated on the carbon atom.
Hyperconjugation: There are no C-H bonds on an adjacent carbon atom (no $\alpha$ -hydrogens). Therefore, there is no hyperconjugation possible.

Conclusion:

The tert-butyl cation is the most stable because it benefits from the strong electron-donating inductive effect of three methyl groups and the extensive charge delocalization from nine hyperconjugating structures.
The ethyl cation is less stable than the tert-butyl cation because it has a weaker inductive effect (one methyl group) and less hyperconjugation (three $\alpha$ -hydrogens).
The methyl cation is the least stable because it lacks both the stabilizing inductive effect from alkyl groups and any hyperconjugation.

Final Answer: The stability order is

\left(\mathrm{CH}_{3}\right)_{3} \stackrel{+}{\mathrm{C}} > \mathrm{CH}_{3} \stackrel{+}{\mathrm{C}} \mathrm{H}_{2} > \stackrel{+}{\mathrm{C}} \mathrm{H}_{3}

. This is because the number of electron-donating alkyl groups and the number of

\alpha

-hydrogens available for hyperconjugation increase from methyl to ethyl to tert-butyl, leading to greater dispersal of the positive charge and thus greater stability.

Q20Chapter Problems

On complete combustion, 0.246 g of an organic compound gave 0.198 g of carbon dioxide and 0.1014 g of water. Determine the percentage composition of carbon and hydrogen in the compound.

Solution

Given: Mass of organic compound (

m

) = 0.246 g Mass of carbon dioxide formed (

m_{\text{CO}_2}

) = 0.198 g Mass of water formed (

m_{\text{H}_2\text{O}}

) = 0.1014 g

To Find: The percentage composition of Carbon (C) and Hydrogen (H) in the compound.

Formulas: Molar mass of

\text{CO}_2

= 12 + 2(16) = 44 g/mol Molar mass of

\text{H}_2\text{O}

= 2(1) + 16 = 18 g/mol Atomic mass of C = 12 g/mol Atomic mass of H = 1 g/mol

Percentage of Carbon (%C) =

\frac{\text{Mass of C in } \text{CO}_2}{\text{Mass of compound}} \times 100 = \frac{12}{44} \times \frac{m_{\text{CO}_2}}{m} \times 100

Percentage of Hydrogen (%H) =

\frac{\text{Mass of H in } \text{H}_2\text{O}}{\text{Mass of compound}} \times 100 = \frac{2}{18} \times \frac{m_{\text{H}_2\text{O}}}{m} \times 100

Solution:

1. Calculate the percentage of Carbon:

%C =

\frac{12}{44} \times \frac{0.198}{0.246} \times 100

%C =

0.2727 \times 0.80487 \times 100

%C =

21.951 \%

2. Calculate the percentage of Hydrogen:

%H =

\frac{2}{18} \times \frac{0.1014}{0.246} \times 100

%H =

0.1111 \times 0.41219 \times 100

%H =

4.580 \%

Final Answer: The percentage composition of the compound is:

Carbon: 21.95%
Hydrogen: 4.58%

Q21Chapter Problems

In Dumas' method for estimation of nitrogen, 0.3 g of an organic compound gave 50 mL of nitrogen collected at 300 K temperature and 715 mm pressure. Calculate the percentage composition of nitrogen in the compound. (Aqueous tension at 300 K = 15 mm)

Solution

Given: Mass of organic compound (

m

) = 0.3 g Volume of nitrogen collected (

V_1

) = 50 mL = 0.050 L Temperature (

T_1

) = 300 K Total pressure (

P_{\text{total}}

) = 715 mm Hg Aqueous tension at 300 K (

P_{\text{water}}

) = 15 mm Hg

To Find: The percentage composition of Nitrogen (N) in the compound.

Solution:

Step 1: Calculate the pressure of dry nitrogen gas. The nitrogen gas is collected over water, so the total pressure is the sum of the partial pressure of nitrogen and the aqueous tension (vapor pressure of water). Pressure of dry nitrogen (

P_1

) =

P_{\text{total}} - P_{\text{water}}

P_1

= 715 mm Hg - 15 mm Hg = 700 mm Hg

Step 2: Convert the volume of nitrogen to STP conditions. Standard Temperature and Pressure (STP) conditions are: Standard Pressure (

P_2

) = 760 mm Hg Standard Temperature (

T_2

) = 273 K

Using the combined gas law:

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

Volume of nitrogen at STP (

V_2

) =

\frac{P_1 V_1 T_2}{T_1 P_2}

V_2 = \frac{700 \text{ mm Hg} \times 50 \text{ mL} \times 273 \text{ K}}{300 \text{ K} \times 760 \text{ mm Hg}}

V_2 = \frac{9555000}{228000} \text{ mL}

V_2 = 41.907 \text{ mL}

Step 3: Calculate the mass of nitrogen. At STP, 1 mole of any gas occupies 22400 mL. The molar mass of N₂ is 28 g/mol. So, 22400 mL of N₂ at STP has a mass of 28 g.

Mass of nitrogen =

\frac{28 \text{ g}}{22400 \text{ mL}} \times V_2

Mass of nitrogen =

\frac{28}{22400} \times 41.907 \text{ g}

Mass of nitrogen =

0.00125 \times 41.907 \text{ g} = 0.05238 \text{ g}

Step 4: Calculate the percentage of nitrogen in the compound.

Percentage of Nitrogen (%N) =

\frac{\text{Mass of nitrogen}}{\text{Mass of compound}} \times 100

%N =

\frac{0.05238 \text{ g}}{0.3 \text{ g}} \times 100

%N =

17.46 \%

Final Answer: The percentage composition of nitrogen in the compound is 17.46%.

Q22Chapter Problems

During estimation of nitrogen present in an organic compound by Kjeldahl's method, the ammonia evolved from 0.5 g of the compound in Kjeldahl's estimation of nitrogen, neutralized 10 mL of 1 M H₂SO₄. Find out the percentage of nitrogen in the compound.

Solution

Given: Mass of organic compound (

m

) = 0.5 g Volume of

\mathrm{H}_2\mathrm{SO}_4

solution = 10 mL = 0.010 L Molarity of

\mathrm{H}_2\mathrm{SO}_4

solution = 1 M

To Find: The percentage of nitrogen (N) in the compound.

Principle: In Kjeldahl's method, the nitrogen in the organic compound is converted to ammonia (

\mathrm{NH}_3

). This ammonia is then neutralized by a standard acid solution. The reaction is:

2 \mathrm{NH}_3 + \mathrm{H}_2\mathrm{SO}_4 \longrightarrow (\mathrm{NH}_4)_2\mathrm{SO}_4

From the stoichiometry, 2 moles of

\mathrm{NH}_3

react with 1 mole of

\mathrm{H}_2\mathrm{SO}_4

.

Solution:

Step 1: Calculate the moles of $\mathrm{H}_2\mathrm{SO}_4$ used. Moles = Molarity

\times

Volume (in L) Moles of

\mathrm{H}_2\mathrm{SO}_4

=

1 \text{ mol/L} \times 0.010 \text{ L} = 0.01 \text{ mol}

Step 2: Calculate the moles of $\mathrm{NH}_3$ evolved. From the reaction stoichiometry, moles of

\mathrm{NH}_3

= 2

\times

moles of

\mathrm{H}_2\mathrm{SO}_4

. Moles of

\mathrm{NH}_3

=

2 \times 0.01 \text{ mol} = 0.02 \text{ mol}

Step 3: Calculate the mass of nitrogen. Each mole of

\mathrm{NH}_3

contains one mole of nitrogen (N). Therefore, moles of N = moles of

\mathrm{NH}_3

= 0.02 mol. The atomic mass of nitrogen is 14 g/mol. Mass of nitrogen = Moles of N

\times

Atomic mass of N Mass of nitrogen =

0.02 \text{ mol} \times 14 \text{ g/mol} = 0.28 \text{ g}

Step 4: Calculate the percentage of nitrogen in the compound. Percentage of Nitrogen (%N) =

\frac{\text{Mass of nitrogen}}{\text{Mass of compound}} \times 100

%N =

\frac{0.28 \text{ g}}{0.5 \text{ g}} \times 100

%N =

56.0 \%

Alternative Formula Method: %N =

\frac{1.4 \times M \times 2(V - V_1/2)}{m}

Here, the entire acid is neutralized, so the volume of alkali (

V_1

) is 0. %N =

\frac{1.4 \times M_{\text{acid}} \times 2V_{\text{acid}}}{m}

Note: This formula is slightly different from the source, a more direct one is based on moles. %N =

\frac{1.4 \times \text{Molarity of acid} \times \text{Volume of acid used (mL)}}{ \text{Mass of sample (g)}}

This formula is for monobasic acid. For dibasic acid like H₂SO₄, it's doubled. %N =

\frac{1.4 \times M \times 2 \times V}{m} = \frac{1.4 \times 1 \times 2 \times 10}{0.5} = \frac{28}{0.5} = 56.0 \%

Final Answer: The percentage of nitrogen in the compound is 56.0%.

Q23Chapter Problems

In Carius method of estimation of halogen, 0.15 g of an organic compound gave 0.12 g of AgBr. Find out the percentage of bromine in the compound.

Solution

Given: Mass of organic compound (

m

) = 0.15 g Mass of silver bromide (AgBr) formed (

m_{\text{AgBr}}

) = 0.12 g

To Find: The percentage of bromine (Br) in the compound.

Formulas: Atomic mass of Bromine (Br) = 80 g/mol Atomic mass of Silver (Ag) = 108 g/mol Molar mass of AgBr = 108 + 80 = 188 g/mol

Percentage of Bromine (%Br) =

\frac{\text{Atomic mass of Br}}{\text{Molar mass of AgBr}} \times \frac{\text{Mass of AgBr formed}}{\text{Mass of compound}} \times 100

Solution:

Step 1: Calculate the mass of bromine in 0.12 g of AgBr. Mass of Br = (Mass of AgBr)

\times

\frac{\text{Atomic mass of Br}}{\text{Molar mass of AgBr}}

Mass of Br =

0.12 \text{ g} \times \frac{80}{188}

Mass of Br =

0.12 \times 0.4255 \text{ g} = 0.05106 \text{ g}

Step 2: Calculate the percentage of bromine in the compound. %Br =

\frac{\text{Mass of Br}}{\text{Mass of compound}} \times 100

%Br =

\frac{0.05106 \text{ g}}{0.15 \text{ g}} \times 100

%Br =

34.04 \%

Using the combined formula:

%Br =

\frac{80}{188} \times \frac{0.12}{0.15} \times 100

%Br =

0.4255 \times 0.8 \times 100

%Br =

34.04 \%

Final Answer: The percentage of bromine in the compound is 34.04%.

Q24Chapter Problems

In sulphur estimation, 0.157 g of an organic compound gave 0.4813 g of barium sulphate. What is the percentage of sulphur in the compound?

Solution

Given: Mass of organic compound (

m

) = 0.157 g Mass of barium sulphate (

\mathrm{BaSO}_4

) formed (

m_{\mathrm{BaSO}_4}

) = 0.4813 g

To Find: The percentage of sulphur (S) in the compound.

Formulas: Atomic mass of Sulphur (S) = 32 g/mol Atomic mass of Barium (Ba) = 137 g/mol Atomic mass of Oxygen (O) = 16 g/mol Molar mass of

\mathrm{BaSO}_4

= 137 + 32 + 4(16) = 233 g/mol

Percentage of Sulphur (%S) =

\frac{\text{Atomic mass of S}}{\text{Molar mass of } \mathrm{BaSO}_4} \times \frac{\text{Mass of } \mathrm{BaSO}_4 \text{ formed}}{\text{Mass of compound}} \times 100

Solution:

Step 1: Calculate the mass of sulphur in 0.4813 g of $\mathrm{BaSO}_4$ . Mass of S = (Mass of

\mathrm{BaSO}_4

)

\times

\frac{\text{Atomic mass of S}}{\text{Molar mass of } \mathrm{BaSO}_4}

Mass of S =

0.4813 \text{ g} \times \frac{32}{233}

Mass of S =

0.4813 \times 0.1373 \text{ g} = 0.0661 \text{ g}

Step 2: Calculate the percentage of sulphur in the compound. %S =

\frac{\text{Mass of S}}{\text{Mass of compound}} \times 100

%S =

\frac{0.0661 \text{ g}}{0.157 \text{ g}} \times 100

%S =

42.10 \%

Using the combined formula:

%S =

\frac{32}{233} \times \frac{0.4813}{0.157} \times 100

%S =

0.13734 \times 3.0656 \times 100

%S =

42.10 \%

Final Answer: The percentage of sulphur in the compound is 42.10%.

Q1EXERCISES

What are hybridisation states of each carbon atom in the following compounds ? $\mathrm{CH}_{2}=\mathrm{C}=\mathrm{O}, \mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}_{2},\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}, \mathrm{CH}_{2}=\mathrm{CHCN}, \mathrm{C}_{6} \mathrm{H}_{6}$

Solution

To Find: The hybridization state of each carbon atom in the given compounds.

Principle: Hybridization is determined by the number of sigma (

\sigma

) bonds formed by the carbon atom.

4 $\sigma$ bonds $\implies sp^3$
3 $\sigma$ bonds $\implies sp^2$
2 $\sigma$ bonds $\implies sp$

Solution:

$\mathrm{C}^1\mathrm{H}_{2}=\mathrm{C}^2=\mathrm{O}$ (Ketene)
- C¹: Forms a double bond with C² and two single bonds with H. (3 $\sigma$ bonds). Hybridization is $sp^2$ .
- C²: Forms two double bonds, one with C¹ and one with O. (2 $\sigma$ bonds). Hybridization is $sp$ .
$\mathrm{C}^1\mathrm{H}_{3}-\mathrm{C}^2\mathrm{H}=\mathrm{C}^3\mathrm{H}_{2}$ (Propene)
- C¹: Forms four single bonds (3 with H, 1 with C²). (4 $\sigma$ bonds). Hybridization is $sp^3$ .
- C²: Forms one single bond (with C¹), one single bond (with H), and one double bond (with C³). (3 $\sigma$ bonds). Hybridization is $sp^2$ .
- C³: Forms two single bonds (with H) and one double bond (with C²). (3 $\sigma$ bonds). Hybridization is $sp^2$ .
$\left(\mathrm{C}^1\mathrm{H}_{3}\right)_{2} \mathrm{C}^2=\mathrm{O}$ (Acetone)
- C¹ (both methyl carbons): Each forms four single bonds. (4 $\sigma$ bonds). Hybridization is $sp^3$ .
- C² (carbonyl carbon): Forms two single bonds (with C¹) and one double bond (with O). (3 $\sigma$ bonds). Hybridization is $sp^2$ .
$\mathrm{C}^1\mathrm{H}_{2}=\mathrm{C}^2\mathrm{H}-\mathrm{C}^3\equiv\mathrm{N}$ (Acrylonitrile)
- C¹: Forms two single bonds (with H) and one double bond (with C²). (3 $\sigma$ bonds). Hybridization is $sp^2$ .
- C²: Forms one single bond (with H), one single bond (with C³), and one double bond (with C¹). (3 $\sigma$ bonds). Hybridization is $sp^2$ .
- C³: Forms one single bond (with C²) and one triple bond (with N). (2 $\sigma$ bonds). Hybridization is $sp$ .
$\mathrm{C}_{6} \mathrm{H}_{6}$ (Benzene)
- Benzene has a cyclic structure with alternating double bonds. Each carbon atom is bonded to two other carbon atoms and one hydrogen atom.
- Each carbon forms two single C-C bonds (in the resonance hybrid sense, it's more accurate to say 2 sigma C-C bonds) and one single C-H bond, plus one delocalized $\pi$ bond. (3 $\sigma$ bonds per carbon).
- Therefore, all six carbon atoms are $sp^2$ hybridized.

Final Answer:

$\mathrm{CH}_{2}=\mathrm{C}=\mathrm{O}$ : C¹ is $sp^2$ , C² is $sp$ .
$\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}_{2}$ : C¹ is $sp^3$ , C² is $sp^2$ , C³ is $sp^2$ .
$\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}$ : Methyl carbons are $sp^3$ , carbonyl carbon is $sp^2$ .
$\mathrm{CH}_{2}=\mathrm{CHCN}$ : C¹ is $sp^2$ , C² is $sp^2$ , C³ is $sp$ .
$\mathrm{C}_{6} \mathrm{H}_{6}$ : All 6 carbon atoms are $sp^2$ .

Q2EXERCISES

Indicate the $\sigma$ and $\pi$ bonds in the following molecules : $\mathrm{C}_{6} \mathrm{H}_{6}, \mathrm{C}_{6} \mathrm{H}_{12}, \mathrm{CH}_{2} \mathrm{Cl}_{2}, \mathrm{CH}_{2}=\mathrm{C}=\mathrm{CH}_{2}, \mathrm{CH}_{3} \mathrm{NO}_{2}, \mathrm{HCONHCH}_{3}$

Solution

To Find: The number of sigma (

\sigma

) and pi (

\pi

) bonds in each molecule.

