Haloalkanes and Haloarenes Class 12 Chemistry NCERT Solutions

Q1Exercises

Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:

(i)

$(CH_3)_2CHCH(Cl)CH_3$

(ii)

$CH_3CH_2CH(CH_3)CH(C_2H_5)Cl$

(iii)

$CH_3CH_2C(CH_3)_2CH_2I$

(iv)

$(CH_3)_3CCH_2CH(Br)C_6H_5$

(v)

$CH_3CH(CH_3)CH(Br)CH_3$

(vi)

$CH_3C(C_2H_5)_2CH_2Br$

(vii)

$CH_3C(Cl)(C_2H_5)CH_2CH_3$

(viii)

$CH_3CH=C(Cl)CH_2CH(CH_3)_2$

(ix)

$CH_3CH=CHC(Br)(CH_3)_2$

(x)

$p-ClC_6H_4CH_2CH(CH_3)_2$ (xi) $m-ClCH_2C_6H_4CH_2C(CH_3)_3$ (xii) $o-Br-C_6H_4CH(CH_3)CH_2CH_3$

Solution

(i)

IUPAC Name: 2-Chloro-3-methylbutane Classification: Secondary alkyl halide

(ii)

IUPAC Name: 3-Chloro-4-methylhexane Classification: Secondary alkyl halide

(iii)

IUPAC Name: 1-Iodo-2,2-dimethylbutane Classification: Primary alkyl halide

(iv)

IUPAC Name: 1-Bromo-3,3-dimethyl-1-phenylbutane Classification: Secondary benzylic halide

(v)

IUPAC Name: 2-Bromo-3-methylbutane Classification: Secondary alkyl halide

(vi)

IUPAC Name: 1-Bromo-2-ethyl-2-methylbutane Classification: Primary alkyl halide

(vii)

IUPAC Name: 3-Chloro-3-methylpentane Classification: Tertiary alkyl halide

(viii)

IUPAC Name: 3-Chloro-5-methylhex-2-ene Classification: Vinylic halide

(ix)

IUPAC Name: 4-Bromo-4-methylpent-2-ene Classification: Allylic halide

(x)

IUPAC Name: 1-Chloro-4-(2-methylpropyl)benzene Classification: Aryl halide

(xi) IUPAC Name: 1-Chloromethyl-3-(2,2-dimethylpropyl)benzene Classification: Primary benzylic halide

(xii) IUPAC Name: 1-Bromo-2-(1-methylpropyl)benzene Classification: Aryl halide

Q2Exercises

Give the IUPAC names of the following compounds:

(i)

$CH_3CH(Cl)CH(Br)CH_3$

(ii)

$CHF_2CBrClF$

(iii)

$ClCH_2C\equiv CCH_2Br$

(iv)

$(CCl_3)_3CCl$

(v)

$CH_3C(p-ClC_6H_4)_2CH(Br)CH_3$

(vi)

$(CH_3)_3CCH=CClC_6H_4I-p$

Solution

(i)

IUPAC Name: 2-Bromo-3-chlorobutane

(ii)

IUPAC Name: 1-Bromo-1-chloro-1,2,2-trifluoroethane

(iii)

IUPAC Name: 1-Bromo-4-chlorobut-2-yne

(iv)

IUPAC Name: 1,1,1,2,3,3,3-Heptachloro-2-(trichloromethyl)propane. A more common name is Perchloroparacyclophane, but the systematic name is as given. Alternatively, it can be named as 2-(Trichloromethyl)-1,1,1,2,3,3,3-heptachloropropane.

(v)

IUPAC Name: 2-Bromo-3,3-bis(4-chlorophenyl)butane

(vi)

IUPAC Name: 1-Chloro-1-(4-iodophenyl)-3,3-dimethylbut-1-ene

Q3Exercises

Write the structures of the following organic halogen compounds.

(i)

2-Chloro-3-methylpentane

(ii)

$p$ -Bromochlorobenzene

(iii)

1-Chloro-4-ethylcyclohexane

(iv)

2-(2-Chlorophenyl)-1-iodooctane

(v)

2-Bromobutane

(vi)

4-tert-Butyl-3-iodoheptane

(vii)

1-Bromo-4-sec-butyl-2-methylbenzene

(viii)

1,4-Dibromobut-2-ene

Solution

(i)

2-Chloro-3-methylpentane

CH_3-CH_2-CH(CH_3)-CH(Cl)-CH_3

(ii)

$p$ -Bromochlorobenzene Structure: A benzene ring with a Br atom at position 1 and a Cl atom at position 4.

(iii)

1-Chloro-4-ethylcyclohexane Structure: A cyclohexane ring with a Cl atom at position 1 and an ethyl group (

-CH_2CH_3

) at position 4.

(iv)

2-(2-Chlorophenyl)-1-iodooctane

CH_3-(CH_2)_5-CH(C_6H_4Cl)-CH_2I

(The chlorophenyl group is attached to the second carbon of the octane chain, and the chlorine is at the ortho position of that phenyl ring).

(v)

2-Bromobutane

CH_3-CH_2-CH(Br)-CH_3

(vi)

4-tert-Butyl-3-iodoheptane

CH_3-CH_2-CH_2-CH(C(CH_3)_3)-CH(I)-CH_2-CH_3

(vii)

1-Bromo-4-sec-butyl-2-methylbenzene Structure: A benzene ring with a Br at position 1, a methyl group at position 2, and a sec-butyl group (

-CH(CH_3)CH_2CH_3

) at position 4.

(viii)

1,4-Dibromobut-2-ene

Br-CH_2-CH=CH-CH_2-Br

Q4Exercises

Which one of the following has the highest dipole moment?

(i)

$CH_2Cl_2$

(ii)

$CHCl_3$

(iii)

$CCl_4$

Solution

Answer: (i)

CH_2Cl_2

has the highest dipole moment.

Explanation:

$CCl_4$ (Carbon tetrachloride): This molecule has a symmetrical tetrahedral geometry. The four C-Cl bond dipoles are equal in magnitude and are oriented at $109.5^\circ$ to each other. The vector sum of these dipoles cancels out completely, resulting in a net dipole moment of zero.
$CHCl_3$ (Chloroform): This molecule has a tetrahedral geometry. It has three polar C-Cl bonds and one less polar C-H bond. The three C-Cl bond dipoles point towards the vertices of the tetrahedron, and their resultant vector is partially cancelled by the opposing dipole of the C-H bond. This results in a net dipole moment.
$CH_2Cl_2$ (Dichloromethane): This molecule also has a tetrahedral geometry. It has two polar C-Cl bonds and two less polar C-H bonds. The two C-Cl bond dipoles are oriented such that their resultant vector is large. The resultant of the two C-H bond dipoles adds to the resultant of the C-Cl bonds. The angle between the two C-Cl bonds is approximately $109.5^\circ$ , leading to a significant net dipole moment that is greater than that of $CHCl_3$ .

The experimental dipole moments are:

$CCl_4$ : 0 D
$CHCl_3$ : 1.08 D
$CH_2Cl_2$ : 1.60 D

Therefore,

CH_2Cl_2

has the highest dipole moment.

Q5Exercises

A hydrocarbon $C_5H_{10}$ does not react with chlorine in dark but gives a single monochloro compound $C_5H_9Cl$ in bright sunlight. Identify the hydrocarbon.

