NEET 2025 Chemistry Question 49 Solution
Question
Predict the major product 'P' in the following sequence of reactions - 1-methylcyclopentene → (i) HBr, benzoyl peroxide (ii) KCN (iii) Na(Hg)/C₂H₅OH → P (Major)
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Predict the major product 'P' in the following sequence of reactions - 1-methylcyclopentene → (i) HBr, benzoyl peroxide (ii) KCN (iii) Na(Hg)/C₂H₅OH → P (Major)
Step 1: HBr + peroxide gives Anti-Markovnikov addition: Br attaches to the LESS substituted C2 (adjacent to C1-methyl). Step 2: KCN replaces Br by CN at C2. Step 3: Reduction converts -CN into -CH₂NH₂ at C2 ⟹ 1,2-disubstituted product (Option 1)!
In the presence of benzoyl peroxide, HBr undergoes homolytic cleavage to produce bromine free radicals (Br•). The Br• radical adds regioselectively to the less substituted double-bond carbon (C2) to generate the more stable 3° tertiary radical at C1:
Potassium cyanide (KCN) is an ionic cyanide source. The cyanide ion (CN⁻) acts as an ambident nucleophile and attacks through the carbon atom, displacing bromide in an S_N2 reaction:
Sodium amalgam in ethanol (Na(Hg) / C₂H₅OH) provides nascent hydrogen to reduce the nitrile group completely to a primary aminomethyl group (-CH₂NH₂):
Diagnose every choice: Know exactly why the right option works and why distractors fail.
100% correct. Anti-Markovnikov addition places Br at C2, giving 1,2-substitution.
Incorrect. Corresponds to Markovnikov addition (without peroxide) where Br attaches to C1.
Incorrect. Contains an isonitrile group (-NC) and omits the reduction step.
Incorrect. Combines Markovnikov regiochemistry with isonitrile formation without reduction.
Without peroxide, HBr adds according to Markovnikov’s rule, placing Br at C1 (tertiary carbon), leading to 1,1-substitution (Option 2). The presence of benzoyl peroxide completely reverses regiochemistry to 1,2-substitution (Option 1)!
"HBr + peroxide undergoes free radical Anti-Markovnikov addition."
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