IIT-JAM Spectroscopy Basics — UV-Vis, IR and NMR in One Place

By Aniket Bhardwaj · 23 September 2026 · IIT-JAM Chemistry

Spectroscopy sits at the join between physical and organic chemistry, which is why students who study those two units separately often find it hardest. It is really one idea applied three times: a molecule absorbs radiation whose photon energy exactly matches a gap between two of its own energy levels, and the position, intensity and shape of that absorption tell you something structural. UV-Vis probes electrons, IR probes bonds vibrating, NMR probes nuclei flipping in a magnetic field. This guide covers all three at the level JAM expects, with the arithmetic done in full, and ends with a combined structure problem of the type that carries the most marks.

The equations you must be able to use both ways

Photon energy: E = hν = hc/λ = hcν̄ (ν̄ = 1/λ, the wavenumber)
Beer–Lambert law: A = εcl and A = −log₁₀T, where T = I/I₀
Vibrational wavenumber: ν̄ = (1/2πc) √(k/μ), with reduced mass μ = m₁m₂/(m₁ + m₂)
Chemical shift: δ = (ν_sample − ν_TMS) × 10⁶ ÷ ν_spectrometer (in ppm)
Degree of unsaturation: DBE = C − H/2 − X/2 + N/2 + 1

Two unit facts save time in the exam. First, 1 cm⁻¹ is equivalent to 11.96 J mol⁻¹, which converts any wavenumber into an energy per mole in one step. Second, δ is defined as a ratio, so it is the same number on any spectrometer, whereas a coupling constant J is a real frequency in hertz and is also field-independent — but the separation between two signals measured in hertz is not, because it scales with the operating frequency.

The three techniques on one energy scale

Technique	Typical region	What is excited	Order of energy
UV-Visible	200–800 nm	Valence electrons: π→π, n→π, charge transfer, d–d	hundreds of kJ mol⁻¹
Infrared	4000–400 cm⁻¹	Bond stretching and bending	tens of kJ mol⁻¹
NMR	radiofrequency, in a strong magnetic field	Nuclear spin flip (¹H, ¹³C)	fractions of a J mol⁻¹

Put numbers on the first two rows and the scale becomes obvious. A 500 nm photon is 20 000 cm⁻¹, which is 20 000 × 11.96 = 239 200 J mol⁻¹ ≈ 239 kJ mol⁻¹. A carbonyl stretch at 1700 cm⁻¹ is 1700 × 11.96 = 20 332 J mol⁻¹ ≈ 20.3 kJ mol⁻¹, about twelve times smaller. NMR transitions are smaller again by many orders of magnitude, which is exactly why NMR needs a strong external magnet to create a gap worth probing at all.

Worked example 1 — Beer–Lambert, both directions

A coloured complex has ε = 1.55 × 10³ L mol⁻¹ cm⁻¹ at its λ_max. In a 1.00 cm cell the absorbance is 0.62. Find the concentration and the percentage transmittance.

Concentration. From A = εcl,
c = A ÷ (ε l) = 0.62 ÷ (1.55 × 10³ × 1.00) = 0.62 ÷ 1550 = 4.0 × 10⁻⁴ mol L⁻¹
(check: 1550 × 4.0 × 10⁻⁴ = 0.62 ✓)

Transmittance. A = −log₁₀T, so T = 10^−A = 10^−0.62.
10^−0.62 = 10^−0.6 × 10^−0.02 = 0.2512 × 0.9550 = 0.2399, i.e. 24.0 % transmittance.

Two things follow that are worth remembering. Absorbance is additive for a mixture at a given wavelength, so a two-component analysis is two simultaneous equations at two wavelengths. And the linear relation fails at high concentration — association, refractive-index change and stray light all bend the plot — which is why calibration curves are kept in the low-A range and why "deviations from Beer's law" is a standard short question.

Infrared — where the absorptions sit and why

A vibration is IR active only if it changes the dipole moment of the molecule. That single rule explains why N₂, O₂ and Cl₂ show no IR spectrum at all, and why the symmetric stretch of CO₂ is IR inactive while its asymmetric stretch and its bend are active. Raman spectroscopy has the complementary rule — it needs a change in polarisability — and for a centrosymmetric molecule the rule of mutual exclusion says no vibration can be active in both.

Group	Approximate wavenumber (cm⁻¹)	Appearance
O–H, alcohol (H-bonded)	3200–3600	Strong, very broad
O–H, carboxylic acid	2500–3300	Extremely broad, overlaps C–H
N–H, amine or amide	3300–3500	Two bands for a primary, one for a secondary
C–H, sp³ / sp² / sp	2850–3000 / 3000–3100 / ≈3300	The 3000 line is a useful divider
C≡N	2220–2260	Sharp, medium
C≡C	2100–2260	Weak; absent if the alkyne is symmetrical
C=O	1650–1820	Very strong and sharp — the most diagnostic band in IR
C=C, alkene / aromatic	1620–1680 / around 1450–1600	Medium; aromatic gives several

Within the carbonyl range the exact position is itself the information, and it moves for reasons you can reason out rather than memorise: conjugation lowers the wavenumber (partial single-bond character weakens the C=O), ring strain raises it (cyclopentanone above cyclohexanone, cyclobutanone higher again), and hydrogen bonding lowers it. So the sequence anhydride > ester > aldehyde ≈ ketone > amide is not arbitrary: the amide nitrogen donates electron density into the carbonyl and weakens it most.

