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Question
CH3−CH3Br2/hv−−−−→ALiAlH4−−−−→BBr2/hv−−−−→CNa/ether−−−−−→D
The compound D is:
CH3−CH3Br2/hv−−−−→ALiAlH4−−−−→BBr2/hv−−−−→CNa/ether−−−−−→D
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Solution
The correct option is A CH3−CH2−CH2−CH3
By this process, first alkyl halide forms which reduced by LiAlH4 to form alkane and again bromination takes place to give alkyl halide which in presence of Na/dryether forms alkane as product (Wurtz reaction).
CH3CH3Br2/hv−−−−→CH3CH2BrLiAlH4−−−−→CH3CH3Br2/hv−−−−→CH3−CH2−BrNa/dryether−−−−−−−−→CH3−CH2−CH2−CH3
Option A is correct.
By this process, first alkyl halide forms which reduced by LiAlH4 to form alkane and again bromination takes place to give alkyl halide which in presence of Na/dryether forms alkane as product (Wurtz reaction).
CH3CH3Br2/hv−−−−→CH3CH2BrLiAlH4−−−−→CH3CH3Br2/hv−−−−→CH3−CH2−BrNa/dryether−−−−−−−−→CH3−CH2−CH2−CH3
Option A is correct.
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