The major product in the given reaction is

$\underset{\left(2\right) \mathrm{HBr}}{\overset{\left(1\right) {\mathrm{H}}^{+}, \mathrm{heat}}{\to }}$

Let's analyze the reaction stepwise:

Step 1: The compound is a tertiary alcohol with a double bond in the structure.

• First reagent: H+,heatH+,heat
  • This causes acid-catalyzed dehydration (loss of water), forming the most stable alkene product via the E1 mechanism.

Step 2: The product alkene is then treated with HBr.

• This step is a Markovnikov addition of HBr to the alkene, where Br adds to the more substituted carbon.

Major Product:

1. Dehydration (E1 elimination):
   • The alcohol group leaves, generating a double bond such that the most substituted (and thus most stable) alkene is formed.
   • Here, the major alkene will be at the more substituted position (Zaitsev's rule).
2. Addition of HBr:
   • HBr adds to the double bond produced above.
   • Br attaches to the more substituted carbon (Markovnikov addition) due to carbocation stability.

Thus, the major product is a bromoalkane where Br is attached to the carbon that was tertiary in the double bond, resulting from dehydration followed by Markovnikov addition.

Summary:

• Alcohol undergoes acid-catalyzed dehydration to yield the most substituted alkene.
• HBr then adds to the alkene, with bromine attaching to the more substituted carbon.

Final structure:
The major product is 2-bromo-2,4,4-trimethylpentane (if the double bond forms between C-2 and C-3, Br ends up at the tertiary carbon).