GATE NMR Spectroscopy — The Problem Types That Actually Appear

By Aniket Bhardwaj · 21 September 2026 · GATE Chemistry

NMR questions in GATE Chemistry are not really "spectroscopy" questions. They are small, well-defined puzzles built on four ideas — chemical shift, integration, spin–spin coupling and molecular symmetry — and once you can see which of the four a question is testing, the answer usually falls out in under a minute. This guide works through each recurring type with the arithmetic done in full, including the ones students get wrong for reasons that have nothing to do with chemistry: converting ppm to hertz, and deciding whether a pattern is even allowed to be read by the n + 1 rule.

Because images are switched off on this knowledge base, every spectrum here is written out as a table of (δ, multiplicity, integral). That is exactly how GATE prints them anyway, so it is good practice.

The two relations everything is built on

δ (ppm) = Δν (Hz) / ν₀ (MHz) ⇔ Δν (Hz) = δ (ppm) × ν₀ (MHz)

Here Δν is the separation between the signal and the reference (TMS, defined as δ = 0), and ν₀ is the spectrometer's operating frequency for the nucleus being observed. Because ν₀ is quoted in MHz and Δν in Hz, the ratio comes out directly in parts per million — no extra factor of 10⁶ is needed. This is the single most common source of a factor-of-a-million slip.

Quantity	Meaning	Depends on field?
δ (ppm)	Chemical shift — electronic environment of the nucleus	No
Δν (Hz)	Same shift expressed as a frequency	Yes — scales with ν₀
J (Hz)	Coupling constant — through-bond nuclear interaction	No
Integral	Area under the signal ∝ number of equivalent nuclei	No

The asymmetry in that table — δ fixed in ppm, J fixed in Hz — is the source of most "why does a higher-field magnet help?" questions, and we use it below.

Type 1 — converting between ppm and Hz

(a) A proton resonates 1 250 Hz downfield of TMS on a 500 MHz spectrometer. Give its chemical shift.

δ = 1 250 / 500 = 2.50 ppm

(b) The same compound is run on a 300 MHz instrument. Where is that signal now, in Hz from TMS?

Δν = 2.50 × 300 = 750 Hz

The chemical shift is unchanged at 2.50 ppm — that is the whole point of the ppm scale. Only the hertz separation shrank, in exact proportion to the field.

(c) Two signals are 0.40 ppm apart. How far apart are they in Hz at 400 MHz?

Δν = 0.40 × 400 = 160 Hz

Type 2 — multiplicity and the limits of the n + 1 rule

For first-order spectra, a nucleus coupled to n equivalent neighbours gives n + 1 lines with intensities from Pascal's triangle. The word "equivalent" is doing all the work.

n equivalent neighbours	Lines	Name	Relative intensities
0	1	singlet	1
1	2	doublet	1 : 1
2	3	triplet	1 : 2 : 1
3	4	quartet	1 : 3 : 3 : 1
4	5	quintet	1 : 4 : 6 : 4 : 1
6	7	septet	1 : 6 : 15 : 20 : 15 : 6 : 1

When the neighbours are not equivalent you must apply each coupling separately. Two different couplings give a doublet of doublets (dd), four lines of equal intensity.

A proton H_X is coupled to H_A with J = 10.0 Hz and to H_B with J = 4.0 Hz. Where are its four lines relative to the centre?

First splitting: ± 10.0/2 = ± 5.0 Hz → lines at −5.0 and +5.0.
Second splitting applied to each: ± 4.0/2 = ± 2.0 Hz.

−5.0 − 2.0 = −7.0 Hz · −5.0 + 2.0 = −3.0 Hz · +5.0 − 2.0 = +3.0 Hz · +5.0 + 2.0 = +7.0 Hz

Check: the outer spacing 7.0 − 3.0 = 4.0 Hz is the small J, and the distance between line 1 and line 3 is (−7.0) → (+3.0) = 10.0 Hz, the large J. A dd always lets you read both coupling constants directly off the spacings, which is a favourite one-mark question.

