NMR Spectroscopy

NMR Spectroscopy — Foundations

Part 1 of 7 — $^1\text{H}$ NMR Basics

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 — 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 $^1\text{H}$ (a single proton) and $^{13}\text{C}$, possess a property called nuclear spin ($I = \tfrac{1}{2}$). 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 $B_0$, each nuclear magnet can adopt one of two orientations:

• aligned with $B_0$ — the lower-energy ($\alpha$) state
• aligned against $B_0$ — the higher-energy ($\beta$) state

The energy gap between these two states, $\Delta E$, is directly proportional to the field strength: $\Delta E = \gamma\, h\, B_0 / 2\pi$, where $\gamma$ is the gyromagnetic ratio of the nucleus. Irradiate the sample with a radiofrequency (RF) pulse whose photon energy exactly matches $\Delta E$, and nuclei in the $\alpha$ state absorb that energy and flip to the $\beta$ state. That absorption — the nucleus coming into resonance — is what the spectrometer detects.

Why a stronger magnet is "higher resolution": because $\Delta E \propto B_0$, 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 $^1\text{H}$ 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 — 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 $B_0$ applied by the magnet.

Every proton is surrounded by electrons. When the sample sits in $B_0$, those electrons circulate and generate their own small induced magnetic field that, near the nucleus, opposes $B_0$. The proton therefore experiences a slightly weaker effective field:

$B_{\text{eff}} = B_0 - B_{\text{induced}}$

This is called shielding: the electrons "shield" the nucleus from the full applied field. A more heavily shielded proton feels a weaker $B_{\text{eff}}$, 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 — it is deshielded, feels a stronger $B_{\text{eff}}$, 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 $\rightarrow$ shielded $\rightarrow$ upfield (small $\delta$); electron-poor environment $\rightarrow$ deshielded $\rightarrow$ downfield (large $\delta$).

Terminology that trips students up: "upfield" and "downfield" are historical terms from old instruments that swept the field. Today "upfield" simply means right / smaller $\delta$ and "downfield" means left / larger $\delta$. Upfield = more shielded; downfield = more deshielded.

Checkpoint — 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 — 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 $^1\text{H}$ 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 — count the signals:

• Methane, $\text{CH}_4$: all four H's are equivalent by tetrahedral symmetry $\rightarrow$ 1 signal.
• Ethanol, $\text{CH}_3\text{CH}_2\text{OH}$: three sets of inequivalent H's — the $\text{CH}_3$, the $\text{CH}_2$, and the $\text{OH}$ — $\rightarrow$ 3 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) $\rightarrow$ 2 signals, even though there are 10 hydrogens.
• Acetone, $(\text{CH}_3)_2\text{C=O}$: the two methyl groups are mirror images across the carbonyl, hence equivalent $\rightarrow$ 1 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.

Equivalence, More Carefully: Homotopic and Enantiotopic vs. Diastereotopic

For everyday structure problems, "related by symmetry $\rightarrow$ 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 $C_n$ axis). They are equivalent in every environment. Example: the three H's of a freely rotating $\text{CH}_3$.
• 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 $\text{CH}_2\text{Cl}_2$, or the two $\text{CH}_2$ 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 $\text{CH}_2$ when the molecule contains a stereocenter or a nearby double bond — for example the two methylene H's of an $\text{R*-CH(X)-CH}_2\text{-Y}$ fragment, or the two $=\text{CH}_2$ 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 — and they are common around stereocenters and double bonds. When a $\text{CH}_2$ unexpectedly shows two signals (often with a large mutual coupling), diastereotopicity is the usual culprit.

Checkpoint — Counting Signals

Chemical Shift ($\delta$): A Field-Independent Address

Each signal needs an address on the horizontal axis. We could report the raw resonance frequency in hertz — 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, $\delta$, in parts per million ($\text{ppm}$):

$\delta = \dfrac{\nu_{\text{sample}} - \nu_{\text{reference}}}{\nu_{\text{spectrometer}}} \times 10^6$

Dividing the frequency offset by the spectrometer frequency cancels the field dependence, so $\delta$ for a given proton is the same number on every instrument. A proton at $\delta = 2\ \text{ppm}$ on a 300 MHz magnet sits 600 Hz downfield of the reference; on a 500 MHz magnet it sits 1000 Hz downfield — but $\delta$ is $2\ \text{ppm}$ in both cases.

The reference compound is tetramethylsilane, $\text{Si(CH}_3)_4$ (TMS), defined as $\delta = 0$. Silicon is more electropositive than carbon, so the 12 equivalent TMS protons are unusually shielded and resonate upfield of virtually all organic protons — a clean, single peak at the right edge of the scale. The $^1\text{H}$ scale then runs from about $0$ to $12\ \text{ppm}$, with deshielded protons at large $\delta$ on the left and shielded protons at small $\delta$ 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 — and that is exactly the table we build next.

