Alcohols, Phenols and Ethers Class 12 Chemistry NCERT Solutions

Q1Exercises

Write IUPAC names of the following compounds:

(i)

$CH_3-CH(CH_3)-CH(OH)-C(CH_3)_3$

(ii)

$CH_3CH_2CH(CH_2CH_3)CH(OH)CH(CH_3)CH_2OH$

(iii)

$CH_3CH(OH)CH(OH)CH_3$

(iv)

$HOCH_2CH(OH)CH_2OH$

(v)

o-Cresol (structure with -OH and -CH3 on adjacent carbons of a benzene ring)

(vi)

p-Cresol (structure with -OH and -CH3 on opposite carbons of a benzene ring)

(vii)

2,5-Dimethylphenol

(viii)

2,6-Dimethylphenol

(ix)

$CH_3-O-CH_2-CH(CH_3)_2$

(x)

$C_6H_5-O-C_2H_5$ (xi) $C_6H_5-O-C_7H_{15}(n-)$ (xii) $CH_3CH_2-O-CH(CH_3)CH_2CH_3$

Solution

(i)

The longest carbon chain containing the -OH group has 5 carbons. Numbering from the right gives the -OH group position 3. The substituents are methyl groups at positions 2, 4, and 4. IUPAC Name: 2,4,4-Trimethylpentan-3-ol

(ii)

The longest carbon chain containing a hydroxyl group has 7 carbons. The parent chain is heptane. There are two -OH groups at positions 1 and 3. The substituents are an ethyl group at position 4 and a methyl group at position 2. IUPAC Name: 4-Ethyl-2-methylheptane-1,3-diol

(iii)

The parent chain is butane. There are two -OH groups at positions 2 and 3. IUPAC Name: Butane-2,3-diol

(iv)

The parent chain is propane. There are three -OH groups at positions 1, 2, and 3. IUPAC Name: Propane-1,2,3-triol

(v)

The -OH group is on a benzene ring, so the parent name is phenol. A methyl group is at the ortho (position 2) position. IUPAC Name: 2-Methylphenol

(vi)

The parent name is phenol. A methyl group is at the para (position 4) position. IUPAC Name: 4-Methylphenol

(vii)

The parent name is phenol. Methyl groups are at positions 2 and 5. IUPAC Name: 2,5-Dimethylphenol

(viii)

The parent name is phenol. Methyl groups are at positions 2 and 6. IUPAC Name: 2,6-Dimethylphenol

(ix)

This is an ether. The smaller alkyl group is methyl, so the alkoxy group is methoxy. The parent alkane is isobutane (2-methylpropane). The methoxy group is attached to C-1. IUPAC Name: 1-Methoxy-2-methylpropane

(x)

The alkoxy group is ethoxy. The parent hydrocarbon is benzene. IUPAC Name: Ethoxybenzene

(xi) The alkoxy group is phenoxy. The parent alkane is heptane. IUPAC Name: 1-Phenoxyheptane

(xii) The alkoxy group is ethoxy. The parent alkane is pentane. The ethoxy group is attached at position 3. IUPAC Name: 3-Ethoxypentane

Q2Exercises

Write structures of the compounds whose IUPAC names are as follows:

(i)

2-Methylbutan-2-ol

(ii)

1-Phenylpropan-2-ol

(iii)

3,5-Dimethylhexane-1,3,5-triol

(iv)

2,3 - Diethylphenol

(v)

1 - Ethoxypropane

(vi)

2-Ethoxy-3-methylpentane

(vii)

Cyclohexylmethanol

(viii)

3-Cyclohexylpentan-3-ol

(ix)

Cyclopent-3-en-1-ol

(x)

4-Chloro-3-ethylbutan-1-ol.

Solution

(i)

2-Methylbutan-2-ol:

CH_3-C(OH)(CH_3)-CH_2-CH_3

(ii)

1-Phenylpropan-2-ol:

C_6H_5-CH_2-CH(OH)-CH_3

(iii)

3,5-Dimethylhexane-1,3,5-triol:

HO-CH_2-CH_2-C(OH)(CH_3)-CH_2-C(OH)(CH_3)-CH_3

(iv)

2,3-Diethylphenol: A benzene ring with an -OH group, an ethyl group at position 2, and an ethyl group at position 3.

(v)

1-Ethoxypropane:

CH_3CH_2-O-CH_2CH_2CH_3

(vi)

2-Ethoxy-3-methylpentane:

CH_3-CH(OCH_2CH_3)-CH(CH_3)-CH_2-CH_3

(vii)

Cyclohexylmethanol: A cyclohexyl ring attached to a

-CH_2OH

group.

(viii)

3-Cyclohexylpentan-3-ol:

CH_3-CH_2-C(OH)(C_6H_{11})-CH_2-CH_3

(ix)

Cyclopent-3-en-1-ol: A five-membered ring with a double bond between C3 and C4, and an -OH group on C1.

(x)

4-Chloro-3-ethylbutan-1-ol:

Cl-CH_2-CH(CH_2CH_3)-CH_2-CH_2-OH

Q3Exercises

(i) Draw the structures of all isomeric alcohols of molecular formula C_5H_12O and give their IUPAC names.…

(i)

Draw the structures of all isomeric alcohols of molecular formula $C_5H_{12}O$ and give their IUPAC names.

(ii)

Classify the isomers of alcohols in question 7.3 (i) as primary, secondary and tertiary alcohols.

Solution

(i)

There are 8 isomeric alcohols with the molecular formula

C_5H_{12}O

.

Pentan-1-ol: $CH_3CH_2CH_2CH_2CH_2OH$
Pentan-2-ol: $CH_3CH_2CH_2CH(OH)CH_3$
Pentan-3-ol: $CH_3CH_2CH(OH)CH_2CH_3$
2-Methylbutan-1-ol: $(CH_3)_2CHCH_2CH_2OH$
3-Methylbutan-1-ol: $CH_3CH(CH_3)CH_2CH_2OH$
2-Methylbutan-2-ol: $(CH_3)_2C(OH)CH_2CH_3$
3-Methylbutan-2-ol: $CH_3CH(OH)CH(CH_3)_2$
2,2-Dimethylpropan-1-ol: $(CH_3)_3CCH_2OH$

(ii)

Classification of the isomers:

Primary (1°) alcohols: The -OH group is attached to a primary carbon atom (a carbon attached to only one other carbon atom).
- Pentan-1-ol
- 2-Methylbutan-1-ol
- 3-Methylbutan-1-ol
- 2,2-Dimethylpropan-1-ol
Secondary (2°) alcohols: The -OH group is attached to a secondary carbon atom (a carbon attached to two other carbon atoms).
- Pentan-2-ol
- Pentan-3-ol
- 3-Methylbutan-2-ol
Tertiary (3°) alcohols: The -OH group is attached to a tertiary carbon atom (a carbon attached to three other carbon atoms).
- 2-Methylbutan-2-ol

Q4Exercises

Explain why propanol has higher boiling point than that of the hydrocarbon, butane?

Solution

Propanol (

CH_3CH_2CH_2OH

) and butane (

CH_3CH_2CH_2CH_3

) have comparable molecular masses (Propanol: 60 g/mol, Butane: 58 g/mol). However, propanol has a much higher boiling point (370 K) than butane (273 K).

The reason for this difference is the presence of intermolecular hydrogen bonding in propanol. The hydroxyl (-OH) group in propanol is highly polar, allowing one propanol molecule to form strong hydrogen bonds with other propanol molecules. A significant amount of energy is required to break these hydrogen bonds, resulting in a high boiling point.

In contrast, butane is a nonpolar hydrocarbon. The only intermolecular forces present are weak van der Waals forces. These forces require less energy to overcome, leading to a much lower boiling point.

Q5Exercises

Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.

Solution

Alcohols are significantly more soluble in water than hydrocarbons of comparable molecular masses due to their ability to form hydrogen bonds with water molecules.

The hydroxyl (-OH) group in an alcohol molecule is polar. The oxygen atom has a partial negative charge, and the hydrogen atom has a partial positive charge. This allows the alcohol molecule to form hydrogen bonds with polar water molecules. The oxygen of the alcohol can form a hydrogen bond with a hydrogen of water, and the hydrogen of the alcohol's -OH group can form a hydrogen bond with the oxygen of a water molecule.

