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JEE CHEMISTRY MCQ: HYDROCARBONS

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Application +4 | -1

\mathop {C{H_3} - C \equiv CH}\limits_{{\text{Propyne}}} \xrightarrow[{\,\,\,\,\,}]{{\,\,\,\,\,}}\mathop {C{H_3} - C \equiv C - C{H_2} - C{H_3}}\limits_{{\text{Pent}}\,\,2 - {\text{yne}}}

The most suitable reagents required for above
conversion?

View answer and explanation

Correct option: B

NaN{H_2},\,C{H_3}C{H_2}I

Option B (

NaNH_2, CH_3CH_2I

) is correct because it uses a strong base to deprotonate the alkyne, followed by alkylation with ethyl iodide.

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