Nuclear magnetic resonance (NMR) is the single most powerful tool the organic chemist has for determining molecular structure. Where a melting point or a combustion analysis tells you what atoms are present, NMR tells you how those atoms are connected โ which carbons bear how many hydrogens, which groups sit next to which, and how the whole skeleton is assembled.
The physics behind it: certain nuclei, including 1H (a single proton) and 13C, possess a property called nuclear spin (I=21โ). A spinning charged nucleus behaves like a tiny bar magnet. In the absence of an external field these nuclear magnets point in random directions and all have the same energy. But when the sample is placed inside a strong, uniform magnetic field B0โ, each nuclear magnet can adopt one of two orientations:
aligned withB0โ โ the lower-energy (ฮฑ) state
aligned againstB0โ โ the higher-energy () state
The energy gap between these two states, ฮE, is directly proportional to the field strength: ฮE=ฮณhB0โ/2ฯ, where is the gyromagnetic ratio of the nucleus. Irradiate the sample with a radiofrequency (RF) pulse whose photon energy exactly matches , and nuclei in the state absorb that energy and flip to the state. That absorption โ the nucleus coming into โ is what the spectrometer detects.
Why a stronger magnet is "higher resolution": because ฮEโB0โ, a bigger field spreads the absorptions farther apart in frequency, so signals that overlap on a weak instrument separate cleanly on a strong one. A "300 MHz" or "500 MHz" instrument is named for the RF frequency at which 1H resonates in its field.
Why Different Protons Absorb at Different Frequencies
If every proton in a molecule resonated at exactly the same frequency, NMR would be useless โ it would give one peak for every organic compound. The reason NMR is informative is that each proton feels a slightly different magnetic field than the bare B0โ applied by the magnet.
Every proton is surrounded by electrons. When the sample sits in B0โ, those electrons circulate and generate their own small induced magnetic field that, near the nucleus, opposes . The proton therefore experiences a slightly weaker :
Checkpoint โ The Resonance Phenomenon
Equivalent Protons Give One Signal
Here is the rule that lets you predict how many peaks a molecule produces: chemically equivalent protons resonate at the same frequency and produce a single signal. Two protons are chemically equivalent if they are in identical electronic environments โ most reliably, if a symmetry operation of the molecule (a rotation, a mirror plane, or the molecule's own rapid bond rotation) interchanges them.
The number of signals in a 1H NMR spectrum therefore equals the number of distinct proton environments, not the number of hydrogen atoms. Counting environments is a skill built on recognizing symmetry.
Worked example โ count the signals:
Methane, CH4: all four H's are equivalent by tetrahedral symmetry
Equivalence, More Carefully: Homotopic and Enantiotopic vs. Diastereotopic
For everyday structure problems, "related by symmetry โ equivalent" is enough. But a rigorous course distinguishes how two protons are related, because one case quietly breaks the rule.
Homotopic protons are interchanged by a rotation of the molecule (a Cnโ axis). They are equivalent in every environment. Example: the three H's of a freely rotating CH.
Checkpoint โ Counting Signals
Chemical Shift (ฮด): A Field-Independent Address
Each signal needs an address on the horizontal axis. We could report the raw resonance frequency in hertz โ but that number depends on the magnet, so a peak at one frequency on a 300 MHz instrument lands at a different frequency on a 500 MHz instrument. To make spectra comparable, chemists report position as a dimensionless chemical shift, ฮด, in parts per million (ppm):
ฮด
Exit Ticket โ Part 1 Synthesis
Part 2: Chemical Shift
Chemical Shift in Depth
Part 2 of 7 โ Chemical Shift (ฮด)
In Part 1 we established that ฮด is a field-independent address measured in ppm relative to TMS at ฮด=0, and that electron density around a proton sets its position: shielded (electron-rich) protons sit upfield at small , deshielded (electron-poor) protons sit downfield at large . This part turns that qualitative idea into a working chemical-shift table and explains the three structural factors โ electronegativity, hybridization, and magnetic anisotropy โ that move a proton along the scale.
Part 3: Integration & Splitting
Integration and SpinโSpin Splitting
Part 3 of 7 โ Integration & Splitting
Chemical shift (Parts 1โ2) tells you what kind of proton each signal represents. This part adds the two pieces of information that make 1H NMR genuinely structure-determining:
Integration โ the area under each signal, which is proportional to the number of protons giving rise to it. This counts H's in each environment.
Spinโspin splitting (coupling) โ the way a signal is split into multiple lines by neighboring protons, which reveals how many protons sit next door and therefore what is bonded to what.
Together with shift, these turn a spectrum into a connectivity map. We will build the n+1 rule, read , define the , and draw the for the two patterns you must recognize on sight: the ethyl group and the isopropyl group.
Part 4: ยนยณC NMR
Carbon-13 NMR
Part 4 of 7 โ 13C NMR
So far we have watched protons. But the carbon skeleton itself can be observed directly with 13C NMR, and it is wonderfully complementary to : it counts environments, spreads them over a much wider scale, and is often easier to read because the spectra are usually a clean set of single lines.
Part 5: Structure Determination
Putting It Together: Structure Determination
Part 5 of 7 โ Structure Determination
Individually, a chemical shift or a multiplicity is a clue. The power of spectroscopy comes from combining them in a disciplined order so that each piece of data constrains the next. This part lays out the systematic workflow that turns a molecular formula plus a few spectra into a unique structure, and walks one unknown all the way through.
The four-step routine:
Molecular formula โ degrees of unsaturation. Establish how many rings and ฯ bonds the molecule must contain.
IR โ functional groups. Identify the major functional groups (C=O, OโH, NโH, CโกN, aromatic).
NMR environments, counts, neighbors. Count signals, read integration, and decode splitting to find fragments.
Part 6: Problem-Solving Workshop
Problem-Solving Workshop
Part 6 of 7 โ Problem-Solving Workshop
This part is all application. We take the systematic pipeline from Part 5 and run it on a series of unknowns of increasing subtlety โ predicting spectra from structures, deducing structures from data, and combining IR, mass spectrometry, and NMR. Work each example with paper in front of you: write the fragments down as you read each signal, keep a running proton tally, and only commit to a structure when every datum is explained.
