¹H NMR — chemical shift, multiplicity (n + 1 rule), integration; common ranges: Visual walkthrough

4.8.4 · D2 · Chemistry › Spectroscopy & Analysis (Intro) › ¹H NMR — chemical shift, multiplicity (n + 1 rule), integrat

Prerequisites jin par hum depend karte hain: Spin-spin coupling (J), Pascal's triangle, aur parent ¹H NMR topic note.

Step 1 — Ek proton ek tiny bar magnet hai jiske paas do choices hain

KYA. Kisi bhi splitting se pehle, humein raw object chahiye. Ek hydrogen nucleus (ek proton) ek microscopic bar magnet ki tarah behave karta hai — isme ek property hoti hai jise spin kehte hain. Ek bade external magnetic field $B_{0}$ mein yeh chota magnet sirf do tarike se baith sakta hai.

YE DO HI KYUN, smooth range kyun nahi? Kyunki nuclear spin quantised hoti hai — proton ek "spin-½" particle hai, jo physics ki bhaasha mein hai "exactly do allowed orientations, kuch beech mein nahi." Hum unhe label karte hain:

==spin-up ( $↑$ )==: $B_{0}$ ke saath aligned → thodi lower energy (comfortable).
==spin-down ( $↓$ )==: $B_{0}$ ke against aligned → thodi higher energy (strained).

Dono ke beech energy gap hai $Δ E = fixed constants \frac{γ h}{2 π} \cdot field jo proton actually feel karta hai B_{local}$

$γ$ — gyromagnetic ratio, proton ke liye ek fixed number (kitna strongly uska magnet respond karta hai).
$h$ — Planck's constant, ek aur fixed number.
$B_{local}$ — woh cheez jo badlati hai proton se proton mein. Yahi poore show ka star hai.

PICTURE. Ek magnet ek field mein, do allowed orientations, do energy rungs.

Step 2 — Ek proton, koi neighbour nahi = ek peak (singlet)

KYA. Ek aisa isolated proton lo jiske neighbouring carbons par koi hydrogen nahi hai. Woh exactly ek value of $B_{local}$ feel karta hai, ek radio frequency absorb karta hai, aur ek line deta hai: ek singlet.

YAHAN SE KYUN SHURU KAREIN? Yeh humara "zero neighbours" baseline hai, $n = 0$ . $n + 1$ rule predict karta hai $0 + 1 = 1$ line — hum us ek line ko dekhna chahte hain pehle koi perturbation kiye bina.

$ν = \frac{γ}{2 π} B_{local} ⟹ single frequency ⟹ single peak$

$ν$ — woh frequency jo proton absorb karta hai.
ek $B_{local}$ value $\to$ ek $ν$ $\to$ spectrum par ek vertical line.

PICTURE. Ek clean single spike apne chemical shift $δ$ par baithe hue.

Step 3 — EK neighbour laao: peak do mein split ho jaati hai (doublet)

KYA. Ab adjacent carbon par ek single hydrogen rakho. Woh neighbour khud ek tiny magnet hai (Step 1), toh woh bhi $↑$ ya $↓$ point karta hai — aur uska field hamare proton ke feel kiye field mein add ya subtract hota hai.

YEH LINE SPLIT KYUN KARTA HAI? Kyunki haara proton ab ek $B_{local}$ feel nahi karta. Tube ke millions of molecules mein:

unke aadhe mein neighbour $↑$ hai → haara proton $B_{local} + \frac{1}{2} J^{'}$ feel karta hai → thodi higher frequency par absorb karta hai.
baaki aadhe mein neighbour $↓$ hai → haara proton $B_{local} - \frac{1}{2} J^{'}$ feel karta hai → thoda lower frequency par absorb karta hai.

Do sub-populations → do lines of equal height → ek doublet. Dono ke beech gap ==coupling constant $J$ == hai (dekho Spin-spin coupling (J)) — ek fixed splitting jo Hz mein measure hoti hai aur spectrometer ke saath nahi badlati.

$1 neighbour \to {↑, ↓} \to 2 equally likely fields \to 2 lines, ratio 1 : 1$

PICTURE. Step 2 ki single line do barabar lines mein phatt jaati hai, $J$ se split hokar.

