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Learn/ 12th Class Chemistry/ Unit 30 /Short Questions

Unit 30: Spectroscopy-2 NMR — Short Questions

12th Class Chemistry · Unit 30: Spectroscopy-2 NMR

Q2.a.State two reasons why TMS is a standard for ¹H NMR.

(1) All 12 protons in TMS, Si(CH₃)₄, are chemically equivalent, so it gives a single sharp reference peak (assigned δ = 0.0 ppm). (2) Because silicon has low electronegativity, the TMS peak appears at the extreme right (upfield) end of the spectrum and does not overlap with most organic proton signals; it is also chemically inert and volatile, making it easy to remove after use.

Q2.b.Why does deuterium (D) not appear in a ¹H NMR spectrum?

Deuterium requires a different frequency of radio waves (different energy) to flip its spin compared to ¹H. Hence it does not absorb the radiofrequency used for ¹H NMR and gives no peak — it is effectively 'invisible' in a ¹H NMR spectrum.

Q2.c.A compound (C₄H₁₀O) has a ¹H NMR peak area ratio of 1:3:6. How many protons does each signal represent?

C₄H₁₀O contains 10 hydrogen atoms in total. Since 1 : 3 : 6 adds up to 10, the ratio already represents the actual proton count: the three signals correspond to 1H, 3H and 6H respectively.

Q2.d.What is the purpose of the D₂O shake test in ¹H NMR?

The D₂O shake test is used to confirm whether a singlet peak in a ¹H NMR spectrum is due to an exchangeable O-H or N-H proton. After adding D₂O and re-running the spectrum, the O-H/N-H peak disappears (its proton is exchanged for deuterium), while all other peaks remain, allowing that peak to be identified.

Q2.e.List the key features used to interpret a ¹H NMR spectrum.

(1) Number of peaks — gives the number of different proton environments. (2) Position/chemical shift (δ) of each peak — identifies the type of environment (e.g. alkyl, vinylic, aromatic, O-H) using reference data. (3) Area under each peak — gives the relative number of protons in that environment. (4) Splitting pattern of each peak (n+1 rule) — gives the number of protons on adjacent carbon atoms.

Q2.f.What does the number of peaks in a ¹³C NMR spectrum represent?

The number of peaks in a ¹³C NMR spectrum equals the number of different (chemically non-equivalent) carbon environments present in the molecule.

Q3.a.A -CH₂- group is lying between a -CH₃ and a -CH< group. What is the splitting pattern for the -CH₂- signal? Show your calculation.

By the n+1 rule, n = number of protons on carbons adjacent to -CH₂- = 3 (from -CH₃) + 1 (from -CH<) = 4. Number of sub-peaks = n + 1 = 4 + 1 = 5. So the -CH₂- signal will split into a quintet (5 sub-peaks).

Q4.a.A chemist suspects a sample contains an O-H group. [A ¹H NMR spectrum of the sample is shown, δ (ppm) axis running from 5 to 0, with one cluster of split peaks near δ ≈ 3.5–4 ppm and another cluster of split peaks near δ ≈ 1–1.5 ppm.] Describe the "D₂O shake" procedure.

First, record the normal ¹H NMR spectrum of the sample, which shows all the proton signals including the suspected O-H peak. Then add a few drops of D₂O to the NMR sample tube, shake the solution well, and record the spectrum again. Finally, compare the two spectra.

Q4.b.What is the expected observation in the ¹H spectrum after the shake?

The peak due to the O-H (or N-H) proton disappears from the spectrum after the D₂O shake, while all the other proton signals remain unchanged, confirming the presence and identity of the O-H group.

Q4.c.Write the equilibrium equation for the exchange and explain why it causes this observation.

R-O-H + D₂O ⇌ R-O-D + HOD. Protons attached to highly electronegative atoms like O or N exchange readily; when D₂O is added, the O-H proton is replaced by deuterium. Since deuterium requires a different (non-¹H-NMR) energy to flip its spin, R-O-D gives no signal in the ¹H NMR spectrum, so the O-H peak disappears (becomes 'invisible').

Q5.a.Predict the number of ¹³C NMR peaks for the following two isomers, justifying each answer by identifying the sets of equivalent carbons: Pentane (CH₃CH₂CH₂CH₂CH₃). [A ¹H NMR spectrum, δ 8–0 ppm, is also shown alongside this question, with peaks labelled 3H, 2H and 3H, but it is not needed to answer this ¹³C NMR question.]
Q5.b.Predict the number of ¹³C NMR peaks for the following two isomers, justifying each answer by identifying the sets of equivalent carbons: 2,2-dimethylpropane (neopentane, C(CH₃)₄).

2,2-Dimethylpropane, C(CH₃)₄, has a central quaternary carbon bonded to four identical methyl groups. All four -CH₃ carbons are chemically equivalent (one environment), and the central carbon is a separate, unique environment. Hence 2,2-dimethylpropane shows only 2 peaks in its ¹³C NMR spectrum.