Basic Spectroscopy for JEE Main Chemistry

Spectroscopic techniques appear in JEE Main Chemistry as concept-based questions about what absorptions occur in which region of the electromagnetic spectrum. These are not heavy calculation questions — they reward precise concept knowledge. This guide gives you the IR and UV-Vis fundamentals at the level JEE tests, along with mass spectrometry basics for completeness.

Infrared (IR) Spectroscopy

IR radiation (wavelength 2.5–25 μm, wavenumber 4000–400 cm⁻¹) causes molecular vibrations (stretching and bending of bonds). Functional group identification by IR: O-H stretch (broad, 3200–3550 cm⁻¹, hydrogen-bonded); N-H stretch (3300–3500 cm⁻¹, sharper than OH); C-H stretch (2850–3000 cm⁻¹ for sp³, ~3100 cm⁻¹ for sp², ~3300 cm⁻¹ for sp); C≡N stretch (~2200 cm⁻¹); C=O stretch (1650–1750 cm⁻¹, the most diagnostic carbonyl peak — different C=O environments give slightly different frequencies); C=C stretch (~1600–1680 cm⁻¹). The fingerprint region (400–1400 cm⁻¹) is unique to each compound but hard to interpret without reference spectra.

JEE question types: (1) given the IR absorption, identify the functional group present; (2) which compound shows a broad absorption around 3300 cm⁻¹ (answer: carboxylic acid or alcohol — look for O-H); (3) compound A has a carbonyl absorption at 1715 cm⁻¹ — is it an acid, ester, aldehyde, or ketone? (aldehydes ~1725, ketones ~1715, esters ~1735, acids ~1700 — rough trends, not memorisation-heavy for JEE). Take a free organic chemistry mock. For carbonyl chemistry, see our aldehydes and ketones guide.

UV-Visible Spectroscopy

UV-Vis radiation (200–800 nm) promotes electrons from bonding/non-bonding MOs to antibonding MOs. Only compounds with chromophores (π systems or lone pairs conjugated with π systems) absorb in the UV-Vis range. σ→σ* transitions (below 200 nm, far UV, JEE rarely tests). π→π* transitions (conjugated dienes, aromatic rings, around 200–300 nm). n→π* transitions (lone pair to π* in carbonyls, ~280–330 nm). Key principle: the more extensive the conjugation, the longer the wavelength of absorption (λ_max shifts to red — bathochromic shift). Lycopene (tomato) and β-carotene (carrot) absorb visible light because of extensive conjugation (11 and 9 conjugated double bonds respectively) → appear coloured.

Mass Spectrometry Basics

In mass spectrometry, the molecule is ionised (loses one electron) to give a radical cation M⁺•, whose mass/charge ratio gives the molecular weight. Fragments of M⁺• give the fragmentation pattern. Molecular ion peak (M⁺) appears at mass = molecular weight. Base peak = the most abundant fragment. Loss of common fragments: lose 15 (CH₃•), lose 17 (OH•), lose 18 (H₂O), lose 29 (CHO•), lose 77 (C₆H₅•, phenyl). Odd-nitrogen rule: if the molecular ion has an odd mass number, the compound contains an odd number of nitrogen atoms. JEE uses mass spectra in multi-correct questions about which molecular formula is consistent with a given M⁺ peak and fragment peaks.

Spectroscopy in Structure Elucidation

JEE occasionally gives a compound with a molecular formula, an IR absorption, and asks you to identify the structure among the given options. Approach: (1) degree of unsaturation (DoU) = (2C + 2 + N − H − X)/2 — tells you how many rings and/or π bonds are present; (2) IR: identify key functional groups (C=O, O-H, N-H); (3) UV: if absorbs above 300 nm, likely has aromatic ring or extended conjugation; (4) narrow down to the consistent option. Degree of unsaturation calculation: for C₈H₈: DoU = (16 + 2 − 8)/2 = 5 → likely benzene ring (4 DoU) + one C=C or ring. For the broader organic chemistry strategy see our organic chemistry reactions guide and our organic isomerism guide.