Solution

Analysis:

The molecular formula $C_5H_{10}$ corresponds to the general formula $C_nH_{2n}$ , which indicates the hydrocarbon could be an alkene or a cycloalkane.
The hydrocarbon does not react with chlorine in the dark. This rules out the possibility of it being an alkene, as alkenes undergo addition reactions with halogens even in the dark.
Therefore, the hydrocarbon must be a cycloalkane.
The hydrocarbon reacts with chlorine in bright sunlight (UV light) to give a single monochloro compound, $C_5H_9Cl$ . This is a free-radical substitution reaction.
The formation of a single monochloro product implies that all the hydrogen atoms in the hydrocarbon are chemically equivalent.

Identification: Let's consider the possible cycloalkanes with the formula

C_5H_{10}

:

Cyclopentane
Methylcyclobutane
Ethylcyclopropane
1,1-Dimethylcyclopropane
1,2-Dimethylcyclopropane (cis and trans isomers)

Now, let's check for the equivalence of hydrogen atoms:

In cyclopentane, all 10 hydrogen atoms are equivalent due to the symmetry of the ring. Substitution of any hydrogen atom will result in the same product, chlorocyclopentane.
In methylcyclobutane and other substituted cycloalkanes, there are different types of hydrogen atoms, which would lead to a mixture of isomeric monochloro products.

Conclusion: The only structure that fits all the conditions is cyclopentane. All its 10 hydrogen atoms are equivalent, so its free-radical chlorination yields only one product.

Hydrocarbon: Cyclopentane

Reaction:

C_5H_{10} \text{ (Cyclopentane)} + Cl_2 \xrightarrow{\text{Sunlight}} C_5H_9Cl \text{ (Chlorocyclopentane)} + HCl

Q6Exercises

Write the isomers of the compound having formula $C_4H_9Br$ .

Solution

The compound with the formula

C_4H_9Br

has the following four structural isomers:

1-Bromobutane (n-Butyl bromide)
- Structure: $CH_3-CH_2-CH_2-CH_2-Br$
- Type: Primary ( $1^\circ$ ) alkyl halide
2-Bromobutane (sec-Butyl bromide)
- Structure: $CH_3-CH_2-CH(Br)-CH_3$
- Type: Secondary ( $2^\circ$ ) alkyl halide
- Note: This compound is chiral and exists as a pair of enantiomers.
1-Bromo-2-methylpropane (Isobutyl bromide)
- Structure: $(CH_3)_2CH-CH_2-Br$
- Type: Primary ( $1^\circ$ ) alkyl halide
2-Bromo-2-methylpropane (tert-Butyl bromide)
- Structure: $(CH_3)_3C-Br$
- Type: Tertiary ( $3^\circ$ ) alkyl halide

Q7Exercises

Write the equations for the preparation of 1-iodobutane from

(i)

1-butanol

(ii)

1-chlorobutane

(iii)

but-1-ene.

Solution

(i)

From 1-butanol 1-Iodobutane can be prepared from 1-butanol by reacting it with sodium or potassium iodide in the presence of 95% orthophosphoric acid.

CH_3CH_2CH_2CH_2OH + KI + H_3PO_4 \xrightarrow{\Delta} CH_3CH_2CH_2CH_2I + KH_2PO_4 + H_2O

Alternatively, it can be prepared by reacting 1-butanol with red phosphorus and iodine.

3CH_3CH_2CH_2CH_2OH + P + \frac{3}{2}I_2 \rightarrow 3CH_3CH_2CH_2CH_2I + H_3PO_3

(ii)

From 1-chlorobutane 1-Iodobutane is prepared from 1-chlorobutane by the Finkelstein reaction. This involves treating 1-chlorobutane with sodium iodide in dry acetone.

CH_3CH_2CH_2CH_2Cl + NaI \xrightarrow{\text{dry acetone}} CH_3CH_2CH_2CH_2I + NaCl(s)

The sodium chloride formed is insoluble in acetone and precipitates, driving the reaction forward according to Le Chatelier's principle.

(iii)

From but-1-ene This is a two-step conversion. First, but-1-ene is converted to 1-bromobutane using the anti-Markovnikov addition of HBr in the presence of peroxide. Then, 1-bromobutane is converted to 1-iodobutane using the Finkelstein reaction.

Step 1: Anti-Markovnikov addition of HBr

CH_3CH_2CH=CH_2 + HBr \xrightarrow{\text{Peroxide}} CH_3CH_2CH_2CH_2Br

Step 2: Finkelstein Reaction

CH_3CH_2CH_2CH_2Br + NaI \xrightarrow{\text{dry acetone}} CH_3CH_2CH_2CH_2I + NaBr(s)

Q8Exercises

What are ambident nucleophiles? Explain with an example.

Solution

Ambident Nucleophiles: Ambident nucleophiles are nucleophiles that have two nucleophilic centers, meaning they can attack an electrophilic center from two different atoms. Although they have two sites for attack, they attack from only one site at a time.

Explanation with an Example: The cyanide ion ( $CN^−$ ) is a classic example of an ambident nucleophile. It is a hybrid of two contributing resonance structures:

[^\ominus C \equiv N \longleftrightarrow :C = N^\ominus]

Attack through Carbon: It can attack an electrophile (like the carbon atom in an alkyl halide) through the carbon atom, forming an alkyl cyanide (nitrile). This happens when the reaction is carried out with an ionic cyanide like KCN. The C-C bond formed is stronger and more stable. $R-X + KCN \rightarrow R-C \equiv N + KX$ Here, the carbon atom acts as the nucleophilic center.
Attack through Nitrogen: It can also attack through the nitrogen atom, forming an alkyl isocyanide (isonitrile). This is the major product when the reaction is carried out with a covalent cyanide like AgCN. In AgCN, the C-Ag bond is covalent, so the nitrogen atom's lone pair is more available for donation. $R-X + AgCN \rightarrow R-N \equiv C + AgX$ Here, the nitrogen atom acts as the nucleophilic center.

Another common example is the nitrite ion ( $NO_2^−$ ), which can attack through the oxygen atom to form alkyl nitrites (R-O-N=O) or through the nitrogen atom to form nitroalkanes (R-NO2).

Q9Exercises

Which compound in each of the following pairs will react faster in $S_N2$ reaction with $^-OH$ ?

(i)

$CH_3Br$ or $CH_3I$

(ii)

$(CH_3)_3CCl$ or $CH_3Cl$

Solution

The rate of an

S_N2

reaction depends on two main factors: the nature of the leaving group and steric hindrance around the carbon atom being attacked.

(i) $CH_3Br$ or $CH_3I$

In this pair, the alkyl group (methyl) is the same, but the leaving groups are different (Br⁻ vs I⁻). The rate of an

S_N2

reaction increases as the leaving group's ability to depart increases. A better leaving group is a weaker base. The basicity of halide ions decreases down the group:

F^− > Cl^− > Br^− > I^−

. Therefore, iodide (

I^−

) is a much better leaving group than bromide (

Br^−

) because the C-I bond is weaker and longer than the C-Br bond, making it easier to break.

Answer:

CH_3I

will react faster than

CH_3Br

.

(ii) $(CH_3)_3CCl$ or $CH_3Cl$

In this pair, the leaving group (Cl⁻) is the same, but the alkyl groups are different.

S_N2

reactions involve a backside attack by the nucleophile on the carbon atom bonded to the leaving group. The presence of bulky groups on or near this carbon atom creates steric hindrance, which slows down or prevents the nucleophile's approach.

$CH_3Cl$ (Methyl chloride) is a primary halide with three small hydrogen atoms, offering very little steric hindrance.
$(CH_3)_3CCl$ (tert-Butyl chloride) is a tertiary halide with three bulky methyl groups attached to the carbon atom. This causes significant steric hindrance, making the backside attack by the nucleophile virtually impossible.

