Question

The major product ' $\mathrm{R}$ ' obtained in the following reaction is: $$2\mathrm{Ph}-\mathrm{CH}=\mathrm{O}\frac{\mathrm{conc}}{\mathrm{NaOH}}⟶\mathrm{P}+\mathrm{Q}\frac{{\mathrm{H}}^{+}}{\mathrm{\Delta }}\mathrm{R}$$ (A) $\mathrm{Ph}-{\mathrm{CH}}_2-\mathrm{OH}$ (B) $\mathrm{Ph}-\mathrm{CHO}$ (C) $\mathrm{Ph}-{\mathrm{COO}}^{\circ }$ (D) O=C(O)Cc1ccccc1

Short answer

The major product 'R' obtained in this reaction sequence is (D) $O=C(O)Cc1ccccc1$.

Step-by-step solution

Reaction with Concentrated Sodium Hydroxide (NaOH)

In the initial step, we have the molecule 2 Ph-CH=O, which consists of two phenylacetaldehyde moieties, reacting with concentrated sodium hydroxide (NaOH). In this reaction, the sodium hydroxide acts as a strong base, promoting the formation of aldol through nucleophilic attack by the deprotonated aldehyde (forming a resonance-stabilized enolate ion) on the other carbonyl group. The overall result of this step is the formation of two products: P and Q.

Acidic Conditions and Heating (H+/Δ)

In the second step, the products P and Q are reacted under acidic conditions and heating, denoted by H+/Δ. Under these conditions, aldol condensation usually occurs. This involves the dehydration of P and/or Q, forming an α,β-unsaturated carbonyl compound as the product R.

Examining the Proposed Answers

Now, we will analyze each of the proposed answers and compare them to the expected characteristics of product R mentioned in the previous step. (A) Ph-CH₂-OH: This compound is an alcohol and does not include the α,β-unsaturated carbonyl group characteristic of an aldol condensation product. Thus, this answer is incorrect. (B) Ph-CHO: This compound is an aldehyde and does not include the α,β-unsaturated carbonyl group characteristic of an aldol condensation product. Thus, this answer is incorrect. (C) Ph-COO⁰: This compound is a carboxylic acid derivative and does not include the α,β-unsaturated carbonyl group characteristic of an aldol condensation product. Thus, this answer is incorrect. (D) O=C(O)Cc1ccccc1: This compound is an ester, which features an α,β-unsaturated carbonyl group. Therefore, this compound is consistent with the expected product of an aldol condensation reaction, making it the correct answer.

Final Answer

The major product 'R' obtained in this reaction sequence is (D) O=C(O)Cc1ccccc1.

Key concepts

Phenylacetaldehyde Reaction

Phenylacetaldehyde is an aromatic organic compound with a formyl group attached to a benzene ring. When it reacts under aldol condensation conditions, it undergoes a transformation that typically leads to the formation of a new carbon-carbon bond.

In the presence of a strong base, such as concentrated sodium hydroxide (NaOH), the alpha hydrogen of phenylacetaldehyde, which is somewhat acidic, is abstracted to form a resonance-stabilized enolate ion. This ion then acts as a nucleophile and can attack another carbonyl carbon, such as that in a second molecule of phenylacetaldehyde, leading to the formation of a larger molecule known as an aldol.

When the aldol product is subjected to acidic conditions and heat (denoted as H+/Δ), it typically undergoes a dehydration reaction, losing water to form an α,β-unsaturated carbonyl compound. This is a common and significant reaction in organic chemistry, as it is used to create complex molecules with high precision.

Enolate Ion Formation

The enolate ion plays a vital role in the aldol condensation reaction process. It is formed when a base removes an alpha hydrogen from an aldehyde or ketone. Considering phenylacetaldehyde, the alpha hydrogen is the hydrogen attached to the carbon next to the carbon of the carbonyl group.

The process happens as follows: the base, generally a hydroxide ion from sodium hydroxide in the aldol condensation, abstracts this slightly acidic hydrogen, leaving behind a pair of electrons. These electrons move towards the carbonyl group, resulting in the formation of an enolate ion. This enolate ion is resonance-stabilized, meaning electrons can be delocalized between the negatively charged oxygen and the carbon-carbon double bond.

Enolate ions are great nucleophiles and can attack electrophilic carbonyl carbons in other molecules, leading to the formation of new carbon-carbon bonds—as occurs in the formation of aldols.

Aldol Condensation Product

An aldol condensation product typically arises from the reaction between an enolate ion of an aldehyde or a ketone and the carbonyl group of another aldehyde or ketone molecule. The term 'aldol' is derived from 'aldehyde' and 'alcohol', which are the two functional groups present in the initial condensation product.

In the JEE exercise, the aldol condensation product is an α,β-unsaturated ester, which is the result of further dehydration of the initial aldol addition product under acidic conditions and heating (denoted as H+/Δ). This elimination of water from the aldol addition leads to the formation of a double bond between the α and β carbons, adjacent to the carbonyl group, hence, it's denoted as α,β-unsaturated.

This unsaturation along with the ester functional group are key structural features that can be identified in option (D): O=C(O)Cc1ccccc1, making it the correct choice for the major product 'R' as delineated by the exercise.