Exit Ticket — Part 1 Synthesis

Chemical Shift in Depth

Part 2 of 7 — Chemical Shift ($\delta$)

In Part 1 we established that $\delta$ is a field-independent address measured in $\text{ppm}$ relative to TMS at $\delta = 0$, and that electron density around a proton sets its position: shielded (electron-rich) protons sit upfield at small $\delta$, deshielded (electron-poor) protons sit downfield at large $\delta$. This part turns that qualitative idea into a working chemical-shift table and explains the three structural factors — electronegativity, hybridization, and magnetic anisotropy — that move a proton along the scale.

A practiced organic chemist reads $\delta$ the way a reader reads words: a peak at $1\ \text{ppm}$ says "alkyl," a peak at $7\ \text{ppm}$ says "aromatic," a peak near $10\ \text{ppm}$ says "aldehyde." The goal of this part is to make those associations automatic and, more importantly, to make them explainable — so you can predict the shift of a proton you have never seen.

The Benchmark Chemical-Shift Table

These ranges are the backbone of $^1\text{H}$ NMR interpretation. Commit the order and the approximate values to memory.

Two columns of "exchangeable" protons — $\text{O-H}$ (alcohol $\sim 1\text{–}5$, broad) and $\text{N-H}$ ($\sim 1\text{–}4$, broad) — 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.

Checkpoint — Reading the Table

Factor 1 — Electronegativity (Inductive Deshielding)

The largest routine influence on $\delta$ is the electronegativity of nearby atoms. An electronegative atom pulls electron density away from the C–H bond through the $\sigma$ 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:

More electronegative neighbor $\rightarrow$ larger $\delta$. Multiple withdrawing groups add up: $\text{CH}_3\text{Cl}$ is at $3.0$, $\text{CH}_2\text{Cl}_2$ at $5.3$, $\text{CHCl}_3$ at $7.3$ — each extra chlorine drags the lone remaining proton further downfield.

Distance matters. In 1-chloropropane, $\text{Cl-CH}_2\text{-CH}_2\text{-CH}_3$, the protons nearest chlorine sit at $\sim 3.5\ \text{ppm}$, the middle $\text{CH}_2$ near $1.8\ \text{ppm}$, and the far $\text{CH}_3$ near $1.0\ \text{ppm}$ — essentially back to the value of a plain alkyl group two bonds away. The inductive pull is short-range.

Factor 2 — Hybridization and Factor 3 — Magnetic Anisotropy

Hybridization. Protons on $sp^2$ carbons (alkenes, aromatics) generally appear downfield of protons on $sp^3$ carbons, partly because $sp^2$ carbon is more electronegative (more s-character). But hybridization alone does not explain the numbers, and the famous exception — the terminal alkyne $\text{C}\equiv\text{C-H}$ at only $2\text{–}3\ \text{ppm}$ despite $sp$ carbon — forces us to invoke the third factor.

Magnetic anisotropy is the deshielding (or shielding) produced by the circulating $\pi$ electrons of a nearby multiple bond, and it is the key to the most diagnostic shifts in all of $^1\text{H}$ NMR. When a $\pi$ system sits in $B_0$, its electrons circulate and set up a secondary field that is not uniform — it reinforces $B_0$ in some regions of space and opposes it in others.

• Aromatic ring current: the six $\pi$ electrons of benzene circulate around the ring, generating an induced field that opposes $B_0$ inside the ring but reinforces it in the plane outside the ring — exactly where the ring protons sit. Aromatic H's are therefore strongly deshielded to $6.5\text{–}8\ \text{ppm}$. (A proton held above a ring would instead be shielded — a real effect in some fused systems.)
• Carbonyl and alkene: the $\pi$ electrons of $\text{C=O}$ and $\text{C=C}$ likewise deshield protons lying in the bond plane, contributing to vinylic ($4.5\text{–}6.5$) and aldehyde ($9\text{–}10$) shifts.
• Alkyne (the exception explained): the cylindrical $\pi$ system of $\text{C}\equiv\text{C}$ circulates around the molecular axis, so the terminal proton lying along that axis falls in the shielding cone. This anisotropic shielding pulls $\equiv\text{C-H}$ back upfield to $2\text{–}3\ \text{ppm}$, opposing the $sp$-electronegativity that would otherwise push it downfield.

The trap to name: "$sp$ carbon is the most electronegative, so the alkyne $\equiv\text{C-H}$ should be the most deshielded." It is not — the anisotropic shielding cone of the triple bond overrides hybridization and places it near $2.5\ \text{ppm}$, upfield of vinylic and aromatic protons. Whenever a shift defies the simple electronegativity argument, suspect anisotropy.

Checkpoint — Why Protons Deshield

Exchangeable Protons: O–H and N–H

Protons on oxygen and nitrogen behave differently from C–H protons in three diagnostic ways.

1. Variable shift. Because $\text{O-H}$ and $\text{N-H}$ participate in hydrogen bonding, their $\delta$ depends on concentration, temperature, and solvent. An alcohol $\text{O-H}$ can appear anywhere from $\sim 1$ to $\sim 5\ \text{ppm}$; a carboxylic acid $\text{O-H}$, locked in strong hydrogen bonding, sits far downfield at $10\text{–}12\ \text{ppm}$.
2. Often broad, often a singlet. In many samples the O–H/N–H proton exchanges rapidly between molecules. This fast exchange averages out spin–spin coupling, so the O–H 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 — but the broad, coupling-free singlet is the common case.)
3. The $\text{D}_2\text{O}$ shake test. Add a drop of deuterium oxide and shake: exchangeable O–H/N–H protons swap for deuterium and their signal disappears (or shrinks dramatically) as a new HDO peak grows in. A signal that vanishes on a $\text{D}_2\text{O}$ shake is diagnostic of an exchangeable proton — a quick way to flag an alcohol, acid, or amine.

