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

4.8.4 · D4 · Chemistry › Spectroscopy & Analysis (Intro) › ¹H NMR — chemical shift, multiplicity (n + 1 rule), integrat

Shuru karne se pehle, teen words jo tumhare paas bilkul crisp hone chahiye (parent note se):

Chemical shift $δ$ — peak KAHAaN baith'ta hai, ppm mein, TMS ke relative $0$ par. Low $δ$ = shielded (electron-rich); high $δ$ = deshielded (electron-poor).
Integration — peak ke neeche ka AREA, equivalent protons ki SANKHYA ke proportional. Sirf ratio matter karta hai.
Multiplicity ( $n + 1$ rule) — ek proton jo adjacent carbons par $n$ equivalent neighbours rakhta hai, woh $n + 1$ lines mein split hota hai, spin–spin coupling se, aur intensities Pascal's triangle se padhte hain.

Level 1 — Recognition

Exercise 1.1

Ek molecule mein teen signals hain jinke integrals $9.0$ , $6.0$ , $3.0$ (arbitrary height units mein) measure hote hain. Smallest whole-number proton ratio kya hai?

Recall Solution

HUM KYA karte hain: integrals sirf relative areas dete hain, isliye har number ko smallest se divide karte hain taaki simplest whole-number ratio saamne aaye. KYUN: NMR absolute counts report nahi kar sakta; sirf ratios reliable hote hain, isliye hum normalise karte hain. $9.0 : 6.0 : 3.0 \div 3.0 3 : 2 : 1$ Answer: $3 : 2 : 1$ . (Un groups mein six protons bhi ho sakte hain — jaise $6 : 4 : 2$ — lekin simplest ratio $3 : 2 : 1$ hai.)

Exercise 1.2

Har proton ko "shielded (low $δ$ )" ya "deshielded (high $δ$ )" se match karo: (a) ek alkane ka $- C H_{3}$ , (b) aromatic $A r - H$ , (c) ek carboxylic acid $- C O O H$ .

Recall Solution

Rule: electron-poor surroundings electron density ko kheench lete hain, toh proton $B_{0}$ ko zyada feel karta hai → deshielded → high $δ$ . Neeche ruler figure dekho.

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

(a) $- C H_{3}$ alkane: sirf C aur H neighbours, electron-rich → shielded, $δ \approx 0.9$ .
(b) $A r - H$ : ring current strongly deshield karta hai → deshielded, $δ \approx 7$ .
(c) $- C O O H$ : extreme electron withdrawal + H-bonding → bahut zyada deshielded, $δ \approx 11$ .

Level 2 — Application

Exercise 2.1

Ek proton ke $4$ equivalent neighbouring protons hain. State karo (a) multiplicity ka naam, (b) intensity pattern.

Recall Solution

KYA/KYUN: $n = 4$ equivalent neighbours → $n + 1$ rule se signal $n + 1$ lines mein split hota hai. Relative intensities count karti hain ki kitne spin combinations har net field dete hain — yeh Pascal's triangle ki row $n$ hai, binomials $(k 4)$ . $n + 1 = 4 + 1 = 5 \Rightarrow quintet$ $(0 4) : (1 4) : (2 4) : (3 4) : (4 4) = 1 : 4 : 6 : 4 : 1$ Sum $= 2^{4} = 16$ (row total).

Exercise 2.2

Chloroethane, $C H_{3} C H_{2} C l$ ka full ¹H NMR (shift ballpark, multiplicity, integral) predict karo.

Recall Solution

Do distinct proton environments: $C H_{3}$ aur $C H_{2}$ .

$C H_{3}$ (3H): iske sirf vicinal neighbours $C H_{2}$ ke 2 protons hain → $n = 2$ → triplet, integral $3$ . Environment: plain alkyl → $δ \approx 1.5$ .
$C H_{2}$ (2H): vicinal neighbours $C H_{3}$ ke 3 protons hain → $n = 3$ → quartet, integral $2$ . Electronegative Cl se attached (Electronegativity) → deshielded → $δ \approx 3.5$ . Spectrum: triplet (3H, ~1.5 ppm) + quartet (2H, ~3.5 ppm) — classic ethyl signature.

