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Which of the following synthetic sequences correctly converts the starting material to propane (CH3-CH2-CH3)?

1. CH3-CH=CH2 treated with (1) B2H6 then (2) NH2-NH2 / CH3COOH gives CH3-CH2-CH3
2. CH3-CH2-CHO treated with Wolff-Kishner reduction (N2H4 / KOH / heat) gives CH3-CH2-CH3
3. CH3-CH2-CH2-OH treated with Red phosphorus and HI at 150 deg C gives CH3-CH2-CH3
4. CH3-CH2-CH2-COONa treated with NaOH / CaO (dry distillation) gives CH3-CH2-CH3

Correct answer: CH3-CH2-CHO treated with Wolff-Kishner reduction (N2H4 / KOH / heat) gives CH3-CH2-CH3

Solution

A) CH3CH=CH2 + B2H6 then acetic acid/hydrazine: hydroboration-oxidation gives propan-1-ol, not propane. This sequence is not a standard reduction to alkane. B) CH3CH2CHO + N2H4/KOH/heat (Wolff-Kishner): the carbonyl C=O is reduced to CH2, giving CH3CH2CH3 (propane). Correct. C) Primary alcohol + Red P/HI gives primary alkyl iodide (CH3CH2CH2I), not propane — would need further reduction. D) CH3CH2CH2COONa has 4 carbons; decarboxylation with NaOH/CaO removes one CO2 to give propane? Actually Kolbe decarboxylation of sodium butanoate gives propane. Wait — CH3CH2CH2COONa is sodium butanoate (4C), and NaOH/CaO decarboxylation gives CH3CH2CH3 (propane, 3C). So D is actually also correct. However, among classical JEE options, B (Wolff-Kishner) is the textbook-standard correct answer. D is also valid chemistry; if single answer expected, B is the cleanest match (propanal -> propane, same carbon count).

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