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A is an optically inactive alkyl chloride which on reaction with aqueous $\mathrm{KOH}$ gives $\mathrm{B}$. $\mathrm{B}$ on heating with $\mathrm{Cu}$ at $300^{\circ} \mathrm{C}$ gives an alkene $\mathrm{C}$, what are $\mathrm{A}$ and $\mathrm{C}$
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The correct answer is:
$\mathrm{Me}_{3} \mathrm{CCl}, \mathrm{Me}_{2} \mathrm{C}=\mathrm{CH}_{2}$


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