Organic Chemistry Questions and Answers – Nuclear Magnetic Resonance – 2

This set of Organic Chemistry Mcqs focuses on “Nuclear Magnetic Resonance – 2”.

1. In NMR spectroscopy, what is the product of Nuclear ‘g’ factor (g_N), the nuclear magneton and the magnetic field strength (B₀)?
a) energy of transition from alpha to beta state
b) chemical shift
c) spin-spin coupling constant
d) magnetogyric ratio
View Answer

Answer: a
Explanation: In NMR spectroscopy. the product of Nuclear ‘g’ factor (g_N), the nuclear magneton (β_N) and the magnetic field strength (B₀) gives the energy of transition from alpha to beta state. As shown below:

2. An organic compound having the molecular formulae C₁₀H₁₄ exhibited two singlets in the ¹H NMR spectrum and three signals in the ¹³C NMR. What is the compound?
a)
b)
c)
d)
View Answer

Answer: a
Explanation: In compound a;

Here only 2 types of Hydrogen are present (two Ha & twelve Hb). So, 2 singlets only. There is no need to check ¹³C NMR. Still, if you check three types carbon are present (two C₁, four C₂ and four C₂ carbon). Hence, there peak for ¹³C NMR.

3. The ¹H NMR spectrum of a dilute solution of a mixture of acetone and dichloromethane in CDCl₃ exhibits two singlets of 1:1 intensity. What will be the molar ratio of acetone to dichloromethane in the solution?
a) 3:1
b) 1:3
c) 1:1
d) 1:2
View Answer

Answer: b
Explanation: If intensity is 1:1, the ratio must be 1:3

4. What will be the strength of coupling between geminal protons in the following molecules?

a) Decrease as the size of ring increase
b) Increase as the size of ring increase
c) Remains same
d) No relation between the size of the ring & coupling
View Answer

Answer: b
Explanation: Geminal proton ²J coupling i.e. coupling of H & H on same carbon.

as ring size increases, bond angle outside decreases i.e. α1 > α2 > α3 > α4 (according % s character). If B.A. decreases, 2J coupling increases.

5. What is the value of gyramagnetic ratio of proton?
a) 41.10 radian/Tesla
b) 42.57 MHz/Tesla
c) 26.75 radian/Tesla
d) 41.10 MHz/Tesla
View Answer

Answer: b
Explanation:

Where, I: nucleus spin=1/2
μ_n: nuclear magneton=5.05×10^-27
μ: magnetic moment by protons=2.79
h: Plank constant= 6.63x 10^-34joule-seconds.

6. What are the number of signals in ¹H NMR in the given molecules?

a) 3, 4, 4, 3 respectively
b) 2, 6, 4, 2 respectively
c) 2, 4, 6, 2 respectively
d) 2, 4, 2, 6 respectively
View Answer

Answer: b
Explanation: Different types of proton give the different type of signals, so the different hydrogen will give different numbers of signals.
As we can see first compound has two type of hydrogen Ha and Hb. Compound second has six type of hydrogen Ha, Hb, Hc, Hd, He and H. Compound third has four type of hydrogen Ha, Hb, Hc, Hd and l forth compound has two type of hydrogen Ha and Hb.

7. How many peaks are expected in low-resolution NMR spectrum of vinyl chloride and ethyl cyclopropane?
a) 3,5
b) 5,3
c) 6,3
d) 3,6
View Answer

Answer: a
Explanation: Vinyl chloride compound has three types of proton as shown below and ethyl cyclopropane has five types of protons so five peaks will be observed.

8. What will be the NMR frequency in MHz of bare ¹H in a magnetic field of intensity 1.4092 tesla (given g_N = 5.585 and μ_N = 5.05 x 10^-27 JT^-1)?
a) 60 MHz
b) 120 MHz
c) 100 MHz
d) 15 MHz
View Answer

Answer: a
Explanation: By using the below formula we can calculate NMR frequency, where h is Plank’s constant.

9. At room temperature, what is the number of singlet resonances observed in the ¹H NMR spectrum of Me₃CC(O)NMe₂ (N, N-Dimethylpivalamide)?
a) 3
b) 4
c) 5
d) 2
View Answer

Answer: a
Explanation: As shown below; three signals are observed. 3Ha will experience shielding due to O^–, so They are different from 3Hb.

10. An organic compound (MF; C₈H₁₀O) exhibited the following ¹H NMR special data: 62.5 (3H, s), 3.8 (314, s), 6.8 (2H, d, J 8 Hz), 7.2 (2H, d, J 8 Hz) ppm. What will be the compound among the choices?
a) 4-ethylphenol
b) 2-ethylphenol
c) 4-methylanisole
d) 4-methylbenzyl alcohol
View Answer

Answer: c
Explanation: C₈H₁₀O

3Ha –> 3.8, singlet (deshielded because of –I of oxygen)
3Hb –> 2.5, singlet (No such –I)
2Hc –> 7.2, doublet (deshielded because of anisotropy effect of benzene as well as because of –I of oxygen)
2Hd –> 6.8, doublet (deshielded because of anisotropy of benzene).

Sanfoundry Global Education & Learning Series – Organic Chemistry.

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