¹³C NMR (overview); DEPT — Hinglish notes

4.8.5 · Chemistry › Spectroscopy & Analysis (Intro)

¹³C NMR kaam kyun karta hai?

Ye problem kyun hai? Sensitivity ki wajah se. Kam abundance (1.1%) ko ek chhote magnetic moment (gyromagnetic ratio $γ_{C} \approx \frac{1}{4} γ_{H}$ ) ke saath combine karo, toh ¹³C signal roughly ~5700× weak hai ¹H se, per nucleus. Isliye humein chahiye:

Bahut saare scans (signal averaging) — signal $N$ ke saath badhta hai, noise $N$ ke saath, toh S/N $\propto N$ .
Fourier-transform (pulsed) NMR.

¹³C spectrum kaisa dikhta hai

Rough chemical-shift map (ppm)

Carbon type	δ (ppm)
Alkyl C–H ( $s p^{3}$ )	5–50
C–N, C–O ( $s p^{3}$ , electronegative ke paas)	40–80
Alkyne ( $s p$ )	65–90
Alkene / aromatic ( $s p^{2}$ )	100–150
C=O ester/acid/amide	160–185
C=O ketone/aldehyde	190–220

DEPT — carbons ko attached H ke hisaab se sort karna

Ye kaam kaise karta hai (conceptually): Magnetization H → C mein transfer hoti hai, phir $1/ J_{C H}$ ke liye tune kiye gaye ek time ke liye evolve hoti hai. Final-pulse angle $θ$ har multiplicity ki sign control karta hai:

Formula DEPT signal sign vs

$θ$ $n$ protons wale carbon ki intensity roughly aise follow karti hai $I_{C H_{n}} \propto sin θ cos^{n - 1} θ$ Quaternary carbons (n=0) koi polarization transfer nahi lete ⇒ DEPT mein hamesha absent hote hain.

Teen standard angles par evaluate karo:

	$θ = 45°$	$θ = 90°$	$θ = 135°$
CH ( $n = 1$ ): $sin θ$	+	+	+
CH₂ ( $n = 2$ ): $sin θ cos θ$	+	0	−
CH₃ ( $n = 3$ ): $sin θ cos^{2} θ$	+	0	+

Ise padhna (80/20 rule): Zyaatar log sirf DEPT-135 run karte hain:

Upar (positive): CH aur CH₃
Neeche (negative): CH₂
Absent: quaternary C (aur C=O, etc.)

Normal ¹³C ke saath combine karo (jo sab carbons dikhata hai): koi bhi peak jo ¹³C mein present hai lekin DEPT mein missing hai wo quaternary hai.

Worked example 1 — Ethanol, CH₃CH₂OH

Step 1 — Unique carbons gino. Do: CH₃ aur CH₂. Kyun? Koi symmetry unhe equivalent nahi banati. ⇒ ¹³C mein 2 peaks.

Step 2 — Shifts assign karo. CH₂ O se bonded hai (electronegative) ⇒ zyada deshielded, ~58 ppm. CH₃ ~18 ppm. Kyun? Oxygen adjacent carbon se electron density withdraw karta hai.

Step 3 — DEPT-135. CH₃ → upar; CH₂ → neeche. Kyun? 135° par $sin θ cos^{n - 1} θ$ se: $n = 3$ gives +, $n = 2$ gives −. (OH mein koi carbon nahi, toh koi peak nahi.)

Worked example 2 — Acetone, (CH₃)₂C=O

Step 1 — Unique carbons gino. Do CH₃ groups symmetry se equivalent hain ⇒ wo ek peak dete hain. C=O ek aur hai. ⇒ 2 peaks. Ye step kyun? Symmetry-equivalent atoms NMR ke liye indistinguishable hote hain.

Step 2 — Shifts. Carbonyl C ~206 ppm (bahut deshielded ketone). CH₃ ~30 ppm.

Step 3 — DEPT-135. CH₃ → upar. C=O quaternary (n=0) hai → absent. Kyun? Koi attached H nahi ⇒ koi polarization transfer nahi.

Conclusion: ~206 ppm ke paas ek peak jo DEPT mein gayab ho jaati hai ek ketone carbonyl confirm karti hai.

Worked example 3 — Para-xylene (1,4-dimethylbenzene)

Step 1 — Symmetry dhundo, phir environments carefully gino. Para-xylene ka ek $C_{2}$ axis hai (aur mirror planes) jo do CH₃ groups ko equivalent banata hai aur do ring-substituted (ipso) carbons ko equivalent banata hai. Lekin chaar un-substituted aromatic CH carbons sab equivalent NAHI hain: symmetry unhe do pairs mein map karti hai.

