Question

(CH3)2C=CHCH3+HBr in presence of peroxide Find the major and minor product [anti markovnikove rule]​

Answer 1

Answer:

compound having unstable carbocation will form the major product.

Explanation:

markonikov is when stable carbocation is formed

this means in anti markonikov unstable carbocation will form and Br will attach to unstable carbocation.

Answer 2

Answer:

The major and minor products are 2- Bromo-3-methyl butane and 2- Bromo-2-methyl butane respectively.

Explanation:

First, let's understand the anti-Markonikoff rule. This is stated as the carbon atom with the most carbon atoms will be the main product when the negative portion of the reagent is added to it. This happens when the peroxide is present as a reagent.

The reaction proceeds as follows,

$\mathrm{(CH_3)_2C=CHCH_3+HBr}\xrightarrow{\mathrm{Peroxide}} \mathrm{(CH_3)_2CH-CH(Br)CH_3+(CH_3)_2C(Br)-CH_2CH_3}$

Where,

(CH₃)₂C=CHCH₃⇒ 2-methyl but-2- ene.

HBr⇒ Hydrogen bromide

(CH₃)₂CH-CH(Br)CH₃⇒ 2- Bromo-3-methyl butane

(CH₃)₂CH(Br)-CH₂CH₃⇒2- Bromo-2-methyl butane

In 2- Bromo-3-methyl butane(Anti-markonikoff product)  secondary carbocation is formed while in 2- Bromo-2-methyl butane(Markonikoff product) tertiary carbocation is formed.

So,

The major product=2- Bromo-3-methyl butane

The minor product=2- Bromo-2-methyl butane

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