Question

The predominant product formed, when 3-methyl-2-pentene reacts with HOCl, is
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: In this question it is important to keep in mind that when alkenes react with hypohalous acids, the result is 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:
We know that hypohalous acids are other halogen containing chemicals, which add to the pi bonds. These reagents are not symmetrical therefore 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 half of the reagent goes to the carbon in the pi bond with 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 combines quickly with the nucleophile to produce the addition product. Because oxygen is more electronegative than halogens, hypochlorous and hypobromous acids operate as electrophiles rather than proton donors. The hydroxide ion, on the other hand, behaves as a nucleophile and attacks the carbocation.

The result of the reaction between 3-methyl-2-pentene and hypochlorous acid is 2-chloro-3-methyl pentanol-3. The hydroxide and chloro group are added across the double bond of two carbons as a result of the reaction according to Markovnikov's addition.
The correct answer is C.

Note: 2-Bromopropane synthesis using Markovnikov addition:
- The addition of HBr to propene (an unsymmetrical alkene) follows the Markovnikov rule, according to which the carbon atom with the fewest hydrogen atoms receives the negative portion of the addition.
- Carbocation is formed as an intermediate in the reaction, which develops through an ionic mechanism. Secondary carbocation is more stable than primary carbocation.
- Compared to primary carbocations, secondary carbocations are more stable. As a result, the secondary carbocation is attacked by bromine in the following step, resulting in the main product, 2-bromopropane.