If instead J_AX = J_BX = 7.0 Hz, the middle two lines coincide and the dd collapses to a 1 : 2 : 1 triplet — which is exactly why a CH₂ flanked by two similar CH groups often looks like a clean quintet even though the neighbours are not strictly equivalent.

Type 3 — is the spectrum even first order?

The n + 1 rule is an approximation valid when the shift difference is large compared with the coupling. The usual working criterion is Δν / J greater than about 6–10; below that the roofing becomes severe and intensities stop following Pascal's triangle.

First-order treatment valid when Δν (Hz) / J (Hz) ≳ 6–10, where Δν = Δδ × ν₀

Two coupled protons appear at δ 7.20 and δ 7.32 with J = 8.0 Hz. Is the spectrum first order at 400 MHz? At 100 MHz?

Δδ = 7.32 − 7.20 = 0.12 ppm.

At 400 MHz: Δν = 0.12 × 400 = 48 Hz, so Δν/J = 48 / 8.0 = 6.0 — just about first order, an AB system on the edge of becoming AX.

At 100 MHz: Δν = 0.12 × 100 = 12 Hz, so Δν/J = 12 / 8.0 = 1.5 — strongly second order. The two "doublets" would lean into each other so heavily that the outer lines almost vanish.

This is the quantitative answer to "why buy a higher-field magnet": J is fixed in Hz, Δν grows with the field, so the ratio improves linearly and the spectrum simplifies itself.

Type 4 — structure from a printed data set

The standard format gives you a molecular formula plus a shift/multiplicity/integral table. Always start with the degree of unsaturation, because it tells you before you read a single shift whether there is a ring or a π bond.

Degree of unsaturation (DBE) = C − H/2 + N/2 + 1 (halogens count as H, O is ignored)

C₄H₈O₂ gives: δ 4.12 (q, 2H, J = 7.1 Hz); δ 2.04 (s, 3H); δ 1.26 (t, 3H, J = 7.1 Hz). Identify it.

DBE = 4 − 8/2 + 1 = 4 − 4 + 1 = 1. With two oxygens and no aromatic signals, that one degree is a C=O.

Integrals 2 : 3 : 3 sum to 8H, matching the formula exactly — so no proton is hidden.

The quartet at 2H and the triplet at 3H share J = 7.1 Hz, so they couple to each other: that is a CH₂CH₃ unit. The CH₂ at δ 4.12 is far downfield, which means it sits on an ester oxygen (O–CH₂), not merely next to a carbonyl.

The 3H singlet at δ 2.04 has no neighbours at all — a methyl on a carbonyl carbon.

Putting it together: CH₃–CO–O–CH₂–CH₃, ethyl acetate. Cross-check the formula: 2 C in the ethyl, 2 C in the acetyl = C₄; 5 + 3 = 8 H; 2 O. ✓

The discriminating detail is which methyl is the singlet. In methyl propanoate (CH₃CH₂–CO–O–CH₃), same formula, the singlet methyl would sit near δ 3.7 on the oxygen and the CH₂ near δ 2.3. The shifts, not the multiplicities, separate the two isomers.

Type 5 — counting signals from symmetry (especially ¹³C)

Proton-decoupled ¹³C spectra are all singlets, so the only information is how many lines there are. That makes them pure symmetry questions, and they are quick marks if you count carefully.

How many ¹³C signals do the three dichlorobenzenes give?

1,2-dichlorobenzene: a mirror plane makes C1 ≡ C2, C3 ≡ C6, C4 ≡ C5 → 3 signals.

1,3-dichlorobenzene: C1 ≡ C3, C4 ≡ C6, while C2 and C5 are each unique → 4 signals.

1,4-dichlorobenzene: C1 ≡ C4, and all four CH carbons are equivalent → 2 signals.

So a single ¹³C count identifies the isomer outright. The same reasoning gives p-xylene three signals: one methyl, one substituted ring carbon, one CH ring carbon.

