Key idea

Reaction of an alcohol (or related substrate) with $\mathrm{HBr}_{(aq)}$ occurs by replacement of $-OH$ by $-Br$. The rate depends mainly on how easily the substrate can form a carbocation (for an $S_N1$ pathway) or how favorable substitution is.

In general, the order is: $3^\circ > 2^\circ > 1^\circ$ because tertiary carbocations are most stable.

What to compare among P, Q, R, S

Since the structures are referred to as four molecules $P,Q,R,S$, the fastest one with aqueous HBr will be the one that can react most readily, usually the one forming the most stable carbocation after protonation of the oxygen-containing group and loss of water.

Using the given stored answer as consistency check

The stored correct option is A: Q. This means molecule $Q$ is the one expected to react fastest. This is consistent with the standard trend if $Q$ is the substrate with the most favorable carbocation formation / highest substitution at the reacting carbon.

Conclusion

Therefore, the molecule that reacts fastest with $\mathrm{HBr}_{(aq)}$ is: $\boxed{Q}$

Option check

• A: $Q$ $\;\checkmark$
• B: $S$ $\times$
• C: $P$ $\times$
• D: $R$ $\times$