CSIR-NET Spectroscopy — Reading IR, NMR and Mass Together

By Aniket Bhardwaj · 5 September 2026 · CSIR-NET Chemistry

Structure elucidation questions in CSIR-NET Chemical Sciences rarely give you enough information in any single spectrum. You get a molecular ion, one or two IR bands and a short NMR listing, and the answer only falls out when you use all three together. The good news is that the reasoning is completely mechanical once you follow a fixed order. This article sets that order out and then works three structures from start to finish.

Each technique answers a different question

Technique	The question it answers
Mass spectrum	How heavy is it? Which heteroatoms are present? What breaks off easily?
Infrared	Which functional groups are present (and which are absent)?
¹H NMR	How many hydrogens of each kind, and who is next to whom?
¹³C NMR	How many chemically distinct carbons, and in what environment?

Work in that order. Never start from the NMR — without the molecular formula you have no way to know when your structure is finished.

Step 1 — molecular formula and degrees of unsaturation

DBE = (2C + 2 + N − H − X) / 2

C = carbons, H = hydrogens, N = nitrogens, X = halogens. Oxygen and sulfur are ignored.

DBE (degrees of unsaturation, also called double-bond equivalents) counts rings plus π bonds. A benzene ring alone is 4 — one ring and three double bonds. So a DBE of 4 or more should make you check for aromatic signals immediately, and a DBE of 0 rules out every C=O, C=C and ring in one line.

Two more free pieces of information come from the molecular ion itself:

Nitrogen rule. An odd nominal molecular mass means an odd number of nitrogen atoms. An even mass means zero or an even number.
Isotope pattern. The relative height of the M+2 peak identifies halogens and sulfur without any other evidence.

Element	Isotopes	M+2 height, as % of M
Chlorine	³⁵Cl 75.8%, ³⁷Cl 24.2%	about 32% (a 3 : 1 look)
Bromine	⁷⁹Br 50.7%, ⁸¹Br 49.3%	about 97% (two peaks of equal height)
Sulfur	³⁴S 4.3%	about 4.4%
Carbon (M+1, not M+2)	¹³C 1.07%	M+1 ≈ 1.1% × number of carbons

Step 2 — the IR bands that actually decide questions

You do not need the whole IR table. Six regions settle almost every exam structure.

Band (cm⁻¹)	Assignment	Note
2500–3300, very broad	O–H of a carboxylic acid	Sits on top of the C–H peaks; unmistakable once seen
3200–3600, broad	O–H of an alcohol	Sharp and weak when the sample is dilute
3300–3500	N–H	Two bands for a primary amine, one for a secondary
2720 and 2820, two weak bands	C–H of an aldehyde	The single most useful confirmation of CHO
2220–2260, sharp	C≡N	Nearly nothing else absorbs there
1630–1820	C=O	The exact position names the carbonyl — see below

Inside the carbonyl region the number matters: amide about 1650, conjugated ketone about 1685, saturated ketone about 1715, aldehyde about 1730, ester about 1740, anhydride two bands near 1760 and 1820. Two rules move these numbers predictably — conjugation lowers a C=O by roughly 20–30 cm⁻¹, and ring strain raises it (cyclopentanone near 1745, cyclobutanone near 1780).

Step 3 — the NMR shift ranges worth memorising

δ (¹H, ppm)	Proton type	δ (¹³C, ppm)	Carbon type
0.9–1.5	Plain alkyl CH₃, CH₂, CH	0–50	Alkyl
2.1–2.6	CH next to C=O	50–90	C–O (alcohol, ether, ester O–C)
3.3–4.5	CH–O	100–150	Alkene
4.5–6.5	Vinyl	110–160	Aromatic
6.5–8.5	Aromatic	160–185	Ester, acid, amide C=O
9.5–10.5	CHO	190–205	Aldehyde C=O
10–13	COOH	195–220	Ketone C=O

Integration gives the ratio of hydrogens, never the absolute number — scale it to the molecular formula. Multiplicity follows the n+1 rule for equivalent neighbours, and the coupling constant J is measured in hertz and does not change with field strength, which is why two mutually coupled signals must show the same J.

Worked example 1 — C₄H₈O₂, M⁺ 88

DBE = (2×4 + 2 + 0 − 8 − 0)/2 = (8 + 2 − 8)/2 = 2/2 = 1. One π bond or one ring only.

IR: strong band at 1740 cm⁻¹, nothing broad above 3000. That is an ester — the one degree of unsaturation is the C=O, and the two oxygens are used up by the ester group.

¹H NMR: δ 4.12 (quartet, 2H, J = 7.1 Hz), δ 2.04 (singlet, 3H), δ 1.26 (triplet, 3H, J = 7.1 Hz).