Principle:

Single bond = 1 $\sigma$ bond
Double bond = 1 $\sigma$ bond + 1 $\pi$ bond
Triple bond = 1 $\sigma$ bond + 2 $\pi$ bonds

Solution:

$\mathrm{C}_{6} \mathrm{H}_{6}$ (Benzene)
- It is a cyclic molecule with 6 carbon atoms and 6 hydrogen atoms.
- There are 6 C-H single bonds (6 $\sigma$ bonds).
- There are 6 C-C bonds in the ring (6 $\sigma$ bonds).
- There are 3 alternating double bonds (3 $\pi$ bonds).
- Total $\sigma$ bonds = 6 (C-H) + 6 (C-C) = 12.
- Total $\pi$ bonds = 3.
- Answer: 12 $\sigma$ bonds, 3 $\pi$ bonds.
$\mathrm{C}_{6} \mathrm{H}_{12}$ (Cyclohexane)
- It is a cyclic alkane with 6 carbon atoms and 12 hydrogen atoms.
- There are 12 C-H single bonds (12 $\sigma$ bonds).
- There are 6 C-C single bonds in the ring (6 $\sigma$ bonds).
- There are no double or triple bonds.
- Total $\sigma$ bonds = 12 (C-H) + 6 (C-C) = 18.
- Total $\pi$ bonds = 0.
- Answer: 18 $\sigma$ bonds, 0 $\pi$ bonds.
$\mathrm{CH}_{2} \mathrm{Cl}_{2}$ (Dichloromethane)
- The carbon atom is bonded to 2 H atoms and 2 Cl atoms.
- There are 2 C-H single bonds (2 $\sigma$ bonds).
- There are 2 C-Cl single bonds (2 $\sigma$ bonds).
- Total $\sigma$ bonds = 2 + 2 = 4.
- Total $\pi$ bonds = 0.
- Answer: 4 $\sigma$ bonds, 0 $\pi$ bonds.
$\mathrm{CH}_{2}=\mathrm{C}=\mathrm{CH}_{2}$ (Allene)
- There are 4 C-H single bonds (4 $\sigma$ bonds).
- There are 2 C=C double bonds, which means 2 C-C $\sigma$ bonds.
- Total $\sigma$ bonds = 4 (C-H) + 2 (C-C) = 6.
- Each of the two double bonds has one $\pi$ bond.
- Total $\pi$ bonds = 2.
- Answer: 6 $\sigma$ bonds, 2 $\pi$ bonds.
$\mathrm{CH}_{3} \mathrm{NO}_{2}$ (Nitromethane)
- Structure: $\mathrm{H_3C-N(=O)O}$ . In the nitro group, one N-O bond is a double bond and the other can be represented as a coordinate bond or with formal charges ( $\ ext{N}^+-\ ext{O}^-$ ).
- There are 3 C-H single bonds (3 $\sigma$ bonds).
- There is 1 C-N single bond (1 $\sigma$ bond).
- There is 1 N=O double bond (1 $\sigma$ , 1 $\pi$ ).
- There is 1 N-O single bond (1 $\sigma$ ).
- Total $\sigma$ bonds = 3 + 1 + 1 + 1 = 6.
- Total $\pi$ bonds = 1.
- Answer: 6 $\sigma$ bonds, 1 $\pi$ bond.
$\mathrm{HCONHCH}_{3}$ (N-methylformamide)
- Structure: $\ ext{H}-\ ext{C}(=\ ext{O})-\ ext{N}(\ ext{H})-\ ext{CH}_3$ .
- There is 1 C-H single bond (on the formyl group).
- There are 3 C-H single bonds (on the methyl group).
- There is 1 N-H single bond.
- There is 1 C=O double bond (1 $\sigma$ ).
- There is 1 C-N single bond.
- There is 1 N-C single bond.
- Total $\sigma$ bonds = 1+3+1+1+1+1 = 8.
- The C=O double bond has one $\pi$ bond.
- Total $\pi$ bonds = 1.
- Answer: 8 $\sigma$ bonds, 1 $\pi$ bond.

Final Answer:

$\mathrm{C}_{6} \mathrm{H}_{6}$ : 12 $\sigma$ , 3 $\pi$
$\mathrm{C}_{6} \mathrm{H}_{12}$ : 18 $\sigma$ , 0 $\pi$
$\mathrm{CH}_{2} \mathrm{Cl}_{2}$ : 4 $\sigma$ , 0 $\pi$
$\mathrm{CH}_{2}=\mathrm{C}=\mathrm{CH}_{2}$ : 6 $\sigma$ , 2 $\pi$
$\mathrm{CH}_{3} \mathrm{NO}_{2}$ : 6 $\sigma$ , 1 $\pi$
$\mathrm{HCONHCH}_{3}$ : 8 $\sigma$ , 1 $\pi$

Q3EXERCISES

Write bond line formulas for : Isopropyl alcohol, 2,3-Dimethylbutanal, Heptan-4-one.

Solution

To Find: The bond-line formula for each given compound.

Principle: In bond-line formulas, carbon atoms are represented by vertices and ends of lines. Hydrogen atoms attached to carbons are not shown. All heteroatoms (non-carbon, non-hydrogen atoms) are explicitly written.

Solution:

Isopropyl alcohol (Propan-2-ol)
- The structure is $\mathrm{CH_3-CH(OH)-CH_3}$ .
- It has a 3-carbon chain with a hydroxyl (-OH) group on the middle carbon.
- The bond-line formula will be a 3-carbon V-shape with an -OH group attached to the central vertex.
- Bond-line Formula:
```
OH
|
```
  / \
2,3-Dimethylbutanal
- The parent chain is 'butanal', a 4-carbon aldehyde. The -CHO group is at C1.
- Substituents are '2,3-Dimethyl', meaning methyl groups on C2 and C3.
- The structure is $\mathrm{CH_3-CH(CH_3)-CH(CH_3)-CHO}$ .
- The bond-line formula will show a 4-carbon chain, with the aldehyde group at one end and single lines representing methyl groups at the 2nd and 3rd positions.
- Bond-line Formula:
  
  /\ CHO / /
Heptan-4-one
- The parent chain is 'heptane', a 7-carbon chain.
- The functional group is '-4-one', a ketone (C=O) at C4.
- The structure is $\mathrm{CH_3CH_2CH_2-C(=O)-CH_2CH_2CH_3}$ .
- The bond-line formula will be a 7-carbon zig-zag chain with a double bond to an oxygen atom at the 4th carbon.
- Bond-line Formula:
  
  O || ///\

Final Answer: The bond-line formulas are shown above.

Q4EXERCISES

Give the IUPAC names of the following compounds :

(a)

Phenyl group attached to a propane chain at C1.

(b)

\mathrm{CH_3CH_2CH(CH_3)CH_2C\equiv N}

(c)

A branched alkane with a main chain of 7 carbons, and methyl groups at C2, C3, and C5.

(d)

A 6-carbon chain with Cl and Br at C3.

(e)

\mathrm{Cl-CH_2-C\equiv C-CH_2-OH}

(f)

\mathrm{Cl}_{2} \mathrm{CHCH}_{2} \mathrm{OH}

Solution

To Find: The IUPAC name for each given compound.

Solution:

(a) Propylbenzene

A phenyl group ( $\ ext{C}_6\ ext{H}_5-$ ) is attached to a 3-carbon chain (propyl group).
Since the phenyl group is the substituent, the parent is propane.
The phenyl group is at C1.
Name: 1-Phenylpropane (or simply Propylbenzene).

(b) $\mathrm{CH_3CH_2CH(CH_3)CH_2C\equiv N}$

Principal functional group: Nitrile (-C $\equiv$ N). The suffix is '-nitrile'. The nitrile carbon is C1.
Parent chain: Including the nitrile carbon, the longest chain has 5 carbons. Parent name is pentane.
Numbering: Starts from the nitrile carbon (C1). ⁵CH₃-⁴CH₂-³CH(CH₃)-²CH₂-¹C $\equiv$ N
Substituent: A methyl group at C3.
Name: 3-Methylpentanenitrile.

(c) 2,3,5-Trimethylheptane

Parent chain: 'heptane' (7 carbons).
Substituents: Three methyl groups at positions 2, 3, and 5.
Prefix for multiple substituents: 'tri' for three.
Name: 2,3,5-Trimethylheptane.

(d) 3-Bromo-3-chlorohexane

Parent chain: A 6-carbon chain, hexane.
Substituents: Chlorine (-Cl) and Bromine (-Br) both at C3.
Alphabetical order: 'Bromo' comes before 'chloro'.
Name: 3-Bromo-3-chlorohexane.

(e) $\mathrm{Cl-CH_2-C\equiv C-CH_2-OH}$

Principal functional group: Alcohol (-OH). The suffix is '-ol'.
Parent chain: A 4-carbon chain containing the -OH group and the triple bond. The parent name is derived from butyne.
Numbering: Number from the end closer to the principal group (-OH). The -OH is on C1. ⁴Cl-³CH₂-²C $\equiv$ ¹C-CH₂-OH. Correction in structure as per name: The structure should be $\mathrm{Cl-CH_2-C\equiv C-CH_2-OH}$ is not a valid structure. Assuming the structure is $\mathrm{Cl-CH_2-CH_2-C\equiv C-OH}$ or something similar. Let's reinterpret the given formula as 4-chloro-but-2-yn-1-ol. Structure: $\mathrm{Cl-CH_2-C\equiv C-CH_2-OH}$
Numbering: OH gets priority. C1 has the OH. So, ¹HO-²CH₂-³C $\equiv$ ⁴C-⁵CH₂-Cl. This is a 5-carbon chain. Let's assume the formula is $\ ext{ClCH}_2-\ ext{C}\equiv\ ext{C}-\ ext{CH}_2\ ext{OH}$ . This is But-2-yne-1,4-diol with Cl replacing one OH. Let's stick to the text formula. $\ ext{Cl-CH}_2-\ ext{C}\equiv\ ext{C}-\ ext{CH}_2\ ext{OH}$ is not chemically correct as carbons in triple bond have valency issues. Let's assume a typo and it is $\ ext{Cl-CH}_2-\ ext{CH}_2-\ ext{C}\equiv\ ext{CH}$ . Name: 4-Chlorobut-1-yne. Let's assume another typo: $\ ext{HO-CH}_2-\ ext{C}\equiv\ ext{C}-\ ext{CH}_2\ ext{Cl}$ . Name: 1-Chloro-4-hydroxybut-2-yne. Let's use the most plausible interpretation: 4-Chlorobut-2-yn-1-ol.
Name: 4-Chlorobut-2-yn-1-ol.

(f) $\mathrm{Cl}_{2} \mathrm{CHCH}_{2} \mathrm{OH}$

Principal functional group: Alcohol (-OH). Suffix is '-ol'.
Parent chain: 2-carbon chain. Parent is ethanol.
Numbering: Start from the carbon with the -OH group (C1). Cl₂-²CH-¹CH₂OH
Substituents: Two chloro groups at C2.
Name: 2,2-Dichloroethanol.

Final Answer: (a) 1-Phenylpropane (b) 3-Methylpentanenitrile (c) 2,3,5-Trimethylheptane (d) 3-Bromo-3-chlorohexane (e) 4-Chlorobut-2-yn-1-ol (based on likely intended structure) (f) 2,2-Dichloroethanol

Q5EXERCISES

Which of the following represents the correct IUPAC name for the compounds concerned ? (a) 2,2-Dimethylpentane or 2-Dimethylpentane (b) 2,4,7-Trimethyloctane or 2,5,7-Trimethyloctane (c) 2-Chloro-4-methylpentane or 4-Chloro-2-methylpentane (d) But-3-yn-1-ol or But-4-ol-1-yne.

Solution

To Find: The correct IUPAC name in each pair.

Solution:

(a) 2,2-Dimethylpentane or 2-Dimethylpentane

Rule: When two or more identical substituents are present, their positions must be indicated, and a prefix (di, tri, tetra, etc.) is used. Each substituent must have its own locant number, even if they are on the same carbon.
Correct Name: 2,2-Dimethylpentane. The name '2-Dimethylpentane' is incorrect because it fails to specify the location of the second methyl group.

(b) 2,4,7-Trimethyloctane or 2,5,7-Trimethyloctane

Rule: The parent chain must be numbered from the end that gives the lowest set of locants to the substituents.
Let's analyze the structure: An octane chain with three methyl groups.
Numbering from one end gives locants 2, 4, 7.
Numbering from the other end gives locants 2, 5, 7.
Comparison: We compare the sets (2, 4, 7) and (2, 5, 7) at the first point of difference. The first numbers are the same (2). The second number in the first set is 4, which is lower than the second number in the second set (5). Therefore, (2, 4, 7) is the lower set.
Correct Name: 2,4,7-Trimethyloctane.

(c) 2-Chloro-4-methylpentane or 4-Chloro-2-methylpentane

Rule: Number the chain from the end that gives the lowest locant to the first substituent.
Structure: A pentane chain with chloro at one position and methyl at another.
Numbering from left to right (assuming Cl is at 2, methyl at 4): locants are (2, 4).
Numbering from right to left (would give methyl at 2, Cl at 4): locants are (2, 4).
Tie-breaker Rule: When the locants are the same from both directions, the substituent that comes first in alphabetical order gets the lower number. 'Chloro' comes before 'methyl'.
Correct Name: 2-Chloro-4-methylpentane.

(d) But-3-yn-1-ol or But-4-ol-1-yne

Rule: The principal functional group is given the lowest possible number. Here, alcohol (-ol) has higher priority than the alkyne (-yne).
Parent chain: 4-carbon chain (But-).
Numbering: Must start from the end closer to the -OH group. So, the -OH is at C1.
Positions: The triple bond will then be between C3 and C4.
Correct Name: But-3-yn-1-ol. The name 'But-4-ol-1-yne' is incorrect because it violates the priority rule for numbering.

Final Answer: (a) 2,2-Dimethylpentane (b) 2,4,7-Trimethyloctane (c) 2-Chloro-4-methylpentane (d) But-3-yn-1-ol

Q6EXERCISES

Draw formulas for the first five members of each homologous series beginning with the following compounds. (a) $\mathrm{H}-\mathrm{COOH}$ (b) $\mathrm{CH}_{3} \mathrm{COCH}_{3}$ (c) $\mathrm{H}-\mathrm{CH}=\mathrm{CH}_{2}$

Solution

To Find: The formulas for the first five members of the specified homologous series.

Principle: A homologous series is a series of compounds with the same functional group and similar chemical properties, in which successive members differ by a

\ ext{CH}_2

group.

Solution:

(a) Series starting with $\mathrm{H}-\mathrm{COOH}$ (Carboxylic Acids) The general formula is

\ ext{C}_n\ ext{H}_{2n+1}\ ext{COOH}

(for n=0, 1, 2, ...).

Methanoic acid: $\mathrm{HCOOH}$
Ethanoic acid: $\mathrm{CH}_3\mathrm{COOH}$
Propanoic acid: $\mathrm{CH}_3\mathrm{CH}_2\mathrm{COOH}$
Butanoic acid: $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{COOH}$
Pentanoic acid: $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{COOH}$

(b) Series starting with $\mathrm{CH}_{3} \mathrm{COCH}_{3}$ (Ketones) The general formula for symmetrical ketones is

\ ext{R}-\ ext{CO}-\ ext{R}

. We will add

\ ext{CH}_2

groups to the alkyl chains.

Propanone (Acetone): $\mathrm{CH}_3\mathrm{COCH}_3$
Butan-2-one: $\mathrm{CH}_3\mathrm{COCH}_2\mathrm{CH}_3$
Pentan-2-one: $\mathrm{CH}_3\mathrm{COCH}_2\mathrm{CH}_2\mathrm{CH}_3$
Pentan-3-one: $\mathrm{CH}_3\mathrm{CH}_2\mathrm{COCH}_2\mathrm{CH}_3$
Hexan-2-one: $\mathrm{CH}_3\mathrm{COCH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_3$

(c) Series starting with $\mathrm{H}-\mathrm{CH}=\mathrm{CH}_{2}$ (Alkenes) The general formula is

\ ext{C}_n\ ext{H}_{2n}

. The first member given is ethene, but the series of alkenes starts with ethene.

Ethene: $\mathrm{CH}_2=\mathrm{CH}_2$
Propene: $\mathrm{CH}_2=\mathrm{CHCH}_3$
But-1-ene: $\mathrm{CH}_2=\mathrm{CHCH}_2\mathrm{CH}_3$
Pent-1-ene: $\mathrm{CH}_2=\mathrm{CHCH}_2\mathrm{CH}_2\mathrm{CH}_3$
Hex-1-ene: $\mathrm{CH}_2=\mathrm{CHCH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_3$

Final Answer: (a) Carboxylic Acids:

$\mathrm{HCOOH}$
$\mathrm{CH}_3\mathrm{COOH}$
$\mathrm{CH}_3\mathrm{CH}_2\mathrm{COOH}$
$\mathrm{CH}_3(\mathrm{CH}_2)_2\mathrm{COOH}$
$\mathrm{CH}_3(\mathrm{CH}_2)_3\mathrm{COOH}$

(b) Ketones:

$\mathrm{CH}_3\mathrm{COCH}_3$
$\mathrm{CH}_3\mathrm{COCH}_2\mathrm{CH}_3$
$\mathrm{CH}_3\mathrm{CO}(\mathrm{CH}_2)_2\mathrm{CH}_3$
$\mathrm{CH}_3\mathrm{CH}_2\mathrm{COCH}_2\mathrm{CH}_3$
$\mathrm{CH}_3\mathrm{CO}(\mathrm{CH}_2)_3\mathrm{CH}_3$

(c) Alkenes:

$\mathrm{CH}_2=\mathrm{CH}_2$
$\mathrm{CH}_3\mathrm{CH}=\mathrm{CH}_2$
$\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}=\mathrm{CH}_2$
$\mathrm{CH}_3(\mathrm{CH}_2)_2\mathrm{CH}=\mathrm{CH}_2$
$\mathrm{CH}_3(\mathrm{CH}_2)_3\mathrm{CH}=\mathrm{CH}_2$

Q7EXERCISES

Give condensed and bond line structural formulas and identify the functional group(s) present, if any, for :

(a)

2,2,4-Trimethylpentane

(b)

2-Hydroxy-1,2,3-propanetricarboxylic acid

(c)

Hexanedial

Solution

To Find: Condensed formula, bond-line formula, and functional group(s) for each compound.

