Q10 · CSIR NET Chemistry, June 2018
Paper: CSIR NET June 2018 · Subject: Organic Chemistry · Chapter: Organic Spectroscopy · Topic: NMR 1H · Marks: 2 · Difficulty: Hard
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 
$\rightarrow$ 
$\rightarrow$ | 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
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).
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