NMR Spectroscopy Short Notes, Solved Examples And Practice Questions Pdf Download Now

 

☢  Chapter 15  |  Klein — Organic Chemistry

Nuclear Magnetic Resonance
Spectroscopy

Complete Study Notes · Less Theory, More Practice · All Examples Included

JEE Advanced NEET IIT-JAM BITSAT GATE CSIR-NET TGT / PGT

📋 Quick Index

1. Why NMR? · Basic Principle · Nuclear Spin
2. Acquiring a Spectrum · FT-NMR · TMS · Deuterated Solvents
3. Number of Signals · Chemical Equivalence · Replacement Test
4. Chemical Shift (δ) · Inductive Effects · Anisotropic Effects · Table 15.2
5. Integration · Step Curves · Worked Examples
6. Multiplicity · n+1 Rule · Coupling Constant J · Complex Splitting
7. Drawing & Analyzing ¹H NMR Spectra · SkillBuilders
8. ¹³C NMR · DEPT · Table 15.4 & 15.5
9. Exam Tips, Tricks & High-Value MCQ Patterns

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01Why NMR? — The Most Powerful Structure Tool

NMR spectroscopy is the #1 technique for determining the carbon-hydrogen framework of organic molecules — often without needing any other data.

• Tells you exactly HOW C and H atoms are connected in a molecule.
• Structure can often be determined from NMR alone (combined with IR + MS in practice).
• Nuclei studied: ¹H, ¹³C, ¹⁵N, ¹⁹F, ³¹P — exam focus is ¹H and ¹³C.
• Works because nuclei with an odd number of protons and/or neutrons possess nuclear spin (a quantum property).
• ¹²C has even protons + even neutrons → no spin, invisible to NMR. ¹³C has odd neutrons → has spin. ✓

The Physics in 3 Lines

• Spinning proton → tiny magnet (magnetic moment).
• Place it in external field B₀ → aligns with (α-state, low energy) or against (β-state, high energy) the field.
• Hit it with radio-frequency (rf) radiation matching ΔE → nucleus flips to β-state → RESONANCE (absorption detected).

⚡ Why Don't All Protons Absorb the Same Frequency?

Diamagnetism! Circulating electron density around each proton creates a tiny induced field opposing B₀. So each chemically distinct proton is shielded differently → absorbs at a unique frequency. Without this, NMR would be useless — all protons would give one signal!

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02Acquiring a ¹H NMR Spectrum

Field Strength & Operating Frequency

• Stronger field → bigger ΔE → better separation of signals.
• 1.41 T → 60 MHz  |  7.04 T → 300 MHz  |  11.7 T → 500 MHz
• Higher frequency = better — signals less likely to overlap (J values look smaller relative to ppm scale).

FT-NMR vs. Old CW (Continuous Wave)

• Old CW: sweeps through frequencies one by one — slow and inefficient.
• Modern FT-NMR: single short rf PULSE excites ALL protons at once → they relax → detector records Free Induction Decay (FID) → Fourier transform → spectrum.
• FIDs can be averaged (hundreds in minutes) → essential for ¹³C NMR sensitivity.

Sample Preparation — Key Points

• Dissolved in deuterated solvent — no ¹H signals from solvent contaminating spectrum.
• Common solvents: CDCl₃, CD₂Cl₂, CD₃CN, C₆D₆, D₂O
• Small amount of TMS [Si(CH₃)₄] added as reference → defined as 0 ppm.
• Why TMS? Its 12 equivalent protons → ONE sharp singlet; Si less electronegative than C → protons very shielded → signal at 0 ppm (upfield of almost all organic signals).

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03Number of Signals & Chemical Equivalence

Each UNIQUE electronic environment = ONE signal. Your first job: count distinct proton types.