The trap to name: treating an O–H like an ordinary C–H and trying to apply the $n+1$ rule to it. Because of rapid chemical exchange, the alcohol O–H typically neither splits the adjacent $\text{CH}_2$ nor is split by it — so do not predict (or expect) coupling to it in a routine spectrum.

Exit Ticket — Part 2 Synthesis

Integration and Spin–Spin Splitting

Part 3 of 7 — Integration & Splitting

Chemical shift (Parts 1–2) tells you what kind of proton each signal represents. This part adds the two pieces of information that make $^1\text{H}$ NMR genuinely structure-determining:

• Integration — the area under each signal, which is proportional to the number of protons giving rise to it. This counts H's in each environment.
• Spin–spin splitting (coupling) — 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 integration ratios, define the coupling constant $J$, and draw the splitting trees for the two patterns you must recognize on sight: the ethyl group and the isopropyl group.

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 — 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 — ethyl acetate, $\text{CH}_3\text{-C(=O)-O-CH}_2\text{-CH}_3$ ($\text{C}_4\text{H}_8\text{O}_2$, 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 ($\times 2$): $2 : 2 : 3$. That sums to 7 — but the formula says 8 H, so we scale to the real counts $2 : 3 : 3$:

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 — always normalize.

Reading tip: $\text{CH}_3$ 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 ($\text{CH}$), an aldehyde, or an exchangeable O–H/N–H.

Checkpoint — Integration

Spin–Spin 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 $B_0$, nudging the observed proton's resonance slightly higher or lighter. The result is that a single signal is split into several lines — 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 — typically three bonds away ($\text{H-C-C-H}$) — not the protons in the signal itself. Equivalent protons do not split one another.

The intensity pattern within a multiplet follows Pascal's triangle — 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 (↑↑), (↑↓ or ↓↑), or (↓↓): the middle, mixed arrangement is twice as likely.

The number-one splitting trap: counting the protons in the signal instead of the neighbors. A $\text{CH}_3$ does not split itself into a quartet; it is split by what is attached to the next carbon. A $\text{CH}_3$ next to a $\text{CH}_2$ appears as a triplet ($n = 2$ neighbors), and that $\text{CH}_2$ appears as a quartet ($n = 3$ neighbors). This reciprocal relationship is the ethyl fingerprint.

Worked Splitting Tree 1 — The Ethyl Group ($\text{CH}_3\text{CH}_2\text{-X}$)

The ethyl group attached to an electron-withdrawing X (as in bromoethane, $\text{CH}_3\text{CH}_2\text{Br}$) is the most important pattern in introductory NMR. Two coupled environments:

The $\text{CH}_3$ (3H): its neighbors are the two $\text{CH}_2$ protons, so $n = 2 \rightarrow 2 + 1 = 3$ lines $\rightarrow$ a triplet, intensities $1:2:1$, near $\delta \approx 1.7$.

The $\text{CH}_2$ (2H): its neighbors are the three $\text{CH}_3$ protons, so $n = 3 \rightarrow 3 + 1 = 4$ lines $\rightarrow$ a quartet, intensities $1:3:3:1$, near $\delta \approx 3.4$ (deshielded by Br).

Building the $\text{CH}_2$ quartet as a tree — 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: $\rightarrow$ 2 lines (doublet)
• Split by H#2: each line splits again $\rightarrow$ 4 lines, but the inner two overlap $\rightarrow$ $1:2:1$
• Split by H#3: $\rightarrow$ the lines overlap into the $1:3:3:1$ 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: $\text{CH}_2 : \text{CH}_3 = 2 : 3$.

Ethyl fingerprint to memorize: a 2H quartet + a 3H triplet sharing the same $J$ ($\approx 7\ \text{Hz}$) almost always means an $\text{-CH}_2\text{CH}_3$ group. Find that pair and you have located an ethyl.

Worked Splitting Tree 2 — The Isopropyl Group $(\text{CH}_3)_2\text{CH-X}$

The isopropyl group is the second must-know pattern. In 2-bromopropane, $(\text{CH}_3)_2\text{CHBr}$:

The two $\text{CH}_3$ groups (6H, equivalent): each methyl's only neighbor is the single methine proton, so $n = 1 \rightarrow 2$ lines $\rightarrow$ a doublet, intensity $1:1$. Because the two methyls are equivalent, they overlap into one 6H doublet near $\delta \approx 1.7$.

The methine $\text{CH}$ (1H): its neighbors are the six equivalent methyl protons, so $n = 6 \rightarrow 6 + 1 = 7$ lines $\rightarrow$ a septet ($1:6:15:20:15:6:1$), near $\delta \approx 4.3$ (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 — 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 $(\text{CH}_3)_2\text{CH-}$ group.

