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

4.8.4 · D3 · Chemistry › Spectroscopy & Analysis (Intro) › ¹H NMR — chemical shift, multiplicity (n + 1 rule), integrat

Scenario Matrix

Har ¹H NMR problem aslam mein kuch case-types ka combination hoti hai. Yahan complete list hai — har worked example neeche us cell(s) ke saath tagged hai jo woh cover karta hai.

Cell	Case class	Tricky kya hai	Covered by
A	Simple ethyl / alkyl splitting	plain $n + 1$ , koi trap nahi	Ex 1
B	Isolated singlet (koi vicinal H nahi)	$n = 0$ → 1 line	Ex 2
C	Equivalent protons split nahi karte	symmetry "degenerate" case	Ex 2, Ex 3
D	Do alag neighbour sets	kaunsa H kisse couple karta hai	Ex 4
E	Integration-only counting	ratios, tert-butyl vs CHO trap	Ex 5
F	Electronegativity → $δ$ ordering	shifts ko environment se rank karo	Ex 6
G	Aromatic ring current (limiting high $δ$ )	$\sim 7$ ppm kyun	Ex 7
H	Exchangeable OH/NH (degenerate splitting)	broad singlet, D₂O trick	Ex 8
I	Real-world / structure-from-data	molecule ko reverse-engineer karo	Ex 9
J	Exam twist (symmetry atoms chhupaati hai)	carbons se kam signals	Ex 10

NMR mein sign/quadrant analogue hai shift ki direction (upfield vs downfield) aur degenerate cases (equivalent protons, $n = 0$ , fast exchange). Hum inhe sab cover karte hain.

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

Ex 1 — Bromoethane $C H_{3} C H_{2} B r$ (Cell A)

Worked example

$C H_{3} C H_{2} B r$ ka full spectrum predict karo. Forecast: Kitne signals honge? Har ek ki multiplicity aur integration kya hogi? Kaun zyada downfield hoga?

Steps

Distinct proton environments identify karo. Do hain: teen $C H_{3}$ protons (sab ek doosre ke equivalent) aur do $C H_{2}$ protons. Yeh step kyun? Signals = chemically alag proton sets ki sankhya. Equivalent H's ek peak share karte hain.
$C H_{3}$ par $n + 1$ apply karo. Iske vicinal neighbours 2 $C H_{2}$ protons hain → $n = 2$ → $2 + 1 = 3$ lines → triplet. Yeh step kyun? Rule sirf adjacent carbons ke protons count karta hai.
$C H_{2}$ par $n + 1$ apply karo. Iske neighbours 3 $C H_{3}$ protons hain → $n = 3$ → quartet. Yeh step kyun? Doosri taraf se wahi rule.
Shifts assign karo. $C H_{2}$ Br se attached hai (electronegative → deshielded) toh yeh downfield baithta hai ( $δ \approx 3.4$ ); $C H_{3}$ sirf carbon ke paas hai ( $δ \approx 1.7$ ). Yeh step kyun? Dekho Electronegativity — electron density kheenchna shielding kam karta hai, $δ$ badhata hai.
Integration: $C H_{3} : C H_{2} = 3 : 2$ .

Verify: Predicted total H $= 3 + 2 = 5$ . Formula $C_{2} H_{5} B r$ mein 5 H hain ✓. Multiplicities $3 + 4 =$ lines Pascal's triangle ke rows $121$ aur $1331$ se match karte hain ✓.

Ex 2 — Neopentane vs acetone (Cells B, C)

Worked example Neopentane

$C (C H_{3})_{4}$ ek single peak kyun deta hai, aur 2,2-dimethylpropane ka spectrum "khali" kyun lagta hai? Forecast: Kitne signals, aur kya multiplicity?

