Identify the product 'C' in the following sequence of reactions.

The correct answer is Option (4) → \(CH_3-CH_2-OH\)

Ozonolysis breaks a C=C bond and puts a C=O group on each of its two carbons. First, ozone adds across the C=C of but-2-ene to give the ozonide A; then Zn/H₂O cleaves it. Both halves are identical, so B is ethanal (two molecules):

    \( \displaystyle \ce{CH3CH=CHCH3} \xrightarrow[\text{(ii) }\ce{Zn/H2O}]{\text{(i) }\ce{O3}} \ce{2CH3CHO} \)

Then lithium aluminium hydride reduces the aldehyde to a primary alcohol, so C is ethanol:

    \( \displaystyle \ce{CH3CHO} \xrightarrow[\text{(ii) }\ce{H2O}]{\text{(i) }\ce{LiAlH4}} \ce{CH3CH2OH} \)

Why the other options are incorrect

• Option (1) \(CH_3-CH_2-CH_3\) — Incorrect. \( \ce{LiAlH4} \) reduces C=O only to CH–OH, not to \( \ce{CH2} \), and ozonolysis gives two-carbon fragments, not a three-carbon one.

• Option (2) \(CH_3-CH_2-CH_2-CH_2-OH\) — Incorrect. The four-carbon chain is broken at the double bond by ozonolysis; a butanol cannot form.

• Option (3) \(CH_3-CH_2-CH=CH_2\) — Incorrect. but-1-ene is an isomer of the starting alkene, not a product of ozonolysis and reduction.

NCERT (2026-27 edition):
By Ozonolysis of Alkenes – “As we know, ozonolysis of alkenes followed by reaction with zinc dust and water gives aldehydes, ketones or a mixture of both depending on the substitution pattern of the alkene (Unit 9, Class XI).”

Reduction to Alcohols – “Aldehydes and ketones are reduced to primary and secondary alcohols respectively by sodium borohydride (NaBH₄) or lithium aluminium hydride (LiAlH₄) as well as by catalytic hydrogenation (Unit 7, Class XII).”