Three Types of Proton Relationships

Relationship	How to Identify	NMR Result
Homotopic	Interchangeable by ROTATIONAL symmetry (axis). Replacement test → same compound.	ONE signal (equivalent)
Enantiotopic	Interchangeable by REFLECTIONAL symmetry (plane). Replacement test → enantiomers.	ONE signal in achiral solvent
Diastereotopic	Cannot be interchanged by ANY symmetry. Replacement test → diastereomers. Common near chiral centers.	DIFFERENT signals (not equivalent)

🔑 Replacement Test — Step by Step

Replace each of the two H atoms with a D atom (one at a time). Draw both structures and compare:
Same compound → Homotopic (equivalent)  |  Enantiomers → Enantiotopic (equivalent in achiral solvent)  |  Diastereomers → Diastereotopic (NOT equivalent)

⚡ Quick Rules for Counting Signals

• All 3 protons of a CH₃ group → ALWAYS equivalent → 1 signal.
• Two protons of a CH₂ group → equivalent IF no chiral center in the molecule.
• Two protons of a CH₂ group → NOT equivalent (diastereotopic) IF chiral center present.
• Two CH₂ groups = equivalent to each other if interchangeable by symmetry (give 4H, 1 signal).
• Aromatic ring: draw resonance hybrid; look for mirror planes to find equivalent positions.

✅ Worked Example — Section 15.4 (Klein)

4-Methoxymethylbenzene analogue — How many ¹H signals?

• OCH₃ group: 3H all on same C → equivalent → 1 signal
• Two CH₃ groups (symmetric): interchangeable → 6H → 1 signal
• Two CH₂ groups (symmetric): interchangeable → 4H → 1 signal
• Two equivalent aromatic H → 1 signal
• One unique aromatic H → 1 signal
• Total: 5 signals ✓

⚠️ Variable-Temperature NMR Trap

Cyclohexane shows ONE ¹H signal at room temperature (ring flip is faster than NMR timescale → axial/equatorial protons average out). Cool to −100°C → ring flip slows → TWO separate signals for axial and equatorial protons. Classic exam concept for CSIR-NET/GATE!

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04Chemical Shift (δ) — Location of the Signal

📐 Formula

δ (ppm) = [observed shift from TMS in Hz] ÷ [operating frequency in Hz]
δ is dimensionless and constant regardless of spectrometer frequency — that is why ppm is used!

• Left of spectrum = DOWNFIELD (deshielded protons, higher frequency, higher δ).
• Right of spectrum = UPFIELD (shielded protons, lower frequency, lower δ).
• Example: Benzene at 300 MHz absorbs 2181 Hz above TMS → δ = 2181 ÷ 300,000,000 = 7.27 ppm. Same result at 60 MHz (436 Hz ÷ 60,000,000 = 7.27 ppm). ✓

Inductive Effects (Electronegativity Deshielding)

• Electronegative atom pulls electrons away → proton is deshielded → signal moves DOWNFIELD (higher ppm).
• Strength: F > Cl > Br > I (matches electronegativity order).
• Effect is ADDITIVE (more halogens = more shift) and TAPERS OFF with distance (α >> β >> γ ≈ 0).

Compound	CH₄	CH₃I	CH₃Br	CH₃Cl	CH₃F
δ (ppm)	1.0	2.2	2.7	3.1	4.3

Distance effect in 1-chloropropane Cl–CH₂–CH₂–CH₃:

• α-CH₂ (next to Cl): 3.3 ppm — large effect
• β-CH₂: 1.6 ppm — small effect
• γ-CH₃: 0.9 ppm — negligible effect

Anisotropic Effects (π-Electron Deshielding)

• Aromatic ring: π electrons circulate in B₀ → protons outside ring are deshielded → ~7 ppm. (Inner protons of annulenes are shielded — e.g., [14]annulene inner H at −1 ppm!)
• Aldehyde C=O: aldehydic H appears at ~10 ppm — very diagnostic.
• Alkene vinylic H: 4.5–6.5 ppm.
• Alkyne terminal H: ~2.5 ppm — surprisingly upfield! H sits along the shielded cylinder axis of C≡C.
• Carboxylic acid O–H: ~12 ppm — most downfield common proton.

📊 TABLE 15.2 — Chemical Shift Reference (Memorize!)

