IR spectroscopy — characteristic group frequencies (O-H, N-H, C=O, C≡N, C=C, etc.); fingerprint region

WHY IR works at all

WHY the dipole rule? Light is an oscillating electric field. To grab energy from it, the molecule must present an oscillating "handle" — a changing dipole. A symmetric stretch of $O_2$ has zero dipole change, so $O_2$ is IR-invisible. A polar bond like $C=O$ has a big dipole swing, so it absorbs strongly.

HOW to predict frequency — derive it from a spring

A bond = harmonic oscillator. Two masses $m_1, m_2$ joined by a spring of stiffness $k$ (the force constant, units N/m).

Step 1 — Newton + Hooke. For a single mass on a spring, $F=-kx$ and $F=ma=m\ddot x$ : $m\ddot x = -kx \;\Rightarrow\; \ddot x = -\frac{k}{m}x$ This is SHM with angular frequency $\omega=\sqrt{k/m}$ . Why this step? The solution of $\ddot x=-\omega^2 x$ is $x=A\cos\omega t$ — oscillation at $\omega$ .

Step 2 — two atoms, not one. Both atoms move, so replace $m$ by the reduced mass: $\mu=\frac{m_1 m_2}{m_1+m_2}$ Why? In the centre-of-mass frame the relative coordinate behaves like a single particle of mass $\mu$ on the spring.

Step 3 — frequency. $\nu=\frac{1}{2\pi}\sqrt{\frac{k}{\mu}}$

Step 4 — chemists use wavenumber $\tilde\nu=1/\lambda=\nu/c$ (units cm⁻¹): $\boxed{\tilde\nu=\frac{1}{2\pi c}\sqrt{\frac{k}{\mu}}}$

The characteristic group frequencies (the 80/20 table)

Memorise this — it solves ~80% of IR problems.

Group	$\tilde\nu$ (cm⁻¹)	Shape / note	Why there
O–H (alcohol)	3200–3550	broad, rounded	H-bonding smears it
O–H (carboxylic acid)	2500–3300	very broad	strong H-bonding
N–H	3300–3500	1–2 sharper spikes	NH₂ shows two
≡C–H	~3300	sharp	high $k$ region
=C–H / C–H	2850–3100	sharp	light H
C≡N	2200–2260	sharp, medium	triple-ish, stiff
C≡C	2100–2260	weak (low dipole)	symmetric ⇒ weak
C=O	1670–1750	strong, sharp	big dipole → tall peak
C=C	1620–1680	weak–medium	small dipole
C–O	1000–1300	strong	fingerprint edge

Figure — IR spectroscopy — characteristic group frequencies (O-H, N-H, C=O, C≡N, C=C, etc.); fingerprint region

The fingerprint region

Worked examples

Common mistakes

Flashcards

What two molecular factors set a bond's IR wavenumber?

Force constant

k

(stiffness) and reduced mass

\mu

— via

\tilde\nu=\frac{1}{2\pi c}\sqrt{k/\mu}

What is the IR selection rule?

The vibration must change the molecule's dipole moment; no dipole change ⇒ IR-inactive.

Define reduced mass.

\mu=\dfrac{m_1m_2}{m_1+m_2}

, the effective single mass for two atoms on a spring.

Why does C–H absorb near 3000 cm⁻¹?

H is very light ⇒ small

\mu

⇒ large

\sqrt{k/\mu}

Typical C=O wavenumber and why it's strong?

~1700 cm⁻¹; strong because the polar carbonyl gives a large dipole change.

Distinguish alcohol O–H from carboxylic-acid O–H.

Alcohol: broad 3200–3550. Acid: very broad 2500–3300 + accompanying C=O ~1710.

Why is C=C often weak in IR?

Small dipole change on stretching (often near-symmetric) ⇒ weak/absent peak.

What is the fingerprint region and its use?

Below ~1500 cm⁻¹; tangled skeletal/bending bands whose unique pattern confirms molecular identity by matching.

Order O–H, C–H, C≡N, C=O, C=C by decreasing wavenumber.

O–H (3300) > C–H (3000) > C≡N (2250) > C=O (1700) > C=C (1650).

Does bond order set peak position or peak height?

Position (via

k

). Height is set by dipole change (polarity).

C≡N vs C=N — which is higher and why?

C≡N higher; larger

k

at same

\mu

⇒ higher

\tilde\nu

Recall Feynman: explain to a 12-year-old

Imagine each bond in a molecule is a little spring holding two balls. If you push it, it bounces back and forth at its own special speed. Tighter springs and lighter balls bounce fast; loose springs and heavy balls bounce slow. Now shine invisible "heat light" (infrared) on the molecule. Each spring grabs only the light whose flicker matches its own bounce. By seeing which lights got swallowed, we figure out which springs (bonds) the molecule has. Springs that are lop-sided (one end pulls electrons more) grab the most light — that's why the C=O spring gives a big dark line. The messy crowd of tiny springs at the bottom is so unique it works like a fingerprint to name the exact molecule.

Connections

Molecular dipole moment — sets which vibrations are IR-active and how intense.
Simple harmonic oscillator — the physics model behind the master equation.
Hydrogen bonding — explains the broadening of O–H and N–H bands.
Bond order and bond strength — links $k$ to single/double/triple bonds.
Mass spectrometry & NMR spectroscopy — complementary structure-elucidation tools.
Electromagnetic spectrum — where IR sits (between visible and microwave).

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, IR spectroscopy ka core idea bahut simple hai: har chemical bond ek chhoti spring jaisa hai jisme do atoms (masses) jude hain. Jab hum molecule par infrared light daalte hain, to har spring sirf apni "favourite" frequency wali light ko absorb karta hai — wahi frequency jis par wo naturally hilta (vibrate karta) hai. Stiff bond (jaise triple bond) aur halke atoms (jaise H) tezi se hilte hain, isliye wo high wavenumber par absorb karte hain. Formula yaad rakho: $\tilde\nu=\frac{1}{2\pi c}\sqrt{k/\mu}$ — yahan $k$ bond ki stiffness hai aur $\mu$ reduced mass.

Ek important rule: bond ko light absorb karne ke liye uska dipole moment change hona chahiye. Isiliye C=O (jo polar hai) bahut strong peak deta hai (~1700 cm⁻¹), jabki C=C (jo zyada symmetric hai) weak peak deta hai. Yaad rakho — bond order se peak ka position decide hota hai, aur polarity (dipole change) se peak ki height. Ye dono alag cheezein hain, confuse mat karna.

Spectrum ko do hisson mein baanto. Upar wala part (1500 cm⁻¹ se zyada) "functional group region" hai — yahan O-H, N-H, C-H, C≡N, C=O ke clean, pehchaan-ne layak peaks aate hain. Niche wala part (1500 se kam) "fingerprint region" hai — yahan itne saare bending vibrations milte hain ki unhe ek-ek karke pehchaanna mushkil hai, lekin poora pattern har molecule ke liye unique hota hai, bilkul fingerprint ki tarah. Identity confirm karne ke liye is region ko reference spectrum se match karte hain.

Exam tip (80/20): bas yeh table ratt lo — broad 3300 = O-H, sharp 1700 = C=O, 2250 = C≡N. Inhi 3-4 peaks se zyadatar questions solve ho jaate hain. Agar broad 3300 + strong 1700 dono dikhein, to wo carboxylic acid hai. Practice karo aur peaks ki "kahani" samjho, sirf numbers ratt-ne se kaam nahi chalega.

Test yourself — Spectroscopy & Analysis (Intro)

Connections