1.

`underset(A)(C_5H_11Br) underset((ii)D_2O)overset((i)Mg//ether)(to)underset(B)(C_5H_11D` only one type of B is formed . Thus , A isA. `CH_3CH_2undersetoverset(|)(Br)( C)HCH_2CH_3`B. `CH_2undersetoverset(|)(Br)(C )Hundersetoverset(|)(CH_3)(C )HCH_3`C. `CH_3-undersetoverset(|)(CH_3)oversetunderset(|)(CH_3)(C)-CH_2Br`D. `CH_3CH_2undersetoverset(|)(CH_3)(C )HCH_2Br`

Answer» Correct Answer - C
If one type of B is formed , it means A is also of one type .
Thus , `underset((A))((CH_(3))_(3))C CHBr overset(Mg) to underset((B)) underset((CH_(3))_(3)C CH_(2)D) underset(darrD_(2)O) ((CH_(3))_(3)C C_(2)MgBr)`


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