Identify the product B in the following sequence of reactions?
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In acid-catalyzed hydration of alkenes, the product is always the more substituted alcohol (Markovnikov addition). The sequence via alkyl hydrogen sulfate is a two-step industrial method for alcohol synthesis.
Step 1: Understand the path.
We start with propan-1-ol. It is first heated with alumina, then treated with concentrated sulfuric acid, then with hot water. We must find the last product B.
Step 2: First reaction, make the alkene.
Hot alumina pulls out water from an alcohol. This is dehydration. Propan-1-ol loses water and becomes propene.
\[ CH_3CH_2CH_2OH \xrightarrow{Al_2O_3,\,623\,K} CH_3CH=CH_2 \]
So A is propene.
Step 3: Second reaction, add sulfuric acid.
Propene reacts with concentrated $H_2SO_4$. The H and the $HSO_4$ group add across the double bond.
Step 4: Apply Markovnikov's rule.
The H goes to the carbon that already has more hydrogens (the end carbon). The $OSO_3H$ group goes to the middle carbon. This gives isopropyl hydrogen sulphate, $(CH_3)_2CH\text{-}OSO_3H$.
Step 5: Third reaction, add hot water.
Hot water breaks the C-O-S bond. The $OSO_3H$ group is swapped for an $OH$ group at the same middle carbon. This gives propan-2-ol, $(CH_3)_2CHOH$.
Step 6: Choose the answer.
The final product B is propan-2-ol, which is option 1.
\[ \boxed{\text{Propan-2-ol}} \]