Level 3 — Analysis

Exercise 3.1

Formula $C_{3} H_{6} O$ ka ek compound dikhata hai: ek singlet (6H, $δ \approx 2.1$ ) aur kuch nahi. Ise identify karo aur singlet explain karo.

Recall Solution

Degree of unsaturation: $\frac{2 ( 3 ) + 2 - 6}{2} = 1$ , toh ek ring ya double bond — O ke saath yeh C=O ka hint hai. Ek single 6H singlet ka matlab hai saare chhe protons equivalent hain aur unke koi vicinal H neighbours nahi hain. Do equivalent $C H_{3}$ groups, har ek kisi aisa carbon se attached jo koi H nahi rakhta, fit karta hai: acetone, $(C H_{3})_{2} C = O$ .

Singlet kyun: dono methyls ke beech ka carbonyl carbon koi proton nahi rakhta, toh $n = 0$ → $0 + 1 = 1$ line.
$δ \approx 2.1$ kyun: methyls C=O ke paas hain (mild deshielding), 2.0–2.5 window mein aate hain.

Exercise 3.2

Isopropyl chloride $(C H_{3})_{2} C H C l$ : multiplicities aur integral ratio predict karo.

Recall Solution

Do environments: chhe equivalent methyl H's aur ek akela methine (CH) H.

$(C H_{3})_{2}$ (6H): har methyl ka vicinal neighbour sirf 1 methine H hai → $n = 1$ → doublet. $δ \approx 1.5$ .
$C H$ (1H): iske vicinal neighbours 6 methyl H's hain (sab equivalent) → $n = 6$ → septet (7 lines, $1 : 6 : 15 : 20 : 15 : 6 : 1$ ). Cl se attached → deshielded → $δ \approx 4.1$ .
Integral ratio $= 6 : 1$ . Coupling ki do-taraf symmetry note karo: methyls 1 neighbour dekhte hain, CH 6 dekhta hai.

Level 4 — Synthesis

Exercise 4.1

Ek unknown $C_{2} H_{4} C l_{2}$ sirf ek singlet deta hai. Kaun sa isomer hai — 1,1-dichloroethane ya 1,2-dichloroethane — aur kyun?

Recall Solution

1,1-dichloroethane, $C H C l_{2} C H_{3}$ : do environments ( $C H C l_{2}$ aur $C H_{3}$ ), toh yeh quartet (1H) + doublet (3H) dikhayega — singlet NAHI.
1,2-dichloroethane, $C l C H_{2} C H_{2} C l$ : symmetry se saare 4 protons equivalent hain (har $C H_{2}$ identical hai). Equivalent protons ek doosre ko split nahi karte → ek singlet. Answer: 1,2-dichloroethane, singlet (4H) $δ \approx 3.7$ par (Cl se deshielded).

Exercise 4.2

Ethyl acetate, $C H_{3} - C O - O - C H_{2} - C H_{3}$ ka ¹H NMR fingerprint design karo: har signal shift, multiplicity, aur integral ke saath list karo.

Recall Solution

Teen distinct environments (acetyl $C H_{3}$ isolated hai; ethyl $C H_{2}$ aur $C H_{3}$ ek doosre se couple karte hain):

Group	Neighbours ( $n$ )	Multiplicity	Integral	$δ$ (ppm)	Shift kyun
$C H_{3} - C O -$ (acetyl)	0	singlet	3	~2.0	C=O ke paas, koi vicinal H nahi
$- O - C H_{2} -$	3 (ethyl $C H_{3}$ ke)	quartet	2	~4.1	O se attached → strongly deshielded
$- C H_{2} - C H_{3}$	2 ( $C H_{2}$ ke)	triplet	3	~1.3	plain alkyl
Fingerprint: singlet(3H, 2.0) + quartet(2H, 4.1) + triplet(3H, 1.3). Isolated acetyl singlet tell-tale hai; quartet+triplet ethyl pair hai.