Contrast the two fingerprints — 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 $\text{O-CH}_2$ or another environment. Always check the integration (2H vs 6H, 3H vs 1H) and the partner multiplicity together — the pair, not a single multiplet, identifies the group.

Checkpoint — 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:

1. $J$ is field-independent. Unlike chemical shift (which we deliberately report in field-independent $\delta$), 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 — a useful way to distinguish a true multiplet from two separate signals.)
2. Coupled partners share the same $J$. Two protons that couple to each other split each other by the identical $J$. This is how you pair up multiplets: the $\text{CH}_2$ quartet and the $\text{CH}_3$ triplet of an ethyl group have matching $J \approx 7\ \text{Hz}$, confirming they are neighbors.

Typical magnitudes are themselves diagnostic of geometry:

The trans > cis relationship across a double bond is especially useful: a $J \approx 16\ \text{Hz}$ between two vinyl protons is strong evidence for a trans (E) alkene, while $J \approx 8\ \text{Hz}$ 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 — it was two distinct chemical shifts.

Exit Ticket — Part 3 Synthesis

Carbon-13 NMR

Part 4 of 7 — $^{13}\text{C}$ NMR

So far we have watched protons. But the carbon skeleton itself can be observed directly with $^{13}\text{C}$ NMR, and it is wonderfully complementary to $^1\text{H}$: it counts carbon 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.

The catch is sensitivity. The dominant carbon isotope, $^{12}\text{C}$, has $I = 0$ and is NMR-silent. Only $^{13}\text{C}$ ($I = \tfrac{1}{2}$) gives a signal, and it makes up just $1.1\%$ of natural carbon. Combined with a gyromagnetic ratio about a quarter that of $^1\text{H}$, $^{13}\text{C}$ NMR is intrinsically thousands of times less sensitive than $^1\text{H}$ NMR — 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 $^{13}\text{C}$ nuclei being adjacent in one molecule are tiny ($\sim 0.01\%$), so carbon–carbon coupling essentially never complicates the spectrum.

One Signal per Unique Carbon Environment

The counting rule mirrors $^1\text{H}$ NMR: the number of signals equals the number of chemically distinct carbon environments, set by molecular symmetry. The huge advantage is dispersion — the $^{13}\text{C}$ scale runs from roughly $0$ to $220\ \text{ppm}$, about twenty times wider than the $^1\text{H}$ window — 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 — counting carbon signals:

• Acetone, $(\text{CH}_3)_2\text{C=O}$: the two methyls are equivalent, the carbonyl is unique $\rightarrow$ 2 signals (the $\text{C=O}$ near $206\ \text{ppm}$, the two equivalent $\text{CH}_3$ near $30\ \text{ppm}$).
• Benzene, $\text{C}_6\text{H}_6$: all six carbons equivalent by symmetry $\rightarrow$ 1 signal (~$128\ \text{ppm}$).
• Toluene, $\text{C}_6\text{H}_5\text{CH}_3$: symmetry makes the ring carbons fall into four sets (ipso, ortho, meta, para) plus the methyl $\rightarrow$ 5 signals.
• 1,4-dimethylbenzene (para-xylene): two unique ring carbons + one methyl $\rightarrow$ 3 signals.

Notice how molecular symmetry can make the $^1\text{H}$ and $^{13}\text{C}$ 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 $^{13}\text{C}$ signals equals the number of $^1\text{H}$ signals. They count different nuclei. Quaternary carbons (bearing no hydrogen) appear in $^{13}\text{C}$ but contribute no $^1\text{H}$ signal at all — so $^{13}\text{C}$ often reveals carbons that are invisible in the proton spectrum.

Checkpoint — Carbon Counting and Sensitivity

Broadband Decoupling: Why $^{13}\text{C}$ Signals Are Singlets

A raw $^{13}\text{C}$ nucleus is coupled — strongly — to the protons attached to it ($^1J_{\text{CH}}$ can be $125\text{–}250\ \text{Hz}$). Left alone, every carbon would be split by its hydrogens (a $\text{CH}_3$ carbon into a quartet, a $\text{CH}_2$ 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 $^{13}\text{C}$, the spectrometer simultaneously irradiates all the protons across their entire frequency range. This rapid irradiation averages the C–H coupling to zero, so each carbon collapses to a single sharp line. The familiar "$^{13}\text{C}$ spectrum" — one singlet per carbon environment — is really a proton-decoupled spectrum.

Two important consequences:

• No C–C splitting either. Because adjacent $^{13}\text{C}$–$^{13}\text{C}$ pairs are vanishingly rare ($1.1\% \times 1.1\%$), carbon–carbon coupling is not observed in routine spectra. With protons decoupled too, every line is a singlet.
• Integration is usually unreliable. A side effect of decoupling (the nuclear Overhauser effect) and the slow relaxation of carbons — especially quaternary carbons — mean that $^{13}\text{C}$ peak areas are not reliably proportional to the number of carbons. Unlike $^1\text{H}$ integration, you normally do not read carbon counts from $^{13}\text{C}$ peak heights or areas without special acquisition conditions.

