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JEE CHEMISTRY MCQ: HALOALKANES AND HALO ARENES

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Application +4 | -1
CH3CH2−CH=CH2→ NBS (X) One of the product → alc. KOH(Y)CH _3 CH _2- CH = CH _2 \xrightarrow{\text { NBS }} \underset{\text { One of the product }}{( X )} \xrightarrow{\text { alc. } KOH }( Y ) The compound (Y)( Y ) is
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Correct option: C
1,3-Butadiene
Option C (1,3-Butadiene) is correct because NBS promotes allylic bromination followed by elimination with alc. KOH to form a conjugated diene.

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