A reliable mental checklist for any unknown:
DoU from the formula โ how many rings/ฯ bonds?
IR โ which functional groups (C=O? OโH? NโH? CโกN? aromatic)?
1H: for each signal note shift (what kind), integration (how many), multiplicity (how many neighbors) a fragment.
Part 7: Synthesis & Review
Synthesis and Review
Part 7 of 7 โ Synthesis & Review
NMR is the most powerful structural tool in the organic chemist's kit because a single pair of experiments (1H and 13C) reports on almost every atom in a molecule at once. This closing part consolidates the whole suite into one coherent picture: the four kinds of information a spectrum delivers, how each maps onto structure, and the decision flow that ties shift, integration, splitting, and the carbon spectrum into a single answer.
ฮฒ
ฮณ
ฮE
ฮฑ
ฮฒ
resonance
B0โ
effective field
Beffโ=B0โโBinducedโ
This is called shielding: the electrons "shield" the nucleus from the full applied field. A more heavily shielded proton feels a weaker Beffโ, needs a slightly lower frequency to reach resonance, and appears toward the right (upfield) of the spectrum. A proton near an electron-withdrawing group has less electron density around it โ it is deshielded, feels a stronger Beffโ, and appears toward the left (downfield).
The take-home idea, which Part 2 develops quantitatively: electron density around a proton determines its position in the spectrum. Electron-rich environment โ shielded โ upfield (small ฮด); electron-poor environment โ deshielded โ downfield (large ฮด).
Terminology that trips students up: "upfield" and "downfield" are historical terms from old instruments that swept the field. Today "upfield" simply means right / smallerฮด and "downfield" means left / largerฮด. Upfield = more shielded; downfield = more deshielded.
โ
โ
1 signal.
Ethanol, CH3โCH2โOH: three sets of inequivalent H's โ the CH3โ, the CH2โ, and the OH โ โ3 signals.
1,4-dimethylbenzene (para-xylene): the two methyl groups are related by symmetry (one signal), and all four aromatic H's are equivalent by symmetry (one signal) โ2 signals, even though there are 10 hydrogens.
Acetone, (CH3โ)2โC=O: the two methyl groups are mirror images across the carbonyl, hence equivalent โ1 signal for all six H's.
The single most common counting mistake: confusing the number of hydrogens with the number of signals. para-Xylene has 10 H's but only 2 signals; acetone has 6 H's but only 1 signal. Always look for symmetry first.
3
โ
Enantiotopic protons are interchanged only by a mirror plane. In an ordinary (achiral) solvent they are equivalent and give one signal. Example: the two H's of CH2โCl2โ, or the two CH2โ protons of ethanol.
Diastereotopic protons cannot be interchanged by any symmetry operation. They are genuinely inequivalent and can give separate signals even though they are on the same carbon. This happens on a CH2โ when the molecule contains a stereocenter or a nearby double bond โ for example the two methylene H's of an R*-CH(X)-CH2โ-Y fragment, or the two =CH2โ protons of a terminal alkene (one is cis, one is trans to the substituent).
A quick operational test for any pair of H's on one carbon: replace each in turn with a test group "Z." If the two products are identical, the H's are homotopic; if they are enantiomers, enantiotopic (equivalent in normal NMR); if they are diastereomers, diastereotopic (inequivalent).
The subtle trap: students often assume "two H's on the same carbon must be equivalent." Diastereotopic protons are the exception โ and they are common around stereocenters and double bonds. When a CH2โ unexpectedly shows two signals (often with a large mutual coupling), diastereotopicity is the usual culprit.
Dividing the frequency offset by the spectrometer frequency cancels the field dependence, so ฮด for a given proton is the same number on every instrument. A proton at ฮด=2ย ppm on a 300 MHz magnet sits 600 Hz downfield of the reference; on a 500 MHz magnet it sits 1000 Hz downfield โ but ฮด is 2ย ppm in both cases.
The reference compound is tetramethylsilane, Si(CH3โ)4โ (TMS), defined as ฮด=0. Silicon is more electropositive than carbon, so the 12 equivalent TMS protons are unusually shielded and resonate upfield of virtually all organic protons โ a clean, single peak at the right edge of the scale. The 1H scale then runs from about 0 to 12ย ppm, with deshielded protons at large ฮด on the left and shielded protons at small ฮด on the right.
Key relationship to carry into Part 2: chemical shift is a map of electron density. Memorizing the shift of a few benchmark environments lets you read a spectrum like an address book โ and that is exactly the table we build next.
ฮด
ฮด
A practiced organic chemist reads ฮด the way a reader reads words: a peak at 1ย ppmsays "alkyl," a peak at 7ย ppmsays "aromatic," a peak near 10ย ppmsays "aldehyde." The goal of this part is to make those associations automatic and, more importantly, to make them explainable โ so you can predict the shift of a proton you have never seen.
The Benchmark Chemical-Shift Table
These ranges are the backbone of 1H NMR interpretation. Commit the order and the approximate values to memory.
Proton environment
Typical ฮด (ppm)
Si(CH3โ)4โ (TMS, reference)
0
Alkyl, R-CH3โ / R-CH2โ-R
0.8โ1.5
Allylic / ฮฑ to a carbonyl, C=C-CH3โ or O=C-CH3โ
CโกC-H (terminal alkyne)
2.0โ3.0
H-C-N (amine ฮฑ)
2.2โ2.9
H-C-Clย /ย Br (halide ฮฑ)
3.0โ4.0
H-C-O (ether / alcohol / ester ฮฑ)
3.3โ4.5
Vinylic, C=C-H
4.5โ6.5
Aromatic, Ar-H
6.5โ8.0
Aldehyde, R-CHO
9.0โ10.0
Carboxylic acid, R-COOH
10โ12
Two columns of "exchangeable" protons โ O-H (alcohol โผ1โ5, broad) and N-H (โผ1โ4, broad) โ are deliberately listed as wide ranges because their shifts depend on concentration, temperature, and hydrogen bonding. We return to them at the end of this part.
The largest routine influence on ฮด is the electronegativity of nearby atoms. An electronegative atom pulls electron density away from the CโH bond through the ฯ framework, reducing the shielding at the proton and shifting it downfield. The effect is strongest on the directly attached carbon and falls off rapidly with distance.