The quartet–triplet pair with matching J is an ethyl group. Its CH₂ at δ 4.12 is far downfield, so the ethyl is bonded to oxygen: –O–CH₂CH₃. The 3H singlet at δ 2.04 has no neighbours and sits in the "next to C=O" window, so it is CH₃–C=O.

Answer: ethyl acetate, CH₃COOCH₂CH₃.

Confirming against the mass spectrum: m/z 43 is the acetyl cation CH₃CO⁺ (base peak), and m/z 29 is C₂H₅⁺. Losing the CH₃CO fragment (43) from 88 leaves m/z 45, ⁺OCH₂CH₃.

The trap here is methyl propanoate, CH₃CH₂COOCH₃ — same formula, same mass, same IR band. One number separates them: its 3H singlet would appear at about δ 3.67 (an OCH₃), not δ 2.04, and its mass spectrum would show loss of 31 (OCH₃) to give m/z 57. The position of the singlet, not its existence, is the answer.

Worked example 2 — C₈H₈O, M⁺ 120

DBE = (16 + 2 − 8)/2 = 10/2 = 5. Four of those are a benzene ring; one is left over.

IR: 1685 cm⁻¹. Lower than a plain ketone at 1715 — that drop is exactly the conjugation shift, so the C=O is attached to the ring. No aldehyde C–H bands at 2720/2820, so it is not an aldehyde.

¹H NMR: δ 7.95 (2H), δ 7.45 (3H), δ 2.60 (singlet, 3H). The 2H : 3H aromatic split is the classic mono-substituted benzene pattern (two ortho, then meta plus para). A clean 3H singlet at δ 2.60 is CH₃ on a carbonyl.

¹³C NMR: δ 198 (in the ketone window, not the ester window), plus four aromatic carbons and one at δ 26.6.

Answer: acetophenone, C₆H₅COCH₃.

Mass spectrum check: loss of CH₃ (15) gives m/z 105, the benzoyl cation C₆H₅CO⁺, usually the base peak; that loses CO (28) to give m/z 77, C₆H₅⁺, which loses acetylene to give m/z 51.

Worked example 3 — C₇H₆O₂, M⁺ 122

This one is set beside example 2 deliberately, because the two share their most obvious fragments.

DBE = (14 + 2 − 6)/2 = 10/2 = 5. Again a ring plus one more.

IR: a very broad absorption from about 2500 to 3300 cm⁻¹ running through the C–H region, plus C=O near 1690. That broad envelope is the hydrogen-bonded dimer O–H of a carboxylic acid; an alcohol O–H is never that wide.

¹H NMR: δ 12.5 (broad singlet, 1H, disappears on D₂O shake), δ 8.1 (2H), δ 7.6 (1H), δ 7.5 (2H). Total 6H, matching the formula.

Answer: benzoic acid, C₆H₅COOH.

Why the mass spectrum alone would mislead you: benzoic acid loses OH (17) to give m/z 105 and then CO to give m/z 77 — the same two ions acetophenone gave. The fragments are identical; only the molecular ion (122 against 120) and the broad IR band separate the two compounds.

Mistakes that cost marks

Ignoring an absent band. No absorption above 3000 cm⁻¹ is a positive result: it rules out OH and NH in one step and often halves the possibilities.
Reading integration as an atom count. An integral of 2 : 3 in a compound with 10 hydrogens means 4H and 6H. Always scale to the molecular formula.
Forgetting the nitrogen rule. An odd M⁺ with no nitrogen in your proposed structure means the structure is wrong, not the spectrum.
Treating m/z 105 and 77 as proof of a phenyl ketone. Benzoic acid, benzamide and benzoyl chloride all give them. Fragments suggest; the molecular ion and IR decide.
Quoting a coupling constant in ppm. J is always in hertz. Two coupled protons must share the same J — that is how you pair up a quartet with its triplet in a crowded spectrum.
Skipping the DBE. It is one line of arithmetic and it eliminates whole families of wrong answers before you look at any spectrum.

Where this appears in the exam

Exam	Typical demand
CSIR-NET Chemical Sciences	Full structure from combined IR + NMR + MS data; predicting a spectrum from a given structure
GATE Chemistry	Assigning a single band or shift; counting distinct ¹³C signals by symmetry
IIT-JAM / CUET-PG	DBE calculation, splitting patterns, identifying a functional group from one IR band
MSc coursework	Interpreting real spectra of a synthesised product

One habit that pays for itself: after you propose a structure, predict its spectra backwards and check every listed signal is accounted for. A structure that explains four of five signals is not the answer.

Do the arithmetic without slips. DBE counts, isotope ratios, integration scaling and mass differences are all small calculations that go wrong under exam pressure. The ABC Chemistry Calculator Suite keeps the molar mass, scientific-constants and general calculation tools together in one page beside your problem set.

Open the ABC Chemistry Calculator Suite →

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