Do CH₃ groups → 1 environment.
Do ipso carbons (ring C jo CH₃ carry karta hai, quaternary) → 1 environment.
Chaar aromatic C–H carbons do equivalent pairs mein split hote hain → 2 environments.

⇒ 4 ¹³C signals, 3 nahi. Ye step kyun? Equivalence ke liye ek symmetry operation chahiye jo ek carbon ko exactly doosre par map kare; para-xylene mein har aromatic CH aise operation se related nahi hota, isliye wo do distinct sets mein aate hain.

Step 2 — Shifts. CH₃ ~21 ppm; do aromatic CH environments ~126–129 ppm (ek doosre ke paas); ipso quaternary ~134 ppm.

Step 3 — DEPT-135. CH₃ → up (~21); dono aromatic CH environments → up (~126–129); ipso carbon absent (quaternary). Kyun? DEPT turant us ek carbon (ipso) ko flag karta hai jo substituent carry karta hai kyunki uske paas koi H nahi.

Recall Feynman: ek 12-saal ke bachche ko samjhao

Socho tumhara molecule ek tinker-toy hai jo carbon balls se bana hai aur sticks se connected hai. ¹³C NMR ek special camera hai jo har alag carbon ball ko light up karta hai taaki tum unhe count kar sako. Catch ye hai: sirf lagbhag 100 mein 1 carbon ball "glowing" wali hoti hai, toh camera ko hazaaron photos lene padte hain aur unhe add karna padta hai ek clear picture ke liye. Phir DEPT ek clever filter hai: ye count karta hai ki har carbon ball par kitne chhote hydrogen beads chipke hain, aur unhe code mein paint karta hai — 2 hydrogens wali balls neeche point karti hain, 1 ya 3 hydrogens wali balls upar point karti hain, aur ganje balls jinke paas koi hydrogen nahi dikhti hi nahi.

Flashcards

¹²C NMR mein invisible kyun hai lekin ¹³C active kyun hai?

¹²C ka nuclear spin

I = 0

hai (koi magnetic moment nahi); ¹³C ka

I = \frac{1}{2}

hai.

¹³C ki natural abundance kya hai, aur ye kyun matter karta hai?

1.1% — ¹³C NMR ko insensitive banata hai (bahut saare scans chahiye) lekin ¹³C–¹³C coupling bhi negligible ho jaati hai.

¹³C NMR ka approximate chemical-shift range kya hai?

~0–220 ppm (¹H ke liye ~0–12 ppm ke comparison mein).

Routine ¹³C spectra mein har carbon singlet kyun dikhta hai?

Broadband proton decoupling C–H coupling hatata hai (aur NOE intensity boost karta hai).

Kya routine ¹³C peak heights quantitative hoti hain?

Nahi — ye

T_{1}

relaxation aur uneven NOE par depend karti hain; quaternary carbons often weak hote hain.

DEPT ka full form kya hai?

Distortionless Enhancement by Polarization Transfer.

DEPT-135 mein kaun se carbons upar, neeche point karte hain, aur kaun absent hain?

Up: CH & CH₃; Down: CH₂; Absent: quaternary C.

Sab DEPT spectra mein quaternary carbons absent kyun hote hain?

Unke paas koi attached H nahi hota, isliye koi H→C polarization transfer nahi hota.

DEPT use karke CH aur CH₃ mein distinguish kaise karte hain?

DEPT-90 run karo: sirf CH bachta hai; CH₃ (aur CH₂)

θ = 90°

par gayab ho jaate hain.

Pulse angle θ ke against

C H_{n}

carbon ke DEPT signal intensity ka formula kya hai?

I \propto sin θ cos^{n - 1} θ

Normal ¹³C mein present lekin DEPT-135 mein missing peak kaun se type ka carbon hai?

Quaternary (koi attached hydrogen nahi), jaise carbonyl ya ipso aromatic C.

Acetone ke ¹³C signals ki number kitni hai aur kyun?

2 — do CH₃ symmetry se equivalent hain; plus C=O.

Para-xylene kitne ¹³C signals deta hai aur kyun?

4 — CH₃, ipso quaternary, aur DO non-equivalent aromatic CH environments (chaar ring CH's do pairs mein split hoti hain).

Connections

¹H NMR (overview) — complementary hydrogen skeleton, wahan integration quantitative hai.
Chemical Shift & Shielding — same deshielding logic, ¹³C ke liye bada range.
Spin–Spin Coupling (J) — jo decoupling hatata hai.
Nuclear Overhauser Effect (NOE) — intensity enhancement & polarization transfer.
IR Spectroscopy — ¹³C ke saath C=O confirm karne ke liye pair karta hai.
Mass Spectrometry — molecular formula / degrees of unsaturation deta hai carbon count support karne ke liye.
Symmetry & Equivalent Nuclei — kyun peak counts < total carbons.