Compare the methyl shift as the attached halogen gets more electronegative:
Compound
Attached atom electronegativity
ฮด of CH3โ
CH3โ-F
F (4.0)
โผ4.3
CH
More electronegative neighbor โ larger ฮด. Multiple withdrawing groups add up: CH3โCl is at 3.0, at , at โ each extra chlorine drags the lone remaining proton further downfield.
Distance matters. In 1-chloropropane, Cl-CH2โ-CH2โ-CH3โ, the protons nearest chlorine sit at , the middle near , and the far near โ essentially back to the value of a plain alkyl group two bonds away. The inductive pull is short-range.
Factor 2 โ Hybridization and Factor 3 โ Magnetic Anisotropy
Hybridization. Protons on sp2 carbons (alkenes, aromatics) generally appear downfield of protons on sp3 carbons, partly because sp2 carbon is more electronegative (more s-character). But hybridization alone does not explain the numbers, and the famous exception โ the terminal alkyne CโกC-H at only 2โ3ย ppm despite sp carbon โ forces us to invoke the third factor.
Magnetic anisotropy is the deshielding (or shielding) produced by the circulating ฯ electrons of a nearby multiple bond, and it is the key to the most diagnostic shifts in all of 1H NMR. When a ฯ system sits in B, its electrons circulate and set up a secondary field that is uniform โ it reinforces in some regions of space and opposes it in others.
Aromatic ring current: the six ฯ electrons of benzene circulate around the ring, generating an induced field that opposesB0โ inside the ring but reinforces it in the plane outside the ring โ exactly where the ring protons sit. Aromatic H's are therefore strongly deshielded to 6.5โ8ย ppm. (A proton held above a ring would instead be shielded โ a real effect in some fused systems.)
Carbonyl and alkene: the electrons of and likewise deshield protons lying in the bond plane, contributing to vinylic () and aldehyde () shifts.
The trap to name: "sp carbon is the most electronegative, so the alkyne โกC-H should be the most deshielded." It is not โ the anisotropic shielding cone of the triple bond overrides hybridization and places it near 2.5ย ppm, upfield of vinylic and aromatic protons. Whenever a shift defies the simple electronegativity argument, suspect anisotropy.
Checkpoint โ Why Protons Deshield
Exchangeable Protons: OโH and NโH
Protons on oxygen and nitrogen behave differently from CโH protons in three diagnostic ways.
Variable shift. Because O-H and N-H participate in hydrogen bonding, their ฮด depends on concentration, temperature, and solvent. An alcohol O-H can appear anywhere from โผ1 to โผ5ย ppm; a carboxylic acid O-H, locked in strong hydrogen bonding, sits far downfield at 10โ12ย ppm.
Often broad, often a singlet. In many samples the OโH/NโH proton exchanges rapidly between molecules. This fast exchange averages out spinโspin coupling, so the OโH usually appears as a broad singlet that does not split its neighbors and is not split by them. (Under scrupulously dry, acid-free conditions, exchange slows and coupling can reappear โ but the broad, coupling-free singlet is the common case.)
The D2โO shake test. Add a drop of deuterium oxide and shake: exchangeable OโH/NโH protons swap for deuterium and their signal disappears (or shrinks dramatically) as a new HDO peak grows in. A signal that vanishes on a D2โO shake is diagnostic of an exchangeable proton โ a quick way to flag an alcohol, acid, or amine.
The trap to name: treating an OโH like an ordinary CโH and trying to apply the n+1 rule to it. Because of rapid chemical exchange, the alcohol OโH typically neither splits the adjacent CH2โ nor is split by it โ so do not predict (or expect) coupling to it in a routine spectrum.
Exit Ticket โ Part 2 Synthesis
integration ratios
coupling constant J
splitting trees
Integration: Counting Protons by Area
The spectrometer can measure the area under each peak, drawn as a step ("integral trace") or printed as a number. That area is proportional to the number of protons in the environment โ but only relative values are meaningful. Integration gives you a ratio, not an absolute count; you convert the ratio to actual hydrogens using the molecular formula.
Worked example โ ethyl acetate, CH3โ-C(=O)-O-CH2โ-CH3โ (C4โH8โO2โ, 8 H total):
Three environments, with integrals measured (say) as 1.5:1.5:2.25. Divide by the smallest (1.5): 1:1:1.5. Multiply to clear the fraction (): . That sums to 7 โ but the formula says 8 H, so we scale to the real counts :
ฮด (ppm)
Integral ratio
Actual H
Assignment
โผ4.1
2
The lesson: integration delivers a ratio, you reduce it to small whole numbers, and you reconcile the total with the molecular formula. A common slip is reading the raw integral as an absolute proton count โ always normalize.
Reading tip:CH3โ groups give the largest single-environment integrals (3H), so a tall 3H signal in the alkyl region is very often a methyl. A 1H signal is frequently a methine (CH), an aldehyde, or an exchangeable OโH/NโH.
Checkpoint โ Integration
SpinโSpin Splitting and the n+1 Rule
A proton senses not only its own electronic environment but also the tiny magnetic fields of protons on adjacent atoms. Each neighboring proton can be aligned with or against B0โ, nudging the observed proton's resonance slightly higher or lighter. The result is that a single signal is split into several lines โ a multiplet.
For a set of protons with n equivalent neighboring protons, the signal is split into
n+1 lines
This is the n+1 rule (valid for first-order spectra, where coupled protons differ substantially in chemical shift). Crucially, n counts the protons on neighboring atoms โ typically three bonds away (H-C-C-H) โ not the protons in the signal itself. Equivalent protons do not split one another.
Neighbors n
Lines (n+1)
Name
Relative line intensities
0
1
singlet (s)
The intensity pattern within a multiplet follows Pascal's triangle โ it arises from the number of equivalent ways the neighboring spins can combine. A 1:2:1 triplet, for instance, reflects that two neighbors can be (โโ), (โโ or โโ), or (โโ): the middle, mixed arrangement is twice as likely.
The number-one splitting trap: counting the protons in the signal instead of the neighbors. A CH3โ does not split itself into a quartet; it is split by what is attached to the next carbon. A CH3โ next to a appears as a triplet ( neighbors), and that appears as a quartet ( neighbors). This reciprocal relationship is the ethyl fingerprint.
