Which of the following alkyl halide is treate | vedclass

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Question

(A) The Wurtz reaction involves the coupling of two alkyl halide molecules in the presence of sodium metal to form a symmetrical alkane.To obtain $2,2

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$tert-$Butyl bromide

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(A) The Wurtz reaction involves the coupling of two alkyl halide molecules in the presence of sodium metal to form a symmetrical alkane.To obtain $2,2,3,3-$tetramethylbutane,we must identify the alkyl group that,when doubled,forms this structure.The structure of $2,2,3,3-$tetramethylbutane is $(CH_3)_3C-C(CH_3)_3$.Splitting this molecule at the central $C-C$ bond gives two $(CH_3)_3C-$ groups.Therefore,the required alkyl halide is $tert-$butyl bromide,$(CH_3)_3C-Br$.

Which of the following alkyl halide is treated with sodium metal to obtain $2,2,3,3-$tetramethylbutane?

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$III$. $(CH_3)_3COH + HCl \rightarrow$
$IV$. $(CH_3)_2CHOH + HCl \xrightarrow{Anh. ZnCl_2}$

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