Q.23 Compounds P and Q undergo E2 elimination with reaction rate constants of k1 and
k2, respectively, as shown below. Which is/are the CORRECT option(s)?
(A) k1 > k2
(B) k2 > k1
(C) Most stable conformer of P gives the product
(D) Most stable conformer of Q gives the product
The correct options are A and D.
For this E2 elimination on substituted cyclohexanes, k1>k2 and the most stable conformer of Q (not P) is the one that gives the product.
Introduction
E2 elimination in cyclohexane systems strongly depends on chair conformation and the anti‑periplanar requirement between the β‑hydrogen and the leaving group.
In this CSIR‑NET style problem, compounds P and Q undergo E2 elimination with rate constants k1 and k2, and careful conformational analysis is needed to compare rates and judge which conformer actually reacts.
Concept recap: E2 in cyclohexanes
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In cyclohexane, E2 requires the β‑H and leaving group (here Cl) to be trans‑diaxial, i.e., both axial and anti‑periplanar.
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The rate of E2 is governed by how populated the chair conformer is in which this trans‑diaxial arrangement exists; the more stable that conformer, the faster the elimination.
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Large groups (like tert‑butyl or isopropyl in your question) strongly prefer the equatorial orientation and destabilize conformers in which they are axial.
In the given problem, the methyl/tert‑butyl–like bulky groups fix the ring preference and decide which conformer is “most stable” for each substrate.
Option (A): k1>k2 – Correct
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For compound P, to achieve E2, the leaving group (Cl) must adopt an axial position while at least one β‑hydrogen on the adjacent carbon is also axial, giving the required trans‑diaxial geometry.
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In P, the arrangement that places Cl axial can coexist with the bulky substituent in an equatorial position, so the reactive trans‑diaxial conformer is relatively stable and significantly populated, leading to a higher rate k1.
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For compound Q, the stereochemistry forces a situation where, to make Cl axial and obtain the needed β‑axial H, the bulky group must become axial, creating strong 1,3‑diaxial repulsions and a much less stable reactive conformer, so the rate k2 is smaller.
Therefore k1>k2, so option (A) is true and option (B) is false.
Option (B): k2>k1 – Incorrect
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If k2 were greater than k1, Q would have to possess a more stable trans‑diaxial (axial Cl/axial β‑H) conformer than P.
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Because Q puts the large substituent axial in its reactive conformer, this form is high in energy and poorly populated, so E2 is slower and k2 is smaller, not larger.
Hence option (B) does not match the conformational energy profile and is incorrect.
Option (C): “Most stable conformer of P gives the product” – Incorrect
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The most stable chair conformer of P is the one where the bulky group is equatorial and the chlorine is also equatorial, minimizing steric strain.
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However, E2 cannot occur from this conformer because an equatorial Cl cannot be anti‑periplanar to any β‑hydrogen; there is no trans‑diaxial H–Cl pair.
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For P to react, it must first undergo a ring flip to a less stable conformer in which Cl is axial and a β‑H is axial trans to it; this higher‑energy conformer, not the most stable one, is the reactive form that gives the product.
Thus, option (C) is false because the product arises from a less stable, not the most stable, conformer of P.
Option (D): “Most stable conformer of Q gives the product” – Correct
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In Q, due to its stereochemistry, the conformation that keeps the bulky group equatorial while making Cl axial (and aligning it trans‑diaxial to a β‑H) happens to also be the most stable accessible conformer.
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This means that Q does not need an energetically costly ring flip to reach the reactive arrangement; the populated lowest‑energy conformer already contains the axial Cl/axial β‑H pair, so this conformer both dominates the equilibrium and directly undergoes E2 to give the alkene product.
Therefore option (D) is true, explaining why Q reacts only from its most stable conformer, even though its overall rate k2 is still lower than k1 because that conformer is less favorable than P’s reactive conformer.
Summary of correctness
| Option | Statement | Correct? | Reason (conformational/E2 logic) |
|---|---|---|---|
| A | k1>k2 | Yes | P’s reactive conformer with axial Cl is more stable and more populated. |
| B | k2>k1 | No | Q’s reactive conformer is higher in energy, so E2 is slower. |
| C | Most stable conformer of P gives product | No | Most stable P has equatorial Cl, which cannot undergo E2; ring flip needed. |
| D | Most stable conformer of Q gives product | Yes | In Q, the lowest‑energy conformer already has axial Cl and β‑axial H. |
Thus, the correct options for this E2 elimination in substituted cyclohexanes are A and D.
