Question

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? Molecules:

\[ P: \text{Cyclic compound with two O groups attached to the ring.} \] \[ Q: \text{Cyclic compound with one O group and one CH\(_3\) group attached to the ring.} \] \[ R: \text{Cyclic compound with one O group attached to the ring and one CH\(_3\) group attached to the ring.} \] \[ S: \text{Cyclic compound with one CH\(_3\) group attached to the ring.} \]

• A. S
• B. Q
• C. R
• D. P

Hint:

The speed of reaction with H-Br is influenced by the electron-donating or electron-withdrawing groups attached to the molecule. Electron-donating groups stabilize the carbocation intermediate, leading to faster reactions.

Answer 1

The Correct Option is B

The question involves determining which cyclic compound reacts fastest with H-Br(aq). Let's evaluate each option: 

1. Molecule P:
   • Contains two oxygen atoms connected to a ring.
   • The two oxygen atoms can lead to resonance stabilization, but their presence reduces the electron density on the ring, making it less reactive toward nucleophilic attack by H-Br.
2. Molecule Q:
   • Contains one oxygen and one CH\(_3\) group connected to the ring.
   • Oxygen can increase electron density through resonance, enhancing reactivity, while the CH\(_3\) group provides inductive stabilization.
   • This combination makes the ring in molecule Q highly reactive to H-Br, facilitating faster reaction.
3. Molecule R:
   • Contains an oxygen and CH\(_3\) group, similar to Q, but arranged differently.
   • The specific position and resonance compared to Q may influence reactivity but is less favorable due to steric and electronic factors.
4. Molecule S:
   • Contains two CH\(_3\) groups attached to a benzene ring.
   • Lacks an oxygen atom, making it the least reactive to H-Br among the options.

Based on the evaluations, molecule Q will react with H-Br(aq) at the fastest rate due to the presence of an oxygen atom that can assist in reaction via electron donation, and a CH\(_3\) group that stabilizes the transition state via hyperconjugation.

Correct Answer: Q

Answer 2

Given: 

- The reaction follows an electrophilic addition mechanism. - During the rate-determining step, a carbocation intermediate is formed.

Step 1: Electrophilic Addition Mechanism

The addition of \( H^+ \) and \( Br^− \) to the alkene follows the electrophilic addition mechanism, where the alkene reacts with an electrophile (such as \( H^+ \)) to form a carbocation intermediate.

Step 2: Stability of Carbocation Intermediates

The stability of the carbocation intermediate plays a crucial role in determining the course of the reaction. Among the compounds P, Q, R, and S, compound \( Q \) forms the most stable carbocation intermediate since it is resonance-stabilized.

Conclusion:

Compound \( Q \) will form the most stable carbocation intermediate due to its resonance stabilization, making it the most stable product in this electrophilic addition reaction.