Alcohols, Phenols and Ethers

Bourveault–Blanc reduction involves the reduction of esters to primary alcohols in the presence of sodium and alcohol.

The reaction involved is

The acidic hydrolysis is fastest when two phenyl groups are replaced by two para-methoxyphenyl groups.

This is because $-$OCH₃ is an electron donating group which increases the stability of the carbocation.

The reactions involved are

Comparatively phenol is a weaker acidic than carbonic acid (H₂CO₃), and hence does not liberate CO₂ on treatment with aqueous NaHCO₃ solution. The rest are more acidic than H₂CO₃, and so liberate CO₂.

The reaction involved is

(A)

The compound undergoes an intramolecular aldol condensation reaction in the presence of aqueous NaOH.

Steps Involved :

Abstraction of Proton :

The proton attached to the alpha carbon is abstracted, generating a carbanion.

Nucleophilic Addition :

The carbanion attacks the electropositive carbonyl carbon, forming a cyclic intermediate.

Dehydration :

The cyclic intermediate undergoes dehydration to yield the final product.

Option (A) in Column I matches with (R), (S), and (T) in Column II.

(B)

Nucleophilic Addition :

The ketone reacts with methyl magnesium bromide, forming a secondary alcohol. This is a nucleophilic addition reaction.

Nucleophilic Substitution :

The highly nucleophilic oxygen attacks the C_1 carbon of the alkyl chain from the back, displacing chlorine to form the final product.

Option (B) in Column I matches with (P) and (S) in Column II.

(C)

Nucleophilic Addition :

The compound undergoes a nucleophilic addition reaction in the presence of $\mathrm{H}_2 \mathrm{SO}_4$.

Dehydration :

The molecule loses water, forming an alkene.

Option (C) in Column I matches with (R) and (S) in Column II.

(D)

The compound reacts with sulfuric acid to form a carbocation, which undergoes electrophilic substitution.

Reaction :

Option (D) in Column I matches with (Q) and (R) in Column II.

$ \text { The products obtained on nitration of } \mathbf{P}, \mathbf{Q} \text { and } \mathbf{S} \text { are } $

Here strongly activating and ortho-, para directing -OH group determines the site for electrophilic substitution.

Here $-\mathrm{OCH}_3$ is a stronger activator than $-\mathrm{CH}_3$ and both are ortho-, para directing.

Here, the substitution takes place on the activated ring (with substituent $\mathrm{PhCOO}^-$ ) at the sterically unhindered para-position.

Alkyl aryl ethers are cleaved at the alkyl-oxygen bond due to stronger aryl-oxygen bond. The reaction yields phenol and alkyl halide. Thus,

The reaction is

In presence of base, phenol forms phenoxide ion which is $ortho$-$para$ directing because in the resonating structures, the electron density increases at ortho and para positions. The phenoxide ion further undergoes electrophilic substitution with Br$_2$ at $ortho$ and $para$ positions.

conc. $\mathrm{H}_2 \mathrm{SO}_4$ gives $\mathrm{SO}_3$ as a neutral electrophile that undergo electrophilic aromatic substitution.

-OH group increases electron density at ortho and para-position which facilitates attack of in coming $\mathrm{NO}_2{ }^{\oplus}$ electrophile. Besides this, $-\mathrm{SO}_3 \mathrm{H}$ is also a good leaving group which will be substituted by $\mathrm{NO}_2{ }^{\oplus}$ and giving corresponding product i.e. Picric acid, one of the strongest acids.

Compound H (a tertiary alcohol) is formed by the reaction of acetophenone PhCOCH$_3$ with benzyl magnesium bromide PhCH$_2$MgBr.

The structure of compound I is -

The reagent used in the reaction is aq $\mathrm{NaOH}+ \mathrm{CHCl}_{3}$

Reaction is

Mechanism :

In this reaction electrophile involved is dichloro carbene ($\mathrm{CCl_2}$) as it contains a vacant orbital.