Answers

2016-03-15T08:55:51+05:30
In case of compound A , as it is reactive to alcoholic Cl , the Cl group must not be present at the terminal.

As there are only three carbons, the compound will be 2-chloropropane(H_3C-CHCl-CH_3).

In case of compound  B, on seeing the formula we can say it is an alkene.
NOTE :- C_nH_{2n} = C_3H_{2*3}

Therefore the compound B will be Propene (H_2C=CH-CH_3)

According to Markonikov Rule compound C will be 2-bromopropane (H_3C-CHBr-CH_3)

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2016-03-15T09:01:27+05:30
A is isopropyl chloride CH3CHCLCH3 it can also be npropyl alcohol CH3CH2CH2CL
then it reacts with alc. KOH and eliminates HCL to form propene CH3CH=CH2 THIS IS compound B then B reacts with HBr to give isopropyl bromide that is CH3CHBrCH3
HOPE THIS MIGHT BE UNDERSTOOD
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