Worked example 2 — predicting an IR band from a force constant

Estimate the C=O stretching wavenumber, taking the force constant as k = 1200 N m⁻¹. Use m(¹²C) = 12.000 u, m(¹⁶O) = 15.995 u, 1 u = 1.66054 × 10⁻²⁷ kg, c = 2.998 × 10¹⁰ cm s⁻¹.

Step 1 — reduced mass.
μ = (12.000 × 15.995) ÷ (12.000 + 15.995) = 191.94 ÷ 27.995 = 6.856 u
In kilograms: 6.856 × 1.66054 × 10⁻²⁷ = 1.1385 × 10⁻²⁶ kg

Step 2 — angular frequency.
k/μ = 1200 ÷ 1.1385 × 10⁻²⁶ = 1.0541 × 10²⁹ s⁻²
√(k/μ) = 3.247 × 10¹⁴ s⁻¹ (check: (3.247 × 10¹⁴)² = 1.054 × 10²⁹ ✓)

Step 3 — convert to wavenumber.
2πc = 6.2832 × 2.998 × 10¹⁰ = 1.8837 × 10¹¹ cm s⁻¹
ν̄ = 3.247 × 10¹⁴ ÷ 1.8837 × 10¹¹ = 1724 cm⁻¹

That lands squarely in the observed ketone region near 1715 cm⁻¹, which is the point of the exercise: the simple harmonic-oscillator model is good enough to be useful. Notice the structure of the formula — ν̄ rises with the square root of bond strength and falls with the square root of mass. That is why C≡C > C=C > C–C in wavenumber, and why C–H sits above C–C even though the bonds are of comparable strength.

Worked example 3 — the deuterium isotope shift

A C–H stretch appears at 3000 cm⁻¹. Where will the corresponding C–D stretch appear? Take m(H) = 1.008 u, m(D) = 2.014 u, m(C) = 12.000 u.

Replacing hydrogen by deuterium does not change the force constant — the electronic structure is the same — so the entire effect is in the reduced mass.

μ(C–H) = (12.000 × 1.008) ÷ 13.008 = 12.096 ÷ 13.008 = 0.9299 u
μ(C–D) = (12.000 × 2.014) ÷ 14.014 = 24.168 ÷ 14.014 = 1.7246 u

ν̄(C–D) ÷ ν̄(C–H) = √[μ(C–H) ÷ μ(C–D)] = √(0.9299 ÷ 1.7246) = √0.5392 = 0.7343

ν̄(C–D) = 3000 × 0.7343 = 2203 cm⁻¹, close to 2200 cm⁻¹.

The useful takeaway is the shortcut: because μ for C–H is close to 1 u and for C–D close to 2 u, the ratio is near 1/√2 ≈ 0.71, so any X–D stretch falls to roughly 0.7 times its X–H value. This is also why C–D bands land in the otherwise empty 2100–2300 window, which is what makes deuterium labelling so useful.

NMR — the four pieces of information in a proton spectrum

Number of signals = number of chemically non-equivalent proton environments.
Chemical shift (δ) = how deshielded each environment is. Electronegative neighbours pull electron density away, the nucleus feels more of the applied field, and the signal moves downfield (higher δ).
Integration = the relative number of protons in each environment. It gives a ratio, never an absolute count, so 3:2:3 could be 3:2:3 or 6:4:6.
Splitting = the n + 1 rule: a signal is split into n + 1 lines by n equivalent neighbouring protons, with intensities from Pascal's triangle.

Proton environment	Approximate δ (ppm)
TMS reference	0.00 by definition
Alkyl C–H	0.8–1.8
C–H next to a carbonyl	2.0–2.6
Benzylic C–H	2.3–2.7
C–H attached to O (ester or ether)	3.3–4.5
Vinylic C–H	4.5–6.5
Aromatic C–H	6.5–8.5
Aldehyde C–H	9.5–10.5
Carboxylic acid O–H	10–13, broad and exchangeable

Aromatic protons sit far downfield because the ring current induced by the applied field reinforces it outside the ring — a genuinely different mechanism from ordinary inductive deshielding, and a favourite short question. O–H and N–H protons are broad and their position depends on solvent, concentration and temperature; they vanish on shaking with D₂O, which is the standard way to confirm them.

Worked example 4 — a complete structure from formula, IR and NMR

A compound has molecular formula C₄H₈O. Its IR shows a strong sharp band at 1715 cm⁻¹ and nothing above 3100 cm⁻¹. Its ¹H NMR at 400 MHz shows three signals: a singlet near δ 2.1 (3H), a quartet near δ 2.4 (2H) and a triplet near δ 1.0 (3H). Identify it.