Hydrocarbons, on the other hand, are nonpolar molecules. They cannot form hydrogen bonds with water. According to the principle of "like dissolves like," nonpolar hydrocarbons are insoluble or sparingly soluble in polar water. The interaction between alcohol and water molecules is strong enough to overcome the hydrogen bonds between water molecules themselves, leading to solubility.

Q6Exercises

What is meant by hydroboration-oxidation reaction? Illustrate it with an example.

Solution

The hydroboration-oxidation reaction is a two-step process used to convert an alkene into an alcohol. It is a method for the hydration of an alkene that results in the anti-Markovnikov addition of water across the double bond. This means the hydroxyl (-OH) group adds to the carbon atom of the double bond that has the greater number of hydrogen atoms.

The reaction involves:

Hydroboration: Addition of diborane ( $B_2H_6$ or $BH_3$ ) to the alkene to form a trialkylborane.
Oxidation: Oxidation of the trialkylborane with hydrogen peroxide ( $H_2O_2$ ) in the presence of a base (like NaOH) to yield the alcohol.

Example: Preparation of Propan-1-ol from Propene

The reaction of propene with diborane followed by oxidation with hydrogen peroxide gives propan-1-ol.

Step 1: Hydroboration

6 CH_3-CH=CH_2 + (BH_3)_2 \rightarrow 2 (CH_3-CH_2-CH_2)_3B

(Propene) (Diborane) (Tripropylborane)

Step 2: Oxidation

(CH_3-CH_2-CH_2)_3B + 3 H_2O_2 + 3 OH^- \rightarrow 3 CH_3-CH_2-CH_2-OH + B(OH)_4^-

(Tripropylborane) (Propan-1-ol)

Q7Exercises

Give the structures and IUPAC names of monohydric phenols of molecular formula, $C_7H_8O$ .

Solution

Monohydric phenols with the molecular formula

C_7H_8O

consist of a benzene ring with one hydroxyl (-OH) group and one methyl (

-CH_3

) group attached. These are known as cresols. There are three possible isomers depending on the relative positions of the -OH and

-CH_3

groups.

ortho-Cresol (o-Cresol)
- Structure: A benzene ring with -OH and $-CH_3$ groups on adjacent carbons (positions 1 and 2).
- IUPAC Name: 2-Methylphenol
meta-Cresol (m-Cresol)
- Structure: A benzene ring with -OH and $-CH_3$ groups at positions 1 and 3.
- IUPAC Name: 3-Methylphenol
para-Cresol (p-Cresol)
- Structure: A benzene ring with -OH and $-CH_3$ groups on opposite carbons (positions 1 and 4).
- IUPAC Name: 4-Methylphenol

Q8Exercises

While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason.

Solution

The isomer which will be steam volatile is ortho-nitrophenol.

Reason:

o-Nitrophenol: In o-nitrophenol, the nitro group ( $-NO_2$ ) and the hydroxyl group (-OH) are in close proximity. This allows for the formation of a strong intramolecular hydrogen bond (a hydrogen bond within the same molecule). This internal bonding prevents the molecule from forming strong hydrogen bonds with other molecules.
p-Nitrophenol: In p-nitrophenol, the -OH and $-NO_2$ groups are far apart, making intramolecular hydrogen bonding impossible. Instead, p-nitrophenol molecules form strong intermolecular hydrogen bonds with each other. This leads to the association of molecules.

Because of the lack of strong intermolecular forces, o-nitrophenol has a lower boiling point and is more volatile than p-nitrophenol. Therefore, when steam is passed through the mixture, the more volatile o-nitrophenol vaporizes along with the steam and can be collected after condensation, while the less volatile p-nitrophenol remains behind.

Q9Exercises

Give the equations of reactions for the preparation of phenol from cumene.

Solution

The preparation of phenol from cumene (isopropylbenzene) is a two-step industrial process.

Step 1: Formation of Cumene Hydroperoxide Cumene is oxidized in the presence of air (oxygen) to form cumene hydroperoxide. This is a free-radical reaction.

C_6H_5CH(CH_3)_2 + O_2 \xrightarrow{\text{air oxidation}} C_6H_5C(CH_3)_2OOH

(Cumene) (Cumene hydroperoxide)

Step 2: Acid-catalyzed cleavage to Phenol and Acetone Cumene hydroperoxide is treated with dilute acid (e.g.,

H_2SO_4

) to undergo rearrangement and cleavage, yielding phenol and acetone as products.

C_6H_5C(CH_3)_2OOH \xrightarrow{H^+/H_2O, \Delta} C_6H_5OH + CH_3COCH_3

(Cumene hydroperoxide) (Phenol) (Acetone)

Q10Exercises

Write chemical reaction for the preparation of phenol from chlorobenzene.

Solution

The preparation of phenol from chlorobenzene is known as the Dow's process. It involves heating chlorobenzene with an aqueous solution of sodium hydroxide (NaOH) under harsh conditions of high temperature and pressure. This is a nucleophilic aromatic substitution reaction.

The process occurs in two steps:

Step 1: Formation of Sodium Phenoxide Chlorobenzene is heated with aqueous NaOH at 623 K and 320 atmospheric pressure. The chlorine atom is replaced by the -ONa group to form sodium phenoxide.

C_6H_5Cl + 2NaOH \xrightarrow{623 K, 320 atm} C_6H_5ONa + NaCl + H_2O

(Chlorobenzene) (Sodium phenoxide)

Step 2: Acidification to form Phenol The resulting sodium phenoxide solution is acidified with a dilute acid, such as HCl or by bubbling

CO_2

through it, to protonate the phenoxide ion and form phenol.

C_6H_5ONa + HCl \rightarrow C_6H_5OH + NaCl

(Sodium phenoxide) (Phenol)

Q11Exercises

Write the mechanism of hydration of ethene to yield ethanol.

Solution

The acid-catalyzed hydration of ethene to ethanol is an electrophilic addition reaction that proceeds through a three-step mechanism involving a carbocation intermediate.

Step 1: Protonation of ethene to form a carbocation The reaction is initiated by the electrophilic attack of a hydronium ion (

H_3O^+

), formed from the acid catalyst and water, on the ethene molecule. The pi bond of ethene breaks, and a new C-H sigma bond is formed, resulting in a primary carbocation.

CH_2=CH_2 + H_3O^+ \rightleftharpoons CH_3-\stackrel{+}{C}H_2 + H_2O

(Ethene) (Hydronium ion) (Ethyl carbocation)

Step 2: Nucleophilic attack of water on the carbocation A water molecule acts as a nucleophile and attacks the electron-deficient carbocation, forming a protonated alcohol (oxonium ion).

CH_3-\stackrel{+}{C}H_2 + H_2O \rightleftharpoons CH_3-CH_2-\stackrel{+}{O}H_2

(Ethyl carbocation) (Protonated ethanol)

Step 3: Deprotonation to form ethanol Another water molecule acts as a base and removes a proton from the protonated alcohol to form the final product, ethanol, and regenerates the hydronium ion catalyst.

CH_3-CH_2-\stackrel{+}{O}H_2 + H_2O \rightleftharpoons CH_3-CH_2-OH + H_3O^+

(Protonated ethanol) (Ethanol)

Q12Exercises

You are given benzene, conc. $H_2SO_4$ and NaOH. Write the equations for the preparation of phenol using these reagents.

Solution

The preparation of phenol from benzene using the given reagents involves three main steps.

Step 1: Sulphonation of Benzene Benzene is heated with concentrated sulphuric acid (

H_2SO_4

) or oleum (fuming sulphuric acid) to produce benzenesulphonic acid. This is an electrophilic aromatic substitution reaction.

C_6H_6 + H_2SO_4 (conc.) \xrightarrow{\Delta} C_6H_5SO_3H + H_2O

(Benzene) (Benzenesulphonic acid)

Step 2: Conversion to Sodium Phenoxide The benzenesulphonic acid is neutralized with sodium hydroxide (NaOH) to form sodium benzenesulphonate. This salt is then fused with excess solid NaOH at high temperature (around 623 K) to produce sodium phenoxide.

