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. $H_A$
1. -0.3
Q. $H_B$
2. 5.1
R. $H_1$ and $H_7$
3. 6.4
S. $H_{2-6}$
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
In the homotropylium ion the ring current shields the endo (inside) methylene proton to δ −0.3 and leaves the exo proton at 5.1; H1/H7 appear at 6.4 and the aromatic H2–H6 at 8.5. With HA exo and HB endo: P-2, Q-1, R-3, S-4 (a).