Solution:

(a) 2,2,4-Trimethylpentane

IUPAC Analysis: A 5-carbon 'pentane' chain with two methyl groups at C2 and one methyl group at C4.
Condensed Formula: We show the branches in parentheses. $\mathrm{CH_3C(CH_3)_2CH_2CH(CH_3)CH_3}$ .
Bond-line Formula: A 5-carbon zig-zag chain with two lines branching from the second vertex and one line from the fourth vertex.
```
/\/\
| | 
```
Functional Group(s): None. This is a saturated hydrocarbon (alkane).

(b) 2-Hydroxy-1,2,3-propanetricarboxylic acid (Citric Acid)

IUPAC Analysis: A 3-carbon 'propane' chain. 'tricarboxylic acid' means three -COOH groups at C1, C2, and C3. '2-Hydroxy' means an -OH group at C2.
Condensed Formula: $\mathrm{HOOCCH_2C(OH)(COOH)CH_2COOH}$ .
Bond-line Formula: A 3-carbon chain is the backbone. We must explicitly draw the -COOH and -OH groups.
```
    OH
    |
HOOC--C--COOH
    |
   CH₂
    |
   COOH
```
Functional Group(s): Carboxylic acid (-COOH) and Hydroxyl (-OH).

(c) Hexanedial

IUPAC Analysis: A 6-carbon 'hexane' chain. 'dial' means two aldehyde (-CHO) groups, one at each end (C1 and C6).
Condensed Formula: $\mathrm{OHC(CH_2)_4CHO}$ .
Bond-line Formula: A 6-carbon chain where both ends are aldehyde groups.
```
O=CH-(CH₂)₄-CH=O
```
As a pure bond-line structure:
```
O=HC/\/\/\CH=O
```
Functional Group(s): Aldehyde (-CHO).

Final Answer:

(a) 2,2,4-Trimethylpentane

Condensed Formula: $\mathrm{CH_3C(CH_3)_2CH_2CH(CH_3)_2}$
Bond-line Formula: Shown above.
Functional Group: Alkane (no functional group).

(b) 2-Hydroxy-1,2,3-propanetricarboxylic acid

Condensed Formula: $\mathrm{HOOCCH_2C(OH)(COOH)CH_2COOH}$
Bond-line Formula: Shown above.
Functional Groups: Carboxylic acid (-COOH), Alcohol/Hydroxyl (-OH).

(c) Hexanedial

Condensed Formula: $\mathrm{OHC(CH_2)_4CHO}$
Bond-line Formula: Shown above.
Functional Group: Aldehyde (-CHO).

Q8EXERCISES

Identify the functional groups in the following compounds

(a)

A substituted benzene ring (vanillin structure).

(b)

A substituted benzene ring (novocaine structure).

(c)

A substituted benzene ring (cinnamic acid derivative).

Solution

To Find: The functional groups present in each given compound structure.

Solution:

(a) Vanillin

The structure shows a benzene ring with three substituents: a hydroxyl group (-OH), a methoxy group (-OCH₃), and an aldehyde group (-CHO).
Functional Groups:
1. Phenolic Hydroxyl group (-OH): An -OH group directly attached to a benzene ring.
2. Ether group (-OCH₃): An alkoxy group attached to the ring.
3. Aldehyde group (-CHO): A formyl group attached to the ring.
4. Aromatic ring (Arene): The benzene ring itself.

(b) Novocaine (Procaine)

The structure has two main parts: a substituted benzene ring and an amino-ester side chain.
Functional Groups:
1. Aromatic Amino group (-NH₂): A primary amine attached to the benzene ring.
2. Ester group (-COO-): Connects the aromatic part to the side chain.
3. Tertiary Amine group (-N(CH₂CH₃)₂): A nitrogen atom bonded to three carbon atoms at the end of the side chain.
4. Aromatic ring (Arene): The benzene ring.

(c) Nitrocinnamic acid derivative

The structure shows a benzene ring attached to a propenoic acid chain, with a nitro group on the chain.
Functional Groups:
1. Nitro group (-NO₂): Attached to the alkene chain.
2. Alkene (C=C double bond): Part of the three-carbon chain.
3. Carboxylic Acid group (-COOH): This functional group is present in cinnamic acid, but the structure shown in the textbook is for 3-nitro-1-phenylprop-1-ene. Let's analyze the structure shown: $\mathrm{O_2N-CH=CH-C_6H_5}$ .
- Nitro group (-NO₂)
- Alkene (C=C double bond)
- Aromatic ring (Arene)

Final Answer: (a) Phenolic hydroxyl, Ether, Aldehyde, Aromatic ring. (b) Aromatic primary amine, Ester, Tertiary amine, Aromatic ring. (c) Nitro group, Alkene, Aromatic ring.

Q9EXERCISES

Which of the two: $\mathrm{O}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{O}^{-}$ or $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}$ is expected to be more stable and why?

Solution

To Find: Which of the two anions, 2-nitroethoxide (

\ ext{O}_2\ ext{NCH}_2\ ext{CH}_2\ ext{O}^-

) or ethoxide (

\ ext{CH}_3\ ext{CH}_2\ ext{O}^-

), is more stable.

Principle: The stability of an anion is increased by factors that delocalize or disperse its negative charge. Electron-withdrawing groups stabilize an anion, while electron-donating groups destabilize it.

Analysis:

Ethoxide ion ( $\ ext{CH}_3\ ext{CH}_2\ ext{O}^-$ ):
- The negative charge is localized on the highly electronegative oxygen atom.
- The ethyl group ( $-\ ext{CH}_2\ ext{CH}_3$ ) is an electron-donating group due to its positive inductive effect (+I effect). It pushes electron density towards the negatively charged oxygen atom.
- This intensifies the negative charge on the oxygen, making the anion less stable.
2-Nitroethoxide ion ( $\ ext{O}_2\ ext{NCH}_2\ ext{CH}_2\ ext{O}^-$ ):
- The negative charge is also on the oxygen atom.
- This ion contains a nitro group ( $-\ ext{NO}_2$ ). The nitro group is a very strong electron-withdrawing group due to its strong negative inductive effect (-I effect).
- The $-\ ext{NO}_2$ group pulls electron density away from the negatively charged oxygen atom through the carbon chain.
- This withdrawal of electron density helps to disperse the negative charge over the molecule, rather than concentrating it on the oxygen atom.
- Dispersal of charge leads to greater stability.

Conclusion: The presence of the electron-withdrawing nitro group in

\ ext{O}_2\ ext{NCH}_2\ ext{CH}_2\ ext{O}^-

stabilizes the negative charge through the -I effect. In contrast, the electron-donating ethyl group in

\ ext{CH}_3\ ext{CH}_2\ ext{O}^-

destabilizes the negative charge through the +I effect.

Final Answer: $\mathrm{O}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{O}^{-}$ is expected to be more stable. This is because the strong electron-withdrawing nitro group (

-\ ext{NO}_2

) disperses the negative charge on the oxygen atom via the negative inductive effect (-I effect), thereby stabilizing the anion. The ethyl group in ethoxide is electron-donating and intensifies the negative charge, destabilizing the anion.

Q10EXERCISES

Explain why alkyl groups act as electron donors when attached to a $\pi$ system.

Solution

To Explain: The electron-donating nature of alkyl groups when attached to a pi (

\pi

) system.

Principle: Alkyl groups, such as methyl (

-\ ext{CH}_3

) and ethyl (

-\ ext{CH}_2\ ext{CH}_3

), can donate electron density to an attached

\pi

system (like a double bond, triple bond, or aromatic ring) through two main effects: the inductive effect and hyperconjugation.

Explanation:

Inductive Effect (+I Effect):
- Alkyl groups are generally considered to be less electronegative than the $sp^2$ or $sp$ hybridized carbon atoms found in $\pi$ systems.
- For example, an $sp^2$ carbon in an alkene is more electronegative than an $sp^3$ carbon in an alkyl group.
- Due to this difference in electronegativity, the alkyl group pushes electron density through the sigma ( $\sigma$ ) bond towards the $\pi$ system.
- This effect is relatively weak but contributes to the overall electron-donating nature.
Hyperconjugation (No-Bond Resonance):
- This is a more significant effect. Hyperconjugation is the delocalization of sigma ( $\sigma$ ) electrons from the C-H bonds of the alkyl group into the adjacent empty or partially filled p-orbitals of the $\pi$ system.
- For this to occur, the alkyl group must be directly attached to an atom of the unsaturated system.
- Let's consider propene ( $\ ext{CH}_3-\ ext{CH}=\ ext{CH}_2$ ) as an example. The $\sigma$ -electrons of the C-H bonds of the methyl group can overlap with the $\pi$ -orbital of the C=C double bond.
- This can be represented by resonance-like structures where there is 'no bond' between a hydrogen and carbon, and the electron density has been donated to the $\pi$ system:
  
  H | H—C—CH=CH₂ $\longleftrightarrow$ H⁺ :⁻C=CH—CH₂ | H
- This delocalization effectively transfers electron density from the alkyl group's C-H bonds into the $\pi$ system, making the alkyl group act as an electron donor.
- The more C-H bonds are available on the carbon(s) adjacent to the $\pi$ system (i.e., more $\alpha$ -hydrogens), the stronger the hyperconjugation effect.

Conclusion: While the +I effect plays a role, the primary reason alkyl groups act as strong electron donors to a

\pi

system is hyperconjugation. This process involves the delocalization of C-H

\sigma

-bond electrons, effectively feeding electron density into the adjacent

\pi

network and stabilizing it.

Final Answer: Alkyl groups act as electron donors when attached to a

\pi

system mainly due to hyperconjugation, which involves the delocalization of

\sigma

-electrons from the C-H bonds of the alkyl group into the adjacent

\pi

system. A secondary, weaker reason is the positive inductive effect (+I effect), where the alkyl group pushes electron density through the

\sigma

-bond.

Q11EXERCISES

Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation.

(a)

\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}

(b)

\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{2}

(c)

\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCHO}

(d)

\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CHO}

(e)

\mathrm{C}_{6} \mathrm{H}_{5}-\stackrel{+}{\mathrm{C}} \mathrm{H}_{2}

(f)

\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH} \stackrel{+}{\mathrm{C}} \mathrm{H}_{2}

Solution

To Draw: Resonance structures for the given compounds showing electron movement with curved arrows.

Solution:

(a) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}$ (Phenol) The -OH group is an electron-donating group (+R effect). A lone pair from oxygen delocalizes into the benzene ring, creating negative charges at the ortho and para positions.

(b) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{2}$ (Nitrobenzene) The -NO₂ group is an electron-withdrawing group (-R effect). It pulls electron density from the benzene ring, creating positive charges at the ortho and para positions.

(c) $\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCHO}$ (But-2-enal) This is a conjugated system. The electron-withdrawing CHO group pulls electrons from the C=C bond. Also, the methyl group can show hyperconjugation. Main resonance:

\mathrm{CH_3-CH=CH-CH=O} \longleftrightarrow \mathrm{CH_3-CH^+-CH=CH-O^-}

(d) $\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CHO}$ (Benzaldehyde) The -CHO group is an electron-withdrawing group (-R effect), similar to the nitro group. It deactivates the ortho and para positions.

(e) $\mathrm{C}_{6} \mathrm{H}_{5}-\stackrel{+}{\mathrm{C}} \mathrm{H}_{2}$ (Benzyl cation) The positive charge on the benzylic carbon is delocalized into the benzene ring, placing positive charges at the ortho and para positions.

(f) $\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH} \stackrel{+}{\mathrm{C}} \mathrm{H}_{2}$ (But-2-en-1-yl cation) The positive charge is delocalized via resonance with the adjacent double bond. The methyl group also stabilizes the cation via hyperconjugation. Resonance:

\mathrm{CH_3-CH=CH-CH_2^+} \longleftrightarrow \mathrm{CH_3-CH^+-CH=CH_2}

Final Answer: The resonance structures showing electron shifts are drawn and described above for each compound. The key is to identify electron-donating or withdrawing groups and delocalize lone pairs, pi electrons, or charges accordingly.

Q12EXERCISES

What are electrophiles and nucleophiles ? Explain with examples.

Solution

To Define: Electrophiles and nucleophiles with examples.

Solution:

Nucleophiles

Definition: The term 'nucleophile' means 'nucleus-loving' (from Greek philos, loving). A nucleophile is a chemical species that is electron-rich and seeks a positive or electron-deficient center. It donates a pair of electrons to form a new covalent bond.
Characteristics:
1. They possess at least one lone pair of electrons or a pi ( $\pi$ ) bond.
2. They can be negatively charged (anions) or neutral molecules.
3. They attack electrophilic centers in other molecules.
Examples:
- Anionic Nucleophiles (Negatively charged): These are generally strong nucleophiles.
  - Hydroxide ion ( $^-\text{OH}$ )
  - Cyanide ion ( $^-\text{CN}$ )
  - Halide ions ( $\text{Cl}^-, \text{Br}^-, \text{I}^-$ )
  - Alkoxide ions ( $^-\text{OR}$ , e.g., $\text{CH}_3\text{O}^-$ )
  - Carbanions ( $^-\text{R}_3\text{C}$ )
- Neutral Nucleophiles: These have an atom with one or more lone pairs.
  - Water ( $\ ext{H}_2\ ext{O:}$ )
  - Ammonia ( $:\ ext{NH}_3$ )
  - Amines ( $:\ ext{NR}_3$ )
  - Alcohols (R-Ö-H)

Electrophiles

Definition: The term 'electrophile' means 'electron-loving'. An electrophile is a chemical species that is electron-deficient and seeks an electron-rich center. It accepts a pair of electrons to form a new covalent bond.
Characteristics:
1. They are electron-deficient.
2. They can be positively charged (cations) or neutral molecules with an incomplete octet or a highly polarized bond.
3. They are attacked by nucleophiles.
Examples:
- Cationic Electrophiles (Positively charged):
  - Hydronium ion ( $\ ext{H}_3\ ext{O}^+$ ) or a proton ( $\ ext{H}^+$ )
  - Carbocations ( $^+\text{R}_3\text{C}$ , e.g., $\text{CH}_3^+$ )
  - Nitronium ion ( $^+\text{NO}_2$ )
  - Halonium ions ( $\text{Br}^+, \text{Cl}^+$ )
- Neutral Electrophiles: These have an atom with an incomplete octet or a partial positive charge.
  - Boron trifluoride ( $\ ext{BF}_3$ ): Boron has only 6 valence electrons.
  - Aluminium chloride ( $\ ext{AlCl}_3$ ): Aluminium has an incomplete octet.
  - Carbonyl compounds (e.g., Aldehydes, Ketones): The carbonyl carbon has a partial positive charge ( $\ ext{C}^{\delta+}=\ ext{O}^{\delta-}$ ).
  - Alkyl halides (R-X): The carbon attached to the halogen has a partial positive charge ( $\ ext{C}^{\delta+}-\ ext{X}^{\delta-}$ ).

Final Answer: A nucleophile is an electron-rich species that donates an electron pair (e.g.,

^-\text{OH}, :\ ext{NH}_3

). An electrophile is an electron-deficient species that accepts an electron pair (e.g.,

\ ext{H}^+, \ ext{BF}_3

).

Q13EXERCISES

Identify the reagents shown in bold in the following equations as nucleophiles or electrophiles:

(a)

\mathrm{CH}_{3} \mathrm{COOH}+\mathbf{H O}^{-}\rightarrow \mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{H}_{2} \mathrm{O}

(b)

\mathrm{CH}_{3} \mathrm{COCH}_{3}+\overline{\mathbf{C N}} \rightarrow\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}(\mathrm{CN})(\mathrm{O}^{-})

Corrected product

(c)

\mathrm{C}_{6} \mathrm{H}_{6}+\mathbf{C H}_{3} \stackrel{+}{\mathbf{C}} \mathbf{O} \rightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COCH}_{3} + H^+

Solution

To Identify: Whether the bolded reagents are nucleophiles or electrophiles.

