Question

Identify the A and B in the following reaction:
$C{H_3} - CH = C{H_2}\xrightarrow{{HBr}}A\xrightarrow{{KOH{\text{ }}acl}}B$

Answer 1

Hint: Markovnikov rule states that in addition of protic acid $\left( {HX} \right)$ to an unsymmetrical alkene, acidic hydrogen gets attached to that carbon atom of double bond which has more hydrogen atoms and halide group gets attached to that carbon atom of double bond which is attached with more alkene groups.

Complete step by step answer:
Reaction given in this question is $C{H_3} - CH = C{H_2}\xrightarrow{{HBr}}A\xrightarrow{{KOH{\text{ }}acl}}B$ and we have to identify A and B in this reaction.
In the first part of the reaction, that is, $C{H_3} - CH = C{H_2}\xrightarrow{{HBr}}A$ anti markovnikov rule will be followed. Markonikov rule states that in addition of protic acid $\left( {HX} \right)$ to an unsymmetrical alkene, acidic hydrogen gets attached to that carbon atom of double bond which has more hydrogen atoms and halide group $\left( X \right)$ gets attached to that carbon atom of double bond which is attached with more alkene groups. In this reactant, the carbon atom of $C{H_2}$ has more hydrogen atoms as compared to $CH$. So, according to anti markovnikov rule hydrogen atom will be attached to $C{H_2}$ group and bromine atom will be attached to $CH$ group. So, the resulting compound that is A will be:

The chemical formula of this compound is ${C_3}{H_7}Br$.
So the first step of reaction is:
$C{H_3} - CH = C{H_2}\xrightarrow{{HBr}}C{H_3} - CHBr - C{H_3}$
Second step of reaction is:
$C{H_3} - CHBr - C{H_3}\xrightarrow{{KOH{\text{ }}acl}}B$
In this reaction halogen of the reactant will be replaced with $OH$ group of $KOH$ and $KBr$ will be released. So, B is ${C_3}{H_7}OH$ and the complete reaction is:
$C{H_3} - CH = C{H_2}\xrightarrow{{HBr}}C{H_3} - CHBr - C{H_3}\xrightarrow{{KOH{\text{ }}acl}}C{H_3} - CHOH - C{H_3}$
Hence A and B are ${C_3}{H_7}Br$ and ${C_3}{H_7}OH$ respectively.


Note:
Anti-markovnikov rule is opposite to the markovnikov rule. According to Anti-markovnikov rule hydrogen atom of $HX$ gets attached to that carbon atom of double bonded carbon which has less number of hydrogen atoms and halogen gets attached to that carbon atom of double bonded carbon atom which has more number of hydrogen atoms.