The order of reactivity for alkyl halides in

S_N2

reactions is: Methyl > Primary (

1^\circ

) > Secondary (

2^\circ

) > Tertiary (

3^\circ

).

Answer:

CH_3Cl

will react much faster than

(CH_3)_3CCl

.

Q10Exercises

Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene:

(i)

1-Bromo-1-methylcyclohexane

(ii)

2-Chloro-2-methylbutane

(iii)

2,2,3-Trimethyl-3-bromopentane.

Solution

Dehydrohalogenation with a strong base like sodium ethoxide follows an elimination (E2) mechanism. The major product is typically the more substituted alkene, as predicted by Zaitsev's rule.

(i) 1-Bromo-1-methylcyclohexane This is a tertiary halide. There are two types of β-hydrogens: on the methyl group and on the C2/C6 positions of the ring.

Elimination of H from the methyl group: gives 1-methylenecyclohexane (less substituted).
Elimination of H from C2 or C6: gives 1-methylcyclohex-1-ene (more substituted).

Products:

1-Methylcyclohex-1-ene (Major)
Methylenecyclohexane (Minor)

According to Zaitsev's rule, the more substituted alkene, 1-methylcyclohex-1-ene, is the major product.

(ii) 2-Chloro-2-methylbutane This is a tertiary halide,

CH_3-C(Cl)(CH_3)-CH_2-CH_3

. There are two types of β-hydrogens: on the C1 methyl groups and on the C3 methylene group.

Elimination of H from C1 ( $CH_3$ ): gives 2-methylbut-1-ene (disubstituted alkene).
Elimination of H from C3 ( $-CH_2-$ ): gives 2-methylbut-2-ene (trisubstituted alkene).

Products:

2-Methylbut-2-ene (Major)
2-Methylbut-1-ene (Minor)

According to Zaitsev's rule, the more substituted (trisubstituted) alkene, 2-methylbut-2-ene, is the major product.

(iii) 2,2,3-Trimethyl-3-bromopentane This is a tertiary halide,

(CH_3)_3C-CH(Br)-CH_2-CH_3

. There are two types of β-hydrogens: on the C2 carbon (no hydrogens) and on the C4 methylene group. Wait, let's re-draw the structure:

CH_3-CH_2-C(Br)(CH_3)-C(CH_3)_3

. There are β-hydrogens on the C2 methylene group and C4 carbon (part of tert-butyl group, no hydrogens). Let me correct the structure from the name: 2,2,3-Trimethyl-3-bromopentane is

CH_3CH_2-C(Br)(CH_3)-CH(CH_3)_2

. No, that's not right. The name implies a pentane chain. Structure:

CH_3-CH_2-C(Br)(CH_3)-C(CH_3)_2-CH_3

. This is 3-bromo-3,4,4-trimethylpentane. Let's assume the question meant 3-Bromo-2,2,3-trimethylpentane:

(CH_3)_3C-C(Br)(CH_3)-CH_2CH_3

. The β-hydrogens are on the C4 methylene group. There are no β-hydrogens on the C2 carbon.

Elimination of H from C4 ( $-CH_2-$ ): gives 3,4,4-trimethylpent-2-ene (tetrasubstituted alkene).

Products:

3,4,4-Trimethylpent-2-ene (Major)
2-ethyl-3,3-dimethylbut-1-ene (Minor) (This would be the Hofmann product, formed by abstracting a proton from the less hindered ethyl group). Wait, the structure is $CH_3-CH(CH_3)-C(Br)(CH_3)-CH_2-CH_3$ . This is 3-Bromo-3,4-dimethylhexane. The question is 2,2,3-Trimethyl-3-bromopentane. Let's draw it correctly: Pentane chain: C-C-C-C-C 3-bromo: Br on C3 2,2,3-trimethyl: two $CH_3$ on C2, one $CH_3$ on C3 Structure: $CH_3-C(CH_3)_2-C(Br)(CH_3)-CH_2-CH_3$ β-hydrogens are at C2 (none) and C4 (two hydrogens on $-CH_2-$ ).

Elimination of H from C4: gives 3,4,4-trimethylpent-2-ene.

Product: Only one alkene is possible: 3,4,4-trimethylpent-2-ene. Since it is the only product, it is also the major product.

Q11Exercises

How will you bring about the following conversions?

(i)

Ethanol to but-1-yne

(ii)

Ethane to bromoethene

(iii)

Propene to 1-nitropropane

(iv)

Toluene to benzyl alcohol

(v)

Propene to propyne

(vi)

Ethanol to ethyl fluoride

(vii)

Bromomethane to propanone

(viii)

But-1-ene to but-2-ene

(ix)

1-Chlorobutane to n-octane

(x)

Benzene to biphenyl.

Solution

(i)

Ethanol to but-1-yne

CH_3CH_2OH \xrightarrow{SOCl_2} CH_3CH_2Cl \xrightarrow{HC\equiv CNa, \text{liq. } NH_3} CH_3CH_2C\equiv CH

(ii)

Ethane to bromoethene

CH_3CH_3 \xrightarrow{Br_2, \text{UV light}} CH_3CH_2Br \xrightarrow{\text{alc. } KOH} CH_2=CH_2 \xrightarrow{Br_2, CCl_4} BrCH_2CH_2Br \xrightarrow{\text{alc. } KOH, \Delta} CH_2=CHBr

(iii)

Propene to 1-nitropropane

CH_3CH=CH_2 \xrightarrow{HBr, \text{Peroxide}} CH_3CH_2CH_2Br \xrightarrow{AgNO_2, \text{ethanol}} CH_3CH_2CH_2NO_2

(iv)

Toluene to benzyl alcohol

C_6H_5CH_3 \xrightarrow{Cl_2, \text{UV light}} C_6H_5CH_2Cl \xrightarrow{\text{aq. } KOH} C_6H_5CH_2OH

(v)

Propene to propyne

CH_3CH=CH_2 \xrightarrow{Br_2, CCl_4} CH_3CH(Br)CH_2Br \xrightarrow{\text{alc. } KOH, \Delta} CH_3C\equiv CH \text{ (via } CH_3C(Br)=CH_2) \xrightarrow{NaNH_2, \text{liq. } NH_3} \text{ (if needed for further steps)}

The double elimination with alcoholic KOH is sufficient. A stronger base like

NaNH_2

ensures complete conversion.

CH_3CH(Br)CH_2Br \xrightarrow{NaNH_2, \text{liq. } NH_3} CH_3C\equiv CH

(vi)

Ethanol to ethyl fluoride

CH_3CH_2OH \xrightarrow{PBr_3} CH_3CH_2Br \xrightarrow{AgF, \text{heat (Swarts Reaction)}} CH_3CH_2F

(vii)

Bromomethane to propanone

CH_3Br \xrightarrow{KCN} CH_3CN \xrightarrow{CH_3MgBr, \text{ether}} CH_3C(CH_3)=NMgBr \xrightarrow{H_3O^+} CH_3COCH_3

(viii)

But-1-ene to but-2-ene

CH_3CH_2CH=CH_2 \xrightarrow{HBr, \text{(Markovnikov)}} CH_3CH_2CH(Br)CH_3 \xrightarrow{\text{alc. } KOH, \Delta} CH_3CH=CHCH_3 \text{ (Major product)}

(ix)

1-Chlorobutane to n-octane This is the Wurtz reaction.

2 CH_3CH_2CH_2CH_2Cl + 2Na \xrightarrow{\text{dry ether}} CH_3(CH_2)_6CH_3 + 2NaCl

(x)

Benzene to biphenyl This is the Fittig reaction.