C_6H_5SO_3H + NaOH \rightarrow C_6H_5SO_3Na + H_2O

(Benzenesulphonic acid) (Sodium benzenesulphonate)

C_6H_5SO_3Na + 2NaOH \xrightarrow{623 K, fuse} C_6H_5ONa + Na_2SO_3 + H_2O

(Sodium benzenesulphonate) (Sodium phenoxide)

Step 3: Acidification to form Phenol The sodium phenoxide is then acidified to produce phenol. Since we are only given

H_2SO_4

, we can use it for acidification.

2 C_6H_5ONa + H_2SO_4 \rightarrow 2 C_6H_5OH + Na_2SO_4

(Sodium phenoxide) (Phenol)

Q13Exercises

Show how will you synthesise:

(i)

1-phenylethanol from a suitable alkene.

(ii)

cyclohexylmethanol using an alkyl halide by an $S_N2$ reaction.

(iii)

pentan-1-ol using a suitable alkyl halide?

Solution

(i)

1-phenylethanol from a suitable alkene: 1-phenylethanol can be synthesized by the acid-catalyzed hydration of styrene (phenylethene). The addition of water follows Markovnikov's rule, where the -OH group adds to the more substituted carbon atom of the double bond.

C_6H_5-CH=CH_2 + H_2O \xrightarrow{H^+} C_6H_5-CH(OH)-CH_3

(Styrene) (1-phenylethanol)

(ii)

cyclohexylmethanol using an alkyl halide by an $S_N2$ reaction: Cyclohexylmethanol can be synthesized by treating cyclohexylmethyl bromide (or chloride) with aqueous sodium hydroxide or potassium hydroxide. The hydroxide ion (

OH^-

) acts as a nucleophile and displaces the bromide ion in an

S_N2

reaction.

C_6H_{11}-CH_2-Br + NaOH(aq) \xrightarrow{S_N2} C_6H_{11}-CH_2-OH + NaBr

(Cyclohexylmethyl bromide) (Cyclohexylmethanol)

(iii)

pentan-1-ol using a suitable alkyl halide: Pentan-1-ol can be synthesized by the nucleophilic substitution of a suitable 1-halopentane, such as 1-chloropentane or 1-bromopentane, with aqueous sodium hydroxide or potassium hydroxide. This is an

S_N2

reaction.

CH_3(CH_2)_3CH_2-Cl + KOH(aq) \xrightarrow{S_N2} CH_3(CH_2)_3CH_2-OH + KCl

(1-Chloropentane) (Pentan-1-ol)

Q14Exercises

Give two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of ethanol.

Solution

Two reactions showing the acidic nature of phenol: Phenol is acidic because it can donate a proton (

H^+

) from its hydroxyl group.

Reaction with active metals: Phenol reacts with active metals like sodium (Na) to liberate hydrogen gas and form sodium phenoxide. $2 C_6H_5OH + 2 Na \rightarrow 2 C_6H_5ONa + H_2 \uparrow$ (Phenol) (Sodium phenoxide)
Reaction with sodium hydroxide: Unlike alcohols, phenol is acidic enough to react with a strong base like sodium hydroxide (NaOH) to form sodium phenoxide and water. This is a neutralization reaction. $C_6H_5OH + NaOH \rightarrow C_6H_5ONa + H_2O$ (Phenol) (Sodium phenoxide)

Comparison of acidity of phenol and ethanol: Phenol is a much stronger acid than ethanol.

Phenol: The acidity of phenol is due to the stability of its conjugate base, the phenoxide ion ( $C_6H_5O^-$ ). The negative charge on the oxygen atom in the phenoxide ion is delocalized over the benzene ring through resonance. This delocalization stabilizes the phenoxide ion and favors the release of the proton.
Ethanol: When ethanol ( $C_2H_5OH$ ) loses a proton, it forms the ethoxide ion ( $C_2H_5O^-$ ). In the ethoxide ion, the negative charge is localized on the oxygen atom. Furthermore, the ethyl group ( $-C_2H_5$ ) is an electron-releasing group (+I effect), which intensifies the negative charge on the oxygen, destabilizing the ethoxide ion and making the release of the proton less favorable.

Because the phenoxide ion is significantly more stable than the ethoxide ion, phenol is a stronger acid than ethanol.

Q15Exercises

Explain why is ortho nitrophenol more acidic than ortho methoxyphenol ?

Solution

Ortho-nitrophenol is significantly more acidic than ortho-methoxyphenol due to the electronic effects of the nitro (

-NO_2

) and methoxy (

-OCH_3

) groups on the stability of their respective conjugate bases (phenoxide ions).

Acidity of a phenol depends on the stability of the phenoxide ion formed after the loss of a proton. Any factor that stabilizes the phenoxide ion will increase the acidity.

o-Nitrophenol: The nitro group ( $-NO_2$ ) is a strong electron-withdrawing group. It exerts both a negative inductive effect (-I effect) and a negative resonance effect (-R effect). Both effects withdraw electron density from the benzene ring and from the oxygen atom of the phenoxide ion. This delocalizes and stabilizes the negative charge on the o-nitrophenoxide ion, making it easier for the parent phenol to donate a proton.
o-Methoxyphenol: The methoxy group ( $-OCH_3$ ) has a dual electronic effect. It has an electron-withdrawing inductive effect (-I effect) due to the high electronegativity of oxygen. However, it has a much stronger electron-donating resonance effect (+R effect) because the lone pairs on the oxygen atom can be delocalized into the benzene ring. The +R effect dominates and increases the electron density on the ring and on the oxygen of the phenoxide ion. This destabilizes the negative charge on the o-methoxyphenoxide ion, making it harder for the parent phenol to donate a proton.

Conclusion: The electron-withdrawing nitro group stabilizes the conjugate base, increasing acidity. The electron-donating methoxy group destabilizes the conjugate base, decreasing acidity. Therefore, o-nitrophenol is more acidic than o-methoxyphenol.

Q16Exercises

Explain how does the -OH group attached to a carbon of benzene ring activate it towards electrophilic substitution?

Solution

The hydroxyl (-OH) group attached to a benzene ring (as in phenol) strongly activates the ring towards electrophilic substitution. This activation occurs because the -OH group is an electron-donating group, primarily through its positive resonance effect (+R effect).

The oxygen atom of the -OH group has two lone pairs of electrons. One of these lone pairs can participate in resonance with the pi electron system of the benzene ring. This delocalization of electrons from the oxygen atom into the ring increases the overall electron density of the ring, especially at the ortho and para positions.

The resonance structures of phenol show the buildup of negative charge at the ortho and para positions:

(See resonance structures in the textbook chapter under Section 7.4.4, Acidity of phenols)

This increased electron density makes the benzene ring more nucleophilic and thus more susceptible to attack by electrophiles (electron-seeking species). Since the electron density is highest at the ortho and para positions, incoming electrophiles are directed to these positions.

In summary, the -OH group activates the benzene ring by increasing its electron density through the +R effect, making it more reactive towards electrophiles.

Q17Exercises

Give equations of the following reactions:

(i)

Oxidation of propan-1-ol with alkaline $KMnO_4$ solution.

(ii)

Bromine in $CS_2$ with phenol.

(iii)

Dilute $HNO_3$ with phenol.

(iv)

Treating phenol wih chloroform in presence of aqueous NaOH.

Solution

(i)

Oxidation of propan-1-ol with alkaline $KMnO_4$ solution: Alkaline potassium permanganate (

KMnO_4

) is a strong oxidizing agent. It oxidizes a primary alcohol like propan-1-ol first to an aldehyde and then to a carboxylic acid. The final product is the salt of the carboxylic acid, which upon acidification gives the acid.

CH_3CH_2CH_2OH \xrightarrow{KMnO_4, OH^-, \Delta} CH_3CH_2COOH

(Propan-1-ol) (Propanoic acid)

(ii)

Bromine in $CS_2$ with phenol: When phenol reacts with bromine in a non-polar solvent like carbon disulphide (

CS_2

) at low temperature, monobromination occurs. The -OH group is ortho, para-directing, so a mixture of o-bromophenol and p-bromophenol is formed. The para isomer is the major product.