Step 4 — DO equivalent neighbours: 1 : 2 : 1 triplet paida hota hai

KYA. Adjacent carbon par do equivalent H's hain. Har ek independently $↑$ ya $↓$ point karta hai, toh hum saari combinations list karte hain aur unke field contributions add karte hain.

COMBINATIONS ENUMERATE KYUN KAREIN? Kyunki haare proton jo field feel karta hai woh do neighbours ke net tilt par depend karta hai. Alag-alag combinations same net field de sakti hain — aur jab deti hain, toh un molecules ka yog same line par ho jaata hai, use taller banaa deta hai.

Neighbour states	Net field shift	Line
$↑↑$	$+ 1$ unit	high-frequency line
$↑↓$ ya $↓↑$	$0$ (ye cancel ho jaate hain)	centre line
$↓↓$	$- 1$ unit	low-frequency line

Top aur bottom rows mein 1-1 way hai.
Middle field ( $0$ ) tak 2 ways se pahuncha jaata hai ( $↑↓$ aur $↓↑$ ) → double height.

Result: 3 lines ratio $1 : 2 : 1$ mein — ek triplet. Teen alag net fields = $2 + 1 = 3$ lines. $n = 2$ ke liye $n + 1$ rule yahan appear ho raha hai.

${↑↑} ∣ {↑↓, ↓↑} ∣ {↓↓} \Rightarrow 1 : 2 : 1$

PICTURE. Chaar spin arrangements teen field bins mein sort kiye hue — middle bin mein double molecules hain.

Step 5 — Pattern exactly Pascal's triangle hai (aur kyun)

KYA. Generalise karo. $n$ equivalent spin-½ neighbours ke saath, $k$ unhe $↑$ point karte hue paane ke ways ki sankhya binomial coefficient hai $(k n) = \frac{n !}{k ! ( n - k )!}, k = 0, 1, \dots, n$

$n$ — kitne neighbours hain.
$k$ — unme se kitne $↑$ choose karte hain.
$(k n)$ — us particular net tilt wale arrangements ki count = ek line ki height.

YEH PASCAL'S TRIANGLE KYUN HAI? Kyunki $(k n)$ is hi Pascal's triangle ki row $n$ mein entry hai — wahi "apne upar ke do numbers add karo" object. Har naya neighbour ek aur coin flip add karne jaisa hai; counts exactly Pascal's rule ki tarah propagate hote hain. Toh:

$lines ki sankhya = values jo k le sakta hai: 0 \dots n n + 1, total intensity = \sum_{k = 0}^{n} (k n) = 2^{n}$

$n + 1$ — kyunki $k$ $n + 1$ values par run karta hai → $n + 1$ rule, derive hua.
$2^{n}$ — spin arrangements ki total sankhya ( $2$ choices, $n$ neighbours) = Pascal's triangle ki row sum.

$n$	lines	Pascal row
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

PICTURE. Pascal's triangle NMR peaks ki stack ki tarah draw kiya — har row ek multiplet hai.

Step 6 — Edge case: equivalent protons ek doosre ko split KYUN NAHI KARTE

KYA. Ek akele $CH_{3}$ ke teen H's kabhi ek doosre ko split nahi karte, chahe har ek ek neighbour wala magnet ho. Ek otherwise-H-free carbon par CH₃ ek singlet hai, khud-ka-quartet nahi.

KYUN? Splitting tabhi visible hoti hai jab spin flip tumhare observe kiye transition energy ko change kare. Chemically equivalent protons ke liye, har spin flip saare unke energy levels ko same amount se shift karta hai — up-shift aur down-shift line up ho jaate hain aur observed transition mein cancel ho jaate hain. Formally, report karne ke liye koi net difference $B_{local}$ mein nahi hota, toh coupling real hai par silent hai.

$equivalent partners \Rightarrow Δ (observed frequency) = 0 \Rightarrow koi splitting nahi$

PICTURE. Do equivalent protons: unki coupling dono levels ko identically shift karti hai, observed line single rehti hai.

Step 7 — Edge case: zero neighbours aur "bahut saare neighbours" limit

KYA. Do boundary checks taaki koi bhi scenario tumhe surprise na kare.

$n = 0$ (koi vicinal H nahi): $0 + 1 = 1$ → ek singlet. Step 2 se match karta hai. ✔
$n$ bada ho: $1 : n : (2 n) : \dots$ pattern bahut lopsided ho jaata hai — ek tall middle, tiny wings. Outermost lines (har end par $1$ ) centre ke relative itni short ho jaati hain ki woh noise mein gaayab ho sakti hain, toh ek septet ( $n = 6$ , jaise isopropyl ka CH) strong centre aur faint edges dikhata hai.