C_6H_6 \xrightarrow{Br_2, FeBr_3} C_6H_5Br \xrightarrow{2Na, \text{dry ether}} C_6H_5-C_6H_5 + 2NaBr

Q12Exercises

Explain why

(i)

the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride?

(ii)

alkyl halides, though polar, are immiscible with water?

(iii)

Grignard reagents should be prepared under anhydrous conditions?

Solution

(i)

Dipole moment of chlorobenzene is lower than that of cyclohexyl chloride. There are two main reasons for this:

Hybridization: In chlorobenzene, the carbon atom attached to chlorine is $sp^2$ -hybridized, while in cyclohexyl chloride, it is $sp^3$ -hybridized. An $sp^2$ carbon has more s-character (33.3%) than an $sp^3$ carbon (25%). This makes the $sp^2$ carbon more electronegative. As a result, it has a reduced tendency to release electrons to the chlorine atom, which decreases the polarity of the C-Cl bond and hence the dipole moment.
Resonance: In chlorobenzene, the lone pair of electrons on the chlorine atom is in conjugation with the $\pi$ -electrons of the benzene ring. This delocalization (resonance) gives the C-Cl bond a partial double-bond character. This movement of electrons from chlorine to the ring opposes the normal inductive electron withdrawal by chlorine, further reducing the bond's polarity. $C_6H_5-Cl \leftrightarrow C_6H_5^+=Cl^-$ In contrast, cyclohexyl chloride has no such resonance effect. Due to these combined effects, the dipole moment of chlorobenzene (1.69 D) is lower than that of cyclohexyl chloride (2.15 D).

(ii)

Alkyl halides, though polar, are immiscible with water. For a substance to dissolve in water, the energy released from forming new solute-water attractions must be sufficient to overcome the energy required to break the original solute-solute attractions and the strong hydrogen bonds between water molecules.

Alkyl halides are polar molecules, so they have dipole-dipole attractions between them.
Water molecules are held together by strong intermolecular hydrogen bonds. When an alkyl halide is mixed with water, new attractions are formed between the alkyl halide and water molecules. However, these attractions are not as strong as the original hydrogen bonds in water. A significant amount of energy is needed to break the hydrogen bonds in water, and not enough energy is released by forming the new, weaker alkyl halide-water attractions. Consequently, alkyl halides are not able to break the hydrogen bond network of water and are therefore immiscible (or very slightly soluble) in it.

(iii)

Grignard reagents should be prepared under anhydrous conditions. Grignard reagents (R-Mg-X) are organometallic compounds containing a highly polar carbon-magnesium bond (

C^{\delta-}-Mg^{\delta+}

). This makes the carbon atom strongly nucleophilic and also a very strong base. Grignard reagents react readily with any source of protons (protic compounds) to form hydrocarbons. Water is a protic solvent and is acidic enough to destroy the Grignard reagent.

The reaction with water is as follows:

R-Mg-X + H-OH \rightarrow R-H + Mg(OH)X

(Grignard reagent) + (Water)

\rightarrow

(Alkane) + (Hydroxy magnesium halide)

Even trace amounts of moisture from the apparatus, solvent (ether), or alkyl halide will react with and consume the Grignard reagent as it forms. Therefore, to ensure the successful preparation and use of Grignard reagents, all reactants and apparatus must be scrupulously dry (anhydrous).

Q13Exercises

Give the uses of freon 12, DDT, carbon tetrachloride and iodoform.

Solution

Uses of Freon 12 ( $CCl_2F_2$ )

Refrigerant: Used extensively in refrigeration and air conditioning systems due to its low boiling point, non-toxic, and non-corrosive nature.
Aerosol Propellant: Used as a propellant in aerosol cans for products like deodorants, hairsprays, and insecticides.
Foam Blowing Agent: Used in the production of foam insulation. (Note: Its use is now heavily restricted due to its role in ozone layer depletion).

Uses of DDT ( $p,p'$ -Dichlorodiphenyltrichloroethane)

Insecticide: It was a highly effective and widely used insecticide, particularly after World War II.
Disease Control: It was instrumental in controlling insect-borne diseases like malaria (by killing mosquitoes) and typhus (by killing lice). (Note: Its use is banned in many countries due to its environmental persistence, bioaccumulation, and toxicity to wildlife).

Uses of Carbon Tetrachloride ( $CCl_4$ )

Solvent: Used as an industrial solvent for oils, fats, resins, and as a degreasing agent.
Fire Extinguisher: Previously used in fire extinguishers under the name Pyrene, especially for electrical fires.
Chemical Synthesis: Used as a feedstock in the synthesis of refrigerants (like freons) and other chemicals.
Spot Remover: Formerly used as a cleaning fluid and spot remover in households. (Note: Its use is now greatly reduced due to its toxicity, particularly its carcinogenic effects on the liver, and its role in ozone depletion).

Uses of Iodoform ( $CHI_3$ )

Antiseptic: It was formerly used as an antiseptic and disinfectant for dressing wounds. Its antiseptic properties are not due to iodoform itself but due to the slow liberation of free iodine upon contact with skin.
Medical Use: It is now largely replaced by other formulations containing iodine that have a less objectionable smell.

Q14Exercises

Write the structure of the major organic product in each of the following reactions:

(i)

$CH_3CH_2CH_2Cl + NaI \xrightarrow{acetone, heat}$

(ii)

$(CH_3)_3CBr + KOH \xrightarrow{ethanol, heat}$

(iii)

$CH_3CH(Br)CH_2CH_3 + NaOH \xrightarrow{water}$

(iv)

$CH_3CH_2Br + KCN \xrightarrow{aq. ethanol}$

(v)

$C_6H_5ONa + C_2H_5Cl \rightarrow$

(vi)

$CH_3CH_2CH_2OH + SOCl_2 \rightarrow$

(vii)

$CH_3CH_2CH=CH_2 + HBr \xrightarrow{peroxide}$

(viii)

$CH_3CH=C(CH_3)_2 + HBr \rightarrow$

Solution

(i)

Reaction: Finkelstein reaction (Halogen exchange). Product: 1-Iodopropane Structure:

CH_3CH_2CH_2I

(ii)

Reaction: Elimination (E2) reaction of a tertiary halide with a strong base in ethanol. Product: 2-Methylpropene (Isobutylene) Structure:

(CH_3)_2C=CH_2

(iii)

Reaction: Nucleophilic substitution (

S_N2

) of a secondary halide with a strong nucleophile in a polar protic solvent (water). Substitution is favored over elimination at lower temperatures with aqueous NaOH. Product: Butan-2-ol Structure:

CH_3CH(OH)CH_2CH_3

(iv)

Reaction: Nucleophilic substitution (

S_N2

) with cyanide ion. Product: Propanenitrile (Ethyl cyanide) Structure:

CH_3CH_2CN

(v)

Reaction: Williamson ether synthesis. Sodium phenoxide acts as a nucleophile. Product: Phenetole (Ethoxybenzene) Structure:

C_6H_5OC_2H_5

(vi)

Reaction: Conversion of an alcohol to an alkyl chloride using thionyl chloride. Product: 1-Chloropropane Structure:

CH_3CH_2CH_2Cl

(vii)

Reaction: Anti-Markovnikov addition of HBr to an alkene in the presence of peroxide. Product: 1-Bromobutane Structure:

CH_3CH_2CH_2CH_2Br

(viii)

Reaction: Markovnikov addition of HBr to an unsymmetrical alkene. The H adds to the carbon with more hydrogens, and Br adds to the more substituted carbon (which forms a more stable carbocation intermediate). Product: 2-Bromo-2-methylbutane Structure:

CH_3CH_2C(Br)(CH_3)_2

Q15Exercises

Write the mechanism of the following reaction: $nBuBr + KCN \xrightarrow{EtOH-H_2O} nBuCN$

Solution

The reaction is a nucleophilic substitution of 1-bromobutane (a primary alkyl halide) with cyanide ion. This reaction proceeds via the $S_N2$ (Substitution Nucleophilic Bimolecular) mechanism.

