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Question: Anti \( - \)Markovnikov addition of \(HBr\) is not observed in: (A) Propene (B) But \( - 2 - \)e...

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

Explanation

Solution

In an addition reaction of an electrophile HXHX(hydrogen halide) to an alkene or alkyne, when the hydrogen atom of HXHXgets 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:
H2C=CH−CH3{H_2}C = CH - C{H_3}
In the above structure of propene with chemical formula C3H6{C_3}{H_6}, the anti−- Markovnikov addition can be observed as follows:
H2C=CH−CH3+HX→CH2X−CH2−CH3{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−- 2 -ene can be shown as
H3C−CH=CH−CH3{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−- 1 -ene is shown below:
H2C=CH−CH2−CH3{H_2}C = CH - C{H_2} - C{H_3}
The anti−- Markovnikov addition with hydrogen halide on this compound can be written as:
H2C=CH−CH2−CH3+HX→CH2X−CH2−CH2−CH3{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−- 2 -ene structure is: H3C−CH=CH−CH2−CH3{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:
CH2−CH=CH−CH2−CH3+HX→CH3−CH2−CHX−CH2−CH3C{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, HBrHBr will give anti−- Markovnikov’s product only in presence of any peroxide.