Worked Splitting Tree 1 โ The Ethyl Group (CH3โCH2โ-X)
The ethyl group attached to an electron-withdrawing X (as in bromoethane, CH3โCH2โBr) is the most important pattern in introductory NMR. Two coupled environments:
The CH3โ (3H): its neighbors are the two CH2โ protons, so n= lines a , intensities , near .
The CH2โ (2H): its neighbors are the three CH3โ protons, so n= lines a , intensities , near (deshielded by Br).
Building the CH2โ quartet as a tree โ start with one line, then split successively by each of the three equivalent methyl protons (each splitting doubles the lines with spacing J):
Split by H#1: โ 2 lines (doublet)
Split by H#2: each line splits again โ 4 lines, but the inner two overlap โ1:2:1
Split by H#3: โ the lines overlap into the quartet
Because all three methyl protons are equivalent, they share one coupling constant J, so the four lines are evenly spaced and collapse to the Pascal pattern. The integral ratio confirms the assignment: CH2โ:CH3โ=.
Ethyl fingerprint to memorize: a 2H quartet + a 3H triplet sharing the same J (โ7ย Hz) almost always means an -CH2โCH group. Find that pair and you have located an ethyl.
Worked Splitting Tree 2 โ The Isopropyl Group (CH3โ)2โCH-X
The isopropyl group is the second must-know pattern. In 2-bromopropane, (CH3โ)2โCHBr:
The two CH3โ groups (6H, equivalent): each methyl's only neighbor is the single methine proton, so n=1โ2 lines โ a doublet, intensity . Because the two methyls are equivalent, they overlap into one near .
The methine CH (1H): its neighbors are the six equivalent methyl protons, so n=6โ6+1=7 lines โ a septet (), near (deshielded by Br).
The outer lines of a septet are only 1/20 the height of the center line, so a septet often looks like a weak picket fence flanking a tall central peak โ easy to miss if you do not expect it.
Isopropyl fingerprint to memorize: a 6H doublet + a 1H septet sharing the same J means an (CH3โ)2โCH- group.
Contrast the two fingerprints โ a frequent exam trap: ethyl = (3H triplet + 2H quartet); isopropyl = (6H doublet + 1H septet). Students who memorize "quartet = ethyl" get caught when the quartet really belongs to an O-CH2โ or another environment. Always check the integration (2H vs 6H, 3H vs 1H) and the partner multiplicity together โ the pair, not a single multiplet, identifies the group.
Checkpoint โ Multiplicity
The Coupling Constant J
The horizontal spacing between adjacent lines of a multiplet is the coupling constant, J, reported in hertz (Hz). J measures the strength of the magnetic interaction between two coupled protons, and it carries two crucial properties:
J is field-independent. Unlike chemical shift (which we deliberately report in field-independent ฮด), the splitting itself is an intrinsic interaction, so J measured in Hz is the same on a 300 MHz and a 500 MHz instrument. (This is also why a multiplet looks narrower in ppm on a higher-field magnet, even though its J in Hz is unchanged โ a useful way to distinguish a true multiplet from two separate signals.)
Coupled partners share the same J. Two protons that couple to each other split each other by the identicalJ. This is how you pair up multiplets: the CH2โ quartet and the CH triplet of an ethyl group have matching , confirming they are neighbors.
Typical magnitudes are themselves diagnostic of geometry:
Coupling type
Typical J (Hz)
Free rotation, H-C-C-H (e.g., ethyl)
6โ8
Vinyl cis, C=C
The trans > cis relationship across a double bond is especially useful: a Jโ16ย Hz between two vinyl protons is strong evidence for a trans (E) alkene, while Jโ8ย Hz points to cis (Z).
The trap to name: confusing J (the line spacing, in Hz, field-independent) with chemical shift difference (the separation between two signals, which scales with field in Hz but is fixed in ppm). If a "splitting" gets wider in Hz when you change magnets, it was never a coupling โ it was two distinct chemical shifts.
Exit Ticket โ Part 3 Synthesis
1
H
carbon
The catch is sensitivity. The dominant carbon isotope, 12C, has I=0 and is NMR-silent. Only 13C (I=21โ) gives a signal, and it makes up just 1.1% of natural carbon. Combined with a gyromagnetic ratio about a quarter that of 1H, 13C NMR is intrinsically thousands of times less sensitive than 1H NMR โ which is why early instruments could barely record it and why modern spectra are acquired by signal-averaging many scans on a pulsed FT spectrometer.
That same low abundance has a happy consequence for splitting, as we will see: the odds of two 13C nuclei being adjacent in one molecule are tiny (โผ0.01%), so carbonโcarbon coupling essentially never complicates the spectrum.
One Signal per Unique Carbon Environment
The counting rule mirrors 1H NMR: the number of signals equals the number of chemically distinct carbon environments, set by molecular symmetry. The huge advantage is dispersion โ the 13C scale runs from roughly 0 to 220ย ppm, about twenty times wider than the 1H window โ so carbons that would overlap as protons usually appear as cleanly separated lines. Even complex molecules often give a fully resolved line for every carbon.
Worked example โ counting carbon signals:
Acetone, (CH3โ)2โC=O: the two methyls are equivalent, the carbonyl is unique โ2 signals (the near , the two equivalent near ).
Notice how molecular symmetry can make the 1H and 13C counts differ: para-xylene gives 2 proton signals but 3 carbon signals, because the substituted and unsubstituted ring carbons are distinct even though only one kind of aromatic hydrogen exists.
The trap to name: assuming the number of 13C signals equals the number of 1H signals. They count different nuclei. Quaternary carbons (bearing no hydrogen) appear in but contribute signal at all โ so often reveals carbons that are invisible in the proton spectrum.
Checkpoint โ Carbon Counting and Sensitivity
Broadband Decoupling: Why 13C Signals Are Singlets
A raw 13C nucleus is coupled โ strongly โ to the protons attached to it (1JCHโ can be 125โ250ย Hz). Left alone, every carbon would be split by its hydrogens (a CH3โ carbon into a quartet, a CH2โ into a triplet, and so on), and the already weak signals would be divided into multiple lines, crippling sensitivity and crowding the spectrum.
