Question

Anti $ - $Markovnikov addition of $HBr$ is not observed in:
(A) Propene
(B) But $ - 2 - $ene
(C) But $ - 1 - $ene
(D) Pent $ - 2 - $ene

Answer 1

Hint:
In an addition reaction of an electrophile $HX$(hydrogen halide) to an alkene or alkyne, when the hydrogen atom of $HX$gets bonded to the carbon atom of alkene that had the less number of hydrogen atoms in starting is called anti$ - $ Markovnikov addition. Both the Markovnikov and the anti$ - $ Markovnikov addition occur on asymmetrical components only.

Complete step by step answer:
(A) The structure of propene can be shown as:
${H_2}C = CH - C{H_3}$
In the above structure of propene with chemical formula ${C_3}{H_6}$, the anti$ - $ Markovnikov addition can be observed as follows:
${H_2}C = CH - C{H_3} + HX \to C{H_2}X - C{H_2} - C{H_3}$
As propene is an asymmetrical compound, the anti$ - $ Markovnikov addition has been done.
(B) But $ - 2 - $ene can be shown as
${H_3}C - CH = CH - C{H_3}$
Since, this compound is symmetrical across the double bond, as shown in the above structure. Thus, anti$ - $ Markovnikov addition is not possible.
(C) The structure of But $ - 1 - $ene is shown below:
${H_2}C = CH - C{H_2} - C{H_3}$
The anti$ - $ Markovnikov addition with hydrogen halide on this compound can be written as:
${H_2}C = CH - C{H_2} - C{H_3} + HX \to C{H_2}X - C{H_2} - C{H_2} - C{H_3}$
Due to the asymmetric structure, the hydrogen atom of hydrogen halide is attached to the carbon which had lesser hydrogen atoms in starting.
(D) Pent $ - 2 - $ene structure is: ${H_3}C - CH = CH - C{H_2} - C{H_3}$
This compound is also not symmetric across the double bond, so its anti$ - $ Markovnikov addition can be shown as:
$C{H_2} - CH = CH - C{H_2} - C{H_3} + HX \to C{H_3} - C{H_2} - CHX - C{H_2} - C{H_3}$

Note: Hydrogen halide reacts with alkene to form markoff products. Whereas, $HBr$ will give anti$ - $ Markovnikov’s product only in presence of any peroxide.