C_6H_5OH + Br_2 \xrightarrow{CS_2, 273 K} o-BrC_6H_4OH + p-BrC_6H_4OH

(Phenol) (o-Bromophenol) (p-Bromophenol)

(iii)

Dilute $HNO_3$ with phenol: Nitration of phenol with dilute nitric acid at low temperature (298 K) gives a mixture of ortho-nitrophenol and para-nitrophenol.

C_6H_5OH + HNO_3 (dil.) \xrightarrow{298 K} o-NO_2C_6H_4OH + p-NO_2C_6H_4OH

(Phenol) (o-Nitrophenol) (p-Nitrophenol)

(iv)

Treating phenol with chloroform in presence of aqueous NaOH: This reaction is the Reimer-Tiemann reaction. It introduces a formyl group (-CHO) onto the benzene ring, primarily at the ortho position, to form salicylaldehyde (2-hydroxybenzaldehyde).

C_6H_5OH + CHCl_3 + 3NaOH \xrightarrow{\Delta} o-HOC_6H_4CHO + 3NaCl + 2H_2O

(Phenol) (Chloroform) (Salicylaldehyde)

Q18Exercises

Explain the following with an example.

(i)

Kolbe's reaction.

(ii)

Reimer-Tiemann reaction.

(iii)

Williamson ether synthesis.

(iv)

Unsymmetrical ether.

Solution

(i)

Kolbe's reaction (or Kolbe-Schmitt reaction): This reaction is used for the synthesis of ortho-hydroxybenzoic acid (salicylic acid). In this reaction, sodium phenoxide is heated with carbon dioxide (

CO_2

) under pressure (4-7 atm) at about 400 K. The product is then acidified to yield salicylic acid.

Example: $C_6H_5ONa + CO_2 \xrightarrow[400 K, 4-7 atm]{(i)} \rightarrow \xrightarrow{(ii) H^+} o-HOC_6H_4COOH$ (Sodium phenoxide) (Salicylic acid)

(ii)

Reimer-Tiemann reaction: This reaction is used to introduce a formyl group (-CHO) onto an aromatic ring, usually at the ortho position to a hydroxyl group. It involves treating phenol with chloroform (

CHCl_3

) in the presence of an aqueous base like sodium hydroxide.

Example: The synthesis of salicylaldehyde from phenol. $C_6H_5OH + CHCl_3 + 3NaOH \rightarrow o-HOC_6H_4CHO + 3NaCl + 2H_2O$ (Phenol) (Salicylaldehyde)

(iii)

Williamson ether synthesis: This is a versatile laboratory method for preparing both symmetrical and unsymmetrical ethers. It involves the reaction of an alkyl halide with a sodium or potassium alkoxide (or phenoxide). The reaction proceeds via an

S_N2

mechanism. For good yields, the alkyl halide should be primary.

Example: The synthesis of ethyl methyl ether. $CH_3CH_2ONa + CH_3Br \rightarrow CH_3CH_2-O-CH_3 + NaBr$ (Sodium ethoxide) (Bromomethane) (Ethyl methyl ether)

(iv)

Unsymmetrical ether (or Mixed ether): An unsymmetrical ether is an ether in which the two alkyl or aryl groups attached to the oxygen atom are different.

Examples:
- Ethyl methyl ether: $CH_3CH_2-O-CH_3$
- Anisole (Methyl phenyl ether): $C_6H_5-O-CH_3$
- tert-Butyl ethyl ether: $(CH_3)_3C-O-CH_2CH_3$

Q19Exercises

Write the mechanism of acid dehydration of ethanol to yield ethene.

Solution

The acid-catalyzed dehydration of ethanol to ethene occurs at 443 K with concentrated sulphuric acid. It follows a three-step mechanism involving the formation of a carbocation.

Step 1: Formation of protonated alcohol (oxonium ion) The ethanol molecule, acting as a Lewis base, accepts a proton from the strong acid catalyst (e.g.,

H_2SO_4

, represented as

H^+

) to form a protonated alcohol, also known as an ethyloxonium ion.

CH_3-CH_2-OH + H^+ \rightleftharpoons CH_3-CH_2-\stackrel{+}{O}H_2

(Ethanol) (Ethyloxonium ion)

Step 2: Formation of a carbocation The protonated alcohol is unstable. The C-O bond breaks heterolytically, and a molecule of water (a good leaving group) is eliminated. This is the slow, rate-determining step and results in the formation of an ethyl carbocation.

CH_3-CH_2-\stackrel{+}{O}H_2 \xrightarrow{slow} CH_3-\stackrel{+}{C}H_2 + H_2O

(Ethyloxonium ion) (Ethyl carbocation)

Step 3: Elimination of a proton to form ethene A base (water in this case) removes a proton from the carbon atom adjacent to the positively charged carbon. This results in the formation of a carbon-carbon double bond (ethene) and regenerates the acid catalyst.

CH_3-\stackrel{+}{C}H_2 \xrightarrow{-H^+} CH_2=CH_2

(Ethyl carbocation) (Ethene)

Q20Exercises

How are the following conversions carried out?

(i)

Propene → Propan-2-ol.

(ii)

Benzyl chloride → Benzyl alcohol.

(iii)

Ethyl magnesium chloride → Propan-1-ol.

(iv)

Methyl magnesium bromide → 2-Methylpropan-2-ol.

Solution

(i)

Propene → Propan-2-ol: This conversion is achieved by the acid-catalyzed hydration of propene. The addition of water follows Markovnikov's rule.

CH_3-CH=CH_2 + H_2O \xrightarrow{H^+} CH_3-CH(OH)-CH_3

(Propene) (Propan-2-ol)

(ii)

Benzyl chloride → Benzyl alcohol: This is a nucleophilic substitution reaction (

S_N2

). Benzyl chloride is heated with an aqueous solution of a base like sodium hydroxide (NaOH) or potassium hydroxide (KOH).

C_6H_5-CH_2-Cl + KOH(aq) \xrightarrow{\Delta} C_6H_5-CH_2-OH + KCl

(Benzyl chloride) (Benzyl alcohol)

(iii)

Ethyl magnesium chloride → Propan-1-ol: This involves the reaction of a Grignard reagent (ethyl magnesium chloride) with formaldehyde (methanal), followed by hydrolysis. This reaction adds one carbon atom and produces a primary alcohol.

Step 1: Reaction with formaldehyde

CH_3CH_2MgCl + HCHO \rightarrow CH_3CH_2CH_2OMgCl

(Ethyl magnesium chloride) (Formaldehyde) (Adduct)

Step 2: Hydrolysis

CH_3CH_2CH_2OMgCl + H_2O \rightarrow CH_3CH_2CH_2OH + Mg(OH)Cl

(Adduct) (Propan-1-ol)

(iv)

Methyl magnesium bromide → 2-Methylpropan-2-ol: This conversion requires the reaction of a Grignard reagent (methyl magnesium bromide) with a ketone, specifically acetone (propanone), followed by hydrolysis. This reaction produces a tertiary alcohol.

Step 1: Reaction with acetone

CH_3MgBr + CH_3COCH_3 \rightarrow (CH_3)_3C-OMgBr

(Methyl magnesium bromide) (Acetone) (Adduct)

Step 2: Hydrolysis

(CH_3)_3C-OMgBr + H_2O \rightarrow (CH_3)_3C-OH + Mg(OH)Br

(Adduct) (2-Methylpropan-2-ol)

Q21Exercises

Name the reagents used in the following reactions:

(i)

Oxidation of a primary alcohol to carboxylic acid.

(ii)

Oxidation of a primary alcohol to aldehyde.

(iii)

Bromination of phenol to 2,4,6-tribromophenol.

(iv)

Benzyl alcohol to benzoic acid.

(v)

Dehydration of propan-2-ol to propene.

(vi)

Butan-2-one to butan-2-ol.