Reactants:

Substrate: n-Butyl bromide ( $CH_3CH_2CH_2CH_2Br$ )
Nucleophile: Cyanide ion ( $CN^−$ ), provided by the ionic compound KCN.

Mechanism: The

S_N2

mechanism is a single-step concerted process. The attack of the nucleophile and the departure of the leaving group occur simultaneously.

Step 1: Backside Attack The cyanide ion (

CN^−

), a strong nucleophile, attacks the electrophilic carbon atom (the one bonded to bromine) from the side opposite to the leaving group (bromide ion). This is called a backside attack.

Step 2: Transition State Formation As the nucleophile approaches, a transition state is formed where the C-CN bond is partially formed and the C-Br bond is partially broken. The carbon atom is momentarily bonded to five groups. The three non-reacting groups (two H atoms and one ethyl group) on the carbon are in a planar arrangement, perpendicular to the attacking and leaving groups.

Step 3: Inversion of Configuration and Product Formation The C-Br bond breaks completely, and the bromide ion departs as the leaving group. The C-CN bond forms completely. The configuration of the carbon atom is inverted, much like an umbrella turning inside out in the wind. This is known as Walden inversion.

Diagrammatic Representation of the Mechanism:

Attack of Nucleophile: The nucleophile $N\equiv C^−$ approaches the n-butyl bromide molecule.
Transition State: A transition state is formed: $[N\equiv C^{...}CH_2(CH_2CH_3)^{...}Br]^{\delta-}$ The dotted lines represent partial bonds.
Product Formation: The final product, n-butyl cyanide (pentanenitrile), is formed along with the bromide ion. $CH_3CH_2CH_2CH_2CN + Br^−$

The overall reaction is:

CH_3CH_2CH_2CH_2Br + CN^− \xrightarrow{S_N2} CH_3CH_2CH_2CH_2CN + Br^−

Q16Exercises

Arrange the compounds of each set in order of reactivity towards $S_N2$ displacement:

(i)

2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane

(ii)

1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 2-Bromo-3-methylbutane

(iii)

1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methylbutane, 1-Bromo-3-methylbutane.

Solution

The reactivity towards an

S_N2

reaction is primarily governed by steric hindrance. The less sterically hindered the carbon atom bearing the halogen, the faster the reaction. The general order of reactivity is: Methyl > Primary (

1^\circ

) > Secondary (

2^\circ

) > Tertiary (

3^\circ

).

(i) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane

1-Bromopentane: Primary halide ( $1^\circ$ ), least steric hindrance.
2-Bromopentane: Secondary halide ( $2^\circ$ ), more steric hindrance than primary.
2-Bromo-2-methylbutane: Tertiary halide ( $3^\circ$ ), most steric hindrance, essentially unreactive in $S_N2$ .

Order of reactivity: 1-Bromopentane > 2-Bromopentane > 2-Bromo-2-methylbutane

(ii) 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 2-Bromo-3-methylbutane

1-Bromo-3-methylbutane: Primary halide ( $1^\circ$ ).
2-Bromo-3-methylbutane: Secondary halide ( $2^\circ$ ).
2-Bromo-2-methylbutane: Tertiary halide ( $3^\circ$ ).

Order of reactivity: 1-Bromo-3-methylbutane > 2-Bromo-3-methylbutane > 2-Bromo-2-methylbutane

(iii) 1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methylbutane, 1-Bromo-3-methylbutane. All are primary (

1^\circ

) alkyl halides. In this case, we compare the steric hindrance caused by branching at the

\beta

-carbon (the carbon adjacent to the one with the halogen) and further down the chain.

1-Bromobutane: Unbranched chain, least steric hindrance.
1-Bromo-3-methylbutane: Branching is on the $\gamma$ -carbon (C3), far from the reaction center. Steric hindrance is minimal, slightly more than 1-bromobutane.
1-Bromo-2-methylbutane: Branching is on the $\beta$ -carbon (C2) with one methyl group. This creates more steric hindrance.
1-Bromo-2,2-dimethylpropane (Neopentyl bromide): Branching is on the $\beta$ -carbon with two methyl groups (a quaternary carbon). This creates very high steric hindrance, making it the least reactive.

Order of reactivity: 1-Bromobutane > 1-Bromo-3-methylbutane > 1-Bromo-2-methylbutane > 1-Bromo-2,2-dimethylpropane

Q17Exercises

Out of $C_6H_5CH_2Cl$ and $C_6H_5CHClC_6H_5$ , which is more easily hydrolysed by aqueous KOH.

Solution

Answer:

C_6H_5CHClC_6H_5

is more easily hydrolysed.

Explanation: Hydrolysis by aqueous KOH is a nucleophilic substitution reaction. The reactivity can be explained by considering the stability of the carbocation intermediate formed if the reaction proceeds via an

S_N1

mechanism. The stability of the carbocation is the key factor for benzylic halides.

$C_6H_5CH_2Cl$ (Benzyl chloride): Upon hydrolysis, it would form a benzyl carbocation ( $C_6H_5CH_2^+$ ). $C_6H_5CH_2Cl \rightarrow C_6H_5CH_2^+ + Cl^-$ This is a primary carbocation, but it is highly stabilized by resonance with the benzene ring. The positive charge is delocalized over the entire ring.
$C_6H_5CHClC_6H_5$ (Diphenylmethyl chloride or Benzhydryl chloride): Upon hydrolysis, it would form a diphenylmethyl carbocation ( $(C_6H_5)_2CH^+$ ). $C_6H_5CHClC_6H_5 \rightarrow (C_6H_5)_2CH^+ + Cl^-$ This is a secondary carbocation, and the positive charge is delocalized over two benzene rings. The extent of resonance stabilization is much greater than in the benzyl carbocation.

Comparison: The diphenylmethyl carbocation is significantly more stable than the benzyl carbocation because the charge is dispersed over two phenyl rings instead of one. A more stable carbocation intermediate means a lower activation energy for its formation, leading to a faster reaction rate.

Therefore,

C_6H_5CHClC_6H_5

will be hydrolysed more easily (faster) than

C_6H_5CH_2Cl

via the

S_N1

mechanism.

Q18Exercises

$p$ -Dichlorobenzene has higher m.p. than those of $o$ - and $m$ -isomers. Discuss.

Solution

The boiling points of ortho-, meta-, and para-dichlorobenzene are very close to each other (453 K, 446 K, and 448 K, respectively) because they have similar molecular masses and polarities.

However, their melting points show a significant difference:

o-Dichlorobenzene: 256 K
m-Dichlorobenzene: 249 K
p-Dichlorobenzene: 323 K

The melting point of a crystalline solid depends not only on the strength of intermolecular forces but also on how well the molecules fit into a crystal lattice.

Structure of p-Dichlorobenzene: The para-isomer is highly symmetrical. The two chlorine atoms are directly opposite each other on the benzene ring. This linear and symmetrical shape allows the molecules to pack very closely and efficiently in the crystal lattice.
Structure of o- and m-Dichlorobenzene: The ortho- and meta-isomers are less symmetrical. Their bent or 'kinked' shapes prevent them from packing as tightly in the crystal lattice.

Conclusion: Because the molecules of p-dichlorobenzene can pack more closely in the solid state, the intermolecular forces of attraction are stronger and more effective. More energy is required to break down this well-ordered crystal lattice, resulting in a significantly higher melting point compared to its less symmetrical ortho- and meta-isomers.

