Question:
An alkene ‘X’ (C\(_4\)H\(_8\)) does not exhibit cis/trans isomerism. Reaction of ‘X’ with Br\(_2\)/CCL\(_4\), followed by reaction with reagent ‘Y’ gave ‘Z’ (C\(_6\)H\(_6\)). What are ‘Y’ and ‘Z’ respectively?
An alkene ‘X’ (C\(_4\)H\(_8\)) does not exhibit cis/trans isomerism. Reaction of ‘X’ with Br\(_2\)/CCL\(_4\), followed by reaction with reagent ‘Y’ gave ‘Z’ (C\(_6\)H\(_6\)). What are ‘Y’ and ‘Z’ respectively?
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Dehydrohalogenation reactions with NaNH\(_2\) lead to the formation of alkynes from vicinal dibromides.
Updated On: Mar 11, 2025
- aq.KOH ; CH\(_3\)CH\(_2\)C = CH
- alc.KOH ; CH\(_3\)C = CCH\(_3\)
- (i) alc.KOH, (ii) NaNH\(_2\) ; CH\(_3\)CH\(_2\)C = CH
- (i) alc.KOH, (ii) NaNH\(_2\) ; CH\(_3\)C = CCH\(_3\) \textbf{Correct Answer:} (3) (i) alc.KOH, (ii) NaNH\(_2\) ; CH\(_3\)CH\(_2\)C = CH
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The Correct Option is C
Solution and Explanation
The reaction of ‘X’ with Br\(_2\)/CCL\(_4\) produces a vicinal dibromide. The subsequent treatment with alcoholic KOH causes elimination to form an alkene, which can undergo dehydrohalogenation with NaNH\(_2\) to give the required alkyne.
Thus, the answer is (i) alc.KOH and (ii) NaNH\(_2\).
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