Solution

(i)

Oxidation of a primary alcohol to carboxylic acid: Strong oxidizing agents are used, such as:

Acidified potassium permanganate ( $KMnO_4 / H^+$ )
Alkaline potassium permanganate ( $KMnO_4 / OH^-$ ) followed by acidification
Acidified potassium dichromate ( $K_2Cr_2O_7 / H^+$ )

(ii)

Oxidation of a primary alcohol to aldehyde: Mild oxidizing agents are used to stop the oxidation at the aldehyde stage, such as:

Pyridinium chlorochromate (PCC)
Anhydrous chromium trioxide ( $CrO_3$ )
Heated copper at 573 K ( $Cu / 573 K$ )

(iii)

Bromination of phenol to 2,4,6-tribromophenol: Aqueous solution of bromine, known as Bromine water ( $Br_2(aq)$ ).

(iv)

Benzyl alcohol to benzoic acid: Strong oxidizing agents, similar to (i), such as alkaline potassium permanganate ( $KMnO_4 / OH^-$ ) followed by acidification.

(v)

Dehydration of propan-2-ol to propene: Acid catalysts are used, such as:

Concentrated sulphuric acid ( $conc. H_2SO_4$ ) with heat (e.g., 85% $H_3PO_4$ at 440 K).

(vi)

Butan-2-one to butan-2-ol: This is a reduction reaction. Reducing agents are used, such as:

Sodium borohydride ( $NaBH_4$ )
Lithium aluminium hydride ( $LiAlH_4$ )
Catalytic hydrogenation ( $H_2$ with Ni, Pt, or Pd catalyst)

Q22Exercises

Give reason for the higher boiling point of ethanol in comparison to methoxymethane.

Solution

Ethanol (

C_2H_5OH

) and methoxymethane (

CH_3OCH_3

) are isomers and have the same molecular mass (46 g/mol). However, ethanol has a much higher boiling point (351 K) than methoxymethane (248 K).

Reason: The significant difference in boiling points is due to the presence of intermolecular hydrogen bonding in ethanol. The ethanol molecule has a highly polar hydroxyl (-OH) group, which allows it to form strong hydrogen bonds with neighboring ethanol molecules. A large amount of energy is required to break these strong intermolecular forces, resulting in a high boiling point.

Methoxymethane, an ether, does not have a hydrogen atom directly bonded to the oxygen atom. Therefore, it cannot form hydrogen bonds with other methoxymethane molecules. The only intermolecular forces present are weaker dipole-dipole interactions and van der Waals forces. These forces are much easier to overcome, leading to a much lower boiling point.

Q23Exercises

Give IUPAC names of the following ethers:

(i)

$CH_3CH_2-O-CH_2-CH(CH_3)_2$

(ii)

$CH_3OCH_2CH_2Cl$

(iii)

$O_2N-C_6H_4-OCH_3(p)$

(iv)

$CH_3CH_2CH_2OCH_3$

(v)

Structure with an ethoxy group and a 1,1-dimethyl group on a cyclohexane ring, with ethoxy at position 2.

(vi)

$C_2H_5-O-C_6H_5$

Solution

(i)

The alkoxy group is ethoxy. The parent alkane chain has 4 carbons (butane) with a methyl group at position 2. Numbering from the left gives the ethoxy group position 1. IUPAC Name: 1-Ethoxy-2-methylpropane

(ii)

The alkoxy group is methoxy. The parent alkane is ethane with a chloro substituent. The methoxy group is at position 1 and the chloro group is at position 2. IUPAC Name: 1-Chloro-2-methoxyethane

(iii)

The parent compound is anisole (methoxybenzene). There is a nitro group at the para (position 4) position. IUPAC Name: 4-Nitroanisole or 1-Methoxy-4-nitrobenzene

(iv)

The alkoxy group is methoxy. The parent alkane is propane. The methoxy group is at position 1. IUPAC Name: 1-Methoxypropane

(v)

The parent is cyclohexane. The substituents are ethoxy and two methyl groups. Numbering should give the substituents the lowest possible locants. Let's assume the question implies 2-Ethoxy-1,1-dimethylcyclohexane as given in Table 7.2. IUPAC Name: 2-Ethoxy-1,1-dimethylcyclohexane

(vi)

The alkoxy group is ethoxy. The parent hydrocarbon is benzene. IUPAC Name: Ethoxybenzene

Q24Exercises

Write the names of reagents and equations for the preparation of the following ethers by Williamson's synthesis:

(i)

1-Propoxypropane

(ii)

Ethoxybenzene

(iii)

2-Methoxy-2-methylpropane

(iv)

1-Methoxyethane

Solution

(i)

1-Propoxypropane (

CH_3CH_2CH_2-O-CH_2CH_2CH_3

) This is a symmetrical ether. It is prepared from a primary alkyl halide and its corresponding alkoxide.

Reagents: Sodium propoxide and 1-bromopropane (or 1-chloropropane).
Equation: $CH_3CH_2CH_2ONa + CH_3CH_2CH_2Br \rightarrow CH_3CH_2CH_2-O-CH_2CH_2CH_3 + NaBr$

(ii)

Ethoxybenzene (

C_6H_5-O-C_2H_5

) This is an aryl alkyl ether. It is best prepared by reacting sodium phenoxide with a primary alkyl halide (ethyl bromide).

Reagents: Sodium phenoxide and ethyl bromide (or ethyl chloride).
Equation: $C_6H_5ONa + C_2H_5Br \rightarrow C_6H_5-O-C_2H_5 + NaBr$

(iii)

2-Methoxy-2-methylpropane (tert-Butyl methyl ether) (

(CH_3)_3C-O-CH_3

) To prepare this ether, we must use a primary alkyl halide to avoid elimination. Therefore, the reactants must be sodium tert-butoxide and methyl bromide.

Reagents: Sodium tert-butoxide and methyl bromide (or methyl chloride).
Equation: $(CH_3)_3CONa + CH_3Br \rightarrow (CH_3)_3C-O-CH_3 + NaBr$ (Note: Using tert-butyl bromide and sodium methoxide would result in elimination to form 2-methylpropene.)

(iv)

1-Methoxyethane (Ethyl methyl ether) (

CH_3-O-C_2H_5

) This unsymmetrical ether can be prepared in two ways, but using the less hindered alkyl halide is preferred.

Method 1 (Preferred):
- Reagents: Sodium ethoxide and methyl bromide.
- Equation: $CH_3CH_2ONa + CH_3Br \rightarrow CH_3CH_2-O-CH_3 + NaBr$
Method 2:
- Reagents: Sodium methoxide and ethyl bromide.
- Equation: $CH_3ONa + CH_3CH_2Br \rightarrow CH_3-O-CH_2CH_3 + NaBr$

Q25Exercises

Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of ethers.

Solution

The Williamson synthesis is a very effective method for preparing ethers, but it has significant limitations, particularly when dealing with secondary or tertiary alkyl halides.

The reaction proceeds via an

S_N2

mechanism, which involves the backside attack of a nucleophile (the alkoxide ion) on the alkyl halide. This mechanism is sensitive to steric hindrance.

Limitation: The reaction gives good yields only if the alkyl halide is primary or methyl. If a secondary or tertiary alkyl halide is used, elimination (E2 reaction) becomes the major pathway, and an alkene is formed instead of an ether. This is because alkoxides (like sodium ethoxide,

CH_3CH_2O^-

) are not only strong nucleophiles but also strong bases.

Example: Attempted synthesis of tert-Butyl ethyl ether Let's try to synthesize tert-butyl ethyl ether using a tertiary alkyl halide.

Incorrect approach: Reacting tert-butyl bromide (a tertiary halide) with sodium ethoxide (a strong base). $(CH_3)_3C-Br + CH_3CH_2ONa \rightarrow (CH_3)_2C=CH_2 + CH_3CH_2OH + NaBr$ (tert-Butyl bromide) (Sodium ethoxide) (2-Methylpropene - Major product)

In this case, the sterically hindered tert-butyl bromide prevents the

S_N2

attack. Instead, the strong base (ethoxide) removes a proton from a beta-carbon, leading to an E2 elimination reaction, which produces 2-methylpropene as the major product.

