The correct match for the protons labeled in compound X in Column M with the corresponding chemical shifts ($\delta$, ppm) in Column N is
Column M Column N P. $H_a$ I. 7.00 (ddd, $J = 8.4, 7.3, 1.4$ Hz, 1H) Q. $H_b$ II. 7.17 (dd, $J = 8.4, 1.4$ Hz, 1H) R. $H_c$ III. 7.59 (ddd, $J = 8.4, 7.3, 1.4$ Hz, 1H) S. $H_d$ IV. 8.12 (dd, $J = 8.4, 1.4$ Hz, 1H)
To solve this problem, we need to assign the chemical shifts of the protons labeled in compound X to the correct values in Column N based on their positions in the aromatic ring. The compound X is a substituted benzene ring, so the chemical shifts will depend on the electronic effects of the substituents:
- Analyzing Substituent Effects:
- The -OH group is an electron-donating group, which tends to activate the ortho and para positions to electrophilic aromatic substitution, but it usually doesn't affect the chemical shifts heavily compared to electron-withdrawing groups.
- The -NO2 group is a strong electron-withdrawing group due to its resonance and inductive effects. This group will deshield the ortho and para protons, leading to downfield shifts (higher ppm values).
- Assigning Chemical Shifts:
- Proton Ha: This is ortho to the -OH group and meta to the -NO2 group. The presence of the -OH group should slightly deshield this proton. Thus, a moderate ppm value like 7.17 is likely.
- Proton Hb: This is para to the -NO2 group, resulting in a significant downfield shift. The value 7.59 is suitable here.
- Proton Hc: This proton is also ortho to -NO2, experiencing considerable deshielding by the electron-withdrawing group. It corresponds to a chemical shift of 7.00.
- Proton Hd: This proton is directly ortho to -NO2 and is highly deshielded, resulting in the highest chemical shift, 8.12.
Conclusion: By analyzing the substituents’ effects on chemical shifts, we find the best matches:
- P ($H_a$) → II (7.17 ppm)
- Q ($H_b$) → III (7.59 ppm)
- R ($H_c$) → I (7.00 ppm)
- S ($H_d$) → IV (8.12 ppm)
This matches with the option:
P → II; Q → III; R → I; S → IV
.
Important Questions from NMR Spectroscopy (1H and 13C)
In the $^1H$-NMR spectrum of the following molecule, the signal of proton $H_a$ appears as
The $^1H$ NMR spectrum of the given iridium complex at room temperature gave a single signal at 2.6 ppm, and its $^{31}P$ NMR spectrum gave a single signal at 23.0 ppm. When the spectra were recorded at lower temperatures, both these signals split into a complex pattern. The intra-molecular dynamic processes shown by this molecule are
Compound K displayed a strong band at $1680 \text{ cm}^{-1}$ in its IR spectrum. Its $^1H$-NMR spectral data are as follows: $\delta$ (ppm) 7.30 (d, J = 7.2 Hz, 2H), 6.8 (d, J = 7.2 Hz, 2H), 3.8 (septet, J = 7.0 Hz, 1H), 2.2 (s, 3H), 1.9 (d, J = 7.0 Hz, 6H). The correct structure of compound K is
$^1H$ NMR spectrum of a mixture containing $CH_3Br$ ($x$ mol) and $(CH_3)_3CBr$ ($y$ mol) shows two singlets at 2.7 ppm and 1.8 ppm, with the relative ratio of 3:1 (integration value), respectively. The value of $x/y$ is ____________
(rounded off to the nearest integer)Consider the following $^1H$-NMR ($400$ MHz, DMSO-$d_6$) data of a compound:
$\delta$ in ppm: $3.85$ (s, $6H$), $6.73$ (t, $J = 2.2$ Hz, $1H$), $7.1$ (d, $J = 2.2$ Hz, $2H$), and $13.05$ (brs, $1H$).
The compound is
