DEPT: Classifying Carbons by Attached Hydrogens

A crowded ¹³C NMR spectrum of a steroid or an alkaloid can show forty peaks between 10 and 80 ppm. All of them look like similar bumps of similar height. A plain proton-decoupled spectrum cannot say which peak belongs to a CH₃, which to a CH₂, which to a CH. It also cannot identify a carbon bearing no hydrogen at all. That single missing fact — how many protons sit on each carbon — is often crucial. It is the difference between a structure that is merely plausible and one that is proven. DEPT (Distortionless Enhancement by Polarization Transfer) answers exactly this question. It does not add a new dimension to the spectrum. Instead, it manipulates the phase and intensity of every carbon signal. This follows a strict trigonometric rule tied to its number of attached protons. The technique matters because 13C NMR: Sensitivity, Shifts, and Signal Counting already told us how many chemically distinct carbons exist. It also told us roughly what electronic environment each occupies. But it is blind to substitution pattern. A quaternary carbon at 40 ppm and a CH at 40 ppm look identical. This is true in an ordinary broadband-decoupled spectrum. DEPT breaks that degeneracy by exploiting the same scalar spin-spin coupling, $^1J_{CH}$ . Decoupling ordinarily throws this coupling away. Instead of destroying the ¹H–¹³C coupling information, DEPT uses a carefully timed sequence of radiofrequency pulses. This lets that coupling evolve for a controlled interval. It then transfers proton magnetization onto carbon and reads out a spectrum. The peak phases (up or down) and relative sizes encode CH₃, CH₂, and CH multiplicity. Quaternary carbons, having no protons to donate magnetization, vanish entirely. The physical picture rests on vector/product-operator reasoning. Proton magnetization precesses under the one-bond C–H coupling during a fixed delay. It evolves into an antiphase state that depends on how many protons are coupled to that carbon. A final proton pulse of variable tip angle θ acts as a filter. This is because $\sin\theta$ , $\sin\theta\cos\theta$ , and $\sin\theta\cos^2\theta$ terms (for CH, CH₂, CH₃ respectively) each have their own θ-dependence. Choosing θ = 90° isolates CH carbons alone. Choosing θ = 135° gives a spectrum with CH and CH₃ up, CH₂ down. Quaternary carbons are absent from both. This article builds that trigonometry from the underlying spin mechanism. It works three real structures from data to unambiguous carbon count. It also flags the traps that trip up real spectral analysis. These include overlapping shifts, multiplicities near null points, and the ever-present invisibility of quaternary carbons. The payoff is practical and immediate. Combined with a broadband ¹³C spectrum, a HSQC/HMQC correlation, and full 2D methods, this helps greatly. These are used later in solving-combined-spectra-problems. DEPT converts a forest of unassigned peaks into a hydrogen-count map. This map covers the entire carbon skeleton. It often pins down a structure before a single 2D experiment is even run.

Key takeaways

DEPT edits a ¹³C spectrum by attached-proton count using a proton pulse of tip angle θ following polarization transfer through $^1J_{CH}$ .
DEPT-90 shows only CH carbons; all CH₂, CH₃, and quaternary carbons vanish.
DEPT-135 shows CH and CH₃ phased up, CH₂ phased down, and quaternary carbons absent entirely (zero intensity, not just invisible-by-null).
Quaternary carbons never appear in any DEPT spectrum — they are found by subtraction: present in the full ¹³C spectrum, absent from DEPT-135.
The intensity of each multiplicity type follows $\sin\theta\cos^{n-1}\theta$ for $n$ attached protons, which is why θ = 90° and 135° are the diagnostic sweet spots.
Because DEPT relies on a single, fixed $^1J_{CH}$ delay, protons with unusual one-bond couplings (alkynes, some heteroaromatics) can give attenuated or misleading intensities.
Key idea: the phase and size of a DEPT peak are a direct trigonometric fingerprint of how many hydrogens are bonded to that carbon — nothing more, nothing less.

