1. Identify the alkene

The given alkene is $CH_3CH=CHCH(CH_3)_2$

Let us number the double-bond carbons: $CH_3-CH=CH-CH(CH_3)_2$

So the double bond is between C-2 and C-3.

2. Addition of HBr in absence of peroxide

Since no peroxide is mentioned, addition of HBr proceeds by the ionic (Markovnikov) mechanism.

The first step is protonation of the double bond to form the more stable carbocation.

There are two possibilities:

Case 1: $H^+$ adds to C-2

Then carbocation forms at C-3: $CH_3-CH_2-CH^+-CH(CH_3)_2$ This is a secondary carbocation.

Case 2: $H^+$ adds to C-3

Then carbocation forms at C-2: $CH_3-CH^+-CH_2-CH(CH_3)_2$ This is also initially secondary carbocation.

So at first sight, both seem comparable.

3. Check for rearrangement

Consider the carbocation formed in Case 1: $CH_3-CH_2-CH^+-CH(CH_3)_2$

The adjacent carbon is: $CH(CH_3)_2$ which is attached to two methyl groups and one hydrogen. A 1,2-hydride shift from this carbon to the carbocation center can occur.

After hydride shift: $CH_3-CH_2-CH_2-C^+(CH_3)_2$

Now the carbocation becomes tertiary, which is much more stable.

Then $Br^-$ attacks this tertiary carbocation to give: $CH_3CH_2CH_2C(Br)(CH_3)_2$

This corresponds to Option D.

4. Why other options are minor or not major

• Option B: $CH_3CH(Br)CH_2CH(CH_3)_2$ comes from direct addition via the less favorable pathway without rearrangement.
• Option C: $CH_3CH_2CH(Br)CH(CH_3)_2$ comes from unrearranged attack on the secondary carbocation of Case 1.
• Option A: does not match the skeleton correctly for normal addition to this alkene.

Because carbocation rearrangement leads to a much more stable tertiary carbocation, the major product is D.

5. Final answer

The major product is: $CH_3CH_2CH_2C(Br)(CH_3)_2$

So the correct option is D.

6. Comparison with stored answer

Stored correct answer: D

My derived answer: D

They agree.