A hydrocarbon A (C_{4}H_{8}) on reaction with | vedclass

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(B) The hydrocarbon $A$ is $but-2-ene$ $(CH_{3}-CH=CH-CH_{3})$.Reaction with $HCl$ follows Markovnikov's rule to give $B$,which is $2-chlorobutane$ $(

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butan$-2-$ol

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(B) The hydrocarbon $A$ is $but-2-ene$ $(CH_{3}-CH=CH-CH_{3})$.Reaction with $HCl$ follows Markovnikov's rule to give $B$,which is $2-chlorobutane$ $(CH_{3}-CHCl-CH_{2}-CH_{3})$.Reaction of $B$ with $NH_{3}$ gives $C$,which is $butan-2-amine$ $(CH_{3}-CH(NH_{2})-CH_{2}-CH_{3})$.Reaction of $C$ with $NaNO_{2}/HCl$ forms a diazonium salt,which upon hydrolysis with water yields $D$,which is $butan-2-ol$ $(CH_{3}-CH(OH)-CH_{2}-CH_{3})$.$Butan-2-ol$ contains a chiral carbon atom and is therefore optically active.

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