Molecular formula of a compound ‘A’ is C4H10O. ‘A’ can liberate hydrogen gas with sodium metal, on strong oxidation using KMnO4/H+ it gives Butanoic acid. The compound ‘A’ is

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(CH3)2CH-CH2OH

(CH3)3COH

CH3-CH2-CH2-CH2-OH

CH3-CHOH-CH2-CH3

answer is C.

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Detailed Solution

‘A’ is an alcohol because it contains an active hydrogen. The carboxylic acid obtained on strong oxidation is containing the same number of carbon atoms as that of alcohol. ∴ the alcohol must be 1° alcohol.1° alcohol'n' carbons→(O) Aldehyde'n' carbons→(O)carboxylic acids'n' carbonsThe two possobilities are (CH3)2CHCH2OH and CH3CH2CH2CH2OHA) (CH3)2CHCH2OH→oxidationStrong(CH3)2CHCOOH2-methyl propanoic acid B) CH3CH2CH2CH2OH→strong oxidationCH3CH2CH2COOHButanoic acid ∴ 'A' is n-butyl alcohol

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