A(predominantly ) is

The correct answer is Option (4) → 2-bromo-2-methylbutane, CH₃C(Br)(CH₃)CH₂CH₃

A secondary alcohol reacts with HBr by the S_N1 path: the –OH group is protonated, water leaves and a carbocation is formed, which is then attacked by \( \ce{Br^{-}} \). A carbocation rearranges to a more stable one whenever a hydrogen (or alkyl group) on the next carbon can shift with its electron pair (1,2-shift). Stability: 3° > 2° > 1°.

The alcohol is 3-methylbutan-2-ol, \( \ce{(CH3)2CH-CH(OH)-CH3} \).

Protonation and loss of water give a secondary carbocation:

    \( \ce{(CH3)2CH-CH(OH)-CH3 + H^+ -> (CH3)2CH-CH(OH2^+)-CH3 -> (CH3)2CH-CH^+-CH3 + H2O} \)

A 1,2-hydride shift from the neighbouring carbon turns it into a more stable tertiary carbocation:

    \( \ce{(CH3)2CH-CH^+-CH3 -> (CH3)2C^+-CH2-CH3} \)  (2° → 3°)

\( \ce{Br^{-}} \) attacks the tertiary carbocation:

    \( \ce{(CH3)2C^+-CH2-CH3 + Br^- -> (CH3)2C(Br)-CH2-CH3} \)

So A (predominantly) is 2-bromo-2-methylbutane.

Why the other options are incorrect

• Option (1) 2-bromo-3-methylbutane, CH₃CH(Br)CH(CH₃)CH₃ — Incorrect. This is the product without rearrangement (Br on the original –OH carbon); it is formed only in small amount, because the secondary carbocation rearranges first.

• Option (2) CH₃CH(CH₃)CH(Br)CH₃ — Incorrect. This is the same compound as Option (1), 2-bromo-3-methylbutane, drawn from the other end; it is the minor, unrearranged product.

• Option (3) 1-bromo-3-methylbutane, (CH₃)₂CHCH₂CH₂Br — Incorrect. Br on a terminal carbon would need a primary carbocation or the primary alcohol 3-methylbutan-1-ol; neither is involved here.

Note: It seems that there is a mistake in the original NTA paper, which prints the same compound, 2-bromo-3-methylbutane, in Options (1) and (2).