The trap to name: trying to apply the $n+1$ rule to a routine $^{13}\text{C}$ spectrum, or trying to integrate it like a proton spectrum. Standard $^{13}\text{C}$ is broadband-decoupled — every signal is a singlet by design — 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 C–H 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 $\text{CH}$ (methine) carbons appear.
• DEPT-135: $\text{CH}$ and $\text{CH}_3$ carbons point up (positive), $\text{CH}_2$ 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:

Worked use: a carbon that is present in the standard spectrum but missing from DEPT-135 must be quaternary — a carbonyl carbon, a fully substituted ring carbon, or a $\text{C}$ bearing no H. This is how you locate carbonyls and quaternary centers that are silent in $^1\text{H}$ NMR. An older alternative, off-resonance decoupling, achieved a similar classification by leaving only the one-bond C–H coupling so that $\text{CH}_3 \rightarrow$ quartet, $\text{CH}_2 \rightarrow$ triplet, $\text{CH} \rightarrow$ doublet, quaternary $\rightarrow$ singlet — 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 $\text{CH}_3$, $\text{CH}_2$, $\text{CH}$, or quaternary. The "disappearing" peaks in DEPT are exactly the quaternary carbons.

Checkpoint — Decoupling and DEPT

The $^{13}\text{C}$ Chemical-Shift Map ($0\text{–}220\ \text{ppm}$)

Carbon shifts respond to the same factors as proton shifts — electronegativity, hybridization, anisotropy — but over a far broader range, which makes the regions very diagnostic. Memorize these zones:

Two regions resolve ambiguities that proton NMR cannot:

• A line at $160\text{–}220\ \text{ppm}$ is essentially always a carbonyl carbon — and its sub-region distinguishes acid/ester/amide ($160\text{–}185$) from aldehyde/ketone ($190\text{–}220$). Proton NMR cannot see a carbonyl carbon directly; $^{13}\text{C}$ pinpoints it.
• The $100\text{–}150\ \text{ppm}$ window flags $sp^{2}$ carbons (alkene or aromatic), confirming unsaturation.

The trap to name: mapping $^{13}\text{C}$ values onto the $^1\text{H}$ scale. A carbonyl carbon sits near $200\ \text{ppm}$, but no proton ever appears at $200\ \text{ppm}$ — the $^1\text{H}$ scale stops around $12$. Keep the two scales mentally separate; they measure different nuclei and span different ranges.

Exit Ticket — Part 4 Synthesis

Putting It Together: Structure Determination

Part 5 of 7 — 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:

1. Molecular formula $\rightarrow$ degrees of unsaturation. Establish how many rings and $\pi$ bonds the molecule must contain.
2. IR $\rightarrow$ functional groups. Identify the major functional groups (C=O, O–H, N–H, C≡N, aromatic).
3. $^1\text{H}$ NMR $\rightarrow$ environments, counts, neighbors. Count signals, read integration, and decode splitting to find fragments.
4. Assemble the fragments into a structure consistent with every piece of data, then check it against the spectra.

Step 1 — Degrees of Unsaturation

The degree of unsaturation (DoU, also "index of hydrogen deficiency") counts the total number of rings plus $\pi$ bonds in a molecule. Every ring and every $\pi$ 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 $\text{C}_c\text{H}_h\text{N}_n\text{O}_o\text{X}_x$ (X = halogen):

$\text{DoU} = \dfrac{2c + 2 + n - h - x}{2}$

Oxygen does not appear because adding an O (e.g., inserting it into a C–H or C–C bond) does not change the hydrogen count. Each halogen counts like a hydrogen; each nitrogen adds one to the numerator.

Worked calculations:

• $\text{C}_4\text{H}_8\text{O}$: $\text{DoU} = (2\cdot 4 + 2 - 8)/2 = 2/2 = 1$. One ring or one $\pi$ bond — consistent with a ketone, an aldehyde, or a cyclic ether/alcohol.
• $\text{C}_7\text{H}_8\text{O}$: $\text{DoU} = (14 + 2 - 8)/2 = 4$. Four degrees strongly suggests a benzene ring (a benzene ring is exactly 4: three $\pi$ bonds + one ring).
• $\text{C}_8\text{H}_9\text{NO}_2$: $\text{DoU} = (16 + 2 + 1 - 9)/2 = 5$. A benzene ring (4) plus one more $\pi$ bond (e.g., a C=O).

The interpretive habit: a DoU of 4 is the fingerprint of an aromatic ring — see it and immediately look for aromatic signals at $\delta\, 6.5\text{–}8$ in $^1\text{H}$ NMR and $100\text{–}150\ \text{ppm}$ in $^{13}\text{C}$. 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 — Degrees of Unsaturation

Step 2 — 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.

The strategy: if DoU $\geq 1$ and IR shows a strong band near $1715\ \text{cm}^{-1}$, a carbonyl accounts for one degree — now decide which carbonyl using NMR shift and chemistry. If DoU = 4 and IR shows bands at $1600/1500\ \text{cm}^{-1}$, 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 $1715\ \text{cm}^{-1}$ band with no leftover degree of unsaturation would be a contradiction — re-examine the formula. When IR and DoU agree (e.g., "one C=O uses the molecule’s single degree"), you have a firm functional-group assignment to carry into the NMR analysis.