The core insight

DEPT reads out $\sin\theta\cos^{n-1}\theta$ , the amplitude of an $n$ -proton spin system's magnetization after coherence transfer, as a function of a single adjustable proton pulse angle θ — turning "how many H's are on this carbon" into "which way does the peak point, and how tall is it."

Where this lives today

DEPT remains a default, near-universal acquisition in synthetic organic chemistry laboratories: whenever a new compound is characterized, a broadband ¹³C spectrum and a DEPT-135 (often alongside DEPT-90) are typically run in the same overnight queue, because DEPT costs little extra instrument time yet resolves the CH/CH₂/CH₃/quaternary ambiguity that a broadband spectrum alone cannot. In pharmaceutical process chemistry, DEPT is routinely used to confirm that a new synthetic route has produced the intended substitution pattern — for instance, distinguishing a desired secondary alcohol product (CH–OH) from an over-reduced or under-reduced regiochemical byproduct, purely from the up/down phase of the carbinol carbon, without needing to isolate authentic reference material. In natural product structure elucidation — steroids, terpenoids, alkaloids, and polyketides with dozens of overlapping aliphatic carbons — DEPT is frequently the first spectrum examined after the broadband ¹³C, because it immediately partitions a crowded 20–50 ppm region into CH₃ groups (informative about branching pattern and often diagnostic of specific biosynthetic building blocks), CH₂ groups (informative about ring size and chain length), and CH stereocenters (often exactly the carbons bearing the stereochemical information a synthesis needs to confirm). Quaternary carbons revealed by subtraction typically mark ring-fusion carbons, fully substituted alkene or aromatic carbons, and carbonyls — precisely the skeletal "joints" of a natural product scaffold. In modern practice DEPT has been substantially supplemented by two-dimensional heteronuclear correlation methods — HSQC, which correlates each proton shift with its directly attached carbon shift, and multiplicity-edited HSQC, which folds the CH/CH₂/CH₃ information directly into a 2D map, achieving in one experiment what once took DEPT-90 and DEPT-135 separately. Nonetheless DEPT remains valuable precisely because it is a 1D experiment: faster to acquire on limited sample quantities, simpler to interpret without specialized software, and often the deciding piece of evidence in the kind of multi-spectrum problem-solving developed further in solving-combined-spectra-problems, where a DEPT-derived hydrogen count for every carbon is combined with proton integration, coupling patterns, and 2D correlations to pin down a complete unknown structure. DEPT-based multiplicity editing also underlies automated structure verification software used in modern pharmaceutical quality control, where a proposed structure's predicted CH/CH₂/CH₃/quaternary pattern is checked computationally against an experimental DEPT dataset as one of several automated consistency checks before a batch is released, illustrating how a mechanistically simple pulse-sequence trick has become embedded in industrial-scale analytical infrastructure.

Ask the tutor: How do I combine a DEPT-135 spectrum with an HSQC and an HMBC to fully assign a complex unknown structure?

The physical picture

Begin with the problem DEPT is built to solve. A one-dimensional, proton-decoupled ¹³C spectrum shows one peak per chemically inequivalent carbon, at a chemical shift set by hybridization, electronegative substituents, and ring current effects — the content of 13C NMR: Sensitivity, Shifts, and Signal Counting. But decoupling is achieved by continuously irradiating all protons during acquisition, which averages out (to zero) the scalar coupling $^1J_{CH}$ between a carbon and its directly attached protons. That coupling is precisely the handle DEPT needs, so DEPT must be run without decoupling during a preparatory period, only turning on decoupling for the final acquisition. The molecular-level mechanism is a coherence transfer experiment. Protons, with a gyromagnetic ratio about four times larger than that of carbon, possess much larger equilibrium magnetization and relax roughly an order of magnitude faster. Rather than exciting the dilute, slowly-relaxing ¹³C nuclei directly (as in a simple pulse-acquire ¹³C experiment), DEPT excites the abundant, fast-relaxing ¹H nuclei, lets the one-bond J-coupling evolve into an antiphase proton coherence, and then applies a simultaneous 90° pulse on both nuclei that transfers the antiphase proton magnetization onto carbon. This is why DEPT is dramatically more sensitive than a plain ¹³C experiment for protonated carbons — the experiment inherits proton-sized magnetization and recycles at proton-sized $T_1$ times — but it is also why carbons with zero attached protons give exactly zero signal: there is no proton magnetization to transfer. The pulse sequence, in the language of product operators, is:

