Question

If HCl is added over $ \left [ CH_2=C{(CH_3)}_2r \right ]$ then what is formed
A. $CH_2Cl=C{(CH_3)}_2$
B. $CH_3-C{(CH_3)}_2Cl$
C. $CH_2=C(CH_3)(CH_2)Cl$
D. None of these

Answer 1

Hint: Addition of alkenes follows markovnikov’s rule according to markovnikov’s rule the attacking nucleophile gets attached to the carbon that is containing less number of hydrogen. The reaction will proceed with the mechanism of the classical carbocation.

Complete Step by Step Answer:
As we know during the addition reaction the formation of carbocation takes place as the carbocation is formed the positive charge move to the carbon at which the most stable carbocation will be formed or we can say that the positive charge will stay on the carbon that will have more no of substituent attached that will have +I effect. So, the stability of primary, secondary, and tertiary carbocation will be-
${CH_3}^+

So here we can see that the most stable carbocation is tertiary carbocation because it has three methyl groups attached with it.

The following reaction will be:
$CH_2=C{(CH_3)}_2+H-Cl\longrightarrow CH_3-C{(CH_3)}_2Cl$
The mechanism of reaction will be:

So, we can see that the product formed from the given reactants is 1-chloro,1-methylpropane. This reaction follows markovnikov’s rule for addition in the alkenes.

In addition to the reaction of alkene the pi(π) bond between the two carbons of alkene breaks and they form two sigma bonds with two separate chemical species. The geometry of the carbon is changed because it changes its hybridization from sp² to sp³ so the shape changes from trigonal planar to tetrahedral.
Thus, Option (A) is correct

Note: Not all reactions follow markovnikov’s rule. In the second reaction 2- bromobutane contains a chiral carbon which means it will have stereoisomers. It will have two stereoisomers; they will be enantiomers that means mirror image of each other.