Q19Exercises

How the following conversions can be carried out?

(i)

Propene to propan-1-ol

(ii)

Ethanol to but-1-yne

(iii)

1-Bromopropane to 2-bromopropane

(iv)

Toluene to benzyl alcohol

(v)

Benzene to 4-bromonitrobenzene

(vi)

Benzyl alcohol to 2-phenylethanoic acid

(vii)

Ethanol to propanenitrile

(viii)

Aniline to chlorobenzene

(ix)

2-Chlorobutane to 3, 4-dimethylhexane

(x)

2-Methyl-1-propene to 2-chloro-2-methylpropane (xi) Ethyl chloride to propanoic acid (xii) But-1-ene to n-butyliodide (xiii) 2-Chloropropane to 1-propanol (xiv) Isopropyl alcohol to iodoform (xv) Chlorobenzene to $p$ -nitrophenol (xvi) 2-Bromopropane to 1-bromopropane (xvii) Chloroethane to butane (xviii) Benzene to diphenyl (xix) tert-Butyl bromide to isobutyl bromide (xx) Aniline to phenylisocyanide

Solution

(i)

Propene to propan-1-ol: (Hydroboration-oxidation)

CH_3CH=CH_2 \xrightarrow{1. B_2H_6, THF} \xrightarrow{2. H_2O_2, OH^-} CH_3CH_2CH_2OH

(ii)

Ethanol to but-1-yne:

CH_3CH_2OH \xrightarrow{SOCl_2} CH_3CH_2Cl \xrightarrow{HC\equiv CNa, \text{liq. } NH_3} CH_3CH_2C\equiv CH

(iii)

1-Bromopropane to 2-bromopropane:

CH_3CH_2CH_2Br \xrightarrow{\text{alc. } KOH, \Delta} CH_3CH=CH_2 \xrightarrow{HBr, \text{(Markovnikov)}} CH_3CH(Br)CH_3

(iv)

Toluene to benzyl alcohol:

C_6H_5CH_3 \xrightarrow{Cl_2, \text{UV light}} C_6H_5CH_2Cl \xrightarrow{\text{aq. } KOH} C_6H_5CH_2OH

(v)

Benzene to 4-bromonitrobenzene:

C_6H_6 \xrightarrow{Br_2, FeBr_3} C_6H_5Br \xrightarrow{\text{conc. } HNO_3 + \text{conc. } H_2SO_4} p-BrC_6H_4NO_2 + o-BrC_6H_4NO_2

(p-isomer is major and can be separated).

(vi)

Benzyl alcohol to 2-phenylethanoic acid:

C_6H_5CH_2OH \xrightarrow{PCl_5} C_6H_5CH_2Cl \xrightarrow{KCN, \text{ethanol}} C_6H_5CH_2CN \xrightarrow{H_3O^+, \Delta} C_6H_5CH_2COOH

(vii)

Ethanol to propanenitrile:

CH_3CH_2OH \xrightarrow{PBr_3} CH_3CH_2Br \xrightarrow{KCN, \text{ethanol}} CH_3CH_2CN

(viii)

Aniline to chlorobenzene: (Sandmeyer reaction)

C_6H_5NH_2 \xrightarrow{NaNO_2 + HCl, 273-278 K} C_6H_5N_2^+Cl^- \xrightarrow{CuCl/HCl} C_6H_5Cl

(ix)

2-Chlorobutane to 3, 4-dimethylhexane: (Wurtz reaction)

2 CH_3CH_2CH(Cl)CH_3 \xrightarrow{2Na, \text{dry ether}} CH_3CH_2CH(CH_3)-CH(CH_3)CH_2CH_3 + 2NaCl

(x)

2-Methyl-1-propene to 2-chloro-2-methylpropane:

(CH_3)_2C=CH_2 \xrightarrow{HCl, \text{(Markovnikov)}} (CH_3)_3CCl

(xi) Ethyl chloride to propanoic acid:

CH_3CH_2Cl \xrightarrow{KCN, \text{ethanol}} CH_3CH_2CN \xrightarrow{H_3O^+, \Delta} CH_3CH_2COOH

(xii) But-1-ene to n-butyliodide:

CH_3CH_2CH=CH_2 \xrightarrow{HBr, \text{Peroxide}} CH_3CH_2CH_2CH_2Br \xrightarrow{NaI, \text{acetone}} CH_3CH_2CH_2CH_2I

(xiii) 2-Chloropropane to 1-propanol:

CH_3CH(Cl)CH_3 \xrightarrow{\text{alc. } KOH, \Delta} CH_3CH=CH_2 \xrightarrow{1. B_2H_6} \xrightarrow{2. H_2O_2, OH^-} CH_3CH_2CH_2OH

(xiv) Isopropyl alcohol to iodoform: (Iodoform test)

CH_3CH(OH)CH_3 + 4I_2 + 6NaOH \rightarrow CHI_3 \downarrow \text{(yellow ppt)} + CH_3COONa + 5NaI + 5H_2O

(xv) Chlorobenzene to p-nitrophenol:

C_6H_5Cl \xrightarrow{\text{conc. } HNO_3 + \text{conc. } H_2SO_4} p-NO_2C_6H_4Cl \xrightarrow{1. \text{aq. } NaOH, 443 K} \xrightarrow{2. \text{dil. } HCl} p-NO_2C_6H_4OH

(xvi) 2-Bromopropane to 1-bromopropane:

CH_3CH(Br)CH_3 \xrightarrow{\text{alc. } KOH, \Delta} CH_3CH=CH_2 \xrightarrow{HBr, \text{Peroxide}} CH_3CH_2CH_2Br

(xvii) Chloroethane to butane: (Wurtz reaction)

2 CH_3CH_2Cl + 2Na \xrightarrow{\text{dry ether}} CH_3CH_2CH_2CH_3 + 2NaCl

(xviii) Benzene to diphenyl: (Fittig reaction)

C_6H_6 \xrightarrow{Br_2, FeBr_3} C_6H_5Br \xrightarrow{2Na, \text{dry ether}} C_6H_5-C_6H_5 + 2NaBr

(xix) tert-Butyl bromide to isobutyl bromide:

(CH_3)_3CBr \xrightarrow{\text{alc. } KOH, \Delta} (CH_3)_2C=CH_2 \xrightarrow{HBr, \text{Peroxide}} (CH_3)_2CHCH_2Br

(xx) Aniline to phenylisocyanide: (Carbylamine reaction)

C_6H_5NH_2 + CHCl_3 + 3KOH \text{(alc.)} \xrightarrow{\Delta} C_6H_5NC + 3KCl + 3H_2O

Q20Exercises

The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH, alkenes are major products. Explain.

Solution

The reaction of an alkyl chloride with KOH can proceed via two competing pathways: substitution and elimination. The choice of solvent (aqueous vs. alcoholic) plays a crucial role in determining which pathway dominates.

