Markovnikov addition of HBr with carbocation rearrangement to tertiary position

Steps:

• Protonate the terminal alkene (3-ethylbut-1-ene, CH2=CHCH(CH2CH3)CH3) at C1, forming a secondary carbocation at C2.
• Hydride shift from adjacent C3 (tertiary carbon) to C2 rearranges to a stable tertiary carbocation at C3.
• Bromide attacks the tertiary carbocation at C3, yielding CH3CH2CBr(CH3)CH2CH3.
• This C6 product is the major due to tertiary stability.

Why A is correct:

• Follows Zaitsev's rule via rearrangement to the more substituted (tertiary) bromide, named 3-bromo-3-methylpentane (equivalent to 3-bromo-3-hexane in option phrasing).

Why the others are wrong:

• B: Lacks bromine; not an addition product.
• C: Incorrect position for Br; would imply secondary without rearrangement.
• D: Wrong carbon numbering and substitution for the tertiary product.

Final answer: A