Correct approach: To synthesize this ether, one must use a primary alkyl halide (ethyl bromide) and a tertiary alkoxide (sodium tert-butoxide).

(CH_3)_3CONa + CH_3CH_2Br \rightarrow (CH_3)_3C-O-CH_2CH_3 + NaBr

(Sodium tert-butoxide) (Ethyl bromide) (tert-Butyl ethyl ether - Major product)

Thus, the Williamson synthesis is limited by the structure of the alkyl halide; it must be primary to favor substitution over elimination.

Q26Exercises

How is 1-propoxypropane synthesised from propan-1-ol? Write mechanism of this reaction.

Solution

The synthesis of 1-propoxypropane from propan-1-ol is a two-step process based on the Williamson ether synthesis.

Step 1: Formation of Sodium Propoxide Propan-1-ol is treated with a reactive metal like sodium (Na) to form the alkoxide, sodium propoxide. This reaction deprotonates the alcohol.

2 CH_3CH_2CH_2OH + 2 Na \rightarrow 2 CH_3CH_2CH_2ONa + H_2

(Propan-1-ol) (Sodium propoxide)

Step 2: Nucleophilic Substitution ( $S_N2$ ) The sodium propoxide is then reacted with a primary alkyl halide, 1-bromopropane (which can also be prepared from propan-1-ol using PBr3). The propoxide ion acts as a nucleophile and displaces the bromide ion.

CH_3CH_2CH_2ONa + CH_3CH_2CH_2Br \rightarrow CH_3CH_2CH_2-O-CH_2CH_2CH_3 + NaBr

(Sodium propoxide) (1-Bromopropane) (1-Propoxypropane)

Mechanism of Step 2 (Williamson Synthesis): The reaction in the second step follows an $S_N2$ (bimolecular nucleophilic substitution) mechanism.

The propoxide ion (

CH_3CH_2CH_2O^-

) acts as the nucleophile. It attacks the carbon atom bonded to the bromine in 1-bromopropane from the side opposite to the leaving group (bromide ion). This is a single, concerted step where the new C-O bond forms at the same time as the C-Br bond breaks. A transition state is formed in which the central carbon is partially bonded to both the incoming nucleophile (propoxide) and the outgoing leaving group (bromide).

CH_3CH_2CH_2O^- + CH_3CH_2CH_2-Br \rightarrow [CH_3CH_2CH_2O \cdots CH_2(CH_2CH_3) \cdots Br]^- \rightarrow CH_3CH_2CH_2-O-CH_2CH_2CH_3 + Br^-

(Nucleophile) (Substrate) (Transition State) (Product) (Leaving group)

Q27Exercises

Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.

Solution

The preparation of ethers by acid-catalyzed dehydration of secondary or tertiary alcohols is not a suitable method because elimination reactions dominate over substitution reactions, leading to the formation of alkenes as the major product.

The mechanism for both ether formation (substitution,

S_N1

) and alkene formation (elimination, E1) proceeds through a carbocation intermediate, especially for secondary and tertiary alcohols.

Protonation of Alcohol: The alcohol is protonated by the acid. $R_2CHOH + H^+ \rightarrow R_2CHOH_2^+$
Formation of Carbocation: Water is eliminated to form a carbocation. Secondary and tertiary alcohols form relatively stable secondary and tertiary carbocations, respectively. $R_2CHOH_2^+ \rightarrow R_2CH^+ + H_2O$
Competition between Substitution and Elimination: The carbocation can then react in two ways:
- Substitution ( $S_N1$ ): It can be attacked by another alcohol molecule to form a protonated ether, which then loses a proton to give the ether. This pathway is sterically hindered for secondary and tertiary alcohols.
- Elimination (E1): It can lose a proton from an adjacent carbon atom to form an alkene. This pathway is highly favored for secondary and tertiary carbocations, especially at the elevated temperatures required for dehydration.

Because the formation of alkenes (elimination) is kinetically and thermodynamically more favorable under these conditions for secondary and tertiary alcohols, it becomes the main reaction pathway. Therefore, attempting to make ethers from these alcohols by acid dehydration results in poor yields of the ether and a high yield of the corresponding alkene.

Q28Exercises

Write the equation of the reaction of hydrogen iodide with:

(i)

1-propoxypropane

(ii)

methoxybenzene

(iii)

benzyl ethyl ether.

Solution

(i)

1-propoxypropane with HI: 1-propoxypropane is a symmetrical ether with two primary alkyl groups. Cleavage by HI will produce one molecule of an alcohol and one molecule of an alkyl iodide. If excess HI is used, the alcohol formed will be further converted to the alkyl iodide.

CH_3CH_2CH_2-O-CH_2CH_2CH_3 + HI \xrightarrow{\Delta} CH_3CH_2CH_2OH + CH_3CH_2CH_2I

(1-propoxypropane) (Propan-1-ol) (1-Iodopropane)

With excess HI:

CH_3CH_2CH_2-O-CH_2CH_2CH_3 + 2HI \xrightarrow{\Delta} 2CH_3CH_2CH_2I + H_2O

(ii)

methoxybenzene (Anisole) with HI: Methoxybenzene is an alkyl aryl ether. The cleavage occurs at the alkyl-oxygen bond because the aryl-oxygen bond has partial double bond character and is stronger. The products are phenol and methyl iodide.

C_6H_5-O-CH_3 + HI \xrightarrow{\Delta} C_6H_5OH + CH_3I

(Methoxybenzene) (Phenol) (Methyl iodide)

(iii)

benzyl ethyl ether with HI: In benzyl ethyl ether, the cleavage of the C-O bond can proceed via an

S_N1

or

S_N2

mechanism. The iodide ion (

I^-

) will attack the less sterically hindered ethyl group in an

S_N2

fashion, or cleavage can occur to form the very stable benzyl carbocation in an

S_N1

fashion. Both pathways lead to the same products: benzyl iodide and ethanol.

C_6H_5CH_2-O-C_2H_5 + HI \xrightarrow{\Delta} C_6H_5CH_2I + C_2H_5OH

(Benzyl ethyl ether) (Benzyl iodide) (Ethanol)

Q29Exercises

Explain the fact that in aryl alkyl ethers (i) the alkoxy group activates the benzene ring towards electrophilic substitution and (ii) it directs the incoming substituents to ortho and para positions in benzene ring.

Solution

In aryl alkyl ethers (e.g., anisole,

C_6H_5OCH_3

), the alkoxy group (-OR) is an activating and ortho, para-directing group due to its strong positive resonance effect (+R effect).

(i) Activation of the benzene ring: The oxygen atom of the alkoxy group has lone pairs of electrons. These lone pairs can be delocalized into the pi-electron system of the benzene ring. This delocalization, known as the +R effect, increases the electron density within the ring.

An increase in electron density makes the ring more nucleophilic and therefore more reactive towards attack by electrophiles (electron-seeking species). This increased reactivity is referred to as activation. Although the oxygen atom also exerts a weak electron-withdrawing inductive effect (-I effect) due to its electronegativity, the +R effect is much stronger and dominates, leading to overall activation of the ring.

(ii) Ortho and para direction: The resonance structures of an aryl alkyl ether show that the delocalization of the lone pair from the oxygen atom results in a buildup of negative charge specifically at the ortho and para positions of the ring.

(See resonance structures of anisole in the textbook chapter under Section 7.6.3, Electrophilic substitution)

Since electrophiles are positively charged or electron-deficient, they are preferentially attracted to these positions of high electron density. Therefore, electrophilic substitution occurs primarily at the ortho and para positions. The alkoxy group thus acts as an ortho, para-director.

Q30Exercises

Write the mechanism of the reaction of HI with methoxymethane.

Solution

The reaction of methoxymethane (

CH_3OCH_3

) with hydrogen iodide (HI) is a nucleophilic substitution reaction that cleaves the C-O ether bond. The mechanism involves two main steps.

Step 1: Protonation of the ether The oxygen atom of the ether has lone pairs of electrons and acts as a Lewis base. It accepts a proton from the strong acid HI to form a protonated ether, a dimethyloxonium ion.

