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When 1,1-dichloroethane is treated with alcoholic KOH and then with sodamide, which of the following reaction sequences leads to the formation of an alkyne?

1. R–CH2–CCl2–R reacts with alcoholic KOH to form R–CH=CCl–R, which on treatment with NaNH2 gives R–C≡C–R.
2. R–CH2–CHCl–R reacts with alcoholic KOH to form R–CH=CH–R, which on treatment with NaNH2 gives R–C≡C–R.
3. R–CH2–CHCl–R reacts with alcoholic KOH to form R–CH=CH–R, which on treatment with NaNH2 gives R–CH=CH–R.
4. R–CH2–CCl2–R reacts with alcoholic KOH to form R–CH=CCl–R, which on treatment with NaNH2 gives R–CH=CH–R.

Correct answer: R–CH2–CHCl–R reacts with alcoholic KOH to form R–CH=CH–R, which on treatment with NaNH2 gives R–C≡C–R.

Solution

Ethylene chloride (1,1-dichloroethane) undergoes dehydrohalogenation with alcoholic KOH to form a vinyl chloride intermediate, which further reacts with sodamide (NaNH2) to yield an alkyne. This matches option B.

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