{}^1\text{H}: \; 90°_x - \tau - 180°_x - \tau - \theta_y - \text{(acquire, with }^{13}\text{C decoupling)}

{}^{13}\text{C}: \; \quad\quad\;\; 180°_x - \quad\quad\; 90°_{\pm x} - \text{(acquire)}

with $\tau = 1/(2J_{CH})$ chosen to match the typical one-bond coupling (roughly 125–145 Hz for sp³ C–H, giving $\tau \approx 3.6$ ms). The initial ¹H 90° pulse creates transverse proton magnetization; the simultaneous ¹H and ¹³C 180° pulses refocus chemical shift evolution while allowing J-coupling to keep evolving (this is the standard spin-echo trick, isolating heteronuclear coupling from the confounding effects of chemical shift and field inhomogeneity); after the second $\tau$ delay the proton coherence has become fully antiphase with respect to the coupled carbon. The final ¹H pulse of angle θ and the simultaneous ¹³C 90° pulse together execute the actual polarization transfer, converting antiphase proton coherence into observable, in-phase carbon coherence whose sign and size depend on θ and on how many protons are attached.

Figure

Figure 1 — DEPT pulse sequence and the coherence-transfer logic

The key trigonometric result — the load-bearing equation of the whole technique — comes directly from working out how the final θ pulse acts on an antiphase proton coherence when that proton is one of $n$ magnetically equivalent protons attached to the same carbon (a CH, a CH₂, or a CH₃ group). Each attached proton contributes one factor of the pulse's action, and the antiphase coherence built during the echo has exactly $n$ orders of multiple-quantum character built in. The general result, derived rigorously in the first mechanism below, is:

I_n(\theta) = n \sin\theta \cos^{n-1}\theta

where $n = 1, 2, 3$ for CH, CH₂, CH₃ respectively (some conventions omit the leading $n$ as a normalization choice; what matters for spectral editing is the sign and the ratio of intensities, both of which are identical either way). Evaluating this function at various θ gives the entire DEPT family:

Figure

Figure 2 — Intensity vs. tip angle θ for CH, CH₂, CH₃

Reading the figure: at $\theta = 90°$ , $\cos\theta = 0$ , so every term with $\cos^{n-1}\theta$ for $n \geq 2$ vanishes, while the CH term $\sin\theta$ is at its maximum. That is DEPT-90 — a spectrum containing only methine (CH) carbons. At $\theta = 135°$ , $\sin\theta = \cos\theta$ in magnitude but $\cos\theta$ is negative (since $135°$ is in the second quadrant), so the CH₃ term ( $\propto \cos^2\theta$ , always positive) stays positive, the CH term ( $\propto \sin\theta$ , always positive for $0 < \theta < 180°$ ) stays positive, but the CH₂ term ( $\propto \sin\theta\cos\theta$ , which flips sign when $\cos\theta$ changes sign) becomes negative. That is DEPT-135 — CH and CH₃ phased up, CH₂ phased down, quaternary carbons absent throughout because they were never fed any proton magnetization to begin with.

The core ideas

The central quantitative relationship governing every DEPT spectrum is the single-proton transfer amplitude as a function of the final pulse angle:

I_1(\theta) = \sin\theta \quad \text{(one attached proton — CH)}

where: θ is the tip angle of the final ¹H pulse (the only adjustable parameter in the sequence after the fixed $\tau = 1/2J$ delays), and $I_1(\theta)$ is the observed carbon signal amplitude for a methine carbon, in arbitrary but internally consistent units. For a carbon bearing two magnetically equivalent protons (a CH₂), the coherence transfer must account for both protons contributing antiphase character, and the algebra (worked in Mechanism 1) yields:

I_2(\theta) = \sin 2\theta = 2\sin\theta\cos\theta \quad \text{(two attached protons — CH}_2\text{)}

where: the factor $\cos\theta$ beyond the single-proton case reflects the fact that the second proton's transverse magnetization must survive being partially rotated back toward $z$ by the very same θ pulse that is doing the transferring — an effect with no analogue in the one-proton case. For three equivalent protons (CH₃), the pattern continues one order further:

I_3(\theta) = 3\sin\theta\cos^2\theta \quad \text{(three attached protons — CH}_3\text{)}

where: each additional proton multiplies the amplitude by another factor of $\cos\theta$ , because each "bystander" proton coherence must be projected back onto the transverse plane by the same pulse, and the combinatorial prefactor $n$ (here 3) counts the number of equivalent spin permutations contributing constructively. Reading these three equations together as one family:

I_n(\theta) = n\sin\theta\cos^{n-1}\theta, \qquad n = 1, 2, 3

the physical content is that a carbon's DEPT intensity is entirely determined by how many protons it carries, not by what those protons are attached to elsewhere in the molecule, nor by the carbon's chemical shift. This is what makes DEPT a clean multiplicity filter rather than a shift-dependent distortion — the price paid is that a carbon with zero attached protons, $n = 0$ , gives $I_0(\theta) \equiv 0$ for every θ, because there is no proton coherence for that carbon to receive in the first place.

Derivation AMechanism 1 — Why

I_n(\theta) = n\sin\theta\cos^{n-1}\theta

: the product-operator walk from CH to CH₃

Goal. Derive, from the pulse sequence and the density-operator (product-operator) formalism, why the observable carbon signal after the DEPT sequence scales as $\sin\theta$ for CH, $\sin2\theta$ for CH₂, and $3\sin\theta\cos^2\theta$ for CH₃, so that θ alone edits the spectrum by proton count.

Step 1 — Initial excitation. The sequence starts at thermal equilibrium with only proton magnetization present along $+z$ : the density operator (ignoring constants) is $\sigma_0 = H_z$ (writing $H_z$ , $C_z$ for the $z$ -magnetization of the relevant proton and carbon spins). The first ¹H 90° $_x$ pulse rotates this into the transverse plane:

H_z \xrightarrow{90°_x(H)} -H_y

No carbon operator appears yet — carbon has not been touched.

Step 2 — Evolution under coupling during the spin echo. During each $\tau = 1/(2J_{CH})$ delay, $-H_y$ evolves under the heteronuclear scalar coupling Hamiltonian $\mathcal{H}_J = 2\pi J_{CH}\, H_zC_z$ (chemical shift evolution is refocused by the simultaneous 180° pulses on both channels, the standard spin-echo cancellation, so only the coupling term survives over the full $2\tau$ ). For a single attached proton (CH), after the full $2\tau = 1/J$ interval:

-H_y \xrightarrow{2\tau,\ J} -H_y\cos(\pi J\cdot 2\tau) + 2H_xC_z\sin(\pi J\cdot 2\tau)

Choosing $2\tau = 1/J$ makes the coupling angle $\pi J(2\tau) = \pi/2$ so $\cos(\pi/2)=0,\ \sin(\pi/2)=1$ , giving purely antiphase coherence:

-H_y \longrightarrow 2H_xC_z

This is the crucial intermediate state: transverse proton magnetization that is antiphase with respect to the coupled carbon spin — it carries carbon-spin information ( $C_z$ ) even though it is still proton coherence.

Step 3 — The final simultaneous pulse pair (CH case, $n=1$ ). Apply θ $_y$ on ¹H and 90° on ¹³C simultaneously. The carbon 90° pulse rotates $C_z \to -C_y$ (or $+C_y$ depending on phase convention; the sign only fixes overall spectrum phase, not the θ-dependence that matters for editing). The proton θ pulse acts on $H_x$ , leaving it as $H_x\cos\theta$ plus a $H_z$ term that does not matter for what follows, and separately rotates part of the coherence into observable carbon antiphase magnetization. Carrying through the operator algebra for the term of interest:

2H_xC_z \xrightarrow{\theta_y(H),\,90°(C)} \sin\theta\,(-2H_zC_y)_{\text{unobserved}} + \sin\theta\,(C_x)_{\text{observable}}

The observable carbon term appears with coefficient exactly $\sin\theta$ : this is the origin of $I_1(\theta) = \sin\theta$ . The unobservable multiple-quantum and antiphase carbon terms are removed by phase cycling before detection.

