Question

Anti-Markovnikov addition of one mole $HBr$ is not observed in:
A.Ethyne
B.Butene
C.Propene
D.Propyne

Answer 1

Hint: Firstly we should know the definition of Markovnikov addition and anti-Markovnikov addition, their difference.
This reaction is addition of alkyl halide to unsaturated hydrocarbons.
The anti-Markovnikov addition of one mole $HBr$ is only observed in unsymmetrical alkenes, which have different numbers of substituents around the double bond.

Complete step by step answer:
The anti-Markovnikov addition of $HBr$ is also called the peroxide effect or Kharasch effect.
Markovnikov addition states that the negative part of addendum is added to the carbon atom of double bond which has a lesser number of hydrogen atoms.
Anti-Markovnikov addition of $HBr$ states that $B{r^ - }$ will be added to double bond that carbon which has a higher number of hydrogen atoms.
But the product of Markovnikov and Anti-Markovnikov will be the same in case of symmetrical unsaturated compounds. Thus let us draw the structures of all the above options given and the one which is symmetrical will not show Anti-Markovnikov addition.
(A)Ethyne
$CH \equiv CH$
This is a symmetrical unsaturated hydrocarbon (Alkyne).
(B)Butene
$C{H_3} - C{H_2} - CH = C{H_2}$
This is unsymmetrical unsaturated hydrocarbon (alkene).
(C)Propene
$C{H_3} - CH = C{H_2}$
This is unsymmetrical unsaturated hydrocarbon (alkene).
(D)Propyne
$CH3 - C \equiv CH$
This is unsymmetrical unsaturated hydrocarbon (alkyne).
After observing all 4 structures, we know that symmetrical unsaturated hydrocarbons will produce the same products for Markovnikov and Anti-Markovnikov.
Thus, the correct option is (A) Ethyne.

Additional information: Anti-Markovnikov addition is possible only in $HBr$ and in presence of peroxide. The mechanism by which this addition takes place is a free radical mechanism. The reaction takes place in 3 steps in presence of sunlight.
Step-1: Initiation
\[R - O - O - R\xrightarrow{{h\nu }}2R - {O^ \bullet }\]
In presence of sunlight, oxygen-oxygen bond breaks, and 2 free radicals are formed. These free radicals will extract Hydrogen from Hydrogen halide.
\[R - {O^ \bullet } + H - Br \to R - O - H + B{r^ \bullet }\]
Step-2: Propagation
Now Bromine free radicals can get attached to any one carbon of double bonded carbon. So we can see 2 cases.
Case 1: $R - CH = C{H_2} + B{r^ \bullet } \to R - C{H^ \bullet } - C{H_2} - Br$
Here Bromine free radical is attached to carbon from double bond, which has more hydrogen, and free radical is formed on carbon in between.
Case 2: $R - CH = C{H_2} + B{r^ \bullet } \to R - CHBr - C{H_2}^ \bullet $
Here Bromine free radical is attached to carbon from double bond, which has less hydrogen, and free radical is formed on carbon at terminal position.
Thus, from the above cases, we can see that a more stable free radical is formed in the first case. So now this will be formed and finally next step will take place:
Step 3: Termination
$R - C{H^ \bullet } - C{H_2} - Br + {H^ \bullet } \to R - C{H_2} - C{H_2} - Br$
Now hydrogen free radical will come and get attached to the intermediate carbon free radical, thus completing the reaction.

Note:
The peroxide used mostly will be Benzoyl peroxide in this reaction.
Anti-Markovnikov addition is possible only in $HBr$ and in the presence of peroxide. No other hydrogen halide will show Anti-Markovnikov addition. The mechanism by which this addition takes place is free radical mechanism, and free radicals are formed by sunlight, thus sunlight is also used in reaction.