Compound A of molecular formula C_4H_{10}O on | vedclass

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(A) $1$. The molecular formula $C_4H_{10}O$ corresponds to an alcohol.
$2$. Lucas reagent $(conc. HCl + ZnCl_2)$ reacts with tertiary alcohols instan

Vedclass · Accepted Answer

$2-$methylpropan$-2-$ol and $2-$chloro$-2-$methylpropane

Vedclass · Answer

(A) $1$. The molecular formula $C_4H_{10}O$ corresponds to an alcohol.
$2$. Lucas reagent $(conc. HCl + ZnCl_2)$ reacts with tertiary alcohols instantaneously at room temperature to form alkyl chlorides.
$3$. Compound $B$ on heating with alcoholic $KOH$ (dehydrohalogenation) gives isobutene $(CH_3-C(CH_3)=CH_2)$.
$4$. The structure of isobutene indicates that the precursor alkyl chloride $(B)$ must be $2-$chloro$-2-$methylpropane.
$5$. Therefore,compound $A$ must be $2-$methylpropan$-2-$ol ($tert-$butyl alcohol).
$6$. The reaction is: $(CH_3)_3COH + HCl \rightarrow (CH_3)_3CCl + H_2O$.
$7$. Thus,$A$ is $2-$methylpropan$-2-$ol and $B$ is $2-$chloro$-2-$methylpropane.

Compound $A$ of molecular formula $C_4H_{10}O$ on treatment with Lucas reagent at room temperature gives compound $B$. When compound $B$ is heated with alcoholic $KOH$,it gives isobutene. Compound $A$ and $B$ are respectively:

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