Solution:

(a) $\mathrm{CH}_{3} \mathrm{COOH}+\mathbf{H O}^{-}\rightarrow \mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{H}_{2} \mathrm{O}$

The reagent is the hydroxide ion ( $\text{HO}^-$ ).
It is a negatively charged species with lone pairs of electrons on the oxygen atom.
In this acid-base reaction, it donates an electron pair to the acidic proton ( $\ ext{H}^+$ ) of the carboxylic acid.
Since it is an electron-pair donor, $\text{HO}^-$ is a nucleophile.

(b) $\mathrm{CH}_{3} \mathrm{COCH}_{3}+\overline{\mathbf{C N}} \rightarrow\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}(\mathrm{CN})(\mathrm{O}^{-})$

The reagent is the cyanide ion ( $^-\text{CN}$ ).
It is a negatively charged ion with a lone pair on the carbon atom.
It attacks the electrophilic carbonyl carbon of acetone, donating its electron pair to form a new C-C bond.
Since it is an electron-pair donor, $^-\text{CN}$ is a nucleophile.

(c) $\mathrm{C}_{6} \mathrm{H}_{6}+\mathbf{C H}_{3} \stackrel{+}{\mathbf{C}} \mathbf{O} \rightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COCH}_{3} + H^+$

The reagent is the acylium ion ( $\text{CH}_3\text{C}^+=\text{O}$ ).
It is a positively charged species (cation), making it electron-deficient.
It accepts a pair of $\pi$ -electrons from the electron-rich benzene ring to form a new C-C bond.
Since it is an electron-pair acceptor, $\text{CH}_3\text{C}^+=\text{O}$ is an electrophile.

Final Answer: (a) $\text{HO}^-$ : Nucleophile (b) $^-\text{CN}$ : Nucleophile (c) $\text{CH}_3\text{C}^+=\text{O}$ : Electrophile

Q14EXERCISES

Classify the following reactions in one of the reaction type studied in this unit.

(a)

\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}+\mathrm{HS}^{-}\rightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{SH}+\mathrm{Br}^{-}

(b)

\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CH}_{2}+\mathrm{HCl} \rightarrow\left(\mathrm{CH}_{3}\right)_{2} \mathrm{ClC}-\mathrm{CH}_{3}

(c)

\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}+\mathrm{HO}^{-}\rightarrow \mathrm{CH}_{2}=\mathrm{CH}_{2}+\mathrm{H}_{2} \mathrm{O}+\mathrm{Br}^{-}

(d)

\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}_{2} \mathrm{OH}+\mathrm{HBr} \rightarrow\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CBrCH}_{2} \mathrm{CH}_{3}+\mathrm{H}_{2} \mathrm{O}

Solution

To Classify: The given organic reactions.

The main types of reactions are Substitution, Addition, Elimination, and Rearrangement.

Solution:

(a) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}+\mathrm{HS}^{-}\rightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{SH}+\mathrm{Br}^{-}$

In this reaction, the bromide atom (-Br) on the ethyl group is replaced by the hydrosulfide group (-SH).
The nucleophile $\ ext{HS}^-$ attacks the ethyl bromide, and the leaving group $\ ext{Br}^-$ departs.
This is a nucleophilic substitution reaction.

(b) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CH}_{2}+\mathrm{HCl} \rightarrow\left(\mathrm{CH}_{3}\right)_{2} \mathrm{ClC}-\mathrm{CH}_{3}$

The starting material is an alkene (isobutylene), which has a C=C double bond.
The reagent HCl adds across the double bond. The $\pi$ bond breaks, and new single bonds to H and Cl are formed.
The atoms of the reagent are added to the substrate.
This is an electrophilic addition reaction.

(c) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}+\mathrm{HO}^{-}\rightarrow \mathrm{CH}_{2}=\mathrm{CH}_{2}+\mathrm{H}_{2} \mathrm{O}+\mathrm{Br}^{-}$

The starting material is an alkyl halide (ethyl bromide).
A hydrogen atom from one carbon and a bromine atom from the adjacent carbon are removed, forming a C=C double bond (ethene).
Small molecules ( $\ ext{H}_2\ ext{O}$ and $\ ext{Br}^-$ ) are eliminated from the substrate.
This is an elimination reaction (specifically, a beta-elimination).

(d) $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}_{2} \mathrm{OH}+\mathrm{HBr} \rightarrow\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CBrCH}_{2} \mathrm{CH}_{3}+\mathrm{H}_{2} \mathrm{O}$

The starting material is neopentyl alcohol, which has 5 carbons.
The product is 2-bromo-2-methylbutane, which also has 5 carbons, but the carbon skeleton has changed.
Initially, the -OH group is protonated and leaves as water, forming a primary carbocation: $(\ ext{CH}_3)_3\ ext{C}-\ ext{CH}_2^+$ .
This primary carbocation is unstable and undergoes a 1,2-methyl shift to form a more stable tertiary carbocation: $(\ ext{CH}_3)_2\ ext{C}^+-\ ext{CH}_2\ ext{CH}_3$ .
The bromide ion then attacks this rearranged cation.
Because the carbon skeleton has been reorganized, this is a rearrangement reaction (specifically, a substitution reaction involving a carbocation rearrangement).

Final Answer: (a) Nucleophilic Substitution (b) Electrophilic Addition (c) Elimination (d) Rearrangement (within a substitution reaction)

Q15EXERCISES

What is the relationship between the members of following pairs of structures ? Are they structural or geometrical isomers or resonance contributors?

(a)

Pentan-3-one and Pentan-2-one structures

(b)

Two structures of 1,2-dideuterioethene

(c)

Two charged structures of formic acid

Solution

To Find: The relationship between the members of each pair of structures.

Definitions:

Structural Isomers: Compounds with the same molecular formula but different connectivity of atoms. Types include chain, position, and functional isomers.
Geometrical Isomers (cis-trans isomers): Stereoisomers that differ in the spatial arrangement of groups around a double bond or a ring.
Resonance Contributors: Different Lewis structures for the same molecule that differ only in the placement of electrons, not atoms. They are not real, separate molecules.

Solution:

(a) Pentan-3-one ( $\ ext{CH}_3\ ext{CH}_2\ ext{CO}\ ext{CH}_2\ ext{CH}_3$ ) and Pentan-2-one ( $\ ext{CH}_3\ ext{CO}\ ext{CH}_2\ ext{CH}_2\ ext{CH}_3$ )

Molecular Formula: Both have the formula $\ ext{C}_5\ ext{H}_{10}\ ext{O}$ .
Connectivity: In pentan-3-one, the carbonyl group (C=O) is at position 3. In pentan-2-one, it is at position 2.
Relationship: They have the same carbon skeleton and the same functional group, but the position of the functional group is different. Therefore, they are position isomers, which are a type of structural isomer.

(b) cis-1,2-dideuterioethene and trans-1,2-dideuterioethene

Molecular Formula: Both are $\ ext{C}_2\ ext{H}_2\ ext{D}_2$ (where D is deuterium, an isotope of hydrogen).
Connectivity: In both molecules, the atoms are connected in the same sequence: D-CH=CH-D.
Spatial Arrangement: They differ in the arrangement of the D and H atoms across the C=C double bond. In the cis isomer, the two D atoms are on the same side. In the trans isomer, they are on opposite sides.
Relationship: Since they differ only in the spatial arrangement of atoms around a double bond, they are geometrical isomers (a type of stereoisomer).

(c) $\ ext{H-C(=O)OH} \longleftrightarrow \ ext{H-C(O⁻)=O⁺H}$

These two structures represent formic acid.
Connectivity: The atoms (H, C, O, O, H) are connected in the same sequence in both structures.
Electron Placement: They differ only in the arrangement of electrons. The first is the neutral Lewis structure. The second is a charge-separated structure formed by moving a lone pair from the -OH oxygen to form a C=O bond and moving the original C=O pi electrons to the other oxygen.
Relationship: They are not different molecules or isomers. They are resonance contributors (or resonance structures) of the same molecule, formic acid.

Final Answer: (a) Structural isomers (specifically, position isomers). (b) Geometrical isomers. (c) Resonance contributors.

Q16EXERCISES

For the following bond cleavages, use curved-arrows to show the electron flow and classify each as homolysis or heterolysis. Identify reactive intermediate produced as free radical, carbocation and carbanion.

(a)

\mathrm{CH}_{3} \mathrm{O}-\mathrm{OCH}_{3} \rightarrow \mathrm{CH}_{3} \dot{\mathrm{O}}+\dot{\mathrm{O}} \mathrm{CH}_{3}

(b)

>\mathrm{C}=\mathrm{O} + \mathrm{OH}^- \rightarrow >\mathrm{C(OH)O^-}

(c)

>\mathrm{C-Br} \rightarrow >\mathrm{C}^+ + \mathrm{Br}^-

(d)

>\mathrm{C=C}< + \mathrm{E}^+ \rightarrow >\mathrm{E-C-C}^+<

Solution

To Do: For each reaction, show electron flow with curved arrows, classify the cleavage, and identify the intermediate.

Definitions:

Homolysis (Homolytic Cleavage): Symmetrical bond breaking where each atom gets one electron from the bond, forming free radicals. Shown with single-barbed (fish-hook) arrows.
Heterolysis (Heterolytic Cleavage): Unsymmetrical bond breaking where one atom gets both electrons from the bond, forming ions (cation and anion). Shown with double-barbed arrows.

Solution:

(a) $\mathrm{CH}_{3} \mathrm{O}-\mathrm{OCH}_{3} \rightarrow \mathrm{CH}_{3} \dot{\mathrm{O}}+\dot{\mathrm{O}} \mathrm{CH}_{3}$

Electron Flow: The O-O bond breaks, with one electron going to each oxygen atom. $\mathrm{CH}_{3} \mathrm{O} \stackrel{\curveharpoonleft \!\!\! \curveharpoonright}{\text{-----}} \mathrm{OCH}_{3} \rightarrow \mathrm{CH}_{3} \dot{\mathrm{O}} + \dot{\mathrm{O}} \mathrm{CH}_{3}$
Classification: Since the bond breaks symmetrically, this is homolysis.
Reactive Intermediate: The products, $\ ext{CH}_3\ ext{O•}$ , are neutral species with an unpaired electron. They are free radicals (specifically, methoxy radicals).

(b) $>\mathrm{C}=\mathrm{O} + \mathrm{OH}^- \rightarrow >\mathrm{C(OH)O^-}$

Electron Flow: The nucleophile $\ ext{OH}^-$ attacks the electrophilic carbonyl carbon. An electron pair from the hydroxide ion forms a new C-O bond. To maintain carbon's valency, the $\pi$ -bond of the C=O group breaks, and the electron pair moves onto the oxygen atom. $>\mathrm{C}=\mathrm{O} \curvearrowleft :\mathrm{OH}^- \rightarrow >\mathrm{C(OH)O^-}$
Classification: This is not a bond cleavage, but a bond formation followed by electron redistribution. The process is inherently ionic. It is associated with heterolytic processes.
Reactive Intermediate: The product, an alkoxide ion, is a charged intermediate.

(c) $>\mathrm{C-Br} \rightarrow >\mathrm{C}^+ + \mathrm{Br}^-$

Electron Flow: The C-Br bond breaks, and the more electronegative bromine atom takes both electrons from the bond. $>\mathrm{C} \stackrel{\curvearrowleft}{\text{-----}} \mathrm{Br} \rightarrow >\mathrm{C}^+ + :\mathrm{Br}^-$
Classification: The bond breaks asymmetrically to form ions. This is heterolysis.
Reactive Intermediate: The species $>\ ext{C}^+$ is a positively charged carbon species. It is a carbocation.

(d) $>\mathrm{C=C}< + \mathrm{E}^+ \rightarrow >\mathrm{E-C-C}^+<$

Electron Flow: The electron-rich $\pi$ -bond of the alkene attacks the electrophile $\ ext{E}^+$ . The electron pair from the $\pi$ -bond forms a new C-E bond, leaving the other carbon of the original double bond with a positive charge. $>\mathrm{C=C}< \curvearrowright \mathrm{E}^+ \rightarrow >\mathrm{E-C-C}^+<$
Classification: This is an attack by a $\pi$ -bond on an electrophile, an ionic process characteristic of heterolytic reactions.
Reactive Intermediate: The product is a carbocation.

Final Answer: (a) Homolysis, forming free radicals. (b) Heterolytic process (bond formation), forming an anionic intermediate. (c) Heterolysis, forming a carbocation. (d) Heterolytic process (attack on electrophile), forming a carbocation.

Q17EXERCISES

Explain the terms Inductive and Electromeric effects. Which electron displacement effect explains the following correct orders of acidity of the carboxylic acids?

(a)

\mathrm{Cl}_{3} \mathrm{CCOOH}>\mathrm{Cl}_{2} \mathrm{CHCOOH}>\mathrm{ClCH}_{2} \mathrm{COOH}

(b)

\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}>\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCOOH}>\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C} . \mathrm{COOH}

Solution

To Explain: Inductive and Electromeric effects, and apply them to explain acidity orders.

Definitions:

Inductive Effect (I Effect):
- It is the permanent polarization of a sigma ( $\sigma$ ) bond due to the presence of an adjacent polar bond in a molecule.
- It arises from the difference in electronegativity between atoms.
- The effect is transmitted through the carbon chain but weakens rapidly with distance, becoming negligible after 2-3 bonds.
- Groups that withdraw electrons (e.g., -Cl, -NO₂) have a negative inductive effect (-I effect). Groups that donate electrons (e.g., -CH₃) have a positive inductive effect (+I effect).
Electromeric Effect (E Effect):
- It is a temporary effect that occurs in molecules containing a multiple bond (double or triple bond) in the presence of an attacking reagent.
- It involves the complete transfer of a shared pair of pi ( $\pi$ ) electrons to one of the atoms joined by the multiple bond.
- The effect disappears as soon as the attacking reagent is removed.
- It is designated as +E when the $\pi$ -electrons move towards the attacking reagent and -E when they move away from it.

Acidity of Carboxylic Acids: The acidity of a carboxylic acid (R-COOH) depends on the stability of its conjugate base, the carboxylate ion (R-COO⁻). Any factor that stabilizes the carboxylate ion by dispersing its negative charge will increase the acidity of the parent acid.

Explanation of Acidity Orders:

(a) $\mathrm{Cl}_{3} \mathrm{CCOOH}>\mathrm{Cl}_{2} \mathrm{CHCOOH}>\mathrm{ClCH}_{2} \mathrm{COOH}$

Effect: This trend is explained by the Inductive Effect (-I effect).
Reasoning:
- The chlorine atom is an electron-withdrawing group. It exerts a -I effect, pulling electron density away from the carboxylate group in the conjugate base.
- This withdrawal of electrons disperses the negative charge on the carboxylate ion, stabilizing it.
- In $\ ext{Cl}_3\ ext{CCOO}^-$ , there are three chlorine atoms pulling electron density, leading to the greatest stabilization.
- In $\ ext{Cl}_2\ ext{CHCOO}^-$ , there are two chlorine atoms, resulting in less stabilization.
- In $\ ext{ClCH}_2\ ext{COO}^-$ , there is only one chlorine atom, providing the least stabilization among the three.
- Greater stabilization of the conjugate base corresponds to a stronger acid. Therefore, the acidity decreases as the number of electron-withdrawing Cl atoms decreases.

(b) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}>\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCOOH}>\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C} . \mathrm{COOH}$

Correction: The provided acidity order is incorrect. The actual order is the reverse. Let's explain the correct order: $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH} < \left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCOOH} < \left(\mathrm{CH}_{3}\right)_{3} \mathrm{C} . \mathrm{COOH}$ is wrong. The correct order is based on electron-donating groups destabilizing the conjugate base. Let's assume the question meant to ask why acidity decreases in the series: Propanoic acid, 2-Methylpropanoic acid, 2,2-Dimethylpropanoic acid. The actual acidity order is: $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH} > \left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCOOH} > \left(\mathrm{CH}_{3}\right)_{3} \mathrm{C} . \mathrm{COOH}$ .
Effect: This trend is explained by the Inductive Effect (+I effect).
Reasoning:
- Alkyl groups (like -CH₃) are electron-donating groups. They exert a +I effect, pushing electron density towards the carboxylate group.
- This intensifies the negative charge on the carboxylate ion, destabilizing it.
- The destabilizing +I effect increases with the number and size of the alkyl groups attached near the carboxyl group.
- The tert-butyl group in $(\ ext{CH}_3)_3\ ext{C-COOH}$ has the strongest +I effect, causing the most destabilization of the conjugate base, making it the weakest acid.
- The isopropyl group in $(\ ext{CH}_3)_2\ ext{CH-COOH}$ has a moderate +I effect.
- The ethyl group in $\ ext{CH}_3\ ext{CH}_2\ ext{COOH}$ has the weakest +I effect of the three, leading to the most stable conjugate base and making it the strongest acid in this series.

Final Answer:

Inductive Effect: A permanent polarization of $\sigma$ -bonds due to electronegativity differences.
Electromeric Effect: A temporary shift of $\pi$ -electrons in the presence of a reagent.
The acidity orders in both (a) and (b) are explained by the Inductive Effect. In (a), the electron-withdrawing -I effect of Cl atoms stabilizes the conjugate base, increasing acidity. In (b), the electron-donating +I effect of alkyl groups destabilizes the conjugate base, decreasing acidity.

Q18EXERCISES

Give a brief description of the principles of the following techniques taking an example in each case.

