Q.72 Chemistry · Section A ⭐ 360° Option Elimination Active

NEET 2024 Chemistry Question 72 Solution

Topic: Aldehydes, Ketones and Carboxylic Acids · Official NTA Booklet Code T1

Question

Identify the correct reagents that would bring about the following transformation. C₆H₁₁−CH₂−CH=CH₂ → C₆H₁₁−CH₂−CH₂−CHO (C₆H₁₁ = cyclohexyl)

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Verified Against NTA Final Answer Key

Option (4) : (i) BH3, (ii) H2O2 / OH⁻, (iii) PCC

📚 Aldehydes, Ketones and Carboxylic Acids Medium

📋 Given Parameters & Problem Setup

Starting material 3-cyclohexylprop-1-ene (Cyclohexyl-CH2-CH=CH2)

Target product 3-cyclohexylpropanal (Cyclohexyl-CH2-CH2-CHO)

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The 30-Second Exam Hall Shortcut: ⚡ Anti-Markovnikov Hydration + Mild Oxidation Route

⏱️ Save ~45 seconds

Terminal Alkene -> Primary Alcohol via Hydroboration-Oxidation (BH3, then H2O2/OH⁻), followed by PCC to halt oxidation at the ALDEHYDE stage!

1 The transformation converts a terminal alkene (-CH=CH2) into an aldehyde (-CH2-CHO).

2 Step 1 must install an -OH group at the TERMINAL carbon (anti-Markovnikov addition of water).

3 Acid-catalyzed hydration (H2O / H⁺) would give a secondary alcohol via Markovnikov rule (ruling out Options 2 and 3).

4 Hydroboration-oxidation: (i) BH3 followed by (ii) alkaline H2O2 produces the primary alcohol 3-cyclohexylpropan-1-ol.

5 Step 2 requires selective oxidation of the 1° alcohol to an aldehyde. Alkaline KMnO4 in Option (1) over-oxidizes it to a carboxylic acid!

6 Pyridinium chlorochromate (PCC) cleanly stops oxidation at the aldehyde.

7 Sequence: (i) BH3, (ii) H2O2/OH⁻, (iii) PCC -> Option (4) in 10 seconds!

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Why You Can 100% Trust This Shortcut: NCERT Chemistry · Aldehydes, Ketones and Carboxylic Acids (Preparation of aldehydes by oxidation of primary alcohols using PCC and hydroboration-oxidation).

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Regioselective Anti-Markovnikov Hydration

Reaction of 3-cyclohexylprop-1-ene with borane (BH3) proceeds with syn-addition across the double bond, with boron attaching to the less hindered terminal carbon. Subsequent oxidation with alkaline hydrogen peroxide (H2O2 / OH⁻) replaces boron with a hydroxyl group with retention of stereochemistry, yielding the primary alcohol 3-cyclohexylpropan-1-ol:

R{-}\text{CH}_2{-}\text{CH}=\text{CH}_2 \xrightarrow{\text{(i) } \text{BH}_3, \; \text{(ii) } \text{H}_2\text{O}_2/\text{OH}^\ominus} R{-}\text{CH}_2{-}\text{CH}_2{-}\text{CH}_2\text{OH}

Step 2

Controlled Oxidation to Aldehyde using PCC

Primary alcohols are easily over-oxidized to carboxylic acids in aqueous acidic or alkaline media. Pyridinium chlorochromate (PCC, C5H5NH⁺ ClCrO3⁻) in anhydrous dichloromethane (CH2Cl2) selectively oxidizes primary alcohols to aldehydes without further oxidation:

R{-}\text{CH}_2{-}\text{CH}_2{-}\text{CH}_2\text{OH} \xrightarrow{\text{PCC}} R{-}\text{CH}_2{-}\text{CH}_2{-}\text{CHO}

Step 3

Evaluating Failure of Distractor Reagent Pathways

Option (1) uses alkaline KMnO4 which oxidizes primary alcohols completely to carboxylic acids (-COOH). Options (2) and (3) use H2O/H⁺ which follows Markovnikov addition to yield a secondary alcohol, which upon oxidation produces a ketone rather than an aldehyde.

\text{H}_2\text{O}/\text{H}^\oplus \implies \text{Markovnikov 2° alcohol} \xrightarrow{\text{PCC}} \text{Ketone (wrong product)}

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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:

Alkaline KMnO4 vigorously oxidizes primary alcohols to carboxylic acids, not aldehydes.

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Student Fallacy (Why candidates pick this trap): Overlooking that KMnO4 is too strong to stop at the aldehyde.

2 (2)

✗ Incorrect

Verdict & Scientific Reason:

Acid-catalyzed hydration gives a secondary alcohol, which oxidizes to a ketone.

⚠️

Student Fallacy (Why candidates pick this trap): Forgetting Markovnikov regioselectivity.

3 (3)

✗ Incorrect

Verdict & Scientific Reason:

Produces a ketone due to Markovnikov hydration.

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Student Fallacy (Why candidates pick this trap): Ignoring carbocation rearrangement and Markovnikov addition.

4 (4)

✓ 100% Correct

Verdict & Scientific Reason:

Anti-Markovnikov hydration yields 1° alcohol; anhydrous PCC mildly oxidizes it to aldehyde.

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Why this is airtight: None.

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NTA Trap Alert: Markovnikov vs Anti-Markovnikov Choice

To obtain an aldehyde from an alkene, the oxygen must attach to the terminal carbon. Only hydroboration-oxidation gives anti-Markovnikov addition of water.

🛡️ How to avoid the -1 mark: Terminal alkene -> Aldehyde requires BH3/H2O2 + PCC. Terminal alkene -> Ketone uses H3O+ + oxidant.

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Core NEET Syllabus Standard

Chemistry Core Organic Chemistry · Aldehydes, Ketones and Carboxylic Acids Standard Benchmark

Core Scientific Principle Tested:

"Hydroboration-oxidation converts terminal alkenes into primary alcohols, which are selectively converted into aldehydes using PCC in non-aqueous medium."

Syllabus Focus: Selective Oxidation of Alcohols to Aldehydes (PCC Reagent)

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

Hydroborate to terminal OH, PCC stops oxidation at the CHO!

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NEET 2024 Chemistry Question 72 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Anti-Markovnikov Hydration + Mild Oxidation Route

Step-by-Step Rigorous Workflow & Mathematical Proof

Regioselective Anti-Markovnikov Hydration

Controlled Oxidation to Aldehyde using PCC

Evaluating Failure of Distractor Reagent Pathways

360° Option Elimination Engine Key Feature

NTA Trap Alert: Markovnikov vs Anti-Markovnikov Choice

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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