Identify product 'B' in following series of reactions
2-methylpropan-2-ol $\xrightarrow[\large{363\ \text{K}}]{\large{20\%\ \text{H}_2\text{SO}_4}}$ A $\xrightarrow[\large{\text{Conc. H}_2\text{SO}_4}]{\large{\text{H}_2\text{O}\ \ \Delta}}$ B

The starting material, 2-methylpropan-2-ol, is a tertiary alcohol. On heating with $20\%$ $H_2SO_4$ at $363\ \text{K}$, it undergoes acid-catalyzed dehydration to form an alkene, 2-methylpropene (product A).

$(CH_3)_3C-OH \xrightarrow{H^+, \Delta} (CH_3)_2C=CH_2 + H_2O$

Product A is 2-methylpropene.

When 2-methylpropene is treated with water in the presence of concentrated $H_2SO_4$ and heat, it undergoes acid-catalyzed hydration. The addition of water follows Markovnikov's rule, proceeding via the formation of a stable tertiary carbocation.

$(CH_3)_2C=CH_2 + H^+ \rightarrow (CH_3)_3C^+$

$(CH_3)_3C^+ + H_2O \rightarrow (CH_3)_3C-OH + H^+$

The final product B is 2-methylpropan-2-ol, which is commonly known as tert-butyl alcohol.

Answer: tert-Butylalcohol