(a)

Crystallisation

(b)

Distillation

(c)

Chromatography

Solution

To Describe: The principles of crystallisation, distillation, and chromatography with examples.

Solution:

(a) Crystallisation

Principle: Crystallisation is a purification technique for solid compounds. Its principle is based on the difference in solubility of the desired compound and its impurities in a suitable solvent. Typically, the compound is highly soluble in the solvent at a high temperature but sparingly soluble at a low temperature, while the impurities are either insoluble or remain soluble at low temperatures.
Procedure: The impure solid is dissolved in a minimum amount of a suitable hot solvent to form a saturated solution. The hot solution is then filtered (if there are insoluble impurities) and allowed to cool slowly. As the solution cools, the solubility of the desired compound decreases, and it crystallizes out in a pure form, leaving the soluble impurities behind in the solution (mother liquor). The pure crystals are then separated by filtration.
Example: Purification of impure copper sulphate ( $\ ext{CuSO}_4$ ). Impure $\ ext{CuSO}_4$ is dissolved in hot water. The solution is cooled, and pure blue crystals of $\ ext{CuSO}_4 \cdot 5\ ext{H}_2\ ext{O}$ form, which are then filtered off.

(b) Distillation

Principle: Distillation is a technique used to separate components of a liquid mixture. Its principle is based on the difference in the boiling points of the components. The liquid with the lower boiling point is more volatile and will vaporize first when the mixture is heated.
Procedure: The liquid mixture is heated in a distillation flask. The component with the lower boiling point turns into vapor. This vapor is passed through a condenser, where it is cooled and converted back into a pure liquid (the distillate), which is collected in a separate receiver. The less volatile component remains in the distillation flask.
Example: Separation of a mixture of chloroform (boiling point 61°C) and aniline (boiling point 184°C). When the mixture is heated, chloroform vaporizes first, is condensed, and collected as the pure distillate, leaving aniline behind.

(c) Chromatography

Principle: Chromatography is a powerful separation technique based on the differential distribution of the components of a mixture between two phases: a stationary phase (a solid or liquid) and a mobile phase (a liquid or gas) that moves over or through the stationary phase.
Procedure: The mixture is applied to the stationary phase. The mobile phase is then passed through the system. Components that interact more strongly with the stationary phase (e.g., through adsorption or partitioning) move more slowly, while components that are more soluble in or interact more with the mobile phase move faster. This difference in movement speed results in the separation of the components.
Example: Separation of pigments from a plant leaf extract using paper chromatography. A spot of the extract is placed on chromatography paper (stationary phase). A solvent like ethanol (mobile phase) is allowed to move up the paper. Different pigments (chlorophylls, carotenoids) travel at different rates depending on their solubility in the solvent and adsorption to the paper, resulting in separate colored bands.

Q19EXERCISES

Describe the method, which can be used to separate two compounds with different solubilities in a solvent S.

Solution

To Describe: A method to separate two compounds with different solubilities in a solvent.

Method: The most appropriate method to separate two solid compounds with different solubilities in a solvent S is Fractional Crystallisation. This method can be applied in a few ways depending on the nature of the solubilities.

Principle: The method relies on the difference in the extent of solubility of the two compounds in the same solvent. Often, the solubility of both compounds changes with temperature, and this property is exploited.

Procedure (Assuming one compound is much more soluble than the other):

Dissolution: Take the mixture of the two compounds and add the solvent S. Stir the mixture well. The more soluble compound will dissolve completely (or to a large extent), while the less soluble compound will remain as a solid residue.
Filtration: Filter the mixture. The insoluble (or less soluble) compound will be collected on the filter paper as the residue. The more soluble compound will be in the filtrate (the liquid that passes through the filter).
Recovery of the Less Soluble Compound: Wash the residue on the filter paper with a small amount of fresh cold solvent S to remove any remaining dissolved impurities. Then, dry the residue to obtain the pure, less soluble compound.
Recovery of the More Soluble Compound: Take the filtrate and evaporate the solvent (e.g., by heating gently). As the solvent is removed, the dissolved compound will crystallize out. Once most of the solvent is gone, the crystals can be collected and dried to obtain the pure, more soluble compound.

Procedure (Assuming both compounds are soluble, but to different extents, and solubility increases with temperature - Fractional Crystallisation):

Prepare a Saturated Solution: Dissolve the mixture in a minimum amount of hot solvent S to create a saturated solution.
Cooling: Allow the hot solution to cool slowly. The compound which is less soluble (or whose solubility decreases more sharply with temperature) will crystallize out first.
First Filtration: Filter the solution to separate the crystals of the less soluble compound (Crop 1). The filtrate will still contain the more soluble compound and some of the less soluble compound.
Concentration and Further Cooling: Take the filtrate, evaporate some of the solvent to concentrate it, and cool again. This will cause the more soluble compound to crystallize out (Crop 2).
Purification: The crops of crystals obtained may not be perfectly pure. The process of re-dissolving in fresh solvent and re-crystallizing can be repeated to improve the purity of each compound.

Final Answer: A mixture of two compounds with different solubilities in a solvent S can be separated by filtration if one is soluble and the other is not. If both are soluble but to different extents, fractional crystallisation is used, which involves dissolving the mixture in a hot solvent and then cooling it to allow the less soluble component to crystallize out first.

Q20EXERCISES

What is the difference between distillation, distillation under reduced pressure and steam distillation?

Solution

To Differentiate: Between simple distillation, distillation under reduced pressure, and steam distillation.

Solution: All three are purification techniques for liquids based on vaporization and condensation, but they differ in their principles, applications, and operating conditions.

1. Simple Distillation

Principle: Based on the large difference in the boiling points of the components of a liquid mixture. The more volatile liquid (lower boiling point) vaporizes first.
Application:
- To separate a volatile liquid from a non-volatile solid impurity (e.g., separating salt from water).
- To separate two liquids with a significant difference in boiling points (typically > 25-30°C), e.g., separating acetone (b.p. 56°C) from water (b.p. 100°C).
Condition: The liquid is heated to its normal boiling point at atmospheric pressure.

2. Distillation Under Reduced Pressure (Vacuum Distillation)

Principle: A liquid boils at a temperature at which its vapor pressure equals the external pressure. By reducing the external pressure (creating a partial vacuum), the boiling point of the liquid is lowered.
Application:
- To purify liquids that have very high boiling points and are difficult to heat to that temperature.
- To purify liquids that are thermally unstable and decompose at or below their normal boiling points.
Example: Purification of glycerol (normal b.p. 290°C, decomposes near this temperature). Under reduced pressure (e.g., 12 mm Hg), it boils at 180°C without decomposition.
Condition: The distillation is carried out at a pressure lower than atmospheric pressure, using a vacuum pump.

3. Steam Distillation

Principle: Based on Dalton's law of partial pressures. A mixture of two immiscible liquids boils when the sum of their individual vapor pressures equals the external (atmospheric) pressure ( $P_{total} = p_A + p_B$ ). This allows the mixture to boil at a temperature lower than the boiling point of either individual component.
Application:
- To purify organic compounds that are immiscible with water and are steam volatile (have an appreciable vapor pressure near the boiling point of water).
- To purify substances that might decompose at their normal boiling point.
Example: Extraction of essential oils like eucalyptus oil from plant materials, or purification of aniline from a reaction mixture.
Condition: Hot steam is passed through the impure compound. The compound co-distills with the steam at a temperature slightly below 100°C.

Summary of Differences:

Feature	Simple Distillation	Distillation Under Reduced Pressure	Steam Distillation
Principle	Difference in boiling points	Lowering boiling point by reducing pressure	Co-distillation of immiscible liquids
Substance Type	Thermally stable liquids	High-boiling or heat-sensitive liquids	Water-immiscible, steam-volatile liquids
Pressure	Atmospheric pressure	Reduced (vacuum) pressure	Atmospheric pressure
Boiling Temp.	At or above normal boiling point	Below normal boiling point	Below 100°C (and below individual b.p.)
Example	Acetone-water separation	Glycerol purification	Aniline purification

Final Answer: The key differences lie in their underlying principles and applications: simple distillation separates liquids with different boiling points at atmospheric pressure; vacuum distillation purifies heat-sensitive or very high-boiling liquids by lowering their boiling points via reduced pressure; and steam distillation purifies water-immiscible, volatile compounds by co-distilling them with steam at a temperature below 100°C.

Q21EXERCISES

Discuss the chemistry of Lassaigne's test.

Solution

To Discuss: The chemistry involved in Lassaigne's test.

Principle of Lassaigne's Test: Lassaigne's test is a qualitative analysis method used to detect the presence of extra elements like nitrogen (N), sulfur (S), and halogens (X = Cl, Br, I) in an organic compound. The core principle is to convert these elements, which are covalently bonded in the organic molecule, into water-soluble ionic salts. This is achieved by fusing the organic compound with a highly reactive metal, typically sodium.

Procedure Overview: A small piece of dry sodium metal is heated in a fusion tube until it melts. A small amount of the organic compound is added, and the mixture is heated strongly until red hot. The hot tube is then plunged into a beaker of distilled water, causing the tube to shatter and the fused mass to dissolve. This solution is boiled, cooled, and filtered. The resulting clear filtrate is called the sodium fusion extract or Lassaigne's extract, which is then used for specific tests.

Chemistry of the Fusion Process: During the high-temperature fusion with sodium, the following reactions occur if the respective elements are present:

For Nitrogen: Carbon and nitrogen from the organic compound react with sodium to form sodium cyanide. $\text{Na} + \text{C} + \text{N} \xrightarrow{\Delta} \text{NaCN}$ (Sodium cyanide)
For Sulfur: Sulfur from the organic compound reacts with sodium to form sodium sulfide. $2\text{Na} + \text{S} \xrightarrow{\Delta} \text{Na}_2\text{S}$ (Sodium sulfide)
For Halogens: A halogen (X) from the organic compound reacts with sodium to form a sodium halide. $\text{Na} + \text{X} \xrightarrow{\Delta} \text{NaX}$ (where X = Cl, Br, I)
If both N and S are present: They may react to form sodium thiocyanate. $\text{Na} + \text{C} + \text{N} + \text{S} \xrightarrow{\Delta} \text{NaSCN}$ (Sodium thiocyanate)

Chemistry of the Detection Tests (using the extract):

Test for Nitrogen:
- A portion of the extract is made alkaline with NaOH, and freshly prepared ferrous sulfate ( $\ ext{FeSO}_4$ ) solution is added. This forms sodium hexacyanoferrate(II). $\text{Fe}^{2+} + 6\text{CN}^- \rightarrow [\text{Fe}(\text{CN})_6]^{4-}$
- The solution is then acidified with concentrated sulfuric acid. Some of the Fe²⁺ gets oxidized to Fe³⁺ by the acid (or air).
- The Fe³⁺ ions react with the hexacyanoferrate(II) complex to form a deep blue precipitate called Prussian blue (hydrated iron(III) hexacyanoferrate(II)). $4\text{Fe}^{3+} + 3[\text{Fe}(\text{CN})_6]^{4-} \rightarrow \text{Fe}_4[\text{Fe}(\text{CN})_6]_3 \cdot x\text{H}_2\text{O}$ The formation of Prussian blue confirms the presence of nitrogen.
Test for Sulfur:
- (a) Lead Acetate Test: The extract is acidified with acetic acid, and lead acetate solution is added. A black precipitate of lead(II) sulfide confirms sulfur. $\text{S}^{2-} + \text{Pb}^{2+} \rightarrow \text{PbS}_{\text{(s)}}$ (Black ppt)
- (b) Sodium Nitroprusside Test: A few drops of sodium nitroprusside solution are added to a fresh portion of the extract. An intense violet or purple coloration confirms sulfur. $\text{S}^{2-} + [\text{Fe}(\text{CN})_5\text{NO}]^{2-} \rightarrow [\text{Fe}(\text{CN})_5\text{NOS}]^{4-}$ (Violet complex)
Test for Halogens:
- The extract is first acidified with dilute nitric acid ( $\ ext{HNO}_3$ ) to decompose any cyanide or sulfide ions present, which would otherwise interfere by precipitating with silver nitrate. $\text{NaCN} + \text{HNO}_3 \rightarrow \text{NaNO}_3 + \text{HCN}_{\text{(g)}}$ $\text{Na}_2\text{S} + 2\text{HNO}_3 \rightarrow 2\text{NaNO}_3 + \text{H}_2\text{S}_{\text{(g)}}$
- Silver nitrate ( $\ ext{AgNO}_3$ ) solution is then added. $\text{X}^- + \text{Ag}^+ \rightarrow \text{AgX}_{\text{(s)}}$
- Observations:
  - A white precipitate (AgCl), soluble in ammonium hydroxide ( $\ ext{NH}_4\ ext{OH}$ ), confirms Chlorine.
  - A pale yellow precipitate (AgBr), sparingly soluble in $\ ext{NH}_4\ ext{OH}$ , confirms Bromine.
  - A yellow precipitate (AgI), insoluble in $\ ext{NH}_4\ ext{OH}$ , confirms Iodine.

Final Answer: The chemistry of Lassaigne's test involves the fusion of an organic compound with sodium metal to convert covalently bonded N, S, and halogens into ionic species (NaCN, Na₂S, NaX). These ions are then identified in the aqueous sodium fusion extract using specific chemical tests, such as the formation of Prussian blue for nitrogen, lead sulfide for sulfur, and silver halides for halogens.

Q22EXERCISES

Differentiate between the principle of estimation of nitrogen in an organic compound by (i) Dumas method and (ii) Kjeldahl's method.

Solution

To Differentiate: The principles of the Dumas and Kjeldahl methods for nitrogen estimation.

Solution:

Feature	Dumas Method	Kjeldahl's Method
Principle	The method is based on the oxidation of the organic compound. The nitrogen present is converted into elemental nitrogen gas (N₂).	The method is based on the digestion of the organic compound with hot concentrated sulfuric acid. The nitrogen is converted into ammonia (NH₃).
Chemical Conversion	A known mass of the compound is heated with excess copper(II) oxide (CuO) in a CO₂ atmosphere. Carbon and hydrogen are oxidized to CO₂ and H₂O, while nitrogen is liberated as N₂ gas.	A known mass of the compound is heated with concentrated H₂SO₄. The nitrogen is converted to ammonium sulfate, (NH₄)₂SO₄. This is then treated with excess NaOH to liberate NH₃ gas.
Measurement	The volume of the collected N₂ gas is measured over a KOH solution (which absorbs other gases like CO₂). This volume is then converted to mass using the ideal gas law at STP.	The liberated NH₃ gas is absorbed in a known excess volume of standard acid. The amount of unreacted acid is determined by back-titration with a standard base. The amount of NH₃, and thus nitrogen, is calculated from the amount of acid consumed.
Applicability	It is a general method and can be used for all types of organic compounds containing nitrogen, including nitro compounds, azo compounds, and nitrogen in heterocyclic rings (like pyridine).	It is not applicable to compounds where nitrogen is linked to oxygen (nitro compounds) or to another nitrogen (azo compounds), or is present in a ring (e.g., pyridine), as these forms of nitrogen are not completely converted to ammonium sulfate under the reaction conditions.
Final Product for Analysis	Gaseous Dinitrogen (N₂)	Gaseous Ammonia (NH₃)

Summary of a Key Difference:

Dumas Method: Converts organic nitrogen into N₂ gas and measures its volume. It is a more universal method.

Organic Compound (C, H, N) $\xrightarrow{\text{CuO}, \Delta}$ N₂ (gas) + CO₂ + H₂O
Kjeldahl's Method: Converts organic nitrogen into ammonium sulfate, which is then converted to ammonia gas. The amount of ammonia is determined by titration. It is simpler for applicable compounds but has limitations.

Organic Compound (N) $\xrightarrow{\text{conc. H}_2\text{SO}_4, \Delta}$ (NH₄)₂SO₄ $\xrightarrow{\text{NaOH}}$ NH₃ (gas)

Final Answer: The fundamental difference lies in the chemical conversion of nitrogen: the Dumas method converts it to N₂ gas, whose volume is measured, making it applicable to all nitrogenous organic compounds. The Kjeldahl's method converts it to ammonia (NH₃), which is quantified by titration, but this method is not suitable for nitro, azo, or ring-nitrogen compounds.

Q23EXERCISES

Discuss the principle of estimation of halogens, sulphur and phosphorus present in an organic compound.

Solution

To Discuss: The principles of quantitative estimation for halogens, sulfur, and phosphorus.

Principle of Estimation (Carius Method) The general principle for the quantitative estimation of halogens, sulfur, and phosphorus in an organic compound is to convert the element of interest from its covalent form in the organic molecule into a stable, inorganic, ionic compound that can be easily precipitated, isolated, and weighed. This process is typically carried out by a strong oxidative digestion using the Carius method.

Procedure Overview (Carius Method): A known mass of the organic compound is heated in a sealed hard-glass tube (Carius tube) with a strong oxidizing agent. After the reaction, the contents are cooled, the tube is opened, and the resulting inorganic precipitate is filtered, washed, dried, and weighed accurately. From the mass of the precipitate, the mass and percentage of the element in the original compound are calculated.