Step 4 — CH₂: two equivalent protons. Now the carbon is coupled to two magnetically equivalent protons, $H^{(1)}$ and $H^{(2)}$ . Each proton independently develops antiphase coherence with the carbon during the echo, as in Step 2, so after $2\tau=1/J$ the relevant term is $2H_x^{(1)}C_z$ and $2H_x^{(2)}C_z$ simultaneously present (plus small multi-spin terms from proton–proton coupling, ignored to first order since $J_{HH} \ll J_{CH}$ over this timescale in most cases). Applying the θ pulse to both protons simultaneously: the transfer from $H^{(1)}$ proceeds as in Step 3, giving a factor $\sin\theta$ , but this transfer is additionally modulated by whatever the θ pulse has done to $H^{(2)}$ 's antiphase contribution, and vice versa. Explicit product-operator expansion of the two-spin system shows the net observable carbon coherence scales as:

I_2(\theta) \propto \sin\theta\cos\theta = \tfrac12\sin2\theta

The extra $\cos\theta$ factor is the signature of the second proton: the θ pulse partially converts the second proton's contribution back toward longitudinal magnetization (a $\cos\theta$ projection) rather than letting it add constructively, and only the geometric mean of the two spins' transfer amplitudes survives as net in-phase carbon signal. Including the standard normalization (so that all three formulas agree at small θ to leading order in $\theta$ , matching the naive "n protons, n first-order transfers" expectation) gives the conventional:

I_2(\theta) = \sin2\theta = 2\sin\theta\cos\theta

Step 5 — CH₃: three equivalent protons. By the identical logic extended to three coupled spins, each additional proton beyond the first contributes one more factor of $\cos\theta$ (from being "along for the ride" while only one antiphase coherence order transfers into observable single-quantum carbon coherence at a time), and the combinatorics of three equivalent protons contributing symmetrically gives an overall factor of 3. The result:

I_3(\theta) = 3\sin\theta\cos^2\theta

Step 6 — Assemble the general rule and check the two diagnostic angles. Collecting Steps 3–5:

I_n(\theta) = n\sin\theta\cos^{n-1}\theta,\qquad n=0,1,2,3

For $n=0$ (quaternary carbon), there is no proton to excite in Step 1 at all — the whole derivation never starts — so $I_0(\theta)=0$ identically, for every θ, not merely at special angles. This is why quaternary carbons never appear in any DEPT spectrum, DEPT-90 or DEPT-135 alike. Evaluate at $\theta = 90°$ : $\sin90°=1,\ \cos90°=0$ , so

I_1(90°)=1,\quad I_2(90°)=2(1)(0)=0,\quad I_3(90°)=3(1)(0)^2=0

Only the CH signal survives — this is DEPT-90. Evaluate at $\theta = 135°$ : $\sin135°=\tfrac{\sqrt2}{2},\ \cos135°=-\tfrac{\sqrt2}{2}$ , so

I_1(135°)=\tfrac{\sqrt2}{2}\approx+0.71,\quad I_2(135°)=2\left(\tfrac{\sqrt2}{2}\right)\left(-\tfrac{\sqrt2}{2}\right)=-1,\quad I_3(135°)=3\left(\tfrac{\sqrt2}{2}\right)\left(\tfrac12\right)\approx+1.06

CH is positive, CH₃ is positive and largest in magnitude, CH₂ is negative — exactly the up/up/down pattern that defines DEPT-135, derived here purely from the trigonometry of a single pulse angle acting on antiphase coherences built from 1, 2, or 3 coupled spins.

Go deeper · Derivation B · 2 minMechanism 2 — Building a full structure elucidation from DEPT-90 and DEPT-135 together

Derivation BMechanism 2 — Building a full structure elucidation from DEPT-90 and DEPT-135 together

Goal. Show, mechanistically and not just by lookup, how a chemist combines a broadband ¹³C spectrum with DEPT-90 and DEPT-135 to assign every carbon's hydrogen count unambiguously, including the quaternary carbons that no DEPT experiment shows directly.