Steps

Neopentane draw karo: ek central carbon jisme koi H nahi, char $C H_{3}$ groups se bonded. Yeh step kyun? Splitting ke liye adjacent carbon par vicinal H's chahiye — central carbon mein koi nahi.
Symmetry se charon $C H_{3}$ groups equivalent hain → ek signal, integration = sab 12 H. Yeh step kyun? Symmetry-equivalent protons ek peak mein collapse ho jaate hain (Cell C).
Multiplicity: neighbours = 0 (central C mein koi H nahi) → $n = 0$ → $0 + 1 = 1$ line → singlet. Yeh step kyun? Yeh degenerate $n = 0$ case hai — "zero input" ka analogue.
Shift: sirf C,H neighbours → $δ \approx 0.9$ (alkane range).

Verify: $C (C H_{3})_{4} = C_{5} H_{12}$ , toh 12 H ek singlet mein. $12$ H, $1$ signal ✓. "Khali" look = ek chota- $δ$ singlet, koi splitting nahi.

Ex 3 — Equivalent protons ek doosre ko kabhi split kyun nahi karte (Cell C)

Ex 4 — 1,1,2-trichloroethane $C H C l_{2} C H_{2} C l$ (Cell D)

Worked example

$C H C l_{2} C H_{2} C l$ ke dono signals assign karo. Forecast: Kaun sa proton doublet hai aur kaun sa triplet? Kaun zyada downfield hai?

Steps

Do environments: akela $C H C l_{2}$ proton aur do $C H_{2} C l$ protons. Yeh step kyun? Alag neighbours + alag substituents → do peaks.
$C H_{2} C l$ : vicinal neighbour ek single $C H C l_{2}$ proton hai → $n = 1$ → doublet. Yeh step kyun? $1 + 1 = 2$ ; ek neighbour ke do spin states do lines dete hain.
$C H C l_{2}$ : vicinal neighbours do $C H_{2}$ protons hain → $n = 2$ → triplet. Yeh step kyun? $2 + 1 = 3$ ; do equivalent spins populations $1 : 2 : 1$ dete hain.
Shifts: $C H C l_{2}$ mein do Cl hain → bahut deshielded, $δ \approx 5.8$ . $C H_{2} C l$ mein ek Cl hai, $δ \approx 3.9$ . Yeh step kyun? Ek hi carbon par zyada electronegative atoms ⇒ zyada $δ$ .

Verify: Total H $= 1 + 2 = 3$ ; $C_{2} H_{3} C l_{3}$ mein 3 H hain ✓. Doublet integrate karta hai 2, triplet integrate karta hai 1, toh 2:1 ratio ✓.

Ex 5 — Integration trap: tert-butyl acetate (Cell E)

Worked example

$C H_{3} - C (= O) - O - C (C H_{3})_{3}$ peaks deta hai raw integral heights 27 : 9 ke saath. Inhe assign karo. Forecast: Tall peak mein zyada protons hain — kya yeh zyada downfield wala hai?

Steps

Ratio reduce karo: $27 : 9$ ko 9 se divide karo → 3 : 1... par ruko, H count karo: acetyl $C H_{3}$ mein 3 H hain; $C (C H_{3})_{3}$ mein 9 H hain. Yeh step kyun? Asli H ratio $3 : 9 = 1 : 3$ hai, toh height $27$ = 9-H group aur $9$ = 3-H group. Integration area = H ki sankhya report karta hai, position nahi.
Toh taller peak (integral 27) $C (C H_{3})_{3}$ hai, aur yeh upfield mein baithta hai $δ \approx 1.4$ par. Yeh step kyun? Yeh "tall = downfield" myth ko khatam karta hai. Height sirf H-count track karta hai.
Acetyl $C H_{3}$ (integral 9, matlab 3 H) $C = O$ ke paas hai, $δ \approx 2.0$ . Yeh step kyun? Carbonyl ke adjacent rehna mild deshielding hai.
Dono singlets hain — kisi bhi taraf koi vicinal H nahi.

Verify: H ratio: tert-butyl $9$ vs acetyl $3 = 3 : 1$ . Given heights $27 : 9 = 3 : 1$ ✓ (9-H group 3× taller hai). General "area = amount" idea ke liye dekho Integration in IR/MS.