CH_3-O-CH_3 + H-I \rightleftharpoons CH_3-\stackrel{+}{O}(H)-CH_3 + I^-

(Methoxymethane) (Dimethyloxonium ion)

Step 2: Nucleophilic attack by iodide ion ( $S_N2$ ) The iodide ion (

I^-

) is a good nucleophile. It attacks one of the carbon atoms of the oxonium ion in an

S_N2

reaction. This attack occurs from the backside, displacing a molecule of methanol (

CH_3OH

), which is a good leaving group.

I^- + CH_3-\stackrel{+}{O}(H)-CH_3 \xrightarrow{S_N2} CH_3-I + CH_3-OH

(Iodide ion) (Dimethyloxonium ion) (Methyl iodide) (Methanol)

Further reaction if HI is in excess: If an excess of HI is used, the methanol produced in Step 2 will react further with HI in a similar two-step mechanism to form another molecule of methyl iodide.

CH_3-OH + HI \rightarrow CH_3-I + H_2O

(Methanol) (Methyl iodide)

Q31Exercises

Write equations of the following reactions:

(i)

Friedel-Crafts reaction - alkylation of anisole.

(ii)

Nitration of anisole.

(iii)

Bromination of anisole in ethanoic acid medium.

(iv)

Friedel-Craft's acetylation of anisole.

Solution

Anisole (

C_6H_5OCH_3

) has an activating methoxy group (-

OCH_3

) which is ortho, para-directing. The para product is usually major due to less steric hindrance.

(i) Friedel-Crafts alkylation of anisole: Anisole reacts with an alkyl halide (e.g., methyl chloride) in the presence of a Lewis acid catalyst (e.g., anhydrous

AlCl_3

) to give a mixture of ortho and para alkylated products.

C_6H_5OCH_3 + CH_3Cl \xrightarrow{Anhyd. AlCl_3} o-CH_3C_6H_4OCH_3 + p-CH_3C_6H_4OCH_3

(Anisole) (2-Methoxytoluene) (4-Methoxytoluene - Major)

(ii) Nitration of anisole: Anisole reacts with a nitrating mixture (conc.

H_2SO_4

and conc.

HNO_3

) to yield a mixture of ortho-nitroanisole and para-nitroanisole.

C_6H_5OCH_3 + conc. HNO_3 + conc. H_2SO_4 \rightarrow o-NO_2C_6H_4OCH_3 + p-NO_2C_6H_4OCH_3

(Anisole) (2-Nitroanisole) (4-Nitroanisole - Major)

(iii) Bromination of anisole in ethanoic acid medium: Anisole undergoes bromination with bromine in a polar solvent like ethanoic acid, even without a Lewis acid catalyst, due to the high activation of the ring by the methoxy group. It gives mainly the para-bromo derivative.

C_6H_5OCH_3 + Br_2 \xrightarrow{CH_3COOH} p-BrC_6H_4OCH_3 + HBr

(Anisole) (4-Bromoanisole - 90% yield)

(iv) Friedel-Crafts acetylation of anisole: Anisole reacts with an acyl halide (e.g., acetyl chloride) in the presence of a Lewis acid catalyst (anhydrous

AlCl_3

) to give a mixture of ortho and para acylated products (ketones).

C_6H_5OCH_3 + CH_3COCl \xrightarrow{Anhyd. AlCl_3} o-(CH_3CO)C_6H_4OCH_3 + p-(CH_3CO)C_6H_4OCH_3

(Anisole) (Acetyl chloride) (2-Methoxyacetophenone) (4-Methoxyacetophenone - Major)

Q32Exercises

Show how would you synthesise the following alcohols from appropriate alkenes?

(i)

1-Methylcyclohexanol

(ii)

4-Methylheptan-4-ol

(iii)

Pentan-2-ol

(iv)

3-Cyclohexylpentan-3-ol

Solution

(i)

1-Methylcyclohexanol This tertiary alcohol can be prepared by the acid-catalyzed hydration of 1-methylcyclohexene. The addition of water follows Markovnikov's rule, with the -OH group adding to the more substituted carbon of the double bond.

Alkene: 1-Methylcyclohexene Reaction: Add water in the presence of an acid catalyst (

H^+

).

(C_6H_9)-CH_3 + H_2O \xrightarrow{H^+} C_6H_{10}(OH)CH_3

(1-Methylcyclohexene) (1-Methylcyclohexanol)

(ii)

4-Methylheptan-4-ol This is a tertiary alcohol that cannot be made by simple hydration of an alkene. It requires a multi-step synthesis using a Grignard reagent and a ketone, both of which can be derived from alkenes. Target:

CH_3(CH_2)_2-C(OH)(CH_3)-(CH_2)_2CH_3

Synthesis plan: React propylmagnesium bromide with butan-2-one.

Preparation of Propylmagnesium bromide from Propene:
1. $CH_3CH=CH_2 + HBr \xrightarrow{Peroxide} CH_3CH_2CH_2Br$ (Anti-Markovnikov addition)
2. $CH_3CH_2CH_2Br + Mg \xrightarrow{Dry Ether} CH_3CH_2CH_2MgBr$
Preparation of Butan-2-one from But-1-ene:
1. $CH_3CH_2CH=CH_2 + H_2O \xrightarrow{H^+} CH_3CH_2CH(OH)CH_3$ (Markovnikov hydration)
2. $CH_3CH_2CH(OH)CH_3 \xrightarrow{PCC} CH_3CH_2COCH_3$
Final Reaction: $CH_3CH_2CH_2MgBr + CH_3CH_2COCH_3 \xrightarrow{(i) Ether} \xrightarrow{(ii) H_3O^+} CH_3(CH_2)_2-C(OH)(CH_3)-(CH_2)_2CH_3$

(iii)

Pentan-2-ol This secondary alcohol can be prepared by the acid-catalyzed hydration of pent-1-ene, following Markovnikov's rule.

Alkene: Pent-1-ene Reaction: Add water in the presence of an acid catalyst (

H^+

).

CH_3CH_2CH_2CH=CH_2 + H_2O \xrightarrow{H^+} CH_3CH_2CH_2CH(OH)CH_3

(Pent-1-ene) (Pentan-2-ol)

(iv)

3-Cyclohexylpentan-3-ol This is a tertiary alcohol requiring a Grignard synthesis. The reactants can be derived from alkenes. Target:

CH_3CH_2-C(OH)(C_6H_{11})-CH_2CH_3

Synthesis plan: React ethylmagnesium bromide with 1-cyclohexylpropan-1-one.

Preparation of Ethylmagnesium bromide from Ethene:
1. $CH_2=CH_2 + HBr \rightarrow CH_3CH_2Br$
2. $CH_3CH_2Br + Mg \xrightarrow{Dry Ether} CH_3CH_2MgBr$
Preparation of 1-Cyclohexylpropan-1-one: This is more complex. A simpler route is to react cyclohexylmagnesium bromide with propanal.
1. Cyclohexene $+ HBr \rightarrow$ Bromocyclohexane
2. Bromocyclohexane $+ Mg \xrightarrow{Dry Ether}$ Cyclohexylmagnesium bromide
3. Propan-1-ol can be made from propene (hydroboration-oxidation). Propan-1-ol can be oxidized to propanal using PCC.
4. Cyclohexylmagnesium bromide + Propanal $\rightarrow$ 1-Cyclohexylpropan-1-ol
5. 1-Cyclohexylpropan-1-ol can be oxidized to 1-cyclohexylpropan-1-one using PCC.
Final Reaction: $CH_3CH_2MgBr + C_6H_{11}COCH_2CH_3 \xrightarrow{(i) Ether} \xrightarrow{(ii) H_3O^+} CH_3CH_2-C(OH)(C_6H_{11})-CH_2CH_3$

Q33Exercises

When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place: $CH_3-CH(OH)-CH(CH_3)_2 + HBr \rightarrow CH_3-C(Br)(CH_3)-CH_2-CH_3$ Give a mechanism for this reaction. (Hint : The secondary carbocation formed in step II rearranges to a more stable tertiary carbocation by a hydride ion shift from 3rd carbon atom.)