Step 1 — Record three spectra on the same sample under identical shift referencing. Acquire (i) the standard proton-decoupled ¹³C spectrum, which shows every carbon including quaternaries, each as a single peak at its true intensity-weighted position (though intensities are unreliable quantitatively due to differential $T_1$ and NOE — this spectrum is used only for peak positions and total peak count); (ii) DEPT-90, in which $I_2=I_3=0$ by the Step-6 result above, so only CH carbons show any peak at all; (iii) DEPT-135, in which every protonated carbon shows a peak, phased according to $I_1>0$ , $I_2<0$ , $I_3>0$ .

Step 2 — Use DEPT-90 to mark every CH position unambiguously. Because $I_2(90°)=I_3(90°)=0$ exactly (not approximately — it is an exact trigonometric zero, robust to moderate mis-calibration of the 90° pulse only to first order), any peak present in DEPT-90 must arise from a methine carbon. Cross-reference these shifts against the broadband spectrum: each DEPT-90 peak identifies one CH carbon's exact chemical shift.

Step 3 — Use DEPT-135 to separate CH₂ from CH₃, now that CH positions are already known. Every peak in DEPT-135 is protonated (CH, CH₂, or CH₃). Peaks phased negative (pointing below baseline, by the sign convention fixed in Step 6 of Mechanism 1) are unambiguously CH₂. Peaks phased positive are either CH or CH₃; since the CH shifts were already flagged in Step 2, any positive DEPT-135 peak not already identified as CH must be CH₃.

Step 4 — Recover quaternary carbons by subtraction, not by direct observation. Compare the full list of peaks in the broadband ¹³C spectrum (Step 1(i)) against the union of all peaks appearing in DEPT-135. Any broadband peak with no corresponding DEPT-135 peak, at the same chemical shift within experimental linewidth, must belong to a carbon bearing zero attached protons — a quaternary carbon (this includes carbonyl carbons of ketones/esters/amides, fully substituted aromatic or olefinic carbons, and nitriles). This is a logical subtraction, not a spectroscopic measurement: quaternary carbons are inferred by their conspicuous absence, precisely because the DEPT mechanism (Step 1 of Mechanism 1) never gives them any proton magnetization to inherit in the first place.

Step 5 — Cross-check multiplicity assignments against expected shift ranges. As an internal consistency check (not a substitute for the trigonometric logic above), confirm that assigned CH₃ signals fall in expected ranges (roughly 10–30 ppm for alkyl methyls, ~55 ppm for OCH₃, ~170–210 ppm never for a methyl), that CH₂ signals for simple alkyl chains fall near 20–40 ppm, and so on — using the shift-correlation knowledge already established for interpreting ordinary ¹³C spectra. Any mismatch (for instance, a peak assigned CH₃ by phase but sitting at 130 ppm, an aromatic region where methyls essentially never occur) signals either an accidental peak overlap or a genuine surprise worth re-examining, most often the former. The result of this five-step process is a complete map: every carbon in the molecule is now tagged as CH₃, CH₂, CH, or C(quaternary), before any 2D correlation experiment has even been run — precisely the input needed for the kind of multi-spectrum structure determination developed in solving-combined-spectra-problems.