Step 3 — $^1\text{H}$ NMR: Environments, Counts, and Neighbors

Now extract fragments from the proton spectrum, reading three properties of every signal in concert:

• Chemical shift $\rightarrow$ what kind of proton (alkyl, $\alpha$-to-O, vinyl, aromatic, aldehyde…).
• Integration $\rightarrow$ how many protons in that environment (reduce the ratio, reconcile with the formula).
• Multiplicity ($n+1$) $\rightarrow$ how many neighbors, which tells you what is attached to the adjacent carbon.

Translate each signal into a fragment and tally the protons:

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 — any shortfall points to an undetected fragment (often an exchangeable proton or a symmetric duplication).

Checkpoint — Reading Fragments

Worked Structure Determination — $\text{C}_9\text{H}_{10}\text{O}_2$

Let us run the full pipeline on an unknown of formula $\text{C}_9\text{H}_{10}\text{O}_2$.

Step 1 — DoU: $(2\cdot 9 + 2 - 10)/2 = (20 - 10)/2 = 5$. Five degrees: a benzene ring (4) plus one more $\pi$ bond — anticipate an aromatic ring and one $\text{C=O}$.

Step 2 — IR: a strong band at $1740\ \text{cm}^{-1}$ confirms an ester-type carbonyl (using the 5th degree); bands at $1600/1500\ \text{cm}^{-1}$ confirm the aromatic ring. No broad O–H, so not a carboxylic acid.

Step 3 — $^1\text{H}$ NMR:

• $\delta\ 7.3$, 5H, multiplet $\rightarrow$ a monosubstituted benzene ($\text{C}_6\text{H}_5\text{-}$).
• $\delta\ 5.1$, 2H, singlet $\rightarrow$ an isolated $\text{O-CH}_2$ deshielded by both ring and oxygen (benzylic, $\text{Ar-CH}_2\text{-O}$).
• $\delta\ 2.1$, 3H, singlet $\rightarrow$ a $\text{CH}_3$ next to C=O with no H neighbors (an acetyl methyl, $\text{CH}_3\text{-C=O}$).

Protons assigned: $5 + 2 + 3 = 10$, matching $\text{H}_{10}$.

Step 4 — Assemble. Fragments: $\text{C}_6\text{H}_5\text{-}$, $\text{-CH}_2\text{-O-}$, $\text{CH}_3\text{-C(=O)-}$, plus the ester carbonyl/oxygen from IR + DoU. Stitching them so the ester links the benzylic $\text{CH}_2$ to the acetyl group gives benzyl acetate, $\text{C}_6\text{H}_5\text{CH}_2\text{-O-C(=O)-CH}_3$. Check: 9 C, 10 H, 2 O ✓; ester C=O at $1740\ \text{cm}^{-1}$ ✓; benzylic $\text{OCH}_2$ singlet at $5.1$ ✓; acetyl singlet at $2.1$ ✓; five aromatic H ✓. Every datum is satisfied — 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; NMR’s singlets (no neighbors) then forced the connectivity. Each step narrowed the field, so the final assembly had essentially one consistent answer.

Exit Ticket — Part 5 Synthesis

Problem-Solving Workshop

Part 6 of 7 — 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 — 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:

1. DoU from the formula — how many rings/$\pi$ bonds?
2. IR — which functional groups (C=O? O–H? N–H? C≡N? aromatic)?
3. $^1\text{H}$: for each signal note shift (what kind), integration (how many), multiplicity (how many neighbors) $\rightarrow$ a fragment.
4. $^{13}\text{C}$ / DEPT — confirm carbon count, find carbonyls and quaternary carbons.
5. Assemble and verify against all data, including symmetry.

Forward Problem — Predicting a Spectrum From a Structure

Before deducing structures, practice the reverse: given a structure, predict its $^1\text{H}$ NMR. This sharpens the same instincts. Take 1,1,2-trichloroethane, $\text{CHCl}_2\text{-CH}_2\text{Cl}$.

• Environments: two — the $\text{CHCl}_2$ methine (1H) and the $\text{CH}_2\text{Cl}$ methylene (2H).
• Shifts: both are deshielded by chlorine. The $\text{CHCl}_2$ proton bears two chlorines on its own carbon $\rightarrow$ very downfield, $\delta \approx 5.8$. The $\text{CH}_2\text{Cl}$ bears one chlorine $\rightarrow \delta \approx 4.0$.
• Multiplicity: the $\text{CHCl}_2$ (1H) has two neighbors (the $\text{CH}_2$) $\rightarrow n+1 = 3 \rightarrow$ triplet. The $\text{CH}_2\text{Cl}$ (2H) has one neighbor (the $\text{CH}$) $\rightarrow n+1 = 2 \rightarrow$ doublet.
• Integration: $\text{CH} : \text{CH}_2 = 1 : 2$.

