The correct match of protons in column A with the ${ }^{1} \mathrm{H}$ NMR chemical shifts in column B for the product of the following reaction is
Column A
Column B
P. $\mathrm{H}_{A}$
1. -0.3
Q. $\mathrm{H}_{B}$
2. 5.1
R. $\mathrm{H}_{1}$ and $\mathrm{H}_{7}$
3. 6.4
S. $\mathrm{H}_{2-6}$
4. 8.5
(a)P-2, Q-1, R-3, S-4
(b)P-1, Q-2, R-4, S-3
(c)P-4, Q-1, R-3, S-2
(d)P-2, Q-4, R-1, S-3
Answer
Answer: A ✓ checked by 4AB · confidence medium
Explanation
Homotropylium ion: the ring current shields the inner methylene proton to δ −0.3 and leaves the outer at 5.1; H1/H7 at 6.4 and H2–H6 at 8.5. With HA the exo and HB the endo proton: P-2, Q-1, R-3, S-4 (a).