Question

3-methyl-2-pentene on reaction with HOCl gives:
A) 3-chloro-3-methyl pentanol-2
B) 2,3-dichloro-3-methylpentane
C) 2-chloro-3-methyl pentanol-3
D) 2,3-dimethyl butanol-2

Answer 1

Hint:We must remember that alkenes in reaction with hypohalous acids gives the product as substituted alcohols. The reaction of 3-methyl-2-pentene with hypochlorous acid follows Markovnikov’s addition to form the product.

Complete step by step answer:
As we know that hypohalous acids are other halogen containing reagents, which add to the pi bonds. These reagents are not symmetrical so they are added to unsymmetrical pi bonds in two ways. These reactions are highly regioselective with one as the major product. The regioselectivity of these reactions are explained in the similar way to Markovnikov rule.
We must know that Markovnikov’s rule predicts the regioselectivity of electrophilic addition reactions of alkenes and alkynes. Addition of hydrogen bromide to an unsymmetrical alkene, the negative part of the reagent goes to the carbon in the pi bond having less hydrogen is called Markovnikov’s rule.
The electrophilic bonding to the pi bonds of an alkene should result in the generation of more highly substituted carbocation and the formed carbocation combine quickly with the nucleophile to produce the addition product. Hypochlorous and hypobromous acid are weak acids and they do not react as proton donors but oxygen is more electronegative than the halogens so the halogens acts as a electrophile while the nucleophile which attacks the carbocation is the hydroxide ion.
When 3-methyl-2-pentene is reacted with hypochlorous acid, the product formed is 2-chloro-3-methyl pentanol-3. The reaction obeys Markovnikov’s addition and the hydroxide and chloro group is added across the double bond of two carbons.
We can write the reaction of 3-methyl-2-pentene with HOCl as,

Therefore, the option (C) is correct.


Note:
- Now we can see an example for Markovnikov addition is the formation of 2-bromopropane
Formation of 2-Bromopropane:
- Addition to HBr to propene (unsymmetrical alkene) follows Markovnikov rule according to which, the negative part of the addition gets attached to that carbon atom, which possesses a lesser number of hydrogen atoms.
- The reaction proceeds via an ionic mechanism and forms carbocation as intermediate. Secondary carbocation is more stable than primary carbocation.
- The secondary carbocations are more stable than primary carbocation. Hence, in the next step, bromine attacks the secondary carbocation to form 2-bromopropane as the major product.