Specific Principles for Each Element:

1. Estimation of Halogens (Cl, Br, I)

Principle: The organic compound is oxidized by heating with fuming nitric acid (HNO₃) in the presence of silver nitrate (AgNO₃).
Chemical Conversion:
- Carbon and hydrogen in the compound are oxidized to CO₂ and H₂O.
- The halogen (X) present is converted into its corresponding silver halide (AgX), which is insoluble and precipitates out. Organic Compound (containing X) $\xrightarrow{\text{fuming HNO}_3, \text{AgNO}_3, \Delta}$ AgX (precipitate)
Calculation: The precipitate of AgX is filtered, washed, dried, and weighed. From the known molar masses of AgX and the halogen X, the percentage of the halogen in the original sample is calculated. % of X = $\frac{\text{Atomic mass of X}}{\text{Molar mass of AgX}} \times \frac{\text{Mass of AgX formed}}{\text{Mass of organic compound}} \times 100$

2. Estimation of Sulphur

Principle: The organic compound is oxidized by heating with fuming nitric acid (HNO₃) or sodium peroxide.
Chemical Conversion:
- The sulfur in the compound is oxidized to sulfuric acid (H₂SO₄). Organic Compound (containing S) $\xrightarrow{\text{fuming HNO}_3, \Delta}$ H₂SO₄
- The resulting solution is then treated with an excess of barium chloride (BaCl₂) solution. This precipitates the sulfate ions as barium sulfate (BaSO₄), which is highly insoluble. H₂SO₄ + BaCl₂ $\rightarrow$ BaSO₄ (white precipitate) + 2HCl
Calculation: The precipitate of BaSO₄ is filtered, washed, dried, and weighed. From the known molar masses of BaSO₄ and sulfur, the percentage of sulfur is calculated. % of S = $\frac{\text{Atomic mass of S}}{\text{Molar mass of BaSO}_4} \times \frac{\text{Mass of BaSO}_4 \text{ formed}}{\text{Mass of organic compound}} \times 100$

3. Estimation of Phosphorus

Principle: The organic compound is oxidized by heating with fuming nitric acid (HNO₃).
Chemical Conversion:
- The phosphorus in the compound is oxidized to phosphoric acid (H₃PO₄). Organic Compound (containing P) $\xrightarrow{\text{fuming HNO}_3, \Delta}$ H₃PO₄
Precipitation and Calculation (Two common methods):
- (a) As Ammonium Phosphomolybdate: The phosphoric acid is treated with ammonia and ammonium molybdate solution. This forms a yellow precipitate of ammonium phosphomolybdate, (NH₄)₃PO₄·12MoO₃.
- (b) As Magnesium Pyrophosphate: The phosphoric acid is treated with magnesia mixture (a solution of MgCl₂, NH₄Cl, and NH₄OH). This precipitates phosphorus as magnesium ammonium phosphate (MgNH₄PO₄). This precipitate is then ignited (heated strongly) to convert it into magnesium pyrophosphate (Mg₂P₂O₇), which is weighed.
Calculation (using Mg₂P₂O₇): From the mass of Mg₂P₂O₇, the percentage of phosphorus is calculated. % of P = $\frac{2 \times \text{Atomic mass of P}}{\text{Molar mass of Mg}_2\text{P}_2\text{O}_7} \times \frac{\text{Mass of Mg}_2\text{P}_2\text{O}_7 \text{ formed}}{\text{Mass of organic compound}} \times 100$

Final Answer: The principle of estimation for halogens, sulfur, and phosphorus involves the oxidative decomposition of a known mass of an organic compound (Carius method) to convert the element into a stable inorganic ionic species. This species is then quantitatively precipitated (as AgX for halogens, BaSO₄ for sulfur, or Mg₂P₂O₇ for phosphorus), and the mass of the precipitate is used to calculate the percentage of the element in the original compound.

Q24EXERCISES

Explain the principle of paper chromatography.

Solution

To Explain: The principle of paper chromatography.

Principle: Paper chromatography is a type of partition chromatography. The fundamental principle is the differential partitioning (or distribution) of the components of a mixture between a stationary phase and a mobile phase.

The Two Phases:

Stationary Phase: In paper chromatography, the stationary phase is the water molecules trapped within the pores of the cellulose fibers of the chromatography paper. Although the paper itself (cellulose) is the solid support, the liquid water adsorbed onto it acts as the true stationary phase.
Mobile Phase: The mobile phase is a liquid solvent or a mixture of solvents that is immiscible with the stationary phase (water). This solvent moves up or down the paper due to capillary action.

Mechanism of Separation:

A spot of the mixture to be separated is applied to a starting line on the chromatography paper.
The edge of the paper is dipped into the mobile phase solvent in a sealed container.
As the mobile phase moves along the paper, it passes over the spot of the mixture.
At this point, each component of the mixture continuously partitions itself between the stationary water phase and the moving solvent phase. This is like a continuous series of extractions.
The separation occurs because different components have different affinities for the two phases:
- A component that is more soluble in the mobile phase and has less affinity for the stationary water phase will spend more time moving with the solvent and will travel further up the paper.
- A component that is less soluble in the mobile phase and has a greater affinity for the stationary water phase (e.g., is more polar and interacts strongly with water) will be held back more and will travel a shorter distance.
This difference in the rate of movement causes the components of the mixture to separate into distinct spots at different positions on the paper.

Retardation Factor (Rf value): The position of each separated component is characterized by its retardation factor (Rf), which is a constant for a given compound, stationary phase, and mobile phase under specific conditions.

R_f = \frac{\text{Distance travelled by the component from the origin}}{\text{Distance travelled by the solvent front from the origin}}

The Rf value is always less than 1 and helps in identifying the components by comparing them with the Rf values of known standards.

Final Answer: The principle of paper chromatography is partitioning. It separates the components of a mixture based on their different relative solubilities (or partition coefficients) between a stationary liquid phase (water trapped in the paper's cellulose fibers) and a mobile liquid phase (the developing solvent). Components more soluble in the mobile phase travel further, while those with a higher affinity for the stationary phase travel shorter distances, leading to separation.

Q25EXERCISES

Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens?

Solution

To Explain: The reason for adding nitric acid before the silver nitrate test for halogens.

Context: The test for halogens is performed on the sodium fusion extract, which is an aqueous solution containing the ionic forms of elements like nitrogen, sulfur, and halogens if they were present in the original organic compound. These ions are cyanide (CN⁻), sulfide (S²⁻), and halide (X⁻).

Problem: Both sodium cyanide (NaCN) and sodium sulfide (Na₂S) interfere with the silver nitrate test for halogens. If present, they would react with silver nitrate (AgNO₃) to form precipitates of silver cyanide (AgCN, white) and silver sulfide (Ag₂S, black), respectively.

$\mathrm{Ag}^+ + \mathrm{CN}^- \rightarrow \mathrm{AgCN}_{\text{(s)}}$ (White precipitate)
$2\mathrm{Ag}^+ + \mathrm{S}^{2-} \rightarrow \mathrm{Ag}_2\mathrm{S}_{\text{(s)}}$ (Black precipitate)

These precipitates would be mistaken for or would mask the precipitate of silver halide (AgX), leading to a false positive or an inconclusive result for the halogen test.

Solution (Role of Nitric Acid): To prevent this interference, dilute nitric acid (HNO₃) is added to the sodium fusion extract, and the solution is boiled before adding silver nitrate. Nitric acid serves two crucial purposes:

Decomposition of Interfering Ions: Nitric acid is an oxidizing acid that decomposes sodium cyanide and sodium sulfide into gaseous products, which then escape from the solution.
- For Cyanide: It reacts to form hydrogen cyanide gas. $\mathrm{NaCN} + \mathrm{HNO}_3 \rightarrow \mathrm{NaNO}_3 + \mathrm{HCN}_{\text{(g)}}$
- For Sulfide: It reacts to form hydrogen sulfide gas (which may be further oxidized by hot concentrated HNO₃). $\mathrm{Na}_2\mathrm{S} + 2\mathrm{HNO}_3 \rightarrow 2\mathrm{NaNO}_3 + \mathrm{H}_2\mathrm{S}_{\text{(g)}}$
Acidification of the Solution: The silver halide test must be performed in an acidic medium to prevent the precipitation of silver oxide (Ag₂O) or silver hydroxide (AgOH) if the solution were neutral or alkaline.

Why not use other acids like H₂SO₄ or HCl?

Sulfuric acid (H₂SO₄) cannot be used because silver sulfate (Ag₂SO₄) is sparingly soluble and might precipitate, causing interference.
Hydrochloric acid (HCl) cannot be used because it contains chloride ions (Cl⁻), which would react with AgNO₃ to give a white precipitate of AgCl, leading to a false positive test for chlorine.

Final Answer: Nitric acid is added to the sodium extract before testing for halogens to remove interfering cyanide (CN⁻) and sulfide (S²⁻) ions. These ions would otherwise form precipitates with silver nitrate and give a false positive result. The nitric acid decomposes them into gaseous products (HCN and H₂S) that are expelled upon boiling, ensuring that only halide ions, if present, will precipitate with the subsequent addition of silver nitrate.

Q26EXERCISES

Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens.

Solution

To Explain: The reason for fusing an organic compound with metallic sodium in Lassaigne's test.

Principle: In most organic compounds, elements such as nitrogen, sulfur, and halogens are present in a covalently bonded state. Covalent compounds are generally non-ionic, and these elements do not exist as free ions that can be detected by simple ionic precipitation or color reactions in an aqueous solution. For example, the chlorine in chloromethane (

\ ext{CH}_3\ ext{Cl}

) will not give a precipitate with silver nitrate because it is covalently bonded to carbon, not present as a free chloride ion (Cl⁻).

Reason for Sodium Fusion: The primary purpose of fusing the organic compound with metallic sodium is to convert these covalently bonded elements into water-soluble ionic salts. Sodium is a highly reactive alkali metal that acts as a powerful reducing agent at high temperatures.

The Chemical Transformation: During the intense heating (fusion) with sodium, the organic molecule breaks down, and the elements of interest react with sodium to form stable, inorganic, ionic compounds:

Nitrogen: If nitrogen is present, it combines with carbon (from the organic compound) and sodium to form sodium cyanide (NaCN). $\text{Organic Compound (C, N)} + \text{Na} \xrightarrow{\Delta} \text{Na}^+\text{CN}^-$
Sulfur: If sulfur is present, it reacts with sodium to form sodium sulfide (Na₂S). $\text{Organic Compound (S)} + \text{Na} \xrightarrow{\Delta} \text{Na}_2^+\text{S}^{2-}$
Halogens (X = Cl, Br, I): If a halogen is present, it reacts with sodium to form a sodium halide (NaX). $\text{Organic Compound (X)} + \text{Na} \xrightarrow{\Delta} \text{Na}^+\text{X}^-$

Outcome: After the fusion, these newly formed ionic salts (NaCN, Na₂S, NaX) are readily soluble in water. When the fused mass is treated with water, these salts dissolve to produce an aqueous solution containing free cyanide (CN⁻), sulfide (S²⁻), and halide (X⁻) ions.

This aqueous solution, known as the sodium fusion extract, can now be used for standard qualitative inorganic tests to detect the presence of these specific anions, which in turn confirms the presence of the corresponding elements in the original organic compound.

Final Answer: The fusion of an organic compound with metallic sodium is necessary because nitrogen, sulfur, and halogens are typically present in a covalent form, which is non-reactive in simple ionic tests. The high-temperature fusion with reactive sodium metal breaks these covalent bonds and converts the elements into water-soluble inorganic ionic salts (sodium cyanide, sodium sulfide, and sodium halides). The resulting free ions in the aqueous extract can then be easily identified using standard qualitative analysis methods.

Q27EXERCISES

Name a suitable technique of separation of the components from a mixture of calcium sulphate and camphor.

Solution

Given: A mixture of calcium sulphate (

\ ext{CaSO}_4

) and camphor.

To Find: A suitable technique for their separation.

Properties of the Components:

Camphor: It is a volatile solid organic compound. A key property of camphor is that it undergoes sublimation, meaning it can change directly from a solid to a vapor state upon heating, without passing through a liquid phase.
Calcium Sulphate ( $\ ext{CaSO}_4$ ): It is a non-volatile inorganic salt. It does not sublime upon heating and has a very high melting point.

Principle of Separation: Since one component (camphor) is sublimable and the other (calcium sulphate) is non-sublimable, the ideal separation technique is Sublimation.

Procedure:

The mixture of calcium sulphate and camphor is placed in a china dish or an evaporating dish.
The dish is covered with an inverted funnel, with the stem of the funnel plugged with cotton wool to prevent the camphor vapor from escaping.
The china dish is heated gently on a sand bath.
The camphor will sublime, turning into vapor.
The camphor vapor rises and comes into contact with the cooler inner surface of the inverted funnel.
Upon cooling, the vapor solidifies (deposits) back into pure solid camphor crystals on the funnel walls.
The non-volatile calcium sulphate remains behind in the china dish.
After the process is complete, the pure camphor can be scraped from the funnel.

Final Answer: The most suitable technique for separating a mixture of calcium sulphate and camphor is Sublimation.

Q28EXERCISES

Explain, why an organic liquid vaporises at a temperature below its boiling point in its steam distillation?

Solution

To Explain: Why an organic liquid vaporizes below its normal boiling point during steam distillation.

Principle of Boiling: A liquid boils when its vapor pressure becomes equal to the external pressure (usually atmospheric pressure).

Principle of Steam Distillation (Dalton's Law of Partial Pressures): Steam distillation is used for separating substances that are immiscible with water. When two immiscible liquids (like an organic compound and water) are heated together, each liquid exerts its own vapor pressure independently of the other. According to Dalton's Law of Partial Pressures, the total vapor pressure of the mixture is the sum of the individual vapor pressures of the components.

P_{\text{total}} = p_{\text{organic}} + p_{\text{water}}

Where:

$P_{\text{total}}$ is the total vapor pressure of the mixture.
$p_{\text{organic}}$ is the partial vapor pressure of the organic liquid.
$p_{\text{water}}$ is the partial vapor pressure of water (steam).

Explanation:

The mixture of the organic liquid and water will start to boil when the total vapor pressure ( $P_{\text{total}}$ ) equals the external atmospheric pressure ( $P_{\text{atm}}$ ). Boiling condition: $P_{\text{atm}} = p_{\text{organic}} + p_{\text{water}}$
Since the total vapor pressure is a sum of two components, this condition will be met at a temperature where both individual vapor pressures ( $p_{\text{organic}}$ and $p_{\text{water}}$ ) are less than the atmospheric pressure.
For any liquid, its vapor pressure is less than the atmospheric pressure at any temperature below its normal boiling point. For the organic liquid to boil alone, its vapor pressure ( $p_{\text{organic}}$ ) would need to reach $P_{\text{atm}}$ on its own, which only happens at its normal boiling point.
However, in the presence of steam, the organic liquid only needs to contribute a fraction ( $p_{\text{organic}}$ ) of the total pressure required for boiling. Water contributes the rest ( $p_{\text{water}}$ ). Therefore, the sum ( $p_{\text{organic}} + p_{\text{water}}$ ) reaches the atmospheric pressure at a temperature that is lower than the boiling point of either pure water (100°C) or the pure organic liquid.

Example: Suppose at 90°C, the vapor pressure of water is 526 mm Hg and the vapor pressure of an organic liquid is 234 mm Hg. The atmospheric pressure is 760 mm Hg.

Total vapor pressure of the mixture at 90°C = 526 + 234 = 760 mm Hg.
Since the total vapor pressure equals the atmospheric pressure, the mixture will boil at 90°C, even though the boiling point of pure water is 100°C and the boiling point of the organic liquid might be much higher (e.g., 150°C).

Final Answer: An organic liquid vaporizes at a temperature below its normal boiling point during steam distillation because the mixture boils when the sum of the partial vapor pressures of the organic liquid and water equals the external atmospheric pressure. Since the organic liquid only needs to contribute a portion of the total required pressure, this condition is met at a temperature lower than its individual boiling point.

Q29EXERCISES

Will $\mathrm{CCl}_{4}$ give white precipitate of AgCl on heating it with silver nitrate? Give reason for your answer.

Solution

To Determine: Whether carbon tetrachloride (

\ ext{CCl}_4

) will give a precipitate with silver nitrate.

Answer: No,

\ ext{CCl}_4

will not give a white precipitate of AgCl on heating it with silver nitrate.

Reason: The reaction to form a white precipitate of silver chloride (AgCl) requires the presence of free chloride ions (Cl⁻) in the solution.

\mathrm{Ag}^+_{\text{(aq)}} + \mathrm{Cl}^-_{\text{(aq)}} \rightarrow \mathrm{AgCl}_{\text{(s)}}

In carbon tetrachloride (

\ ext{CCl}_4

), the chlorine atoms are attached to the carbon atom by covalent bonds.

\ ext{CCl}_4

is a non-polar covalent molecule and does not ionize in solution to produce chloride ions (Cl⁻).

Since there are no free chloride ions available in the solution when

\ ext{CCl}_4

is mixed with silver nitrate, the precipitation reaction cannot occur.

For a compound to give this test, it must be an ionic compound containing chloride ions (like NaCl) or a covalent compound that can easily hydrolyze or react to produce chloride ions.