Ex 6 — Electronegativity se shifts rank karo (Cell F)

Worked example In

$C H_{3}$ -group protons ko lowest se highest $δ$ mein order karo: $C H_{3} - C H_{3}$ , $C H_{3} - C l$ , $C H_{3} - F$ , $C H_{3} - B r$ . Forecast: Kaun sa halogen methyl ko sabse zyada downfield push karta hai?

Steps

Controlling factor yeh hai ki attached atom $C H_{3}$ protons se electron density kitni strongly kheenchta hai — yaani uski Electronegativity. Yeh step kyun? H ke around kam electron density ⇒ kam shielding ⇒ zyada $δ$ .
Electronegativity order: C $(2.5) <$ Br $(2.8) <$ Cl $(3.2) <$ F $(4.0)$ . Yeh step kyun? Yeh directly deshielding strength se map karta hai.
Toh $δ$ order (low → high): $C H_{3} C H_{3} (0.9) < C H_{3} B r (2.7) < C H_{3} C l (3.0) < C H_{3} F (4.3)$ .

Verify: Monotonic: jaise electronegativity badhti hai, $δ$ badhta hai. Pairs check karo: F $>$ Cl $>$ Br $>$ C dono electronegativity aur quoted $δ$ mein ✓.

Ex 7 — Aromatic ring current: benzene (Cell G, limiting high $δ$ )

Worked example Benzene ka single proton signal

$δ \approx 7.3$ par itna far downfield kyun hai, ek normal vinyl $= C H$ ( $δ \approx 5.5$ ) se bahut door? Forecast: Ring current — kya yeh ring H's ko shield karta hai ya deshield?

Steps

Benzene mein $π$ electrons ring ke around circulate karte hain jaise ek loop mein current, $B_{0}$ se driven (dekho Aromaticity & ring current). Yeh step kyun? Ek moving charge apna magnetic field banata hai — yeh plain electronegativity se alag extra effect hai.
Yeh ring current ek aisa field create karta hai jo ring ke bahar (jahan H's baithte hain), $B_{0}$ mein add ho jaata hai. Yeh step kyun? Induced field ring-edge protons par $B_{0}$ reinforce karta hai → extra deshielding → zyada $δ$ .
Result: aromatic H's $δ \approx 6.5$ – $8.0$ par land karte hain, isolated $π$ (vinyl) protons se zyada.

Verify: Benzene: 6 equivalent H → ek singlet, integration 6, $δ \approx 7.3$ par ✓. Yeh aldehyde/acid se kam "limiting" high- $δ$ case hai.

Ex 8 — Exchangeable OH: ethanol revisited (Cell H)

Worked example Ethanol

$C H_{3} C H_{2} O H$ mein CH₃ triplet, CH₂ quartet, aur OH... kya dikhata hai, aur D₂O ke saath kya hota hai? Forecast: Kya OH, CH₂ ko extra lines mein split karta hai? D₂O add karne se kya hota hai?

Steps

OH proton fast chemical exchange karta hai — yeh molecules ke beech NMR timescale se tez jump karta hai. Yeh step kyun? Fast exchange coupling ko average out karta hai, toh OH ek broad singlet ke roop mein dikhta hai aur apne neighbour ko cleanly split nahi karta.
Toh CH₂ ek clean quartet rehta hai (sirf CH₃ ke 3 H se coupling, $n = 3$ ), koi complex pattern nahi. Yeh step kyun? Yeh exchangeable protons ke liye degenerate "no-coupling" case hai.
D₂O add karo: OH proton deuterium (D) ke liye swap ho jaata hai, jise NMR ¹H frequency par detect nahi karta → OH peak gayab ho jaata hai. Yeh step kyun? OH/NH protons identify karne ka ek diagnostic trick.

Verify: Ethanol summary: triplet (3H) + quartet (2H) + singlet (1H) = 6 H total; $C_{2} H_{6} O$ mein 6 H hain ✓. D₂O ke baad: 5 H visible ✓.