The standard solution is broadband proton decoupling: while observing 13C, the spectrometer simultaneously irradiates all the protons across their entire frequency range. This rapid irradiation averages the CโH coupling to zero, so each carbon collapses to a single sharp line. The familiar "13C spectrum" โ one singlet per carbon environment โ is really a proton-decoupled spectrum.
Two important consequences:
No CโC splitting either. Because adjacent 13Cโ13C pairs are vanishingly rare (1.1%ร1.1%), carbonโcarbon coupling is not observed in routine spectra. With protons decoupled too, every line is a singlet.
The trap to name: trying to apply the n+1 rule to a routine 13C spectrum, or trying to integrate it like a proton spectrum. Standard 13C is broadband-decoupled โ every signal is a singlet by design โ and its intensities are not quantitative. Multiplicity information is recovered instead by DEPT.
DEPT: Recovering How Many Hydrogens Each Carbon Bears
Decoupling cleans up the spectrum but discards the CโH information. The DEPT experiment (Distortionless Enhancement by Polarization Transfer) brings that information back in a controlled way, sorting carbons by the number of attached hydrogens. The most useful variants:
DEPT-90: only CH (methine) carbons appear.
DEPT-135:CH and CH3โ carbons point up (positive), CH2โ carbons point down (negative), and quaternary carbons (no attached H) are absent.
Comparing a normal decoupled spectrum (which shows all carbons) with DEPT-135 lets you classify every carbon:
Carbon type
Attached H
DEPT-135
DEPT-90
CH3โ
3
up
absent
CH2โ
2
Worked use: a carbon that is present in the standard spectrum but missing from DEPT-135 must be quaternary โ a carbonyl carbon, a fully substituted ring carbon, or a C bearing no H. This is how you locate carbonyls and quaternary centers that are silent in 1H NMR. An older alternative, off-resonance decoupling, achieved a similar classification by leaving only the one-bond CโH coupling so that CH3โ quartet, triplet, doublet, quaternary singlet โ but DEPT has largely replaced it.
Reading strategy: run the decoupled spectrum to count carbon environments and read their shifts; run DEPT to assign each as CH3โ, CH2โ, CH, or quaternary. The "disappearing" peaks in DEPT are exactly the quaternary carbons.
Checkpoint โ Decoupling and DEPT
The 13C Chemical-Shift Map (0โ220ย ppm)
Carbon shifts respond to the same factors as proton shifts โ electronegativity, hybridization, anisotropy โ but over a far broader range, which makes the regions very diagnostic. Memorize these zones:
Carbon type
Typical ฮด (ppm)
Alkyl C (spยณ, C/H only)
5โ45
C-N (amine)
30โ65
C-O (alcohol, ether, ester )
Two regions resolve ambiguities that proton NMR cannot:
A line at 160โ220ย ppm is essentially always a carbonyl carbon โ and its sub-region distinguishes acid/ester/amide (160โ185) from aldehyde/ketone (190โ220). Proton NMR cannot see a carbonyl carbon directly; pinpoints it.
The trap to name: mapping 13C values onto the 1H scale. A carbonyl carbon sits near 200ย ppm, but no proton ever appears at โ the scale stops around . Keep the two scales mentally separate; they measure different nuclei and span different ranges.
Exit Ticket โ Part 4 Synthesis
1
H
โ
Assemble the fragments into a structure consistent with every piece of data, then check it against the spectra.
Step 1 โ Degrees of Unsaturation
The degree of unsaturation (DoU, also "index of hydrogen deficiency") counts the total number of rings plus ฯ bonds in a molecule. Every ring and every ฯ bond removes two hydrogens relative to the fully saturated formula, so a single number computed from the formula tells you how much unsaturation to look for.
For a compound CcโHhโNnโOoโXxโ (X = halogen):
DoU=22c+2+nโhโxโ
Oxygen does not appear because adding an O (e.g., inserting it into a CโH or CโC bond) does not change the hydrogen count. Each halogen counts like a hydrogen; each nitrogen adds one to the numerator.
Worked calculations:
C4โH8โO: DoU=. One ring one bond โ consistent with a ketone, an aldehyde, or a cyclic ether/alcohol.
The interpretive habit: a DoU of 4 is the fingerprint of an aromatic ring โ see it and immediately look for aromatic signals at ฮด6.5โ8 in 1H NMR and 100โ150ย ppm in . A DoU of 1 with an oxygen says "one C=O or one ring." DoU bounds the whole problem before you read a single peak.
Checkpoint โ Degrees of Unsaturation
Step 2 โ IR for Functional Groups
Infrared spectroscopy answers a question NMR answers only indirectly: which functional groups are present? A few strong, diagnostic IR bands quickly confirm or exclude the major groups, and they pair naturally with the DoU from Step 1.
IR band (cmโ1)
Functional group
Pairs with DoU
1700โ1750, strong, sharp
C=O (carbonyl)
uses 1 degree
3200โ3550, broad
O-H (alcohol/acid)
โ
3300โ3500, medium (1โ2 peaks)
N-H (amine/amide)
โ
โผ2250, sharp
CโกN or CโกC
uses 2 degrees
1600ย andย 1500, medium
aromatic C=C
part of aromatic 4
โผ3300, sharp
terminal alkyne โกC-H
โ
The strategy: if DoU โฅ1 and IR shows a strong band near 1715ย cmโ1, a carbonyl accounts for one degree โ now decide which carbonyl using NMR shift and chemistry. If DoU = 4 and IR shows bands at 1600/1500ย cm, the aromatic ring is confirmed. IR rarely gives the whole structure, but it pins down the functional groups so NMR can focus on connectivity.
Reading habit: treat IR and DoU as cross-checks. A 1715ย cmโ1 band with no leftover degree of unsaturation would be a contradiction โ re-examine the formula. When IR and DoU agree (e.g., "one C=O uses the moleculeโs single degree"), you have a firm functional-group assignment to carry into the NMR analysis.
Step 3 โ 1H NMR: Environments, Counts, and Neighbors
Now extract fragments from the proton spectrum, reading three properties of every signal in concert:
Chemical shiftโwhat kind of proton (alkyl, ฮฑ-to-O, vinyl, aromatic, aldehydeโฆ).
Integrationโhow many protons in that environment (reduce the ratio, reconcile with the formula).