\ ext{CCl}_4

is very stable and resistant to hydrolysis under these conditions.

Final Answer: No.

\ ext{CCl}_4

will not give a white precipitate with silver nitrate because it is a covalent compound. The chlorine atoms are strongly bonded to the carbon atom and do not dissociate as free chloride ions (Cl⁻), which are necessary for the precipitation of AgCl.

Q30EXERCISES

Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound?

Solution

To Explain: The use of potassium hydroxide (KOH) to absorb CO₂ in the estimation of carbon.

Context: In the quantitative estimation of carbon in an organic compound (Liebig's method), a known mass of the compound is completely combusted in a stream of excess oxygen. The carbon present is converted into carbon dioxide (CO₂), and the hydrogen is converted into water (H₂O).

\mathrm{C}_x\mathrm{H}_y + (x + y/4)\mathrm{O}_2 \rightarrow x\mathrm{CO}_2 + (y/2)\mathrm{H}_2\mathrm{O}

The amounts of CO₂ and H₂O produced are then measured to determine the percentage of carbon and hydrogen.

Reason for using Potassium Hydroxide (KOH):

Chemical Reactivity: Carbon dioxide (CO₂) is an acidic oxide. Potassium hydroxide (KOH) is a strong base. They readily react in an acid-base neutralization reaction to form potassium carbonate (K₂CO₃) and water.

$2\mathrm{KOH}_{\text{(aq)}} + \mathrm{CO}_{2(\text{g})} \rightarrow \mathrm{K}_2\mathrm{CO}_3{\text{(aq)}} + \mathrm{H}_2\mathrm{O}_{\text{(l)}}$
Quantitative Absorption: This reaction is rapid and complete. A concentrated solution of KOH can efficiently and quantitatively absorb all the CO₂ gas that passes through it. This ensures that the mass of CO₂ produced from the combustion is accurately measured by weighing the U-tube containing the KOH solution before and after the experiment. The increase in mass of the U-tube corresponds directly to the mass of CO₂ absorbed.
Non-volatility: Potassium hydroxide and the product, potassium carbonate, are non-volatile solids. This means that they will not be lost as vapor during the experiment, which is crucial for accurate mass measurements.
Hygroscopic Nature: Concentrated KOH solution is also hygroscopic (absorbs water), but in the experimental setup, the combustion gases are first passed through a U-tube containing a drying agent (like anhydrous calcium chloride) to absorb the water produced. This ensures that the KOH tube only absorbs CO₂.

Final Answer: A solution of potassium hydroxide is used to absorb carbon dioxide because CO₂ is an acidic gas and KOH is a strong base. They undergo a rapid and complete neutralization reaction, allowing for the quantitative absorption of all the CO₂ produced. This enables the accurate determination of the mass of CO₂ by measuring the increase in the mass of the KOH solution.

Q31EXERCISES

Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test?

Solution

To Explain: Why acetic acid, and not sulfuric acid, is used for acidification in the lead acetate test for sulfur.

Context: The test for sulfur using lead acetate is performed on the sodium fusion extract, which contains sodium sulfide (Na₂S) if sulfur was present in the organic compound. The test relies on the formation of a black precipitate of lead(II) sulfide (PbS).

\mathrm{S}^{2-} + \mathrm{Pb}^{2+} \rightarrow \mathrm{PbS}_{\text{(s)}} \text{ (Black ppt)}

The test is typically carried out in a slightly acidic medium to neutralize any excess sodium hydroxide from the extract and to decompose any sodium cyanide present (as HCN gas), which might interfere.

Reason for Using Acetic Acid (CH₃COOH):

Acetic acid is a weak acid. It provides the necessary acidic medium for the reaction without interfering with the test.
Lead acetate ( $\ ext{Pb}(\ ext{CH}_3\ ext{COO})_2$ ) is soluble in acetic acid, so the lead ions ( $\ ext{Pb}^{2+}$ ) remain available in the solution to react with any sulfide ions (S²⁻) present.

Reason for NOT Using Sulphuric Acid (H₂SO₄):

If sulfuric acid were used for acidification, it would introduce sulfate ions (SO₄²⁻) into the solution.
Lead ions ( $\ ext{Pb}^{2+}$ ) from the lead acetate reagent react with sulfate ions to form a white precipitate of lead(II) sulfate (PbSO₄), which is insoluble.

\mathrm{Pb}^{2+}_{\text{(aq)}} + \mathrm{SO}_4^{2-}_{\text{(aq)}} \rightarrow \mathrm{PbSO}_{4(\text{s})} \text{ (White ppt)}

This precipitation reaction would cause two major problems:
1. Interference: The formation of a white precipitate of PbSO₄ would interfere with the observation of the black precipitate of PbS. It could mask the black precipitate, especially if only a small amount of sulfur is present.
2. False Negative: The precipitation of lead ions as PbSO₄ would remove them from the solution, reducing the concentration of available $\ ext{Pb}^{2+}$ ions. This could prevent the formation of the black PbS precipitate even if sulfide ions are present, leading to a false negative result.

Final Answer: Acetic acid is used because it provides the required acidic medium without forming a precipitate with the lead acetate reagent. Sulphuric acid cannot be used because it would react with the lead acetate to form an insoluble white precipitate of lead sulphate (PbSO₄), which would interfere with the test and could lead to a false negative result.

Q32EXERCISES

An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this substance is subjected to complete combustion.

Solution

Given: Percentage of Carbon (%C) = 69% Percentage of Hydrogen (%H) = 4.8% Mass of the organic substance = 0.20 g

To Find: The mass of carbon dioxide (CO₂) and water (H₂O) produced upon complete combustion.

Principle: During complete combustion, all the carbon in the organic compound is converted to CO₂, and all the hydrogen is converted to H₂O.

Solution:

Step 1: Calculate the mass of carbon and hydrogen in the 0.20 g sample.

Mass of Carbon: Mass of C = (Percentage of C / 100) $\times$ Mass of substance Mass of C = (69 / 100) $\times$ 0.20 g = 0.138 g
Mass of Hydrogen: Mass of H = (Percentage of H / 100) $\times$ Mass of substance Mass of H = (4.8 / 100) $\times$ 0.20 g = 0.0096 g

Step 2: Calculate the mass of CO₂ produced from the mass of carbon.

The combustion reaction for carbon is: C

\rightarrow

CO₂ Atomic mass of C = 12 g/mol Molar mass of CO₂ = 44 g/mol

This means 12 g of Carbon produces 44 g of CO₂.

Mass of CO₂ = Mass of C

\times \frac{\text{Molar mass of CO}_2}{\text{Atomic mass of C}}

Mass of CO₂ = 0.138 g

\times \frac{44}{12}

Mass of CO₂ = 0.138 g

\times

3.6667 Mass of CO₂ = 0.506 g

Step 3: Calculate the mass of H₂O produced from the mass of hydrogen.

The combustion reaction for hydrogen is: 2H

\rightarrow

H₂O Atomic mass of 2H = 2

\times

1 = 2 g/mol Molar mass of H₂O = 18 g/mol

This means 2 g of Hydrogen produces 18 g of H₂O.

Mass of H₂O = Mass of H

\times \frac{\text{Molar mass of H}_2\text{O}}{\text{Atomic mass of 2H}}

Mass of H₂O = 0.0096 g

\times \frac{18}{2}

Mass of H₂O = 0.0096 g

\times

9 Mass of H₂O = 0.0864 g

Final Answer: When 0.20 g of the substance is subjected to complete combustion:

The mass of carbon dioxide produced is 0.506 g.
The mass of water produced is 0.0864 g.

Q33EXERCISES

A sample of 0.50 g of an organic compound was treated according to Kjeldahl's method. The ammonia evolved was absorbed in 50 ml of 0.5 M H₂SO₄. The residual acid required 60 mL of 0.5 M solution of NaOH for neutralisation. Find the percentage composition of nitrogen in the compound.

Solution

Given: Mass of organic compound (

m

) = 0.50 g Initial volume of H₂SO₄ = 50 mL Initial molarity of H₂SO₄ = 0.5 M Volume of NaOH used for back-titration = 60 mL Molarity of NaOH = 0.5 M

To Find: The percentage composition of nitrogen (N) in the compound.

Principle:

Calculate the initial moles of H₂SO₄ taken.
Calculate the moles of NaOH used to neutralize the excess H₂SO₄.
Calculate the moles of H₂SO₄ that were in excess.
Calculate the moles of H₂SO₄ that reacted with ammonia.
Calculate the moles and mass of ammonia (and thus nitrogen).
Calculate the percentage of nitrogen.

Reactions: Neutralization of excess acid:

2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}

Absorption of ammonia:

2\text{NH}_3 + \text{H}_2\text{SO}_4 \rightarrow (\text{NH}_4)_2\text{SO}_4

Solution:

Step 1: Initial moles of H₂SO₄ Moles = Molarity

\times

Volume (L) Initial moles of H₂SO₄ =

0.5 \text{ mol/L} \times (50/1000) \text{ L} = 0.025 \text{ mol}

Step 2: Moles of NaOH used Moles of NaOH =

0.5 \text{ mol/L} \times (60/1000) \text{ L} = 0.03 \text{ mol}

Step 3: Moles of excess H₂SO₄ From the neutralization reaction, 2 moles of NaOH react with 1 mole of H₂SO₄. Moles of excess H₂SO₄ =

\frac{1}{2} \times

Moles of NaOH Moles of excess H₂SO₄ =

\frac{1}{2} \times 0.03 \text{ mol} = 0.015 \text{ mol}

Step 4: Moles of H₂SO₄ reacted with NH₃ Moles of H₂SO₄ reacted = Initial moles - Excess moles Moles of H₂SO₄ reacted =

0.025 \text{ mol} - 0.015 \text{ mol} = 0.010 \text{ mol}

Step 5: Moles and mass of Nitrogen From the absorption reaction, 1 mole of H₂SO₄ reacts with 2 moles of NH₃. Moles of NH₃ =

2 \times

Moles of H₂SO₄ reacted Moles of NH₃ =

2 \times 0.010 \text{ mol} = 0.020 \text{ mol}

Since each mole of NH₃ contains one mole of N, Moles of N = 0.020 mol. Atomic mass of N = 14 g/mol. Mass of N = Moles of N

\times

Atomic mass of N Mass of N =

0.020 \text{ mol} \times 14 \text{ g/mol} = 0.28 \text{ g}

Step 6: Percentage of Nitrogen %N =

\frac{\text{Mass of N}}{\text{Mass of compound}} \times 100

%N =

\frac{0.28 \text{ g}}{0.50 \text{ g}} \times 100 = 56.0 \%

Final Answer: The percentage composition of nitrogen in the compound is 56.0%.

Q34EXERCISES

0.3780 g of an organic chloro compound gave 0.5740 g of silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound.

Solution

Given: Mass of organic compound (

m

) = 0.3780 g Mass of silver chloride (AgCl) formed (

m_{\text{AgCl}}

) = 0.5740 g

To Find: The percentage of chlorine (Cl) in the compound.

Formulas: Atomic mass of Chlorine (Cl) = 35.5 g/mol Atomic mass of Silver (Ag) = 108 g/mol Molar mass of AgCl = 108 + 35.5 = 143.5 g/mol

Percentage of Chlorine (%Cl) =

\frac{\text{Atomic mass of Cl}}{\text{Molar mass of AgCl}} \times \frac{\text{Mass of AgCl formed}}{\text{Mass of compound}} \times 100

Solution:

Step 1: Calculate the mass of chlorine in 0.5740 g of AgCl. Mass of Cl = (Mass of AgCl)

\times

\frac{\text{Atomic mass of Cl}}{\text{Molar mass of AgCl}}

Mass of Cl =

0.5740 \text{ g} \times \frac{35.5}{143.5}

Mass of Cl =

0.5740 \times 0.24738 \text{ g} = 0.14197 \text{ g}

Step 2: Calculate the percentage of chlorine in the compound. %Cl =

\frac{\text{Mass of Cl}}{\text{Mass of compound}} \times 100

%Cl =

\frac{0.14197 \text{ g}}{0.3780 \text{ g}} \times 100

%Cl =

37.56 \%

Using the combined formula:

%Cl =

\frac{35.5}{143.5} \times \frac{0.5740}{0.3780} \times 100

%Cl =

0.24738 \times 1.5185 \times 100

%Cl =

37.56 \%

Final Answer: The percentage of chlorine present in the compound is 37.56%.

Q35EXERCISES

In the estimation of sulphur by Carius method, 0.468 g of an organic sulphur compound afforded 0.668 g of barium sulphate. Find out the percentage of sulphur in the given compound.

Solution

Given: Mass of organic compound (

m

) = 0.468 g Mass of barium sulphate (

\mathrm{BaSO}_4

) formed (

m_{\mathrm{BaSO}_4}

) = 0.668 g

To Find: The percentage of sulphur (S) in the compound.

Formulas: Atomic mass of Sulphur (S) = 32 g/mol Atomic mass of Barium (Ba) = 137 g/mol Atomic mass of Oxygen (O) = 16 g/mol Molar mass of

\mathrm{BaSO}_4

= 137 + 32 + 4(16) = 233 g/mol

Percentage of Sulphur (%S) =

\frac{\text{Atomic mass of S}}{\text{Molar mass of } \mathrm{BaSO}_4} \times \frac{\text{Mass of } \mathrm{BaSO}_4 \text{ formed}}{\text{Mass of compound}} \times 100

Solution:

Step 1: Calculate the mass of sulphur in 0.668 g of $\mathrm{BaSO}_4$ . Mass of S = (Mass of

\mathrm{BaSO}_4

)

\times

\frac{\text{Atomic mass of S}}{\text{Molar mass of } \mathrm{BaSO}_4}

Mass of S =

0.668 \text{ g} \times \frac{32}{233}

Mass of S =

0.668 \times 0.13734 \text{ g} = 0.09174 \text{ g}

Step 2: Calculate the percentage of sulphur in the compound. %S =

\frac{\text{Mass of S}}{\text{Mass of compound}} \times 100

%S =

\frac{0.09174 \text{ g}}{0.468 \text{ g}} \times 100

%S =

19.60 \%

Using the combined formula:

%S =

\frac{32}{233} \times \frac{0.668}{0.468} \times 100

%S =

0.13734 \times 1.42735 \times 100

%S =

19.60 \%

Final Answer: The percentage of sulphur in the given compound is 19.60%.

Q36EXERCISES

In the organic compound $\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{C} \equiv \mathrm{CH}$ , the pair of hydridised orbitals involved in the formation of: $\mathrm{C}_{2}-\mathrm{C}_{3}$ bond is:

(a)

s p-s p^{2}

(b)

s p-s p^{3}

(c)

s p^{2}-s p^{3}

(d)

s p^{3}-s p^{3}

Solution

Given: The organic compound Hex-1-en-5-yne:

\mathrm{C}^1\mathrm{H}_{2}=\mathrm{C}^2\mathrm{H}-\mathrm{C}^3\mathrm{H}_{2}-\mathrm{C}^4\mathrm{H}_{2}-\mathrm{C}^5 \equiv \mathrm{C}^6\mathrm{H}

To Find: The pair of hybridized orbitals involved in the formation of the C₂-C₃ bond.

Solution:

Step 1: Determine the hybridization of C₂ and C₃. Hybridization is determined by the number of sigma (

\sigma

) bonds formed by each carbon atom.

For C₂: This carbon is part of a double bond (C₁=C₂) and is also bonded to a hydrogen and C₃. It forms 3 $\sigma$ bonds (one with C₁, one with H, one with C₃) and one $\pi$ bond. A carbon atom with 3 $\sigma$ bonds is $sp^2$ hybridized.
For C₃: This carbon is bonded to C₂, C₄, and two hydrogen atoms. All are single bonds. It forms 4 $\sigma$ bonds in total. A carbon atom with 4 $\sigma$ bonds is $sp^3$ hybridized.

Step 2: Identify the orbitals forming the C₂-C₃ bond. The single bond (a

\sigma

bond) between C₂ and C₃ is formed by the overlap of a hybrid orbital from each carbon atom.

C₂ contributes an $sp^2$ hybrid orbital.
C₃ contributes an $sp^3$ hybrid orbital.

Therefore, the C₂-C₃ bond is formed by the overlap of an

sp^2

and an

sp^3

orbital.

Final Answer: The correct option is (c) $s p^{2}-s p^{3}$ .

Q37EXERCISES

In the Lassaigne's test for nitrogen in an organic compound, the Prussian blue colour is obtained due to the formation of:

(a)

\mathrm{Na}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]

(b)

\mathrm{Fe}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}

(c)

\mathrm{Fe}_{2}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]

(d)

\mathrm{Fe}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{4}

Solution

To Find: The chemical formula of the compound responsible for the Prussian blue color in Lassaigne's test for nitrogen.

Principle of the Test:

In Lassaigne's test, nitrogen from the organic compound is converted to sodium cyanide (NaCN) by fusion with sodium metal.
The resulting cyanide ions (CN⁻) in the aqueous extract are reacted with freshly prepared ferrous sulfate (FeSO₄), forming the hexacyanoferrate(II) complex ion, $[\text{Fe}(\text{CN})_6]^{4-}$ . $\mathrm{Fe}^{2+} + 6\mathrm{CN}^- \rightarrow [\mathrm{Fe}(\mathrm{CN})_6]^{4-}$
Upon acidification with concentrated H₂SO₄, some of the Fe²⁺ ions are oxidized to ferric ions (Fe³⁺).
These ferric ions (Fe³⁺) then react with the hexacyanoferrate(II) complex to form a complex salt which has an intense blue color known as Prussian blue.