Type of Proton	Group	δ (ppm)
Methyl	R–CH₃	~0.9
Methylene	R–CH₂–R	~1.2
Methine	R₃C–H	~1.7
Allylic	C=C–CH₂–	~2.0
Alkynyl (terminal)	R–C≡C–H	~2.5
Aromatic methyl	Ar–CH₃	~2.5
Alkyl halide	R–CHX–	2–4
Alcohol O–H	R–OH	2–5
Vinylic	C=C–H	4.5–6.5
Aryl (aromatic)	Ar–H	6.5–8
Aldehyde	R–CHO	~10
Carboxylic acid	R–COOH	~12

TABLE 15.1 — Benchmark Offsets for Predicting δ

Neighboring Group	Effect on α-Protons	Effect on β-Protons
–OH or –OR (alcohol/ether)	+2.5 ppm	+0.5 ppm
Ester oxygen (–O–C=O)	+3.0 ppm	+0.6 ppm
Carbonyl C=O (ketone/aldehyde/ester)	+1.0 ppm	+0.2 ppm

✅ SkillBuilder 15.3 — Predict Chemical Shifts (Klein p.698)

Compound: isobutyl propanoate analogue (ester with OCH₂ and isopropyl)

• OCH₂ (α to ester O): 1.2 + 3.0 = 4.2 ppm  (actual ~4.1 ppm ✓)
• CH₃ of ethyl (β to O): 0.9 + 0.2 = 1.1 ppm  (actual ~1.1 ppm ✓)
• CH methine (α to C=O): 1.7 + 1.0 = 2.7 ppm  (actual ~2.6 ppm ✓)
• Two equivalent CH₃ (β to O): 0.9 + 0.6 = 1.5 ppm  (actual ~3.3 for OMe ✓)

⚡ Exam Trick — Predicting δ

Always start from benchmark (CH₃ ≈ 0.9 | CH₂ ≈ 1.2 | CH ≈ 1.7 ppm), then ADD the offset for each nearby functional group. β-group effect ≈ 1/5 of α-group effect.

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05Integration — How Many Protons per Signal?

Integration = area under each signal = RELATIVE number of protons giving rise to that signal. Numbers only have meaning compared to each other.

Steps to Read Integration

1. Write down all raw integration numbers printed under each signal.
2. Divide ALL values by the SMALLEST value → get ratios.
3. Multiply by a whole-number factor so ratios sum to total H from molecular formula.
4. If ratio = 2:3, could be 2H:3H or 4H:6H — the molecular formula tells you which!

✅ SkillBuilder 15.4 — Klein p.703

Molecular formula: C₅H₁₀O₂ → 10 total H

Raw integration values: 6.33 | 19.4 | 37.9

1. Divide by smallest (6.33):   1.00  |  3.06  |  5.99
2. Ratio ≈ 1 : 3 : 6   →   Sum = 10 = total H ✓
3. Exact counts: 1H, 3H, 6H

✅ Symmetry Trap — 3-Pentanone C₅H₁₀O (Klein p.703)

Raw integration: 32.5 | 48.0 → ratio 2:3

Total H = 10 → must be 4H and 6H (not 2H and 3H). The molecule has symmetry — both CH₂ groups equivalent and both CH₃ groups equivalent → signal count doubled!

⚡ Key Point

Integration gives RELATIVE ratios, not absolute numbers. Always use molecular formula to pin down the exact proton count per signal.

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06Multiplicity — Shape of the Signal

Multiplicity = number of peaks in a signal. Caused by spin-spin splitting (coupling) from neighboring non-equivalent protons.

The n+1 Rule

• n = number of neighboring (non-equivalent) protons.
• Multiplicity = n + 1
• TWO conditions required: (1) neighbors must be NON-EQUIVALENT; (2) separated by ≤ 3 bonds.

📊 TABLE 15.3 — Multiplicity Table (Must Memorize!)

n (Neighbors)	Name	Peak Intensity Ratios
0	Singlet (s)	1
1	Doublet (d)	1 : 1
2	Triplet (t)	1 : 2 : 1
3	Quartet (q)	1 : 3 : 3 : 1
4	Quintet	1 : 4 : 6 : 4 : 1
5	Sextet	1 : 5 : 10 : 10 : 5 : 1
6	Septet	1 : 6 : 15 : 20 : 15 : 6 : 1

Two Rules That STOP Splitting

• RULE 1: Equivalent protons do NOT split each other. (e.g., all 4H in ClCH₂CH₂Cl are equivalent → one singlet only.)
• RULE 2: Protons >3 bonds apart generally do not split. (Long-range coupling only in rigid/allylic systems — ignore for most exams.)