Multiplicity (n+1) โhow many neighbors, which tells you what is attached to the adjacent carbon.
Translate each signal into a fragment and tally the protons:
Signal reads as
Likely fragment
3H triplet near 1ย ppm + 2H quartet near 2โ4
ethyl, -CH2โCH
A singlet is especially informative: it means no coupling neighbors, so that group is flanked by carbons bearing no hydrogens (a carbonyl, a quaternary carbon, an oxygen, or the symmetric equivalent). Add up the protons you have assigned and make sure they total the molecular formula โ any shortfall points to an undetected fragment (often an exchangeable proton or a symmetric duplication).
Checkpoint โ Reading Fragments
Worked Structure Determination โ C9โH10โO2โ
Let us run the full pipeline on an unknown of formula C9โH10โO2โ.
Step 1 โ DoU:(2โ 9+2โ10)/2=(20โ10)/2=5. Five degrees: a benzene ring (4) plus one more bond โ anticipate an aromatic ring and one .
Step 2 โ IR: a strong band at 1740ย cmโ1 confirms an ester-type carbonyl (using the 5th degree); bands at 1600/1500ย cmโ1 confirm the aromatic ring. No broad OโH, so not a carboxylic acid.
Step 3 โ 1H NMR:
ฮดย 7.3, 5H, multiplet โ a monosubstituted benzene (C6โH5โ).
Protons assigned: 5+2+3=10, matching H10โ.
Step 4 โ Assemble. Fragments: C6โH5โ-, -CH2โ, , plus the ester carbonyl/oxygen from IR + DoU. Stitching them so the ester links the benzylic to the acetyl group gives , . Check: 9 C, 10 H, 2 O โ; ester C=O at โ; benzylic singlet at โ; acetyl singlet at โ; five aromatic H โ. Every datum is satisfied โ the structure is secure.
Why the order matters: DoU told us to expect a ring + a carbonyl before we touched the NMR; IR confirmed the ester; NMRโs singlets (no neighbors) then forced the connectivity. Each step narrowed the field, so the final assembly had essentially one consistent answer.
Assemble and verify against all data, including symmetry.
Forward Problem โ Predicting a Spectrum From a Structure
Before deducing structures, practice the reverse: given a structure, predict its 1H NMR. This sharpens the same instincts. Take 1,1,2-trichloroethane, CHCl2โ-CH2โCl.
Environments: two โ the CHCl2โ methine (1H) and the CH2โCl methylene (2H).
Shifts: both are deshielded by chlorine. The proton bears two chlorines on its own carbon very downfield, . The bears one chlorine .
Predicted spectrum: a 1H triplet at โผ5.8 and a 2H doublet at โผ4.0, sharing one Jโ6ย Hz. Notice the reciprocity โ the more substituted carbon carries fewer protons but is split into more lines, because multiplicity reflects the neighbors, not the protons in the signal.
Forward-prediction trap: writing "CH2โโ triplet, CHโ doublet" by reflex. Here it is the opposite: the lone CHCl2โ proton is a (two neighbors) and the is a (one neighbor). Always split by what is next door.
Checkpoint โ Forward Prediction
Reverse Problem 1 โ Deduce C3โH6โO From NMR
Given: formula C3โH6โO. 1H NMR: ฮดย 9.8 (1H, triplet, Jโ1.5ย Hz); ฮดย 2.4 (2H, doublet of quartets / multiplet); ฮดย 1.1 (3H, triplet). 13C: a peak near 202ย ppm.
Step 1 โ DoU:(2โ 3+2โ6)/2=1. One ring or one ฯ bond.
Step 2 โ Carbonyl? The 13C line at 202ย ppm is in the aldehyde/ketone region, and the ฮดย 9.8 proton in is the aldehyde signature. So the one degree of unsaturation is a , specifically an (it has a CโH). That consumes : 1 C, 1 H, 1 O.
Step 3 โ Remaining fragments:C3โH6โO minus CHO leaves . The 3H triplet () + 2H signal () is an , and the 2H sits at because it is to the carbonyl.
Step 4 โ Assemble:CH3โCH2โ-CHO = propanal. Verify the fine splitting: the aldehyde proton is a triplet because it couples weakly to the adjacent CH (small ); the is split by both the (3 neighbors) and the (1 neighbor). 3 C, 6 H, 1 O โ.
The diagnostic pivot: a 13C line above 190ย ppm together with a ฮดย 9.8 proton is essentially conclusive for an aldehyde โ neither datum alone is as decisive as the two together. The aldehyde proton being a triplet (not a singlet) is the tell that a is adjacent: propanal, not acetone (which would have no aldehyde proton at all).
Reverse Problem 2 โ Two Isomers of C3โH6โO, and Using MS
A subtle skill is distinguishing isomers that share a formula. C3โH6โO has two common carbonyl isomers:
Their 1H spectra could hardly be more different: propanal has three signals including the tell-tale 9.8 aldehyde; acetone has one singlet. Even without the aldehyde region, the symmetry-driven 6H singlet uniquely marks acetone.
Mass spectrometry as cross-check. The molecular ion gives the molecular weight (both isomers: M=58), but fragmentation distinguishes them:
Acetone loses a methyl radical to give the acylium ionCH3โCO+ at m/z=43 โ a dominant peak (the classic -cleavage of a ketone).
Integrative habit: when two structures fit the NMR, look to MS fragmentation or the 13Ccount. Here, acetoneโs two carbon signals (vs. propanalโs three) and its m/z=43 acylium peak each independently settle the assignment. Cross-confirming with a second technique is what turns a "probably" into a "certainly."
Checkpoint โ Distinguishing Isomers
Reverse Problem 3 โ An Aromatic Unknown, C8โH8โO
Given:C8โH8โO. IR: strong band at 1685ย cmโ1 (conjugated C=O) and 1600/1580ย cmโ1 (aromatic). 1H NMR: ฮดย 7.9 (2H, doublet-like), ฮดย 7.5 (3H, multiplet), ฮดย 2.6 (3H, singlet). 13C: ~198ย ppm (carbonyl), plus aromatic carbons 128โ137.
Step 1 โ DoU:(2โ 8+2โ8)/2=5. Aromatic ring (4) + one C=O (1).
