Single Choice

In the presence of peroxide, HCl and HI do not give anit-Markownikoff's addition to alkenes because:

Aone of the steps is endothermic in HCl and HI
Correct Answer
Bboth HCl and HI are strong acids
CHCl is oxidizing and the HI is reducing
Dall the steps are exothermic in HCl and HI

Solution

In the presence of organic peroxides, bromides will take part in radical based reactions. Fission of the peroxide O-O linkage will induce a Br radical which behaves differently to Bromide, by adding to the less substituted side of the alkene (anti - markovnikov). HI and HCl do not do this for energetic reasons.

The addition of Cl and I radicals to the alkene in an anti-markovnikov fashion is an endothermic reaction and therefore unfavourable. Without the presence of peroxide, all hydrogen halides will add according to the markovnikov rule.

Hence, option A is the correct answer.

SIMILAR QUESTIONS

Hydrocarbons

What is the major product expected from the following reaction? Where $$D$$ is an isotope of hydrogen.

Hydrocarbons

Which of the following will yield a mixture of 2-chlorobutene and 3-chlorobutene on treatment with $$\displaystyle HCl$$?

Hydrocarbons

In the reaction with $$HCl$$, an alkene reacts in accordance with the Markovnikov's rule to give a product 1-chloro-1-methylcyclohexane. The possible alkene is:

Hydrocarbons

In the above reaction, the major product is:

Hydrocarbons

$$CH_3-CH_2-CH = CH_2+HBr\xrightarrow[]{ROOR(peroxide)} \underset{Major}{(X)} + \underset{Minor}{(Y)}$$ X and Y respectively are:

Hydrocarbons

How many structures of $$X$$ is possible?

Hydrocarbons

Arrange the following hydrogen halides in order of their decreasing reactivity with propene.

Hydrocarbons

Provide the structures of compounds A, B, C, and D in the above questions.

Hydrocarbons

Provide the structures of compounds A, B, C, and D in the above questions.

Hydrocarbons

Write the structure of the major organic product in each of the following reactions: $$CH_3CH=C(CH_3)_2+HBr \rightarrow$$

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