Consider the following sequence,where A can b | vedclass

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(A) Let us analyze the reactions for $A$ $(C_{3}H_{6}Cl_{2})$:$1$. Reaction with $aq. \, KOH$ (excess) gives $C$. Geminal dihalides $(R-CH(Cl)_{2})$ o

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$1,1-$dichloropropane

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(A) Let us analyze the reactions for $A$ $(C_{3}H_{6}Cl_{2})$:$1$. Reaction with $aq. \, KOH$ (excess) gives $C$. Geminal dihalides $(R-CH(Cl)_{2})$ on hydrolysis with $aq. \, KOH$ yield aldehydes or ketones,while vicinal dihalides $(R-CH(Cl)-CH_{2}Cl)$ yield glycols.$2$. The sequence $A$ $\xrightarrow{alc. \, KOH / \Delta, NaNH_{2} / \Delta} B$ $\xrightarrow{B_{2}H_{6}, H_{2}O_{2}/OH^{-}} C$ is characteristic of converting a dihalide to an alkyne $(B)$ and then performing hydroboration-oxidation to form an aldehyde or ketone $(C)$.$3$. If $A$ is $1,1$-dichloropropane $(CH_{3}CH_{2}CHCl_{2})$:- $aq. \, KOH$ gives propanal $(CH_{3}CH_{2}CHO)$.- $alc. \, KOH / NaNH_{2}$ gives propyne $(CH_{3}C \equiv CH)$.- Hydroboration-oxidation of propyne $(CH_{3}C \equiv CH)$ gives propanal $(CH_{3}CH_{2}CHO)$.Since both paths lead to the same product $C$ (propanal),$A$ must be $1,1$-dichloropropane.

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