Q.CH3-CH(CH3)-CH(OH)-CH3 $\xrightarrow{H^+/heat}$ (A) major product $\xrightarrow{HBr}$ (B) major product
Identify A and B

Step 1. The starting alcohol, CH3-CH(CH3)-CH(OH)-CH3 (3-methylbutan-2-ol), is protonated by H+ and loses water (E1 dehydration) to give a SECONDARY carbocation at C2: CH3-CH(CH3)-CH+-CH3.

Step 2. This secondary carbocation sits directly next to a carbon (C3) that, if a hydride shifts across from it, would become a TERTIARY carbocation -- a 1,2-hydride shift from C3 to C2 does exactly this, converting the cation to the more stable tertiary carbocation CH3-CH2-C+(CH3)-CH3 (numbering the rearranged skeleton fresh, this is the 2-methylbutan-2-yl cation).

Step 3. Losing a proton (E1, Zaitsev's rule favouring the more substituted alkene) from this rearranged tertiary carbocation gives A = 2-methylbut-2-ene, CH3-CH=C(CH3)-CH3, the major dehydration product (rather than the 'naive', non-rearranged 3-methylbut-2-ene that a simple, no-rearrangement analysis might predict). …

Whenever a dehydration or addition intermediate carbocation sits adjacent to a carbon that could become MORE substituted by a hydride or alkyl shift, check for that rearrangement FIRST -- the major p …

• Skipping the hydride-shift check and predicting A as the non-rearranged elimination product (3-methylbut-2-ene) -- carbocation rearrangement to a more stable cation, when available, generally dominates the product distribution. …