Question

Compounds P and Q undergo E2 elimination with reaction rate constants of \(k_1\) and \(k_2\), respectively, as shown below. Which is/are the CORRECT option(s)? 

• A. \(k_1 \,>\, k_2\)
• B. \(k_2 \,>\, k_1\)
• C. Most stable conformer of P gives the product
• D. Most stable conformer of Q gives the product

Hint:

For E2 on cyclohexane systems, always check whether the leaving group can be {axial} in the most stable conformer.
Bulky substituents like tert-butyl lock the chair conformation and determine whether elimination will be fast or slow.

Answer 1

The Correct Option is A, C

Step 1: Stereoelectronic requirement of E2 in cyclohexane rings.
For E2 elimination in cyclohexanes, the leaving group (\ce{Cl}) and a \(\beta\)-hydrogen must adopt an anti-periplanar geometry.
In a chair conformation, this corresponds to a trans-diaxial arrangement.
Step 2: Conformational locking by tert-butyl group.
The bulky tert-butyl substituent always locks itself in the equatorial position.
This effectively fixes the chair conformation of the cyclohexane ring, preventing ring flipping.
Step 3: Conformation of P.
In compound P, the chlorine atom is axial when the tert-butyl group is equatorial.
Thus, there exists a suitable \(\beta\)-hydrogen that is anti-periplanar to the C–Cl bond.
\(⇒\) Elimination occurs readily in the most stable conformer of P, giving the product without requiring an unfavorable ring flip.
Step 4: Conformation of Q.
In compound Q, the chlorine atom is equatorial when tert-butyl is equatorial.
In this arrangement, there is no anti-periplanar \(\beta\)-H.
A ring flip would place chlorine axial, but this forces tert-butyl into the axial position, which is highly destabilized.
\(⇒\) Elimination is much slower in Q since the reactive conformer is not significantly populated.
Step 5: Comparing rates.
Since P eliminates from its stable conformer and Q requires a rare, high-energy conformer, we conclude:
\[ k_1 \,>\, k_2 \] Hence, (A) and (C) are correct.