Coupling Constant J

• J = distance in Hz between peaks within a split signal.
• Same J for BOTH coupled partners — Ha and Hb always show equal J.
• J range: 0–20 Hz. Typical vicinal (3-bond) J: ~7 Hz.
• J does NOT change with spectrometer frequency → higher field gives better resolved spectra.

Pattern Recognition — Know These!

Group	Signal Pattern	Integration
Ethyl (–CH₂CH₃)	quartet (2H) + triplet (3H) — same J!	2H : 3H
Isopropyl [–CH(CH₃)₂]	septet (1H) + doublet (6H) — same J!	1H : 6H
tert-Butyl [–C(CH₃)₃]	singlet (9H) — quaternary C, no neighbors!	9H singlet

Complex Splitting

• Occurs when a proton has two different types of neighbors with very different J values.
• Example: Hb with 1 Ha (Jab large) and 2 Hc (Jbc small) → doublet of triplets (6 peaks = 2×3).
• When Jab ≈ Jbc (~7 Hz, all single bonds, free rotation): just use n+1 with TOTAL neighbors.
• 1-nitropropane CH₃–CH₂–CH₂–NO₂: middle CH₂ has 3+2=5 neighbors, J values equal → sextet. ✓

Special Cases — Protons With No Observable Coupling

• –OH proton: usually a SINGLET. Rapid proton exchange (catalyzed by trace acid/base) averages J to zero. D₂O shake → OH signal disappears.
• Aldehyde CHO: appears as SINGLET ~10 ppm even though CH₂ neighbors exist — Jaldehyde–CH is tiny (~1–3 Hz), below resolution.
• –COOH, –NH: also exchangeable (labile); often broad singlets.

⚡ D₂O Shake Test

Add D₂O → exchangeable protons (OH, NH, COOH) swap for D → those signals vanish from spectrum. Classic exam question to identify exchangeable protons!

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07Drawing & Analyzing ¹H NMR Spectra

DRAWING a Spectrum — 5-Step Method (SkillBuilder 15.6)

1. Identify all distinct proton types → number of signals.
2. Predict chemical shift (δ) using benchmarks + Table 15.1 offsets.
3. Count protons giving each signal → integration.
4. Apply n+1 rule → multiplicity (singlet/doublet/triplet...).
5. Draw each signal at correct δ with correct peak shape.

✅ SkillBuilder 15.6 — Isopropyl Acetate CH₃COOCH(CH₃)₂ (Klein p.714)

#	Proton Group	δ (ppm)	Integration	Multiplicity
1	CH₃CO– (acetyl methyl)	~1.9	3H	Singlet (0 neighbors — adjacent C has no H)
2	–CH(CH₃)₂ (methine)	~4.7	1H	Septet (6 neighbors = two CH₃)
3	–CH(CH₃)₂ (two equivalent methyls)	~1.5	6H	Doublet (1 neighbor = CH)

ANALYZING an Unknown Spectrum — 5-Step Method (SkillBuilder 15.8)

1. Calculate HDI from molecular formula. HDI ≥ 4 → likely aromatic ring. HDI = 1 → one ring OR one double bond.
2. Count signals + check integration (look for symmetry clues — large integration may indicate symmetry).
3. Analyze each signal: δ = electronic environment | multiplicity = neighbors | integration = proton count.
4. Build fragments from each signal (puzzle pieces).
5. Assemble fragments → verify against ALL spectral data.

