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answer is B, D.

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Detailed Solution

Explanation:

In the given reaction sequence:

Multiple treatments with NaNH₂ (liq) and CH₃I show the chain extension of an alkyne (HC≡CH) by alkylation, forming higher terminal alkynes.
The product is then subjected to Lindlar’s catalyst or Na/NH₃ (liq) reductions, which convert the alkyne to an alkene (cis or trans form depending on reagent).
Subsequent oxidation (OsO₄, NaHSO₃, H₂O or similar) gives a vicinal diol (–CHOH–CHOH–).

The final product Q is a chiral diol (2,3-butanediol) that exists as enantiomers because both hydroxyl-bearing carbons are asymmetric.
Hence, the compound formed is CH₃–CHOH–CHOH–CH₃, i.e., 2,3-butanediol.

Correct Answer: (a)

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