Predicted spectrum: a 1H triplet at $\sim 5.8$ and a 2H doublet at $\sim 4.0$, sharing one $J \approx 6\ \text{Hz}$. Notice the reciprocity — 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 "$\text{CH}_2 \rightarrow$ triplet, $\text{CH} \rightarrow$ doublet" by reflex. Here it is the opposite: the lone $\text{CHCl}_2$ proton is a triplet (two $\text{CH}_2$ neighbors) and the $\text{CH}_2$ is a doublet (one $\text{CH}$ neighbor). Always split by what is next door.

Checkpoint — Forward Prediction

Reverse Problem 1 — Deduce $\text{C}_3\text{H}_6\text{O}$ From NMR

Given: formula $\text{C}_3\text{H}_6\text{O}$. $^1\text{H}$ NMR: $\delta\ 9.8$ (1H, triplet, $J \approx 1.5\ \text{Hz}$); $\delta\ 2.4$ (2H, doublet of quartets / multiplet); $\delta\ 1.1$ (3H, triplet). $^{13}\text{C}$: a peak near $202\ \text{ppm}$.

Step 1 — DoU: $(2\cdot 3 + 2 - 6)/2 = 1$. One ring or one $\pi$ bond.

Step 2 — Carbonyl? The $^{13}\text{C}$ line at $202\ \text{ppm}$ is in the aldehyde/ketone region, and the $\delta\ 9.8$ proton in $^1\text{H}$ is the aldehyde signature. So the one degree of unsaturation is a $\text{C=O}$, specifically an aldehyde (it has a C–H). That consumes $\text{CHO}$: 1 C, 1 H, 1 O.

Step 3 — Remaining fragments: $\text{C}_3\text{H}_6\text{O}$ minus $\text{CHO}$ leaves $\text{C}_2\text{H}_5$. The 3H triplet ($\delta\ 1.1$) + 2H signal ($\delta\ 2.4$) is an ethyl group, and the 2H sits at $2.4$ because it is $\alpha$ to the carbonyl.

Step 4 — Assemble: $\text{CH}_3\text{CH}_2\text{-CHO}$ = propanal. Verify the fine splitting: the aldehyde proton is a triplet because it couples weakly to the adjacent $\text{CH}_2$ (small $J \approx 1.5\ \text{Hz}$); the $\text{CH}_2$ is split by both the $\text{CH}_3$ (3 neighbors) and the $\text{CHO}$ (1 neighbor). 3 C, 6 H, 1 O ✓.

The diagnostic pivot: a $^{13}\text{C}$ line above $190\ \text{ppm}$ together with a $\delta\ 9.8$ proton is essentially conclusive for an aldehyde — neither datum alone is as decisive as the two together. The aldehyde proton being a triplet (not a singlet) is the tell that a $\text{CH}_2$ is adjacent: propanal, not acetone (which would have no aldehyde proton at all).

Reverse Problem 2 — Two Isomers of $\text{C}_3\text{H}_6\text{O}$, and Using MS

A subtle skill is distinguishing isomers that share a formula. $\text{C}_3\text{H}_6\text{O}$ has two common carbonyl isomers:

• Propanal, $\text{CH}_3\text{CH}_2\text{CHO}$ — aldehyde proton at $9.8$ (1H), ethyl pattern, $^{13}\text{C}$ carbonyl ~$202$.
• Acetone, $(\text{CH}_3)_2\text{CO}$ — a single 6H singlet at $\delta\ 2.1$ (two equivalent methyls, no neighbors), no proton downfield of $3$, and a $^{13}\text{C}$ carbonyl ~$206$ with only 2 carbon signals by symmetry.

Their $^1\text{H}$ 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 ion $\text{CH}_3\text{CO}^+$ at $m/z = 43$ — a dominant peak (the classic $\alpha$-cleavage of a ketone).
• Propanal shows loss of $\text{H}$ ($m/z = 57$) and the McLafferty-type and acylium fragments characteristic of an aldehyde, with a strong $m/z = 29$ ($\text{CHO}^+$) often visible.

Integrative habit: when two structures fit the NMR, look to MS fragmentation or the $^{13}\text{C}$ count. Here, acetone’s two carbon signals (vs. propanal’s 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 — Distinguishing Isomers

Reverse Problem 3 — An Aromatic Unknown, $\text{C}_8\text{H}_8\text{O}$

Given: $\text{C}_8\text{H}_8\text{O}$. IR: strong band at $1685\ \text{cm}^{-1}$ (conjugated C=O) and $1600/1580\ \text{cm}^{-1}$ (aromatic). $^1\text{H}$ NMR: $\delta\ 7.9$ (2H, doublet-like), $\delta\ 7.5$ (3H, multiplet), $\delta\ 2.6$ (3H, singlet). $^{13}\text{C}$: ~$198\ \text{ppm}$ (carbonyl), plus aromatic carbons $128\text{–}137$.

Step 1 — DoU: $(2\cdot 8 + 2 - 8)/2 = 5$. Aromatic ring (4) + one $\text{C=O}$ (1).

Step 2 — IR: carbonyl at the lowered $1685\ \text{cm}^{-1}$ signals conjugation with the ring (an aryl ketone, not an aldehyde — there is no aldehyde proton near $9.7$).

