A dihaloalkane 'X',having formula C_{3}H_{6}C | vedclass

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(B) Since the obtained compound reduces Tollen's reagent,it must be an aldehyde. Hydrolysis of a gem-dihalide (where both halogen atoms are on the sam

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$1,1$-dichloropropane

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(B) Since the obtained compound reduces Tollen's reagent,it must be an aldehyde. Hydrolysis of a gem-dihalide (where both halogen atoms are on the same carbon) yields a gem-diol,which is unstable and loses a water molecule to form an aldehyde or ketone. For an aldehyde to be formed,the dihaloalkane must be a terminal gem-dihalide (i.e.,the $-Cl$ atoms are at the $C_{1}$ position). Thus,$CH_{3}CH_{2}CHCl_{2}$ ($1,1$-dichloropropane) on hydrolysis gives propanal $(CH_{3}CH_{2}CHO)$,which reduces Tollen's reagent. The reaction is: $CH_{3}CH_{2}CHCl_{2}$ $\xrightarrow{H_{2}O} CH_{3}CH_{2}CH(OH)_{2}$ $\xrightarrow{-H_{2}O} CH_{3}CH_{2}CHO$ $\xrightarrow{	ext{Tollen's reagent}} CH_{3}CH_{2}COOH + Ag \downarrow$

$A$ dihaloalkane '$X$',having formula $C_{3}H_{6}Cl_{2}$,on hydrolysis gives a compound that can reduce Tollen's reagent. The compound '$X$' is

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