Solution

The reaction of 3-methylbutan-2-ol with HBr is an

S_N1

reaction that involves the formation of a carbocation intermediate, which then undergoes a rearrangement to a more stable carbocation before the final product is formed.

Mechanism:

Step 1: Protonation of the alcohol The lone pair of electrons on the oxygen atom of the alcohol attacks the proton from HBr, forming a protonated alcohol (an oxonium ion). This converts the poor leaving group (-OH) into a good leaving group (

-OH_2

).

CH_3-CH(OH)-CH(CH_3)_2 + H-Br \rightleftharpoons CH_3-CH(\stackrel{+}{O}H_2)-CH(CH_3)_2 + Br^-

(3-Methylbutan-2-ol) (Protonated alcohol)

Step 2: Formation of the secondary carbocation The protonated alcohol is unstable and loses a molecule of water to form a secondary carbocation.

CH_3-CH(\stackrel{+}{O}H_2)-CH(CH_3)_2 \rightarrow CH_3-\stackrel{+}{C}H-CH(CH_3)_2 + H_2O

(Secondary carbocation)

Step 3: Rearrangement via 1,2-Hydride Shift The secondary carbocation is less stable than a tertiary carbocation. It can rearrange to a more stable tertiary carbocation. A hydrogen atom with its pair of electrons (a hydride ion,

H^-

) shifts from the adjacent carbon (C-3) to the positively charged carbon (C-2).

CH_3-\stackrel{+}{C}H-CH(CH_3)_2 \xrightarrow{1,2-Hydride~shift} CH_3-\stackrel{+}{C}(CH_3)-CH_2-CH_3

(Secondary carbocation) (Tertiary carbocation - more stable)

Step 4: Nucleophilic attack by bromide ion The bromide ion (

Br^-

), which is a good nucleophile, attacks the more stable tertiary carbocation to form the final product, 2-bromo-2-methylbutane.

CH_3-\stackrel{+}{C}(CH_3)-CH_2-CH_3 + Br^- \rightarrow CH_3-C(Br)(CH_3)-CH_2-CH_3

(Tertiary carbocation) (2-Bromo-2-methylbutane)

Alcohols, Phenols and EthersClass 12 Chemistry NCERT Solutions

Write IUPAC names of the following compounds:

CH3−CH(CH3)−CH(OH)−C(CH3)3CH_3-CH(CH_3)-CH(OH)-C(CH_3)_3CH3​−CH(CH3​)−CH(OH)−C(CH3​)3​

CH3CH2CH(CH2CH3)CH(OH)CH(CH3)CH2OHCH_3CH_2CH(CH_2CH_3)CH(OH)CH(CH_3)CH_2OHCH3​CH2​CH(CH2​CH3​)CH(OH)CH(CH3​)CH2​OH

CH3CH(OH)CH(OH)CH3CH_3CH(OH)CH(OH)CH_3CH3​CH(OH)CH(OH)CH3​

HOCH2CH(OH)CH2OHHOCH_2CH(OH)CH_2OHHOCH2​CH(OH)CH2​OH

o-Cresol (structure with -OH and -CH3 on adjacent carbons of a benzene ring)

p-Cresol (structure with -OH and -CH3 on opposite carbons of a benzene ring)

2,5-Dimethylphenol

2,6-Dimethylphenol

CH3−O−CH2−CH(CH3)2CH_3-O-CH_2-CH(CH_3)_2CH3​−O−CH2​−CH(CH3​)2​

C6H5−O−C2H5C_6H_5-O-C_2H_5C6​H5​−O−C2​H5​ (xi) C6H5−O−C7H15(n−)C_6H_5-O-C_7H_{15}(n-)C6​H5​−O−C7​H15​(n−) (xii) CH3CH2−O−CH(CH3)CH2CH3CH_3CH_2-O-CH(CH_3)CH_2CH_3CH3​CH2​−O−CH(CH3​)CH2​CH3​

Solution

Write structures of the compounds whose IUPAC names are as follows:

2-Methylbutan-2-ol

1-Phenylpropan-2-ol

3,5-Dimethylhexane-1,3,5-triol

2,3 - Diethylphenol

1 - Ethoxypropane

2-Ethoxy-3-methylpentane

Cyclohexylmethanol

3-Cyclohexylpentan-3-ol

Cyclopent-3-en-1-ol

4-Chloro-3-ethylbutan-1-ol.

Solution

(i) Draw the structures of all isomeric alcohols of molecular formula C_5H_12O and give their IUPAC names.…

Draw the structures of all isomeric alcohols of molecular formula C5H12OC_5H_{12}OC5​H12​O and give their IUPAC names.

Classify the isomers of alcohols in question 7.3 (i) as primary, secondary and tertiary alcohols.

Solution

Explain why propanol has higher boiling point than that of the hydrocarbon, butane?

Solution

Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.

Solution

What is meant by hydroboration-oxidation reaction? Illustrate it with an example.

Solution

Give the structures and IUPAC names of monohydric phenols of molecular formula, C7H8OC_7H_8OC7​H8​O.

Solution

While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason.

Solution

Give the equations of reactions for the preparation of phenol from cumene.

Solution

Write chemical reaction for the preparation of phenol from chlorobenzene.

Solution

Write the mechanism of hydration of ethene to yield ethanol.

Solution

You are given benzene, conc. H2SO4H_2SO_4H2​SO4​ and NaOH. Write the equations for the preparation of phenol using these reagents.

Solution

Show how will you synthesise:

1-phenylethanol from a suitable alkene.

cyclohexylmethanol using an alkyl halide by an SN2S_N2SN​2 reaction.

pentan-1-ol using a suitable alkyl halide?

Solution

Give two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of ethanol.

Solution

Explain why is ortho nitrophenol more acidic than ortho methoxyphenol ?

Solution

Explain how does the -OH group attached to a carbon of benzene ring activate it towards electrophilic substitution?

Solution

Give equations of the following reactions:

Oxidation of propan-1-ol with alkaline KMnO4KMnO_4KMnO4​ solution.

Bromine in CS2CS_2CS2​ with phenol.

Dilute HNO3HNO_3HNO3​ with phenol.

Treating phenol wih chloroform in presence of aqueous NaOH.

Solution

Explain the following with an example.

Kolbe's reaction.

Reimer-Tiemann reaction.

Williamson ether synthesis.

Unsymmetrical ether.

Solution

Write the mechanism of acid dehydration of ethanol to yield ethene.

Solution

How are the following conversions carried out?

Propene → Propan-2-ol.

Benzyl chloride → Benzyl alcohol.

Ethyl magnesium chloride → Propan-1-ol.

Methyl magnesium bromide → 2-Methylpropan-2-ol.

Solution

Name the reagents used in the following reactions:

Oxidation of a primary alcohol to carboxylic acid.

$CH_3-CH(CH_3)-CH(OH)-C(CH_3)_3$

$CH_3CH_2CH(CH_2CH_3)CH(OH)CH(CH_3)CH_2OH$

$CH_3CH(OH)CH(OH)CH_3$

$HOCH_2CH(OH)CH_2OH$

$CH_3-O-CH_2-CH(CH_3)_2$

$C_6H_5-O-C_2H_5$ (xi) $C_6H_5-O-C_7H_{15}(n-)$ (xii) $CH_3CH_2-O-CH(CH_3)CH_2CH_3$

Draw the structures of all isomeric alcohols of molecular formula $C_5H_{12}O$ and give their IUPAC names.

Give the structures and IUPAC names of monohydric phenols of molecular formula, $C_7H_8O$ .

You are given benzene, conc. $H_2SO_4$ and NaOH. Write the equations for the preparation of phenol using these reagents.

cyclohexylmethanol using an alkyl halide by an $S_N2$ reaction.

Oxidation of propan-1-ol with alkaline $KMnO_4$ solution.

Bromine in $CS_2$ with phenol.

Dilute $HNO_3$ with phenol.

$CH_3CH_2-O-CH_2-CH(CH_3)_2$

$CH_3OCH_2CH_2Cl$

$O_2N-C_6H_4-OCH_3(p)$

$CH_3CH_2CH_2OCH_3$

$C_2H_5-O-C_6H_5$