See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 – Identify the reactants. Ethanoic acid is CH3COOH. 3-methyl-1-butanol is (CH3)2CHCH2CH2OH, a primary alcohol where the hydroxyl group is on carbon-1 and there is a methyl branch on carbon-3. Step 2 – Reaction type. Fischer esterification with traces of H2SO4 as acid catalyst converts a carboxylic acid and an alcohol into an ester plus water. The –OH of the acid is replaced by the –O–R group of the alcohol. Step 3 – Form the ester. The product is CH3-C(=O)-O-CH2CH2CH(CH3)2, i.e., 3-methylbutyl ethanoate (isoamyl acetate). The acetyl group (CH3CO–) comes from ethanoic acid, and the 3-methylbutyloxy group (–OCH2CH2CH(CH3)2) comes from 3-methyl-1-butanol. Step 4 – Match to options. Option (b) shows CH3C(=O)–O–CH2CH2CH2CH(CH3)2, which is an acetate ester with a 3-methylbutyl (isopentyl) chain attached through a primary oxygen: –O–CH2–CH2–CH2–CH(CH3)2. This matches the product of esterification of ethanoic acid with 3-methyl-1-butanol. Step 5 – Eliminate other options. - Option (a) shows a branched connection at the carbon directly attached to oxygen (secondary/tertiary type), inconsistent with 3-methyl-1-butanol whose OH is primary (C-1). - Option (c) shows an n-butyl ester, which would come from 1-butanol, not 3-methyl-1-butanol. - Option (d) shows a sec-butyl ester, inconsistent with the primary alcohol 3-methyl-1-butanol. Therefore, the correct answer is B.