See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1: Identify the starting material and first reaction. The starting material is (CH3)2CHCH2C≡N (3-methylbutanenitrile / isoamyl cyanide). Treatment with HCl and H2O hydrolyzes the nitrile (C≡N) to a carboxylic acid. This gives compound A = (CH3)2CHCH2COOH (3-methylbutanoic acid). Step 2: Reaction of compound A with LiAlH4 followed by H2O. LiAlH4 reduces carboxylic acids to primary alcohols. Therefore compound B = (CH3)2CHCH2CH2OH (3-methyl-1-butanol). Step 3: Reaction of compound B with PCC in CH2Cl2. PCC (pyridinium chlorochromate) oxidizes primary alcohols to aldehydes (not further to carboxylic acids). Therefore compound C = (CH3)2CHCH2CHO (3-methylbutanal), which is written as (CH3)2CHCH2C(=O)H. This matches option (c). Why other options fail: (a) (CH3)2CHC(=O)CH3 is a ketone with a different carbon skeleton — not consistent with the reaction sequence. (b) (CH3)2CHCH2C(=O)OH is the carboxylic acid (compound A), not compound C. (d) (CH3)2CHC(=O)CH2OH has a different carbon arrangement and is not the product of these reactions. Therefore, the correct answer is C.