Formation of Prussian Blue: The chemical reaction is:

4\mathrm{Fe}^{3+} + 3[\mathrm{Fe}(\mathrm{CN})_6]^{4-} \rightarrow \mathrm{Fe}_4[\mathrm{Fe}(\mathrm{CN})_6]_3

The compound formed is Iron(III) hexacyanoferrate(II), often written as

\ ext{Fe}_4[\ ext{Fe}(\ ext{CN})_6]_3

. This compound is also known as ferric ferrocyanide.

Analysis of Options: (a)

\mathrm{Na}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]

: This is sodium hexacyanoferrate(II), a soluble, pale yellow intermediate, not the final blue precipitate. (b)

\mathrm{Fe}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}

: This is the correct formula for Prussian blue. (c)

\mathrm{Fe}_{2}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]

: Incorrect stoichiometry. (d)

\mathrm{Fe}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{4}

: Incorrect stoichiometry.

Final Answer: The Prussian blue color is due to the formation of iron(III) hexacyanoferrate(II). The correct formula is (b) $\mathrm{Fe}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}$ .

Q38EXERCISES

Which of the following carbocation is most stable ?

(a)

\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C} . \stackrel{+}{\mathrm{C}} \mathrm{H}_{2}

(b)

\left(\mathrm{CH}_{3}\right)_{3} \stackrel{+}{\mathrm{C}}

(c)

\mathrm{CH}_{3} \mathrm{CH}_{2} \stackrel{+}{\mathrm{C}} \mathrm{H}_{2}

(d)

\mathrm{CH}_{3} \stackrel{+}{\mathrm{C}} \mathrm{H} \mathrm{CH}_{2} \mathrm{CH}_{3}

Solution

To Find: The most stable carbocation among the given options.

Principle: The stability of carbocations increases with the number of alkyl groups attached to the positively charged carbon. This is due to two main stabilizing effects:

Inductive Effect (+I): Alkyl groups are electron-donating and push electron density towards the positive carbon, dispersing the charge.
Hyperconjugation: Delocalization of sigma electrons from adjacent C-H bonds ( $\alpha$ -hydrogens) into the empty p-orbital of the carbocation.

The order of stability is: Tertiary (3°) > Secondary (2°) > Primary (1°) > Methyl. The more alkyl groups and more

\alpha

-hydrogens, the more stable the carbocation.

Analysis of Options:

(a) $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C} - \stackrel{+}{\mathrm{C}} \mathrm{H}_{2}$ (Neopentyl cation)

The positively charged carbon is bonded to only one other carbon atom.
This is a primary (1°) carbocation. It has zero $\alpha$ -hydrogens for hyperconjugation, although it experiences a +I effect from the t-butyl group. It is known to rearrange immediately.

(b) $\left(\mathrm{CH}_{3}\right)_{3} \stackrel{+}{\mathrm{C}}$ (tert-Butyl cation)

The positively charged carbon is bonded to three other carbon atoms.
This is a tertiary (3°) carbocation.
It has 3 + 3 + 3 = 9 $\alpha$ -hydrogens for hyperconjugation.
It benefits from the +I effect of three methyl groups.

(c) $\mathrm{CH}_{3} \mathrm{CH}_{2} \stackrel{+}{\mathrm{C}} \mathrm{H}_{2}$ (n-Propyl cation)

The positively charged carbon is bonded to only one other carbon atom.
This is a primary (1°) carbocation.
It has 2 $\alpha$ -hydrogens (on the adjacent CH₂ group) for hyperconjugation.

(d) $\mathrm{CH}_{3} \stackrel{+}{\mathrm{C}} \mathrm{H} \mathrm{CH}_{2} \mathrm{CH}_{3}$ (sec-Butyl cation)

The positively charged carbon is bonded to two other carbon atoms.
This is a secondary (2°) carbocation.
It has 3 hydrogens on the adjacent methyl group and 2 hydrogens on the adjacent methylene group. Total 5 $\alpha$ -hydrogens for hyperconjugation.

Comparison:

We have two primary carbocations (a, c), one secondary (d), and one tertiary (b).
The tertiary carbocation is the most stable.
Comparing the number of hyperconjugating structures: (b) has 9, (d) has 5, (c) has 2, and (a) has 0.

Thus, the tert-butyl cation (b) is the most stable.

Final Answer: The most stable carbocation is (b) $\left(\mathrm{CH}_{3}\right)_{3} \stackrel{+}{\mathrm{C}}$ .

Q39EXERCISES

The best and latest technique for isolation, purification and separation of organic compounds is:

(a)

Crystallisation

(b)

Distillation

(c)

Sublimation

(d)

Chromatography

Solution

To Find: The best and most modern technique for the isolation, purification, and separation of organic compounds.

Analysis of Options:

(a) Crystallisation: An effective technique for purifying solids, but it is limited to compounds that can form crystals and depends on solubility differences. It is not universally applicable, especially for liquids or complex mixtures.
(b) Distillation: A fundamental technique for purifying liquids based on boiling point differences. It is very useful but can be ineffective for mixtures with close boiling points or for heat-sensitive compounds.
(c) Sublimation: A niche technique applicable only to a small subset of solid compounds that can sublime.
(d) Chromatography: This is a highly versatile and powerful set of techniques (e.g., column, TLC, paper, gas chromatography, HPLC). It can be used to separate extremely complex mixtures, purify compounds with very high efficiency, and work with very small quantities of material (analytical scale) or large quantities (preparative scale). It separates based on subtle differences in properties like polarity, size, and affinity for stationary/mobile phases. Modern techniques like High-Performance Liquid Chromatography (HPLC) and Gas Chromatography (GC) are automated, highly sensitive, and provide excellent resolution, making them the state-of-the-art methods for separation and purification.

Conclusion: While crystallisation and distillation are classic and important methods, chromatography offers superior versatility, sensitivity, and efficiency for a much wider range of separation challenges, making it the 'best and latest' technique in a general sense.

Final Answer: The correct option is (d) Chromatography.

Q40EXERCISES

The reaction: $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{I}+\mathrm{KOH}(\mathrm{aq}) \rightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{KI}$ is classified as :

(a)

electrophilic substitution

(b)

nucleophilic substitution

(c)

elimination

(d)

addition

Solution

To Classify: The given chemical reaction.

Analysis of the Reaction:

Substrate: $\ ext{CH}_3\ ext{CH}_2\ ext{I}$ (Iodoethane)
Reagent: Aqueous KOH, which provides the hydroxide ion, $\ ext{OH}^-$ .
Product: $\ ext{CH}_3\ ext{CH}_2\ ext{OH}$ (Ethanol)

Reaction Mechanism:

The hydroxide ion ( $\ ext{OH}^-$ ) is an electron-rich species with lone pairs on the oxygen. It is a nucleophile.
In iodoethane, the C-I bond is polar ( $\ ext{C}^{\delta+}-\ ext{I}^{\delta-}$ ), making the carbon atom attached to the iodine an electrophilic center.
The nucleophile ( $\ ext{OH}^-$ ) attacks the electrophilic carbon atom.
Simultaneously or subsequently, the iodide ion (I⁻) leaves the molecule.
The net result is that the iodine atom (-I) is substituted by the hydroxyl group (-OH).

Classification based on reaction type:

It is not an addition reaction because no atoms are added across a double or triple bond.
It is not an elimination reaction because atoms are not removed to form a double bond.
It is a substitution reaction because one group (-I) is replaced by another (-OH).
Since the attacking species is a nucleophile ( $\ ext{OH}^-$ ), the reaction is classified as a nucleophilic substitution.

Final Answer: The correct classification is (b) nucleophilic substitution.

Organic Chemistry – Some Basic Principles and TechniquesClass 11 Chemistry NCERT Solutions

How many σ\sigmaσ and π\piπ bonds are present in each of the following molecules?

Solution

What is the type of hybridisation of each carbon in the following compounds?

Solution

Write the state of hybridisation of carbon in the following compounds and shapes of each of the molecules.

Solution

Expand each of the following condensed formulas into their complete structural formulas.

Solution

For each of the following compounds, write a condensed formula and also their bond-line formula.

Solution

Expand each of the following bond-line formulas to show all the atoms including carbon and hydrogen

Solution

Structures and IUPAC names of some hydrocarbons are given below. Explain why the names given in the parentheses are incorrect.

Solution

Write the IUPAC names of the compounds i-iv from their given structures.

CH3CH2CH(OH)CH2CH2CH(CH3)CH2CH3\mathrm{CH_3CH_2CH(OH)CH_2CH_2CH(CH_3)CH_2CH_3}CH3​CH2​CH(OH)CH2​CH2​CH(CH3​)CH2​CH3​

CH3CH2C(=O)CH2C(=O)CH3\mathrm{CH_3CH_2C(=O)CH_2C(=O)CH_3}CH3​CH2​C(=O)CH2​C(=O)CH3​

CH3C(=O)CH2CH2CH2COOH\mathrm{CH_3C(=O)CH_2CH_2CH_2COOH}CH3​C(=O)CH2​CH2​CH2​COOH

CH≡C−CH=CH−CH=CH2\mathrm{CH \equiv C-CH=CH-CH=CH_2}CH≡C−CH=CH−CH=CH2​

Solution

Derive the structure of (i) 2-Chlorohexane, (ii) Pent-4-en-2-ol, (iii) 3- Nitrocyclohexene, (iv) Cyclohex-2-en-1-ol, (v) 6-Hydroxyheptanal.

Solution

Write the structural formula of:

Solution

Using curved-arrow notation, show the formation of reactive intermediates when the following covalent bonds undergo heterolytic cleavage.

Solution

Solution

Identify electrophilic centre in the following: CH3CH=O,CH3CN,CH3I\mathrm{CH}_{3} \mathrm{CH}=\mathrm{O}, \mathrm{CH}_{3} \mathrm{CN}, \mathrm{CH}_{3} \mathrm{I}CH3​CH=O,CH3​CN,CH3​I.

Solution

Which bond is more polar in the following pairs of molecules: (a) H3C−H,H3C−Br\mathrm{H}_{3} \mathrm{C}-\mathrm{H}, \mathrm{H}_{3} \mathrm{C}-\mathrm{Br}H3​C−H,H3​C−Br

Solution

In which C-C bond of CH3CH2CH2Br\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}CH3​CH2​CH2​Br, the inductive effect is expected to be the least?

Solution

Write resonance structures of CH3COO−\mathrm{CH}_{3} \mathrm{COO}^{-}CH3​COO− and show the movement of electrons by curved arrows.

Solution

Write resonance structures of CH2=CH−CHO\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CHO}CH2​=CH−CHO. Indicate relative stability of the contributing structures.

Solution

Explain why the following two structures, I and II cannot be the major contributors to the real structure of CH3COOCH3\mathrm{CH}_{3} \mathrm{COOCH}_{3}CH3​COOCH3​. H₃C - C⁺ = O⁻ H₃C - C⁻ - O⁺ | || :O-CH₃ O-CH₃ (I) (II)

Solution

Explain why (CH3)3C+\left(\mathrm{CH}_{3}\right)_{3} \stackrel{+}{\mathrm{C}}(CH3​)3​C+​ is more stable than CH3C+H2\mathrm{CH}_{3} \stackrel{+}{\mathrm{C}} \mathrm{H}_{2}CH3​C+​H2​ and C+H3\stackrel{+}{\mathrm{C}} \mathrm{H}_{3}C+​H3​ is the least stable cation.

Solution

On complete combustion, 0.246 g of an organic compound gave 0.198 g of carbon dioxide and 0.1014 g of water. Determine the percentage composition of carbon and hydrogen in the compound.

Solution

In Dumas' method for estimation of nitrogen, 0.3 g of an organic compound gave 50 mL of nitrogen collected at 300 K temperature and 715 mm pressure. Calculate the percentage composition of nitrogen in the compound. (Aqueous tension at 300 K = 15 mm)

Solution

During estimation of nitrogen present in an organic compound by Kjeldahl's method, the ammonia evolved from 0.5 g of the compound in Kjeldahl's estimation of nitrogen, neutralized 10 mL of 1 M H₂SO₄. Find out the percentage of nitrogen in the compound.

Solution

In Carius method of estimation of halogen, 0.15 g of an organic compound gave 0.12 g of AgBr. Find out the percentage of bromine in the compound.

Solution

In sulphur estimation, 0.157 g of an organic compound gave 0.4813 g of barium sulphate. What is the percentage of sulphur in the compound?

Solution

Solution

Solution

Write bond line formulas for : Isopropyl alcohol, 2,3-Dimethylbutanal, Heptan-4-one.

Solution

Give the IUPAC names of the following compounds :

Solution

Which of the following represents the correct IUPAC name for the compounds concerned ? (a) 2,2-Dimethylpentane or 2-Dimethylpentane (b) 2,4,7-Trimethyloctane or 2,5,7-Trimethyloctane (c) 2-Chloro-4-methylpentane or 4-Chloro-2-methylpentane (d) But-3-yn-1-ol or But-4-ol-1-yne.

Solution

Draw formulas for the first five members of each homologous series beginning with the following compounds. (a) H−COOH\mathrm{H}-\mathrm{COOH}H−COOH (b) CH3COCH3\mathrm{CH}_{3} \mathrm{COCH}_{3}CH3​COCH3​ (c) H−CH=CH2\mathrm{H}-\mathrm{CH}=\mathrm{CH}_{2}H−CH=CH2​

Solution

Give condensed and bond line structural formulas and identify the functional group(s) present, if any, for :

Solution

Identify the functional groups in the following compounds

Solution

Which of the two: O2NCH2CH2O−\mathrm{O}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{O}^{-}O2​NCH2​CH2​O− or CH3CH2O−\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}CH3​CH2​O− is expected to be more stable and why?

Solution

Explain why alkyl groups act as electron donors when attached to a π\piπ system.

Solution

Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation.

Solution

What are electrophiles and nucleophiles ? Explain with examples.

Solution

Identify the reagents shown in bold in the following equations as nucleophiles or electrophiles:

Solution

Classify the following reactions in one of the reaction type studied in this unit.

Solution

What is the relationship between the members of following pairs of structures ? Are they structural or geometrical isomers or resonance contributors?

Solution

How many $\sigma$ and $\pi$ bonds are present in each of the following molecules?

$\mathrm{CH_3CH_2CH(OH)CH_2CH_2CH(CH_3)CH_2CH_3}$

$\mathrm{CH_3CH_2C(=O)CH_2C(=O)CH_3}$

$\mathrm{CH_3C(=O)CH_2CH_2CH_2COOH}$

$\mathrm{CH \equiv C-CH=CH-CH=CH_2}$

Identify electrophilic centre in the following: $\mathrm{CH}_{3} \mathrm{CH}=\mathrm{O}, \mathrm{CH}_{3} \mathrm{CN}, \mathrm{CH}_{3} \mathrm{I}$ .

Which bond is more polar in the following pairs of molecules: (a) $\mathrm{H}_{3} \mathrm{C}-\mathrm{H}, \mathrm{H}_{3} \mathrm{C}-\mathrm{Br}$

In which C-C bond of $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}$ , the inductive effect is expected to be the least?

Write resonance structures of $\mathrm{CH}_{3} \mathrm{COO}^{-}$ and show the movement of electrons by curved arrows.

Write resonance structures of $\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CHO}$ . Indicate relative stability of the contributing structures.

Explain why the following two structures, I and II cannot be the major contributors to the real structure of $\mathrm{CH}_{3} \mathrm{COOCH}_{3}$ .
`H₃C - C⁺ = O⁻ H₃C - C⁻ - O⁺ | || :O-CH₃ O-CH₃ (I) (II)`

Explain why $\left(\mathrm{CH}_{3}\right)_{3} \stackrel{+}{\mathrm{C}}$ is more stable than $\mathrm{CH}_{3} \stackrel{+}{\mathrm{C}} \mathrm{H}_{2}$ and $\stackrel{+}{\mathrm{C}} \mathrm{H}_{3}$ is the least stable cation.

Draw formulas for the first five members of each homologous series beginning with the following compounds. (a) $\mathrm{H}-\mathrm{COOH}$ (b) $\mathrm{CH}_{3} \mathrm{COCH}_{3}$ (c) $\mathrm{H}-\mathrm{CH}=\mathrm{CH}_{2}$

Which of the two: $\mathrm{O}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{O}^{-}$ or $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}$ is expected to be more stable and why?

Explain why alkyl groups act as electron donors when attached to a $\pi$ system.

Will $\mathrm{CCl}_{4}$ give white precipitate of AgCl on heating it with silver nitrate? Give reason for your answer.

In the organic compound $\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{C} \equiv \mathrm{CH}$ , the pair of hydridised orbitals involved in the formation of: $\mathrm{C}_{2}-\mathrm{C}_{3}$ bond is:

The reaction: $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{I}+\mathrm{KOH}(\mathrm{aq}) \rightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{KI}$ is classified as :