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.
When H-Br(aq) reacts with organic molecules, the rate of reaction typically depends on the stability of the carbocation that forms during the reaction. In general, the presence of electron-donating groups such as \(-{CH3}\) or \(-{OCH3}\) can stabilize the carbocation and speed up the reaction. The order of reactivity with H-Br is governed by the ability of the molecule to stabilize the resulting carbocation.
- Molecule P has two oxygen atoms attached to the ring. Oxygen is an electron-withdrawing group, which decreases the stability of the carbocation and slows the reaction.
- Molecule Q has an oxygen atom and a methyl group attached to the ring. The methyl group is an electron-donating group, which will stabilize the carbocation, making the reaction faster.
- Molecule R has a single oxygen atom and a methyl group attached to the ring, similar to molecule Q. However, the oxygen atom in R is more electron-withdrawing, so the reaction will be slightly slower than Q.
- Molecule S only has a methyl group attached to the ring. Since there is no oxygen or other electron-withdrawing group, the carbocation is less stabilized, resulting in a slower reaction.
Thus, molecule Q will react with H-Br at the fastest rate because the methyl group donates electrons, stabilizing the carbocation.
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