Question

In which of the following, Anti Markovnikov addition of HBr is not observed?
(A) Propene
(B) But-1-ene
(C) But-2-ene
(D) Pent-2-ene

Answer 1

Markovnikov’s rule predicts the major product formed as a result of additional reaction in organic compounds. It states that on reaction of unsaturated hydrocarbons with alkyl halide, the halogen goes to the carbon with lesser number of hydrogens. Anti-Markovnikov’s rule is the opposite of Markovnikov’s rule.

Complete step by step answer:
Let us start this question by understanding the concept of Anti Markovnikov’s addition.
Anti-Markovnikov’s rule is also known as the “peroxide effect’, since this type of addition takes place by a free radical mechanism.
Some points need to be kept in mind to check if peroxide effect takes place –
It happens only in case of HBr.
It needs the presence of peroxide.
Addition of halogen halide across an unsymmetrical alkene.
As a result of this addition, Bromide ion will go to the less substituted carbon of the double bond and proton will go to the more substituted carbon.
Let us now look at the given options and check it for Anti Markovnikov’s rule –
Propene
$C{{H}_{3}}CH=C{{H}_{2}}+HBr\xrightarrow{peroxide}C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Br$
But-1-ene
$C{{H}_{3}}C{{H}_{2}}CH=C{{H}_{2}}+HBr\xrightarrow{peroxide}C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}Br$
But-2-ene
$C{{H}_{3}}CH=C{{H}_{2}}C{{H}_{2}}+HBr\xrightarrow{peroxide}C{{H}_{3}}CH(Br)C{{H}_{2}}C{{H}_{2}}Br$
Since it is a symmetrical alkene, it gives only one product, therefore it does not follow Anti Markovnikov’s rule.
Pent-2-ene
$C{{H}_{3}}C{{H}_{2}}CH=CHC{{H}_{2}}+HBr\xrightarrow{peroxide}C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}BrC{{H}_{3}}$

Therefore, the answer is – option (c) – But-2-ene does not show Anti Markovnikov addition of HBr.

Additional Information: In the absence of a peroxide, hydrogen bromide adds to alkene via an electrophilic addition mechanism and gives a product according to Markovnikov’s rule.

Note: Free radical mechanism takes place in three steps –
(i) Initiation –
In this, the chain is initiated by free radicals produced by breaking an oxygen-oxygen bond in peroxide. These free radicals extract a hydrogen atom from a hydrogen halide molecule to produce free radicals.

& R-O-O-R\xrightarrow{hv}2R-{{O}^{\bullet }} \\\ & R-{{O}^{\bullet }}+H-Br\xrightarrow{fission}R-OH+B{{r}^{\bullet }} \\\ \end{aligned}$$ (ii) Propagation – Free radical attaches to the alkene using one of the electrons in a pi bond. This creates a new radical with the single electron on the other carbon atom. $$C{{H}_{3}}-CH=C{{H}_{2}}+B{{r}^{\bullet }}\to C{{H}_{3}}-{{C}^{\bullet }}H-C{{H}_{2}}-Br$$ (iii) Termination – Eventually two free radicals combine with each other and produce a molecule. The process terminates here, when all free radicals combine. $$C{{H}_{3}}-{{C}^{\bullet }}H-C{{H}_{2}}Br+{{H}^{\bullet }}\to C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}Br$$