With Aqueous KOH: In an aqueous solution, KOH is completely ionized to give hydroxide ions ( $OH^−$ ). Water is a polar protic solvent, which solvates the hydroxide ions effectively through hydrogen bonding. This solvation reduces the basic character of the $OH^−$ ion. While still a strong base, in this environment, it acts primarily as a strong nucleophile. As a nucleophile, it attacks the partially positive carbon atom of the alkyl chloride, leading to a nucleophilic substitution reaction ( $S_N1$ or $S_N2$ ), replacing the chloride ion to form an alcohol. $R-Cl + OH^− (aq) \xrightarrow{\text{Substitution}} R-OH + Cl^−$
With Alcoholic KOH: In an alcoholic solution (e.g., ethanol), KOH reacts with the solvent to form alkoxide ions, such as ethoxide ions ( $C_2H_5O^−$ ). $KOH + C_2H_5OH \rightleftharpoons C_2H_5O^−K^+ + H_2O$ The ethoxide ion is a much stronger base than the hydroxide ion. It is also a stronger nucleophile, but its bulky nature and the less polar nature of the alcoholic solvent favor its role as a base. As a strong base, it preferentially abstracts a $\beta$ -hydrogen atom from the alkyl chloride, leading to an elimination reaction (E2), which forms an alkene. $C_2H_5O^− + H-CHR-CHR-Cl \xrightarrow{\text{Elimination}} CH_2=CH_2 + C_2H_5OH + Cl^−$

Summary:

Aqueous KOH: Provides $OH^−$ ions that are better nucleophiles than bases, favoring substitution to form alcohols.
Alcoholic KOH: Provides alkoxide ions (e.g., $C_2H_5O^−$ ) that are very strong bases, favoring elimination to form alkenes.

Q21Exercises

Primary alkyl halide $C_4H_9Br$ (a) reacted with alcoholic KOH to give compound (b). Compound (b) is reacted with HBr to give (c) which is an isomer of (a). When (a) is reacted with sodium metal it gives compound (d), $C_8H_{18}$ which is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions.

Solution

Step-by-step deduction:

Identify Compound (a):
- (a) is a primary alkyl halide with the formula $C_4H_9Br$ . The possible primary isomers are 1-bromobutane (n-butyl bromide) and 1-bromo-2-methylpropane (isobutyl bromide).
- The Wurtz reaction of (a) with sodium gives compound (d), $C_8H_{18}$ .
- The problem states that (d) is different from the compound formed when n-butyl bromide reacts with sodium. The Wurtz reaction of n-butyl bromide gives n-octane.
- Therefore, compound (a) cannot be n-butyl bromide.
- So, compound (a) must be 1-bromo-2-methylpropane (isobutyl bromide).
Identify Compounds (b), (c), and (d) and write the reactions:
- Structural Formula of (a): $(CH_3)_2CH-CH_2-Br$ (1-Bromo-2-methylpropane)
- Reaction for (b): (a) reacts with alcoholic KOH (elimination). $(CH_3)_2CH-CH_2-Br \xrightarrow{\text{alc. } KOH, \Delta} (CH_3)_2C=CH_2 + KBr + H_2O$ Compound (b) is 2-Methylpropene.
- Reaction for (c): (b) reacts with HBr (Markovnikov addition). $(CH_3)_2C=CH_2 + HBr \rightarrow (CH_3)_3C-Br$ Compound (c) is 2-Bromo-2-methylpropane (tert-butyl bromide). This is a tertiary halide and is an isomer of (a), as stated in the problem.
- Reaction for (d): (a) reacts with sodium metal (Wurtz reaction). $2 (CH_3)_2CH-CH_2-Br + 2Na \xrightarrow{\text{dry ether}} (CH_3)_2CH-CH_2-CH_2-CH(CH_3)_2 + 2NaBr$ Compound (d) is 2,5-Dimethylhexane. This is different from n-octane, which would be formed from n-butyl bromide.

Summary of Structures and Equations:

Compound (a): 1-Bromo-2-methylpropane, $(CH_3)_2CHCH_2Br$
Compound (b): 2-Methylpropene, $(CH_3)_2C=CH_2$
Compound (c): 2-Bromo-2-methylpropane, $(CH_3)_3CBr$
Compound (d): 2,5-Dimethylhexane, $(CH_3)_2CHCH_2CH_2CH(CH_3)_2$

Equations:

$(CH_3)_2CHCH_2Br \text{ (a)} + \text{alc. } KOH \rightarrow (CH_3)_2C=CH_2 \text{ (b)}$
$(CH_3)_2C=CH_2 \text{ (b)} + HBr \rightarrow (CH_3)_3CBr \text{ (c)}$
$2(CH_3)_2CHCH_2Br \text{ (a)} + 2Na \xrightarrow{\text{dry ether}} (CH_3)_2CHCH_2CH_2CH(CH_3)_2 \text{ (d)}$

Q22Exercises

What happens when

(i)

n-butyl chloride is treated with alcoholic KOH,

(ii)

bromobenzene is treated with Mg in the presence of dry ether,

(iii)

chlorobenzene is subjected to hydrolysis,

(iv)

ethyl chloride is treated with aqueous KOH,

(v)

methyl bromide is treated with sodium in the presence of dry ether,

(vi)

methyl chloride is treated with KCN?

Solution

(i)

n-butyl chloride is treated with alcoholic KOH: An elimination reaction (dehydrohalogenation) occurs, forming but-1-ene as the major product.

CH_3CH_2CH_2CH_2Cl + KOH(\text{alc.}) \xrightarrow{\Delta} CH_3CH_2CH=CH_2 + KCl + H_2O

(ii)

bromobenzene is treated with Mg in the presence of dry ether: A Grignard reagent, phenylmagnesium bromide, is formed.

C_6H_5Br + Mg \xrightarrow{\text{dry ether}} C_6H_5MgBr

(iii)

chlorobenzene is subjected to hydrolysis: Chlorobenzene is very resistant to hydrolysis due to the partial double-bond character of the C-Cl bond. No reaction occurs under normal conditions. Hydrolysis to phenol requires drastic conditions (Dow's process): high temperature (623 K) and high pressure (300 atm) with aqueous NaOH.

C_6H_5Cl + H_2O \xrightarrow{\text{No reaction under normal conditions}}

(iv)

ethyl chloride is treated with aqueous KOH: A nucleophilic substitution reaction occurs, forming ethanol.

CH_3CH_2Cl + KOH(\text{aq.}) \rightarrow CH_3CH_2OH + KCl

(v)

methyl bromide is treated with sodium in the presence of dry ether: The Wurtz reaction occurs, forming ethane gas.

2CH_3Br + 2Na \xrightarrow{\text{dry ether}} CH_3-CH_3 + 2NaBr

(vi)

methyl chloride is treated with KCN: A nucleophilic substitution reaction (

S_N2

) occurs, forming ethanenitrile (acetonitrile or methyl cyanide).

CH_3Cl + KCN \rightarrow CH_3CN + KCl

Haloalkanes and HaloarenesClass 12 Chemistry NCERT Solutions

Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:

(CH3)2CHCH(Cl)CH3(CH_3)_2CHCH(Cl)CH_3(CH3​)2​CHCH(Cl)CH3​

CH3CH2CH(CH3)CH(C2H5)ClCH_3CH_2CH(CH_3)CH(C_2H_5)ClCH3​CH2​CH(CH3​)CH(C2​H5​)Cl

CH3CH2C(CH3)2CH2ICH_3CH_2C(CH_3)_2CH_2ICH3​CH2​C(CH3​)2​CH2​I

(CH3)3CCH2CH(Br)C6H5(CH_3)_3CCH_2CH(Br)C_6H_5(CH3​)3​CCH2​CH(Br)C6​H5​

CH3CH(CH3)CH(Br)CH3CH_3CH(CH_3)CH(Br)CH_3CH3​CH(CH3​)CH(Br)CH3​

CH3C(C2H5)2CH2BrCH_3C(C_2H_5)_2CH_2BrCH3​C(C2​H5​)2​CH2​Br

CH3C(Cl)(C2H5)CH2CH3CH_3C(Cl)(C_2H_5)CH_2CH_3CH3​C(Cl)(C2​H5​)CH2​CH3​

CH3CH=C(Cl)CH2CH(CH3)2CH_3CH=C(Cl)CH_2CH(CH_3)_2CH3​CH=C(Cl)CH2​CH(CH3​)2​

CH3CH=CHC(Br)(CH3)2CH_3CH=CHC(Br)(CH_3)_2CH3​CH=CHC(Br)(CH3​)2​

p−ClC6H4CH2CH(CH3)2p-ClC_6H_4CH_2CH(CH_3)_2p−ClC6​H4​CH2​CH(CH3​)2​ (xi) m−ClCH2C6H4CH2C(CH3)3m-ClCH_2C_6H_4CH_2C(CH_3)_3m−ClCH2​C6​H4​CH2​C(CH3​)3​ (xii) o−Br−C6H4CH(CH3)CH2CH3o-Br-C_6H_4CH(CH_3)CH_2CH_3o−Br−C6​H4​CH(CH3​)CH2​CH3​

