Question

Compound, A $C{{H}_{3}}CH=C{{H}_{2}}+HBr$ gives:
A. $C{{H}_{2}}=CH-C{{H}_{2}}-Br$
B. $C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}-Br$
C. $C{{H}_{3}}-CH(Br)-C{{H}_{3}}$
D. $C{{H}_{3}}-C{{H}_{2}}-C{{H}_{3}}$

Answer 1

Hint: Alkene are those compounds which have double bonds present in them. Like in the given question double bond is present in compounds which are alkene in nature and known as propene and all the alkene groups ended with the suffix ene.

Complete answer:
Reaction given in the question is generally based on Markovnikov's rule which is also known by the name Markownikoff’s rule. This rule is used in organic chemistry to know the product formed during the addition reaction of alkenes.
This rule can be defined as when hydrogen halide or protic acid represented by $HX$ is added to an asymmetric alkene where the acid hydrogen gets attached to the carbon with more hydrogen substituents and halide group gets attached to that carbon which have great number of alkyl substituents. This reaction is generally a two-step reaction which can be explained as:
1. First step includes protonation or addition of acidic hydrogen ion: In this step $HX$ dissociates into ${{H}^{+}}$ and ${{X}^{-}}$ ions and ${{H}^{+}}$ gets attached to the carbon which has a great number of hydrogen substituents.
2. Second step includes addition of bromine ion: In this step halide ion attacks on a carbocation and forms a major product, in the given example the major product will be 2-bromo propene. Bromide is attached to the carbon which has greater number of alkyl substituents.
Hence reaction can be shown as:
$C{{H}_{3}}CH=C{{H}_{2}}+HBr\to C{{H}_{3}}-CH(Br)-C{{H}_{3}}$

Option C is the correct answer.

Note:
The main applications of markovnikov’s rule other than this reaction is that it is used in reaction of propene with $HI$ too. This rule was followed by symmetrical and unsymmetrical both types of alkenes and in this protonation step is the rate determining step.