Step 3 — $^1\text{H}$: the aromatic protons total 5H ($2 + 3$) $\rightarrow$ a monosubstituted benzene, with the 2H further downfield ($7.9$) being the ortho protons next to the electron-withdrawing carbonyl. The 3H singlet at $2.6$ is a $\text{CH}_3$ attached to C=O (no neighbors) — an acetyl methyl, shifted slightly downfield of a normal acetyl by the adjacent ring.

Step 4 — Assemble: $\text{C}_6\text{H}_5\text{-C(=O)-CH}_3$ = acetophenone. Check: $5 + 3 = 8$ H ✓; aryl-conjugated C=O at $1685\ \text{cm}^{-1}$ and $^{13}\text{C}\ 198$ ✓; 5 aromatic H as 2H + 3H ✓; methyl singlet ✓.

Two cross-checks worth naming: (1) the lowered carbonyl frequency ($1685$ vs. a normal ketone $\sim 1715\ \text{cm}^{-1}$) reveals conjugation — 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 — Part 6 Synthesis

Synthesis and Review

Part 7 of 7 — Synthesis & Review

NMR is the most powerful structural tool in the organic chemist's kit because a single pair of experiments ($^1\text{H}$ and $^{13}\text{C}$) 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 $^1\text{H}$ spectrum delivers, how each maps onto structure, and the decision flow that ties shift, integration, splitting, and the carbon spectrum into a single answer.

The unifying idea from Parts 1–6: 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 $^1\text{H}$ Spectrum

Each observable is independent, and they constrain one another. A signal that is downfield (large $\delta$), integrates for 1H, and is a singlet might be an aldehyde or an exchangeable O–H; add that it sits at $9.7$ (not broad, not vanishing on $\text{D}_2\text{O}$) and it is an aldehyde. No single observable identifies a fragment — 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 — Mapping Observables to Structure

Consolidated Shift Map — $^1\text{H}$ and $^{13}\text{C}$ 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:

The two scales rhyme — both move downfield as protons/carbons become more deshielded — but they cover very different ranges. A carbonyl carbon lives near $200\ \text{ppm}$; no proton ever does. Use $^{13}\text{C}$ to find carbonyls and quaternary carbons (invisible in $^1\text{H}$), and use DEPT to label each carbon as $\text{CH}_3$, $\text{CH}_2$, $\text{CH}$, or quaternary.

The persistent trap: "a peak at $200$ must be a far-downfield proton." There is no such proton — the $^1\text{H}$ window ends near $12$. A value of $200\ \text{ppm}$ is a carbon shift, almost always a carbonyl. Always check which spectrum you are reading.

Splitting Reveals Connectivity — 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:

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 same $J$, which is how you pair a multiplet with its partner and so chain fragments together. A singlet, by contrast, announces no coupling neighbors — 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 $^{13}\text{C}$ spectrum; (d) expecting an alcohol O–H to couple to its $\text{CH}_2$ (rapid exchange usually erases that coupling). Avoid these four and most interpretation errors vanish.

Checkpoint — Fingerprints and Connectivity

The Master Workflow — Systematic Structure Determination

Pulling Parts 5–6 into one checklist you can run on any unknown:

1. Molecular formula $\rightarrow$ DoU. $\text{DoU} = (2c + 2 + n - h - x)/2$. A DoU of 4 flags an aromatic ring; a DoU of 1 with an O flags one ring or one C=O.
2. IR $\rightarrow$ functional groups. $\sim 1715\ \text{cm}^{-1}$ = C=O (lowered by conjugation); broad $3200\text{–}3550$ = O–H; $\sim 2250$ = nitrile/alkyne; $1600/1500$ = aromatic.
3. $^1\text{H}$ NMR $\rightarrow$ fragments. For each signal: shift (identity), integration (count), multiplicity (neighbors). Recognize the fingerprints; sum the protons against the formula.
4. $^{13}\text{C}$ / DEPT $\rightarrow$ carbon skeleton. Count environments, locate carbonyls ($160\text{–}220$) and $sp^{2}$ carbons ($100\text{–}150$); use DEPT to classify $\text{CH}_3$/$\text{CH}_2$/$\text{CH}$/quaternary.
5. Assemble and verify. Connect fragments consistently with every datum (including symmetry), then read the proposed structure back against each spectrum.

Capstone illustration ($\text{C}_8\text{H}_{10}$, DoU = 4): IR shows aromatic bands only (no C=O, no O–H). $^1\text{H}$ NMR: a 4H aromatic singlet-like region near $7.1$ and a 6H singlet at $2.3$. The 6H singlet = two equivalent ring-attached methyls with no H neighbors; the symmetric 4H aromatic pattern = a para-disubstituted ring. Assemble $\rightarrow$ para-xylene (1,4-dimethylbenzene). The $^{13}\text{C}$ count of just 3 signals confirms the high symmetry. Every observable — DoU, IR, the two $^1\text{H}$ signals, and the carbon count — 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 $\rightarrow$ IR $\rightarrow$ NMR) and pattern recognition (the fingerprints) are what make that convergence fast and reliable.

Exit Ticket — Course Synthesis