Level 5 — Mastery

Exercise 5.1

Ek neutral compound $C_{8} H_{8} O$ exactly deta hai: ek singlet (3H, $δ \approx 2.6$ ), aur aromatic signals jo 5H integrate karte hain $δ \approx 7.5$ – $8.0$ ke paas. Structure deduce karo aur har signal justify karo.

Recall Solution

Step 1 — degrees of unsaturation: $\frac{2 ( 8 ) + 2 - 8}{2} = 5$ . Paanch degrees = benzene ring ka strong hint (4: teen C=C + ek ring) aur ek aur (ek C=O). Step 2 — integrals padhna: total observed H $= 3 + 5 = 8$ , $C_{8} H_{8} O$ se match karta hai. 5H aromatic block ka matlab hai mono-substituted benzene ( $C_{6} H_{5} -$ , paanch ring H's). Step 3 — 3H singlet at 2.6: teen equivalent H, koi vicinal neighbours nahi ( $n = 0$ → singlet), mildly deshielded → ek $C H_{3}$ kisi aise carbon se attached jo koi H nahi rakhta. C=O ke paas shift fit karta hai. Assemble: $C_{6} H_{5} - + - C O - + - C H_{3} = C_{6} H_{5} C O C H_{3}$ = acetophenone.

Aromatic 5H, $δ \approx 7.5$ – $8.0$ : ring current deshields karta hai; ortho H's carbonyl se aur downfield push hote hain.
$C H_{3}$ 3H singlet, $δ \approx 2.6$ : isolated methyl C=O ke adjacent.

Exercise 5.2

Do isomers $C_{3} H_{8} O$ : 1-propanol $C H_{3} C H_{2} C H_{2} O H$ aur 2-propanol $(C H_{3})_{2} C H O H$ . Sirf multiplicity aur integration use karke (exchangeable OH ignore karo), decide karo kaunsa spectrum kiska hai.

Recall Solution

OH ignore karo (broad, exchangeable — neeche trap dekho). 1-propanol $C H_{3} C H_{2} C H_{2} O H$ — teen carbon environments:

$C H_{3}$ (3H): 2 vicinal H → triplet.
middle $C H_{2}$ (2H): 3 + 2 = 5 non-equivalent vicinal H → ek complex multiplet (sextet-like).
$O - C H_{2}$ (2H): 2 vicinal H → triplet, deshielded ( $δ \approx 3.6$ ). Carbon-H integral pattern (OH exclude karke) $= 3 : 2 : 2$ . 2-propanol $(C H_{3})_{2} C H O H$ — do carbon environments:
$(C H_{3})_{2}$ (6H): 1 vicinal H (CH) → doublet.
$C H$ (1H): 6 vicinal H → septet, deshielded ( $δ \approx 4.0$ ). Carbon-H integral pattern (OH exclude karke) $= 6 : 1$ . Decision: jo spectrum 6H doublet + 1H septet dikhata hai woh 2-propanol hai; jo spectrum do triplets ek central multiplet ke around (3:2:2) dikhata hai woh 1-propanol hai.

Active Recall

Recall Rapid-fire (answers cover karo)

Integral heights $12 : 8 : 4$ simplest ratio? ::: $3 : 2 : 1$ 6 equivalent neighbours wala proton → kitni lines? ::: 7 (septet), $1 : 6 : 15 : 20 : 15 : 6 : 1$ Acetone mein kitne ¹H signals hote hain? ::: ek singlet (6H) Ethyl acetate acetyl $C H_{3}$ multiplicity? ::: singlet Kaun sa $C_{2} H_{4} C l_{2}$ isomer lone singlet hai? ::: 1,2-dichloroethane OH peak ko kaise gayab karein? ::: $D_{2} O$ se shake karo (H/D exchange) 6H doublet + 1H septet kaun sa alcohol identify karta hai? ::: 2-propanol

Recall Ek nazar mein cheat map

KAHAN (ppm) = environment · AREA = kitne H · LINES = $n + 1$ neighbours · intensities = Pascal's triangle row.