Step 2 โ IR: carbonyl at the lowered 1685ย cmโ1 signals conjugation with the ring (an aryl ketone, not an aldehyde โ there is no aldehyde proton near 9.7).
Step 3 โ 1H: the aromatic protons total 5H (2+3) โ a monosubstituted benzene, with the 2H further downfield (7.9) being the ortho protons next to the electron-withdrawing carbonyl. The 3H singlet at is a attached to C=O (no neighbors) โ an acetyl methyl, shifted slightly downfield of a normal acetyl by the adjacent ring.
Step 4 โ Assemble:C6โH5โ-C(=O)-CH3โ = acetophenone. Check: H โ; aryl-conjugated C=O at and โ; 5 aromatic H as 2H + 3H โ; methyl singlet โ.
Two cross-checks worth naming: (1) the lowered carbonyl frequency (1685 vs. a normal ketone โผ1715ย cmโ1) reveals conjugation โ a structural clue IR gives that NMR does not state outright. (2) The aromatic protons split into 2H + 3H, the signature of a monosubstituted ring; a para-disubstituted ring would instead give a symmetric 2H + 2H pattern. Reading the aromatic count and shape tells you the substitution pattern.
Exit Ticket โ Part 6 Synthesis
1
H
The unifying idea from Parts 1โ6: every observable is a question about local structure. Where a signal sits asks "what is this proton attached to?" How big it is asks "how many are there?" How it splits asks "who are its neighbors?" Learn to hear those three questions and a spectrum becomes a sentence.
The Four Pieces of Information in a 1H Spectrum
Observable
Question it answers
Governing rule
Number of signals
How many distinct proton environments?
Symmetry / chemical equivalence (Part 1)
Chemical shift ฮด
What is each proton near? (electronic environment)
Each observable is independent, and they constrain one another. A signal that is downfield (large ฮด), integrates for 1H, and is a singlet might be an aldehyde or an exchangeable OโH; add that it sits at 9.7 (not broad, not vanishing on D2โO) and it is an aldehyde. No single observable identifies a fragment โ their combination does.
Mantra:shift = identity, integration = count, splitting = neighbors. If you can state those three for every signal, you have read the spectrum.
Checkpoint โ Mapping Observables to Structure
Consolidated Shift Map โ 1H and 13C Side by Side
Chemical shift indicates electronic environment, and keeping the two scales side by side prevents the most common confusion (mapping carbon values onto the proton axis). Approximate landmarks to know cold:
Environment
1Hย ฮด (ppm)
13Cย ฮด (ppm)
TMS reference
0
0
Alkyl (C/H only)
0.8โ1.5
5โ45
ฮฑ to C=O / allylic
The two scales rhyme โ both move downfield as protons/carbons become more deshielded โ but they cover very different ranges. A carbonyl carbon lives near 200ย ppm; no proton ever does. Use 13C to find carbonyls and quaternary carbons (invisible in 1H), and use DEPT to label each carbon as , , , or quaternary.
The persistent trap: "a peak at 200 must be a far-downfield proton." There is no such proton โ the 1H window ends near 12. A value of 200ย ppm is a carbon shift, almost always a carbonyl. Always check which spectrum you are reading.
Splitting Reveals Connectivity โ The Fingerprints
Splitting pattern reveals connectivity: the multiplets in concert with their integrals fingerprint whole substructures. Re-memorize the canonical pairs, because recognizing them on sight is most of practical spectral interpretation:
Pattern (integration + multiplicity)
Fragment
3H triplet + 2H quartet (shared Jโ7ย Hz)
ethyl, -CH2โCH3โ
6H doublet + 1H septet
isopropyl, (CH3โ)2โCH-
9H singlet (~1ย ppm)
tert-butyl, (CH3โ)3โC-
3H singlet (~2ย ppm)
CH3โ on C=O or ring (no H neighbors)
2H + 2H aromatic (symmetric)
para-disubstituted benzene
2H + 3H aromatic
monosubstituted benzene
Jโ16ย Hz between two vinyl H
trans (E) alkene
Jโ8ย Hz between two vinyl H
cis (Z) alkene
Two rules keep these honest: (1) the n+1 count is over neighbors, never the protons in the signal; and (2) mutually coupled signals share the sameJ, which is how you pair a multiplet with its partner and so chain fragments together. A singlet, by contrast, announces no coupling neighbors โ its group is bounded by carbonyls, oxygens, or quaternary carbons.
Capstone trap roundup: (a) counting a signal's own protons for n+1 instead of its neighbors; (b) confusing a quartet's J (Hz, field-independent) with a chemical-shift gap (ppm-fixed, Hz-scaling); (c) trying to integrate or apply n+1 to a routine decoupled spectrum; (d) expecting an alcohol OโH to couple to its (rapid exchange usually erases that coupling). Avoid these four and most interpretation errors vanish.
Checkpoint โ Fingerprints and Connectivity
The Master Workflow โ Systematic Structure Determination
Pulling Parts 5โ6 into one checklist you can run on any unknown:
Molecular formula โ DoU.DoU=(2c+2+nโhโx)/2. A DoU of 4 flags an aromatic ring; a DoU of 1 with an O flags one ring or one C=O.
IR โ functional groups.โผ1715ย cmโ1 = C=O (lowered by conjugation); broad 3200โ3550 = OโH; โผ = nitrile/alkyne; = aromatic.
1H NMR โ fragments. For each signal: shift (identity), integration (count), multiplicity (neighbors). Recognize the fingerprints; sum the protons against the formula.
13C / DEPT โ carbon skeleton. Count environments, locate carbonyls (160โ220) and spยฒ carbons (100โ150); use DEPT to classify ///quaternary.
Assemble and verify. Connect fragments consistently with every datum (including symmetry), then read the proposed structure back against each spectrum.
Capstone illustration (C8โH10โ, DoU = 4): IR shows aromatic bands only (no C=O, no OโH). 1H NMR: a 4H aromatic singlet-like region near and a 6H singlet at . The 6H singlet = two equivalent ring-attached methyls with no H neighbors; the symmetric 4H aromatic pattern = a -disubstituted ring. Assemble (1,4-dimethylbenzene). The count of just signals confirms the high symmetry. Every observable โ DoU, IR, the two signals, and the carbon count โ agrees.
