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.} \]
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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.
- 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.
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