Q.97 Chemistry ⭐ 360° Option Elimination Active

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NEET 2021 Chemistry Question 97 Solution

Topic: Aldehydes, Ketones and Carboxylic Acids · Official NTA Booklet Code M1

Question

The product formed in the following chemical reaction is:

✓

Verified Against NTA Final Answer Key

Option (4) : Structure (4) (Ketone reduced to secondary alcohol; ester intact)

📚 Aldehydes, Ketones and Carboxylic Acids Medium

📋 Given Parameters & Problem Setup

Substrate Bifunctional compound with cyclohexanone (ketone) and methyl ester (-CH2COOCH3)

Reducing Agent NaBH4 in C2H5OH

⚡

The 30-Second Exam Hall Shortcut: ⚡ NaBH4 Reduces Aldehydes & Ketones ONLY — Esters Are Immune!

⏱️ Save ~20 seconds

Sodium borohydride ( $NaBH_4$ ) is a mild, chemoselective reducing agent that readily reduces ketones to secondary alcohols ( $-OH$ ) but is completely unreactive toward esters ( $-COOCH_3$ ), carboxylic acids, and amides under standard conditions! Thus, the cyclohexanone $C=O$ reduces to $-OH$ , while the $-CH_2COOCH_3$ ester group remains completely intact! This corresponds to Structure (4)! Match Option (4)!

1 Identify functional groups on the substrate: - Ring ketone (

C=O

on cyclohexanone ring) - Side-chain methyl ester (

-CH_2-COOCH_3

2 Identify reagent selectivity:

NaBH_4

in ethanol is a selective reducing agent. - Aldehydes

\to 1^\circ

alcohols - Ketones

\to 2^\circ

alcohols - Esters

\to

NO REACTION.

3 Ring ketone reduces to secondary alcohol (

-OH

4 Ester group (

-CH_2-COOCH_3

) remains completely untouched.

5 Match Structure (4).

🛡️

Why You Can 100% Trust This Shortcut: Canonical NCERT chemoselectivity principle: NaBH4 reduces aldehydes/ketones, not esters.

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Identifying Functional Groups and Their Electrophilicity

The reactant molecule contains two carbonyl-containing functional groups: 1. A cyclic ketone group located on the cyclohexane ring. 2. An aliphatic methyl ester group ( $-CH_2COOCH_3$ ) on the side chain. The carbonyl carbon of a ketone is significantly more electrophilic than the carbonyl carbon of an ester because the ester oxygen atom donates electron density through resonance into the carbonyl $\pi^*$ orbital ( $R-C(=O)-OCH_3 \leftrightarrow R-C(O^-)=O^+CH_3$ ).

Step 2

Reactivity Profile of Sodium Borohydride (NaBH4)

Sodium borohydride ( $NaBH_4$ ) is a mild hydride donor. Due to the lower nucleophilicity of the $[BH_4]^-$ ion compared to $[AlH_4]^-$ ( $LiAlH_4$ ): - $NaBH_4$ readily reduces aldehydes and ketones to primary and secondary alcohols, respectively. - Under standard conditions in alcoholic solvent ( $C_2H_5OH$ ), $NaBH_4$ does not reduce esters, carboxylic acids, or amides.

Step 3

Determining the Structure of the Product

- The cyclic ketone ( $C=O$ ) is smoothly converted to a secondary alcohol group ( $-OH$ ). - The ester group ( $-CH_2COOCH_3$ ) survives untouched. - This chemoselective transformation corresponds precisely to Structure (4).

⭐

360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan

1 1

✗ Incorrect

Verdict & Scientific Reason:

Shows reduction of the ester group into a geminal hydroxy ether / hemiacetal, which does not occur with $NaBH_4$ .

⚠️

Student Fallacy (Why candidates pick this trap): Assuming

NaBH_4

reduces esters partially.

2 2

✗ Incorrect

Verdict & Scientific Reason:

Shows the ketone untouched while the ester is reduced to an alcohol, which is the exact inverse of true chemical selectivity.

⚠️

Student Fallacy (Why candidates pick this trap): Inverting functional group reactivity.

3 3

✗ Incorrect

Verdict & Scientific Reason:

Shows an impossible methylated diol product with loss of oxygen.

⚠️

Student Fallacy (Why candidates pick this trap): Confusing hydride reduction with Grignard alkylation.

4 4

✓ 100% Correct

Verdict & Scientific Reason:

Exact chemoselective product: cyclic ketone reduced to secondary alcohol ( $-OH$ ), ester ( $-COOCH_3$ ) intact.

💡

Why this is airtight: None; rigorous hydride chemoselectivity mastery.

⚠️

NTA Trap Alert: LiAlH4 vs. NaBH4 Chemoselectivity Trap

Lithium aluminium hydride ( $LiAlH_4$ ) is a powerful non-selective reducing agent that reduces BOTH ketones and esters (producing a diol). Sodium borohydride ( $NaBH_4$ ) is milder and strictly SELECTIVE: it reduces the ketone but LEAVES THE ESTER INTACT.

🛡️ How to avoid the -1 mark:

NaBH_4

+ Ester

\implies

NO REACTION! Only the ketone reacts.

📖

Core NEET Syllabus Standard

NCERT Chemistry · Aldehydes, Ketones and Carboxylic Acids Standard Benchmark

Core Scientific Principle Tested:

"NaBH4 selectively reduces aldehydes and ketones to alcohols without affecting esters or carboxylic acids."

Syllabus Focus: Reduction of Carbonyl Compounds (Chemoselectivity of NaBH4 vs LiAlH4)

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2021 Chemistry Question 97 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ NaBH4 Reduces Aldehydes & Ketones ONLY — Esters Are Immune!

Step-by-Step Rigorous Workflow & Mathematical Proof

Identifying Functional Groups and Their Electrophilicity

Reactivity Profile of Sodium Borohydride (NaBH4)

Determining the Structure of the Product

360° Option Elimination Engine Key Feature

NTA Trap Alert: LiAlH4 vs. NaBH4 Chemoselectivity Trap

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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