In the following sequence of reactions, \$CH_3Br\overset{KCN}{\rightarrow}A\overset{H_3O^+}{\rightarrow}B\overset{LiAlH_4}{\rightarrow}C\$The end product C is:

The end product C is ethyl alcohol.\$\displaystyle CH_3Br \overset{KCN}{\rightarrow}\underset {A}{CH_3-CN}\overset{H_3O^+}{\rightarrow}\underset {B}{CH_3-COOH}\overset{LiAlH_4}{\rightarrow}\underset {C}{CH_3-CH_2-OH}\$Methyl bromide reacts with KCN to form acetonitrile. Br atom is replaced with CN group. Hydrolysis of cyano group gives carboxylic group. Reduction of carboxylic group gives hydroxyl group.

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>>Class 12
>>Chemistry
>>Alcohols, Phenols and Ethers
>>Chemical Reactions of Alcohols and Phenols
>> $Electrophilic Aromatic Substitution Reac...$

Electrophilic Aromatic Substitution Reactions of Phenols

Q: Give reasons for the following:(i) Aniline does not undergo Friedal- Crafts reaction.(ii) \$(CH_3)_2NH\$ is more basic than \$(CH_3)_3N\$ in an aqueous solution.(iii) Primary amines have higher boiling point than tertiary amines.

(i) Aniline does not undergo Friedel craft's reactions because the reagent \$AlCl_3\$ (the Lewis acid which is used as a catalyst in friedel crafts reaction), being electron deficient acts as a Lewis base. and attacks on the lone pair of nitrogen present in aniline to form an insoluble complex which precipitates out and the reaction does not proceed.(ii) With increase in alkyl group, the +I effect will increase which will increase the ease of donation of lone pair electron. But in water one other factor is controlling the strength of basicity. Amine will accept a proton and form cation. This cation will be stabilised in water by salvation. Better the salvation by hydrogen bonding, higher will be the basic strength.(refer image)Thus with increase in methyl group, hydrogen bonding and stabilisation by salvation decreases. The net effect is when we move from secondary to tertiary amine basic strength (iii) In a tertiary amine there aren't any hydrogen atoms attached directly to the nitrogen. That means that hydrogen bonding between tertiary amine molecules is impossible. That's why the boiling point is much lower. The small amines of all types are very soluble in water.

Q: Write the following reaction: Reimer-Tiemann reaction.

When phenol reacts with chloroform in the presence of bases \$(NaOH/KOH)\$, an aldehyde \$(-CHO)\$ group gets introduced in the ring at a position ortho to the phenolic group and salicylaldehyde is formed. This reaction is called the Reimer-Tiemann reaction.

Q: Predict the major product of acid catalysed dehydration:(i) 1-methyl cyclohexanol(ii) Butan-1-ol

Due to Saytzeff rule, in \$(i)\$ the product where double bond is formed at the more substituted carbon is the major product. But as Saytzeff rule is only valid for 2° and 3° alcohols, a primary alcohol like \$(ii)\$ will form a terminal alkene as the major product.

Q: Write the equation involved in the following reactions:(i) Reimer-Tiemann reaction (ii) Kolbe's reaction

(i) Reimer-Tiemann reaction \$\rightarrow\$ When phenol is treated with chloroform in the presence of sodium hydroxide, a \$-CHO\$ group is introduced at the ortho position of the benzene ring. This reaction is known as the Reimer-Tiemann reaction. The intermediate is hydrolyzed in presence of alkalis to produce salicylaldehyde.(ii) Kolbe's reaction \$\rightarrow\$ When phenol is treated with sodium hydroxide, sodium phenoxide is produced. This sodium phenoxide when treated with carbon dioxide, followed by acidification, undergoes electrophilic substitution to give ortho-hydroxybenzoic acid as the main product. This reaction is known as Kolbe's reaction.

Q: 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.

(i) Oxidation of primary alcohol to carboxylic acid.Acidified \$ \displaystyle K_2Cr_2O_7\$ or acidified \$ \displaystyle KMnO_4\$(ii) Oxidation of primary alcohol to aldehyde.Pyridinium chlorochromate (PCC) in \$ \displaystyle CH_2Cl_2\$ or Pyridinimium dichromate (PDC) in \$ \displaystyle CH_2Cl_2\$(iii) Bromination of phenol to 2,4,6-tribromophenol.Aqueous bromine ( or bromine water)(iv) Benzyl alcohol to benzoic acid.Acidified or alkaline \$ \displaystyle KMnO_4\$ followed by hydrolysis with dil \$ \displaystyle H_2SO_4\$(v) Dehydration of propan-2-ol to propene.85% phosphoric acid(vi) Butan-2-one to butan-2-ol.\$ \displaystyle NaBH_4\$ or \$ \displaystyle Ni / H_2\$

Q: Explain why is ortho nitrophenol more acidic than ortho methoxyphenol?

In ortho nitro phenol, electron withdrawing nitro group is present in ortho position, the electron density of \$O-H\$ bond is reduced.This makes removal of proton easy. After deprotonation, o-nitrophenoxide ion is resonance stabilized. Hence, ortho nitrophenol is a stronger acid.In ortho methoxyphenol, electron releasing methoxy group is present in ortho position, the electron density of \$O-H\$ bond is increased.This makes removal of proton difficult. Hence, ortho methoxyphenol is less acidic.

Q: The major product obtained on interaction of phenol with sodium hydroxide and carbon dioxide is?

Reaction of phenol with sodium hydroxide and carbon dioxide is Kolbe's Schmidt reaction that is a electrophilic substitution reaction resulting in the formation of salicylic acid as shown in the reaction.The major product obtained on interaction of phenol with sodium hydroxide and carbon dioxide is salicylic acid.

Q: Which one of the following compounds has the most acidic nature?

Acidic nature is directly proportional to the stability of anion when acidic H is removed. Since in Phenol negative charge formed after the removal of acidic H will be delocalized in the whole ring. Hence it will be most stable. So phenol will be most acidic compound.

Q: Which of the following is most acidic?

Option D is the correct answer. A methoxy group is mesomeric donating (+R) and inductive withdrawing (-I). We know that the +R effect is predominant over the -I effect. But +R effect is only valid when the substituent is at the ortho and para position.We also know that addition of Electron donating groups to the Benzene ring in phenol decreases its acidity while the addition of Electron withdrawing groups increase it.By that logic, m-Methoxyphenol is a stronger acid than phenol as the \$-OCH_3\$ group at m-position imparts only -I effect to the ring.