Step 1 — degrees of unsaturation.
DBE = C − H/2 + 1 = 4 − 8/2 + 1 = 4 − 4 + 1 = 1
One degree only: either one ring or one double bond, not both.

Step 2 — read the IR. A strong sharp band at 1715 cm⁻¹ is a carbonyl, and it accounts for the single degree of unsaturation. Nothing above 3100 cm⁻¹ rules out O–H and N–H, so it is not an acid or an alcohol. There is no aldehyde C–H stretch pair near 2720 and 2820 cm⁻¹, and the NMR has no signal near δ 9.5–10.5 — so it is a ketone, not an aldehyde.

Step 3 — read the NMR. Integration 3:2:3 accounts for all eight hydrogens. The quartet-and-triplet pair with a 2:3 ratio is the unmistakable signature of an isolated –CH₂CH₃ group: the CH₂ sees three neighbours (3 + 1 = 4 lines) and the CH₃ sees two (2 + 1 = 3 lines). The remaining 3H singlet has no neighbours at all and sits at δ 2.1, next to the carbonyl — a CH₃–CO– group.

Step 4 — assemble. CH₃CO– plus –CH₂CH₃ gives CH₃COCH₂CH₃, butan-2-one, C₄H₈O ✓. The only other DBE = 1 ketone of this formula would need a different carbon skeleton, and butanal, the aldehyde isomer, is excluded in step 2.

A frequency check, since the field is given. On a 400 MHz instrument, 1 ppm = 400 Hz. The gap between the δ 2.4 and δ 1.0 signals is 1.4 ppm = 1.4 × 400 = 560 Hz. Run the same sample at 100 MHz and that gap becomes 140 Hz — but the coupling constant between the CH₂ and CH₃, roughly 7 Hz, would be unchanged. That difference is the standard test of whether a student really understands what δ means.

UV-Vis in brief — what actually gets asked

π→π* transitions are allowed and intense (ε often 10³–10⁵); n→π* transitions are formally forbidden, so they are weak (ε of order 10–100) and appear at longer wavelength.
Conjugation raises λ_max. Extending a conjugated system narrows the HOMO–LUMO gap, so absorption moves to longer wavelength — a bathochromic or red shift. The opposite is hypsochromic. Intensity changes are hyperchromic and hypochromic.
A chromophore is the group responsible for absorption (C=C, C=O, NO₂, an aromatic ring); an auxochrome (–OH, –NH₂, –OR) does not absorb usefully on its own but shifts and intensifies the chromophore's band by donating a lone pair into it.
Empirical Woodward–Fieser rules predict λ_max for dienes and enones by adding increments to a base value. The increment tables differ between textbooks, so learn them from the book you will be examined against rather than from memory of a different one.
Solvent matters: n→π* bands shift to shorter wavelength in polar hydrogen-bonding solvents because the lone pair is stabilised by hydrogen bonding, widening the gap.

Common mistakes that cost marks

Reading integration as an absolute number of protons. It is a ratio; check it against the molecular formula before assigning.
Applying the n + 1 rule to equivalent protons. The three protons of a CH₃ do not split each other. Only non-equivalent neighbours count.
Saying δ changes with field strength. It does not — that is the whole reason δ was defined. The separation in hertz does.
Assuming every bond gives an IR band. No dipole change, no absorption. Symmetrical alkynes and homonuclear diatomics are silent.
Calling any 1700 cm⁻¹ band "a ketone". Esters, aldehydes, acids and amides all absorb nearby; use the O–H, N–H and aldehyde C–H regions to discriminate.
Forgetting halogen and nitrogen in the DBE formula. Halogens count like hydrogen (−X/2), nitrogen adds +N/2, and oxygen is ignored entirely.
Confusing ε with absorbance. ε is a property of the substance at a given wavelength; A depends on concentration and path length as well.

How to prepare this unit

Theme	What you must be able to do without hesitation
Energy conversions	Move between λ, ν̄, J per photon and kJ mol⁻¹ in either direction
Beer–Lambert	Solve for any of A, ε, c, l, and convert between A and %T
Reduced mass	Compute ν̄ from k, and predict isotope shifts
IR correlation	Assign O–H, N–H, C–H, C≡N, C=O and C=C confidently, and explain carbonyl shifts
Selection rules	State the IR and Raman rules and apply mutual exclusion
NMR reading	Signals, δ, integration and multiplicity — all four, every time
Combined problems	Formula → DBE → IR → NMR → structure, in that order

Treat that as a revision checklist, not as a prediction of the paper. For the syllabus and the current pattern, read the official IIT-JAM notification for your year.

Check your absorbance arithmetic. Worked example 1 is the calculation that appears most often in this unit, and the errors are almost always in rearranging A = εcl or in the log step. The Beer–Lambert calculator solves for whichever quantity you leave out, so you can confirm concentration, ε or path length in seconds and then do the transmittance conversion by hand.

Open the Beer–Lambert Law Calculator →

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