Ex 9 — Structure from data (Cell I, real-world)

Worked example Ek compound

$C_{3} H_{6} O$ dikhata hai: singlet, integral 6, $δ \approx 2.1$ , aur kuch nahi. Yeh kya hai? Forecast: Ek signal, 6 H, carbonyl-ish shift ke paas — structure guess karo.

Steps

Accounted H count karo: ek signal mein 6 H, par formula kehti hai 6 H total → saare H equivalent hain. Yeh step kyun? Sab H ke liye ek signal ke liye high symmetry chahiye.
Molecular formula $C_{3} H_{6} O$ : degrees of unsaturation $= \frac{2 ( 3 ) + 2 - 6}{2} = 1$ → ek ring ya ek double bond. Yeh step kyun? Formula batati hai $C = O$ ya ring present hai; $δ \approx 2.1$ aur carbonyl par missing 6 H $C = O$ ki taraf point karte hain.
Do equivalent $C H_{3}$ groups ek $C = O$ ke dono taraf = acetone, $(C H_{3})_{2} C = O$ . Yeh step kyun? Dono methyls equivalent hain (singlet, 6 H), carbonyl ke paas ( $δ \approx 2.1$ ), koi vicinal H nahi → koi splitting nahi.

Verify: Acetone $C_{3} H_{6} O$ : 6 H, sab equivalent, singlet $δ \approx 2.1$ par ✓. Degrees of unsaturation $= 1$ (ek C=O) ✓.

Ex 10 — Exam twist: symmetry carbons chhupaati hai (Cell J)

Worked example

p-xylene (1,4-dimethylbenzene) mein 8 carbons hain. Kitne ¹H signals honge? Multiplicities aur integrations do. Forecast: Aath carbons — par kitne alag tarah ke protons hain?

Steps

Proton types identify karo. Do $C H_{3}$ groups symmetry se equivalent hain → ek methyl signal (6 H). Chaar aromatic H's sab equivalent hain (para-symmetry) → ek aromatic signal (4 H). Yeh step kyun? Molecular symmetry bahut atoms ko thode signals mein collapse kar deti hai — classic exam trap.
$C H_{3}$ ki multiplicity: aromatic ring carbon se attached (jisme koi H nahi) → $n = 0$ → singlet, $δ \approx 2.3$ . Yeh step kyun? Benzylic methyls ek aisi carbon ke paas hain jisme koi H nahi.
Ar–H ki multiplicity: charon equivalent hain → equivalent protons ek doosre ko split nahi karte → singlet, $δ \approx 7.0$ . Yeh step kyun? Wahi degenerate rule jaise Ex 3 mein, ab ring par.

Verify: Do signals: $C H_{3}$ (6 H, singlet, 2.3) aur Ar–H (4 H, singlet, 7.0). Total H $= 6 + 4 = 10$ ; $C_{8} H_{10}$ mein 10 H hain ✓. Ratio $6 : 4 = 3 : 2$ ✓. ¹³C NMR se compare karo, jo zyada carbon environments dikhaega.

Active Recall

Recall Answers dhako

Neopentane $C (C H_{3})_{4}$ : kitne signals? ::: Ek singlet, 12 H (Cell B/C).
$C H C l_{2} C H_{2} C l$ : kaun sa proton triplet hai? ::: $C H C l_{2}$ proton ( $n = 2$ neighbours) $δ \approx 5.8$ par.
Tert-butyl (9 H) 1.4 par vs aldehyde (1 H) 9.7 par — kaun sa peak taller hai? ::: Tert-butyl (area = H count, position nahi).
Benzene $δ \approx 7.3$ par kyun hai? ::: Ring current ring-edge H's par $B_{0}$ mein add karta hai → extra deshielding.
D₂O ek OH peak ke saath kya karta hai? ::: H ko D se exchange karta hai → OH signal gayab ho jaata hai.
$C_{3} H_{6} O$ , ek singlet 6 H at 2.1 — yeh kya hai? ::: Acetone.