✅ SkillBuilder 15.8 — C₉H₁₀O (Klein p.718)

Full Worked Analysis

1. HDI: [2(9)+2−10]/2 = 5 → aromatic ring (HDI=4) + one C=O (HDI=1).
2. Signals & integration: 4 signals with ratio 1 : 5 : 2 : 2.
3. Signal at ~10 ppm, singlet, 1H → ALDEHYDE (CHO).
   Signal just above 7 ppm, multiplet, 5H → MONOSUBSTITUTED BENZENE RING (5 aromatic H).
   Two triplets at ~2.4 and ~3 ppm, each 2H → two adjacent CH₂ groups coupling each other.
4. Fragments: Ph–  +  –CH₂CH₂–  +  –CHO
5. Assembly: Ph–CH₂–CH₂–CHO (3-phenylpropanal) ✓
   Verify: monosubstituted Ph ✓, two triplets with same J ✓, aldehyde singlet at 10 ppm ✓

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08¹³C NMR Spectroscopy

Key Differences vs. ¹H NMR

• Only 1.1% of C atoms are ¹³C — low natural abundance, needs signal averaging.
• Only CHEMICAL SHIFT reported (range: 0–220 ppm). No routine integration or multiplicity.
• Broadband decoupling: all ¹³C signals collapse to SINGLETS by continuous ¹H irradiation → removes ¹³C–¹H coupling entirely.
• ¹³C–¹³C splitting: negligible (adjacent ¹³C atoms statistically very rare).
• Number of signals = number of UNIQUE carbon environments (accounting for symmetry).

Chemical Shift Regions (Figure 15.22)

Carbon Type	Hybridization	δ Range (ppm)
Alkanes (methyl/methylene)	sp³	10–60
C–O (alcohols, ethers)	sp³	40–80
C–X (halides)	sp³	0–80
Alkynes (C≡C)	sp	65–90
Alkenes (C=C)	sp²	100–150
Aromatic C	sp²	110–170
C=O (esters, acids, amides)	sp²	165–185
C=O (aldehydes, ketones)	sp²	185–220

DEPT ¹³C NMR

DEPT = Distortionless Enhancement by Polarization Transfer. Reveals how many H atoms are attached to each carbon — information lost in broadband decoupling.

📊 TABLE 15.5 — DEPT Signal Patterns (Must Know!)

Spectrum	CH₃	CH₂	CH (methine)	Quat. C (no H)
Broadband-decoupled	✅ Present	✅ Present	✅ Present	✅ Present
DEPT-90	❌ Absent	❌ Absent	✅ Positive (+)	❌ Absent
DEPT-135	✅ Positive (+)	🔻 Negative (–)	✅ Positive (+)	❌ Absent

✅ SkillBuilder 15.10 — DEPT Analysis (Klein p.725)

Formula: C₄H₁₀O (alcohol, HDI = 0) — 3 signals in broadband ¹³C (4 C atoms → 1 signal = 2 equivalent C)

• ~69 ppm: negative in DEPT-135 → CH₂ attached to O (downfield shifted).
• ~30 ppm: positive in ALL spectra including DEPT-90 → CH (methine).
• ~19 ppm: positive in broadband + DEPT-135, absent in DEPT-90 → CH₃.
• Only 3 signals for 4 C → CH₃ signal represents 2 equivalent CH₃ groups.
• Structure: (CH₃)₂CH–CH₂OH → 2-methyl-1-propanol ✓

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09Exam Tips, Tricks & High-Value Patterns

🎯 12 High-Priority Points for All Exams

1. HDI First — Always

Calculate HDI before anything else. HDI ≥ 4 → aromatic ring likely. HDI = 1 → one ring OR one double bond. HDI = 0 → no rings, no π bonds.

2. Replacement Test (Favourite in CSIR-NET, GATE, IIT-JAM)

Diastereotopic = chiral center nearby = DIFFERENT signals. Replace both H with D one at a time — if products are diastereomers → diastereotopic → inequivalent.

3. Aldehyde Singlet Near 10 ppm

Signal at ~10 ppm, singlet, 1H = ALDEHYDE (CHO). J is tiny even though neighbors exist. Instant identification in MCQs.

4. OH / NH / COOH Protons

D₂O shake → exchangeable protons disappear. OH broad singlet 2–5 ppm. COOH ~12 ppm. All appear as singlets due to rapid exchange.

5. Ethyl Group Fingerprint

Quartet (2H) + triplet (3H) with same J value. Instantly recognizable. Both signals must have identical J.

6. Isopropyl Group Fingerprint

Septet (1H, usually shown as inset — it's tiny!) + doublet (6H) with same J. If you see a large 6H doublet upfield, look for a small septet.

