Following are the four molecules "P", "Q", "R" and "S". Which one among the four molecules will react with

$$H-Br_{(aq)}$$ at the fastest rate?

The reaction of an alcohol with aqueous $$HBr$$ proceeds through three elementary steps:
  1. Protonation of the -OH group to give $$R-OH_2^{+}$$.
  2. Loss of water (the rate-determining step) to generate a carbocation $$R^{+}$$.
  3. Nucleophilic attack of $$Br^{-}$$ on the carbocation.

The overall rate therefore depends almost entirely on the ease with which the carbocation is produced in step 2. The more stable the carbocation that can form, the lower the activation energy and the faster the substitution.

Carbocation stability order:
  benzylic ≈ allylic > tertiary > secondary > primary > methyl

Now compare the four given molecules.

P is a primary alcohol. Loss of water would give a primary carbocation, which is least stable; therefore P reacts slowly.

Q is a tertiary benzylic alcohol (the -OH is attached to a carbon that is both tertiary and benzylic). On protonation, water leaves to give a benzylic-tertiary carbocation, the most stabilised among the four molecules because it is resonance-stabilised by the benzene ring and additionally stabilised by three alkyl groups through hyperconjugation. Hence Q reacts extremely fast with $$HBr_{(aq)}$$.

R is a secondary alcohol. Loss of water forms a secondary carbocation, appreciably less stable than the benzylic-tertiary carbocation from Q, so the rate is lower than that of Q.

S is an allylic primary alcohol that can form a resonance-stabilised allylic carbocation. Although allylic stabilisation helps, the carbocation is still less stabilised than the benzylic-tertiary carbocation from Q, so S is slower than Q.

Comparing all four, the fastest rate is shown by the molecule that generates the most stable carbocation, namely Q.

Therefore, the molecule that reacts with $$HBr_{(aq)}$$ at the fastest rate is
Option C which is: Q.