LIMIT KYUN DIKHAYEIN? Taaki jab tum real data mein "5-line-jaisi dikhne wali" septet dekho, tum jaano ki do faint outer lines wahan hain — maths kehta hai $1 : 6 : 15 : 20 : 15 : 6 : 1$ — woh bas beech ke $20$ ke saamne bahut chhoti lag rahi hain.

$n = 6 : 1 : 6 : 15 : 20 : 15 : 6 : 1 (sum = 2^{6} = 64)$

PICTURE. Doublet, triplet, quartet, septet side by side — dekho wings shrink karte hain jaise $n$ badhta hai.

Step 8 — Teeno readouts ko saath mein rakhna (ethanol)

KYA. Chemical shift (Step-1 field idea) + integration + woh splitting jo humne abhi derive ki, ethanol $CH_{3} CH_{2} OH$ par combine karo.

ETHANOL KYUN? Yeh har ek piece exercise karta hai: ek triplet, ek quartet, ek exchange-broadened singlet.

Group	Neighbours ( $n$ )	Multiplicity ( $n + 1$ )	Integral	$δ$ / ppm
$CH_{3}$	2 (woh CH₂)	triplet $1 : 2 : 1$	3	~1.2
$CH_{2}$	3 (woh CH₃)	quartet $1 : 3 : 3 : 1$	2	~3.7 (O ke paas → deshielded, cf. Electronegativity)
$O - H$	fast exchange	singlet (broad)	1	variable

CH₃ $2$ CH₂ protons dekhta hai → Step 4 → triplet.
CH₂ $3$ CH₃ protons dekhta hai → Step 5, $n = 3$ → quartet.
OH: fast proton exchange coupling ko average kar deta hai → Step 6 ki "silence" ek alag reason se → broad singlet.

PICTURE. Poora ethanol spectrum, har peak kyun us shape ka hai yeh label kiya hua.

Ek-picture summary

Recall Feynman retelling — apne words mein wapas bolo

Har hydrogen ek bahut chota magnet hai jo ek strong field mein sirf do taraf face kar sakta hai. Akele hone par woh ek field feel karta hai aur ek spike deta hai. Ab agla-door carbon par ek hydrogen bhi ek magnet hai, aur uski choice — upar face karna ya neeche — hamare hydrogen ke field ko nudge karti hai. Ek neighbour ke paas do choices hain, toh haari spike do barabar lines mein split ho jaati hai. Do neighbours chaar choices dete hain, par unme se do (up-down, down-up) same field tak cancel ho jaate hain, toh middle line double hoti hai: yahi 1-2-1 triplet hai. Neighbours badhate raho aur "is net tilt ko paane ke kitne ways hain" count exactly Pascal's triangle hai — jo turant tumhe bata deta hai ki $n + 1$ lines hain aur unki heights $2^{n}$ tak add hoti hain. Bas ek twist: woh neighbours jo hamare hydrogen se identical hain sab kuch equally shift kar dete hain, toh unki coupling kabhi dikhti hi nahi — same-neighbour matlab koi splitting nahi. Shift padho kahan ke liye, integral padho kitne ke liye, aur yeh line-count padho kitne neighbours ke liye, aur tumne molecule decode kar li.

Recall Rapid self-test

$n = 3$ kitni lines deta hai aur kya pattern? ::: 4 lines, $1 : 3 : 3 : 1$ (quartet) Equivalent protons ek doosre ko kyun split nahi karte? ::: unki coupling saare levels ko equally shift karti hai → koi observed frequency difference nahi $n$ neighbours se multiplet ki total intensity kya hoti hai? ::: $2^{n}$ Akela CH₃ singlet kyun hota hai? ::: $n = 0$ vicinal H (apne 3 H equivalent hain, toh silent) → $0 + 1 = 1$ Septet ki outer lines faint hain — kya woh sach mein wahan hain? ::: haan, ratio $1 : 6 : 15 : 20 : 15 : 6 : 1$ , wings centre ke saamne dabi hui hain

Related: ¹³C NMR · Aromaticity & ring current · Tetramethylsilane (TMS) · Integration in IR/MS.