7. tert-Butyl Fingerprint

Large singlet integrating for 9H (three equivalent CH₃). No neighbors because connected to quaternary C.

8. Equivalent Protons Don't Split

1,2-Dichloroethane: all 4H equivalent → ONE singlet. This is one of the most common wrong-answer traps.

9. Higher Field = Better Resolution

300 MHz vs. 60 MHz: J stays the same in Hz but ppm scale is 5× bigger → peaks less likely to overlap. 300/500 MHz preferred for research.

10. ¹³C Signal Count

Count UNIQUE carbons after applying symmetry. Symmetric molecule → fewer signals than C atoms. Example: 3-pentanone has 3 ¹³C signals for 5 carbons (symmetry).

11. DEPT Trick in MCQs

Negative in DEPT-135 = CH₂.  Absent from both DEPTs = quaternary C.  Positive in DEPT-90 = CH.  Positive in broadband + DEPT-135, absent in DEPT-90 = CH₃.

12. Variable-Temperature NMR

Cyclohexane: ONE signal at RT (fast ring flip) → TWO signals at −100°C (slow ring flip). Rate of dynamic processes can be measured this way.

⚡ Quick-Fire MCQ Patterns

Q: A compound C₅H₁₂ shows only ONE ¹H NMR signal.

→ Neopentane [C(CH₃)₄] — all 12H equivalent (homotopic).

Q: A spectrum shows only two doublets.

→ Two non-equivalent CH groups coupling only with each other (e.g., vinyl protons Hₐ and H_b of a disubstituted alkene).

Q: Signal at ~10 ppm (s, 1H) + quartet (2H) + triplet (3H) in same spectrum.

→ Ethyl ester with adjacent methylene (like propionaldehyde or ethyl formate) — aldehyde + ethyl group pattern.

Q: DEPT-135 shows a negative peak at 35 ppm.

→ CH₂ group at 35 ppm (only CH₂ gives negative signal in DEPT-135).

Q: ¹³C shows 3 signals for a compound with 6 carbon atoms.

→ Molecule has symmetry such that only 3 unique C environments exist (e.g., 3-hexanone or a symmetric diester).

Q: Compound shows 9H singlet + 1H singlet.

→ tert-Butyl group (9H, s) plus an isolated CH proton with no vicinal H [e.g., (CH₃)₃C–CHO].

Q: No signal in DEPT-90 but signal present in broadband ¹³C.

→ Quaternary carbon (no H attached) — invisible in all DEPT spectra.

Q: Signal at δ 4.1 ppm, 2H, triplet in ¹H NMR.

→ –OCH₂CH₃ of an ester: OCH₂ is α to ester O (+3 ppm from 1.2 ppm benchmark ≈ 4.2 ppm) split into triplet by adjacent CH₃ (n=2 neighbors, n+1=3).

Q: Which is more downfield — aromatic H or vinylic H?

→ Aromatic H (~7 ppm) > vinylic H (4.5–6.5 ppm) due to stronger ring current deshielding.

Q: Compound C₁₂H₂₄ gives 1 ¹H signal and 2 ¹³C signals.

→ Cyclododecane (cyclic, one type of CH₂ in ¹H, but ring carbons in two orientations give 2 ¹³C signals) — or similar symmetric cyclic molecule.

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📌 One-Page Summary — Four Characteristics of Every ¹H Signal

Characteristic	What It Tells You	How to Determine It
Number of Signals	Number of chemically distinct proton environments	Count unique H types using symmetry rules + replacement test
Chemical Shift (δ)	Electronic environment — shielded vs. deshielded	Benchmarks (0.9 / 1.2 / 1.7 ppm) + Table 15.1 offsets + Table 15.2
Integration (area)	Relative number of H giving rise to signal	Divide raw values by smallest → ratio; use mol. formula for exact count
Multiplicity (shape)	Number of non-equivalent neighboring H atoms	Apply n+1 rule; look for ethyl/isopropyl/t-Bu patterns; special cases for OH, CHO

Based on: Organic Chemistry, David Klein — Chapter 15  ·  Prepared as exam study notes for JEE Advanced, NEET, IIT-JAM, BITSAT, GATE, CSIR-NET, TGT, PGT

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