Type 6 — DEPT and what each experiment removes

Carbon type	Broadband-decoupled ¹³C	DEPT-90	DEPT-135
Quaternary (C)	present	absent	absent
Methine (CH)	present	present	positive
Methylene (CH₂)	present	absent	negative
Methyl (CH₃)	present	absent	positive

The standard exam move is to give you the ordinary spectrum and the DEPT-135 and ask for the number of quaternary carbons. Subtract: quaternary count = (total lines) − (lines visible in DEPT-135). Nothing more is needed.

Approximate ¹H shift ranges worth carrying in memory

Environment	Typical δ (ppm)	Note
TMS reference	0.00	By definition
Alkane C–H	0.8 – 1.8	CH₃ < CH₂ < CH
C–H next to C=O or C=C	2.0 – 2.7	Allylic / α to carbonyl
C–H on N	2.2 – 3.0	Amines
C–H on O (ether, ester alkoxy)	3.3 – 4.3	Strong deshielding
Vinylic C=C–H	4.5 – 6.5	Terminal alkenes at the low end
Aromatic H	6.5 – 8.0	Ring current deshields
Aldehyde CHO	9.5 – 10.1	Diagnostic singlet (or small d)
Carboxylic acid COOH	10 – 13	Broad; exchanges with D₂O
OH / NH	variable	Concentration- and solvent-dependent; exchangeable

Treat these as ranges, never as fixed numbers. Hydrogen bonding moves OH and NH signals by several ppm, which is precisely why they are quoted as "variable" rather than pinned down.

Common mistakes

Quoting a coupling constant in ppm. J is a molecular property measured in Hz. If a question reports "J = 0.02 ppm" it is testing whether you convert: 0.02 × ν₀.
Expecting J to change with the magnet. It does not. Only the ppm-to-Hz spacing of chemical shifts changes.
Applying n + 1 to inequivalent neighbours. Three neighbours with two different J values give a dd of d (up to 8 lines), not a quartet.
Reading integrals as absolute proton counts. They give a ratio. Scale the ratio so the total matches the molecular formula before drawing conclusions.
Forgetting exchangeable protons. OH, NH and COOH often appear broad, may not couple, and disappear on shaking with D₂O — a standard exam clue.
Assuming coincidental overlap means equivalence. Two chemically distinct protons can happen to fall at the same δ at low field and separate at high field.
Counting ¹³C signals from the drawn structure without checking symmetry. Rotate the molecule mentally and look for mirror planes and axes first.
Treating DEPT-135 intensities as concentrations. The sign encodes the number of attached protons; the height does not reliably count carbons.

Where these types show up in GATE Chemistry

Question type	What is really being tested	Usual form
ppm ⇄ Hz conversion	Field independence of δ, field dependence of Δν	Numerical answer
Multiplicity prediction	Chemical and magnetic equivalence	Multiple choice on a named structure
Reading J from a dd	Line-spacing analysis	Numerical answer in Hz
Δν/J judgement	Limits of first-order analysis	Conceptual multiple choice
Structure from data table	DBE plus shift logic	Choose the correct isomer
¹³C signal counting	Molecular symmetry	Numerical answer (an integer)
DEPT interpretation	CH_n classification	Count quaternary carbons

The syllabus and question pattern are set by the conducting institute and are revised from time to time, so confirm the current scope and marking scheme from the official GATE notification for your exam year rather than from any secondary source, including this one.

Practise the arithmetic, not just the theory. Almost every NMR numerical reduces to one multiplication or one division — δ × ν₀, Δν ÷ ν₀, or a ratio of line spacings. Those are exactly the steps where a rushed decimal costs a mark, so run them through the Scientific Calculator rather than in your head under time pressure. There is no dedicated NMR tool in the suite, so this button opens the calculator home view honestly.

Open the ABC Chemistry Calculator Suite →

Preparing for GATE, IIT-JAM, CSIR-NET or CUET-PG? ABC Chemistry runs dedicated chemical-sciences batches — coaching centre and online classes across India. Details at abcchemistry.in.