Solution

Give the IUPAC names of the following compounds:

CH3CH(Cl)CH(Br)CH3CH_3CH(Cl)CH(Br)CH_3CH3​CH(Cl)CH(Br)CH3​

CHF2CBrClFCHF_2CBrClFCHF2​CBrClF

ClCH2C≡CCH2BrClCH_2C\equiv CCH_2BrClCH2​C≡CCH2​Br

(CCl3)3CCl(CCl_3)_3CCl(CCl3​)3​CCl

CH3C(p−ClC6H4)2CH(Br)CH3CH_3C(p-ClC_6H_4)_2CH(Br)CH_3CH3​C(p−ClC6​H4​)2​CH(Br)CH3​

(CH3)3CCH=CClC6H4I−p(CH_3)_3CCH=CClC_6H_4I-p(CH3​)3​CCH=CClC6​H4​I−p

Solution

Write the structures of the following organic halogen compounds.

2-Chloro-3-methylpentane

ppp-Bromochlorobenzene

1-Chloro-4-ethylcyclohexane

2-(2-Chlorophenyl)-1-iodooctane

2-Bromobutane

4-tert-Butyl-3-iodoheptane

1-Bromo-4-sec-butyl-2-methylbenzene

1,4-Dibromobut-2-ene

Solution

Which one of the following has the highest dipole moment?

CH2Cl2CH_2Cl_2CH2​Cl2​

CHCl3CHCl_3CHCl3​

CCl4CCl_4CCl4​

Solution

A hydrocarbon C5H10C_5H_{10}C5​H10​ does not react with chlorine in dark but gives a single monochloro compound C5H9ClC_5H_9ClC5​H9​Cl in bright sunlight. Identify the hydrocarbon.

Solution

Write the isomers of the compound having formula C4H9BrC_4H_9BrC4​H9​Br.

Solution

Write the equations for the preparation of 1-iodobutane from

1-butanol

1-chlorobutane

but-1-ene.

Solution

What are ambident nucleophiles? Explain with an example.

Solution

Which compound in each of the following pairs will react faster in SN2S_N2SN​2 reaction with −OH^-OH−OH?

CH3BrCH_3BrCH3​Br or CH3ICH_3ICH3​I

(CH3)3CCl(CH_3)_3CCl(CH3​)3​CCl or CH3ClCH_3ClCH3​Cl

Solution

Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene:

1-Bromo-1-methylcyclohexane

2-Chloro-2-methylbutane

2,2,3-Trimethyl-3-bromopentane.

Solution

How will you bring about the following conversions?

Ethanol to but-1-yne

Ethane to bromoethene

Propene to 1-nitropropane

Toluene to benzyl alcohol

Propene to propyne

Ethanol to ethyl fluoride

Bromomethane to propanone

But-1-ene to but-2-ene

1-Chlorobutane to n-octane

Benzene to biphenyl.

Solution

Explain why

the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride?

alkyl halides, though polar, are immiscible with water?

Grignard reagents should be prepared under anhydrous conditions?

Solution

Give the uses of freon 12, DDT, carbon tetrachloride and iodoform.

Solution

Write the structure of the major organic product in each of the following reactions:

CH3CH2CH2Cl+NaI→acetone,heatCH_3CH_2CH_2Cl + NaI \xrightarrow{acetone, heat}CH3​CH2​CH2​Cl+NaIacetone,heat​

(CH3)3CBr+KOH→ethanol,heat(CH_3)_3CBr + KOH \xrightarrow{ethanol, heat}(CH3​)3​CBr+KOHethanol,heat​

CH3CH(Br)CH2CH3+NaOH→waterCH_3CH(Br)CH_2CH_3 + NaOH \xrightarrow{water}CH3​CH(Br)CH2​CH3​+NaOHwater​

CH3CH2Br+KCN→aq.ethanolCH_3CH_2Br + KCN \xrightarrow{aq. ethanol}CH3​CH2​Br+KCNaq.ethanol​

$(CH_3)_2CHCH(Cl)CH_3$

$CH_3CH_2CH(CH_3)CH(C_2H_5)Cl$

$CH_3CH_2C(CH_3)_2CH_2I$

$(CH_3)_3CCH_2CH(Br)C_6H_5$

$CH_3CH(CH_3)CH(Br)CH_3$

$CH_3C(C_2H_5)_2CH_2Br$

$CH_3C(Cl)(C_2H_5)CH_2CH_3$

$CH_3CH=C(Cl)CH_2CH(CH_3)_2$

$CH_3CH=CHC(Br)(CH_3)_2$

$p-ClC_6H_4CH_2CH(CH_3)_2$ (xi) $m-ClCH_2C_6H_4CH_2C(CH_3)_3$ (xii) $o-Br-C_6H_4CH(CH_3)CH_2CH_3$

$CH_3CH(Cl)CH(Br)CH_3$

$CHF_2CBrClF$

$ClCH_2C\equiv CCH_2Br$

$(CCl_3)_3CCl$

$CH_3C(p-ClC_6H_4)_2CH(Br)CH_3$

$(CH_3)_3CCH=CClC_6H_4I-p$

$p$ -Bromochlorobenzene

$CH_2Cl_2$

$CHCl_3$

$CCl_4$

A hydrocarbon $C_5H_{10}$ does not react with chlorine in dark but gives a single monochloro compound $C_5H_9Cl$ in bright sunlight. Identify the hydrocarbon.

Write the isomers of the compound having formula $C_4H_9Br$ .

Which compound in each of the following pairs will react faster in $S_N2$ reaction with $^-OH$ ?

$CH_3Br$ or $CH_3I$

$(CH_3)_3CCl$ or $CH_3Cl$

$CH_3CH_2CH_2Cl + NaI \xrightarrow{acetone, heat}$

$(CH_3)_3CBr + KOH \xrightarrow{ethanol, heat}$

$CH_3CH(Br)CH_2CH_3 + NaOH \xrightarrow{water}$

$CH_3CH_2Br + KCN \xrightarrow{aq. ethanol}$

$C_6H_5ONa + C_2H_5Cl \rightarrow$

$CH_3CH_2CH_2OH + SOCl_2 \rightarrow$

$CH_3CH_2CH=CH_2 + HBr \xrightarrow{peroxide}$

$CH_3CH=C(CH_3)_2 + HBr \rightarrow$

Write the mechanism of the following reaction: $nBuBr + KCN \xrightarrow{EtOH-H_2O} nBuCN$

Arrange the compounds of each set in order of reactivity towards $S_N2$ displacement:

Out of $C_6H_5CH_2Cl$ and $C_6H_5CHClC_6H_5$ , which is more easily hydrolysed by aqueous KOH.

$p$ -Dichlorobenzene has higher m.p. than those of $o$ - and $m$ -isomers. Discuss.