The major products 'A' and 'B', respectively, are:

Step 1: Identify the starting material and reagents

The reaction involves an alkene (2-methylprop-2-ene) treated with $\ce{H2SO4}$ under different temperature conditions.

Step 2: Low temperature reaction (gives A)

At low temperature, $\ce{H2SO4}$ adds to the alkene via Markovnikov addition: $\ce{(CH3)_2C=CH2 + H2SO4 -> (CH3)_3C-OSO3H}$

The product is an alkyl hydrogen sulfate (tert-butyl hydrogen sulfate) = Product A

Step 3: High temperature reaction (gives B)

At high temperature ($\Delta$), elimination (dehydration) of the sulfate ester occurs, regenerating an alkene — but now a more substituted (Saytzeff) alkene forms:

Or the reaction at high temp with $\ce{H2SO4}$ acts as a dehydrating agent to give the more stable, more substituted alkene = 2,3-dimethylbut-2-ene or similar tetrasubstituted alkene = Product B

Answer: Option (a)

Key Points to Remember:

• Low temp + $\ce{H2SO4}$: addition → alkyl hydrogen sulfate
• High temp + $\ce{H2SO4}$: elimination → alkene (more substituted, Saytzeff product)
• The reaction is reversible — used for purifying alkenes (dissolve in cold $\ce{H2SO4}$, then heat to regenerate)