The closing principle: NMR rarely hands you a structure from one number; it hands you constraints, and the answer is the unique structure that satisfies all of them simultaneously. Discipline (coarse-to-fine: DoU โ IR โ NMR) and pattern recognition (the fingerprints) are what make that convergence fast and reliable.
Exit Ticket โ Course Synthesis
2.0โ2.5
3
โ
-Cl
Cl (3.2)
โผ3.0
CH3โ-Br
Br (3.0)
โผ2.7
CH3โ-I
I (2.7)
โผ2.2
CH3โ-H
H (2.2)
โผ0.2
CH2โCl2โ
5.3
CHCl3โ
7.3
โผ
3.5ย ppm
CH2โ
1.8ย ppm
CH3โ
1.0ย ppm
0
โ
not
B0โ
ฯ
C=O
C=C
4.5โ6.5
9โ10
Alkyne (the exception explained): the cylindrical ฯ system of CโกC circulates around the molecular axis, so the terminal proton lying along that axis falls in the shielding cone. This anisotropic shielding pulls โกC-H back upfield to 2โ3ย ppm, opposing the sp-electronegativity that would otherwise push it downfield.
ร
2
2:2:3
2:3:3
2H
O-CH2โ (downfield: next to O)
โผ2.0
3
3H
C(=O)-CH3โ (acetyl methyl)
โผ1.3
3
3H
CH2โ-CH3โ (ethyl methyl)
1
1
2
doublet (d)
1:1
2
3
triplet (t)
1:2:1
3
4
quartet (q)
1:3:3:1
4
5
quintet
1:4:6:4:1
6
7
septet
1:6:15:20:15:6:1
CH2โ
n=2
CH2โ
n=3
2โ
2+
1=
3
โ
triplet
1:2:1
ฮดโ1.7
3โ
3+
1=
4
โ
quartet
1:3:3:1
ฮดโ3.4
1:3:3:1
2
:
3
3
โ
1:1
6H doublet
ฮดโ1.7
1:6:15:20:15:6:1
ฮดโ4.3
3
โ
Jโ7ย Hz
6โ12
Vinyl trans, C=C
12โ18
Geminal, H-C-H (diastereotopic)
0โ3 (often โผ2, or large/negative in spยณ)
Aromatic ortho
7โ10
C=O
206ย ppm
CH3โ
30ย ppm
Benzene, C6โH6โ: all six carbons equivalent by symmetry โ1 signal (~128ย ppm).
Toluene, C6โH5โCH3โ: symmetry makes the ring carbons fall into four sets (ipso, ortho, meta, para) plus the methyl โ5 signals.
1,4-dimethylbenzene (para-xylene): two unique ring carbons + one methyl โ3 signals.
13
C
no
1H
13C
Integration is usually unreliable. A side effect of decoupling (the nuclear Overhauser effect) and the slow relaxation of carbons โ especially quaternary carbons โ mean that 13C peak areas are not reliably proportional to the number of carbons. Unlike 1H integration, you normally do not read carbon counts from 13C peak heights or areas without special acquisition conditions.
down
absent
CH
1
up
up
quaternary (C, C=O)
0
absent
absent
โ
CH2โโ
CHโ
โ
ฮฑ
50โ90
Alkyne CโกC (sp)
65โ90
Alkene / aromatic C (spยฒ)
100โ150
Nitrile CโกN
115โ120
Ester / acid / amide carbonyl C=O
160โ185
Aldehyde / ketone carbonyl C=O
190โ220
13
C
The 100โ150ย ppm window flags spยฒ carbons (alkene or aromatic), confirming unsaturation.
200ย ppm
1H
12
(
2
โ
4+
2โ
8)/2=
2/2=
1
or
ฯ
C7โH8โO: DoU=(14+2โ8)/2=4. Four degrees strongly suggests a benzene ring (a benzene ring is exactly 4: three ฯ bonds + one ring).
C8โH9โNO2โ: DoU=(16+2+1โ9)/2=5. A benzene ring (4) plus one more ฯ bond (e.g., a C=O).
13C
โ1
3โ
6H doublet + 1H septet
isopropyl, (CH3โ)2โCH-
9H singlet near 1ย ppm
tert-butyl, (CH3โ)3โC-
3H singlet near 2ย ppm
CH3โ next to C=O or aromatic (no H neighbors)
2H singlet near 3.5โ5
isolated CH2โ (e.g., Ar-CH2โ-O, benzylic)
1H singlet near 9.7
aldehyde CHO
signals at 6.5โ8 summing to 4โ5 H
aromatic ring (the pattern hints at substitution)
broad 1H that vanishes on D2โO
exchangeable O-H / N-H
ฯ
C=O
-
ฮดย 5.1, 2H, singlet โ an isolated O-CH2โ deshielded by both ring and oxygen (benzylic, Ar-CH2โ-O).
ฮดย 2.1, 3H, singlet โ a CH3โ next to C=O with no H neighbors (an acetyl methyl, CH3โ-C=O).
-O-
CH3โ-C(=O)-
CH2โ
benzyl acetate
C6โH5โCH2โ-O-C(=O)-CH3โ
1740ย cmโ1
OCH2โ
5.1
2.1
CHCl2โ
โ
ฮดโ5.8
CH2โCl
โฮดโ4.0
Multiplicity: the CHCl2โ (1H) has two neighbors (the CH2โ) โn+1=3โtriplet. The CH2โCl (2H) has one neighbor (the CH) โn+1=2โdoublet.
Integration:CH:CH2โ=1:2.
triplet
CH2โ
CH2โ
doublet
CH
1
H
C=O
aldehyde
CHO
C2โ
H5โ
ฮดย 1.1
ฮดย 2.4
ethyl group
2.4
ฮฑ
2
โ
Jโ1.5ย Hz
CH2โ
CH3โ
CHO
CH2โ
13
C
202
Acetone, (CH3โ)2โCO โ a single 6H singlet at ฮดย 2.1 (two equivalent methyls, no neighbors), no proton downfield of 3, and a 13C carbonyl ~206 with only 2 carbon signals by symmetry.
ฮฑ
Propanal shows loss of H (m/z=57) and the McLafferty-type and acylium fragments characteristic of an aldehyde, with a strong m/z=29 (CHO+) often visible.