Worked examples

Example 1 (foundational) — Assigning 2-methylbutan-1-ol, a five-carbon structure with all four carbon types. The structure is $\text{CH}_3\text{CH}_2\text{CH(CH}_3\text{)CH}_2\text{OH}$ , with carbons labeled C1 (the CH₂OH), C2 (the CH stereocenter), C3 (the CH₂ of the ethyl group), C4 (the terminal CH₃ of ethyl), and C5 (the branch CH₃ on C2). There is no quaternary carbon in this molecule, which makes it a clean first test of the CH/CH₂/CH₃ trichotomy before adding the complication of a missing peak. Broadband ¹³C shows five peaks: approximately 65.8 ppm (C1), 40.1 ppm (C2), 26.3 ppm (C3), 16.0 ppm (C5), and 11.4 ppm (C4). Applying the logic of Mechanism 2: DEPT-90 shows exactly one peak, at 40.1 ppm, because C2 is the only carbon in the molecule bonded to exactly one hydrogen ( $I_1(90°) = \sin90° = 1$ , while $I_2(90°)=I_3(90°)=0$ for every other carbon). This immediately confirms C2 = CH. DEPT-135 shows four peaks. The peak at 40.1 ppm is positive (confirming CH, consistent with DEPT-90). The peaks at 16.0 ppm and 11.4 ppm are positive and typically taller than the CH peak (since $I_3(135°) \approx 1.06$ versus $I_1(135°)\approx0.71$ in the idealized trigonometric ratio, though real spectra also carry relaxation and NOE factors that modulate absolute height) — these are assigned CH₃ (C4 and C5). The peak at 65.8 ppm and the peak at 26.3 ppm are negative, phased below baseline — these are CH₂ (C1 and C3), consistent with $I_2(135°) = -1$ . Every peak in the broadband spectrum has a DEPT-135 counterpart, so by Step 4 of Mechanism 2 there is no quaternary carbon — consistent with the structure, since indeed every carbon in 2-methylbutan-1-ol bears at least one hydrogen. As a limiting-case check: if the two DEPT-135 CH₃ peaks were instead mistaken for two CH₂ peaks (a sign error), the total attached-proton count implied would be $1(\text{CH}) + 2(2\times\text{CH}_2) + 2(2\times\text{CH}_2) = 1+4+4=9$ rather than the correct $1+2+2+3+3=11$ , immediately failing to match the known molecular formula $\text{C}_5\text{H}_{12}\text{O}$ — a useful arithmetic cross-check that catches phase-reading errors.

Example 2 (applied) — Distinguishing two isomeric ketones, 3-methylbutan-2-one vs. pentan-3-one, by DEPT alone. Both isomers, $\text{C}_5\text{H}_{10}\text{O}$ , have a broadband ¹³C spectrum with a carbonyl peak near 208–212 ppm and four aliphatic peaks, and in a crowded mixture the raw shift list alone might not instantly distinguish them. Applying the DEPT logic resolves the ambiguity immediately because the two structures have different hydrogen-count "fingerprints." Pentan-3-one, $\text{CH}_3\text{CH}_2\text{C(=O)CH}_2\text{CH}_3$ , is symmetric: it has only three unique carbons (by the signal-counting logic already established) — the carbonyl C (quaternary, ~211 ppm), one CH₂ shift (~35.5 ppm, doubled in intensity by the molecular symmetry), and one CH₃ shift (~7.8 ppm, doubled). In DEPT-135, only two peaks appear in the aliphatic region: one negative (CH₂, 35.5 ppm) and one positive (CH₃, 7.8 ppm). DEPT-90 shows no peaks at all in the aliphatic region, since pentan-3-one contains no methine carbon whatsoever. The carbonyl at ~211 ppm appears in the broadband spectrum but in neither DEPT spectrum, confirmed quaternary by subtraction. 3-Methylbutan-2-one, $\text{CH}_3\text{C(=O)CH(CH}_3)_2$ , is asymmetric with four unique carbons: carbonyl C (quaternary, ~212 ppm), a CH at the isopropyl carbon (~41.6 ppm), a CH₃ on the carbonyl side (~27.5 ppm), and the two equivalent CH₃ groups of the isopropyl unit (~18.4 ppm, doubled). Here DEPT-90 shows one clear peak at 41.6 ppm — instantly diagnostic, because pentan-3-one gives zero DEPT-90 peaks while 3-methylbutan-2-one gives exactly one. DEPT-135 for 3-methylbutan-2-one shows three positive peaks (41.6, 27.5, and 18.4 ppm — CH and two chemically distinct CH₃ environments, one of them double-intensity) and no negative peaks at all, since this isomer has no CH₂ carbon. The diagnostic logic reduces to a single sentence: pentan-3-one shows a DEPT-135 spectrum with one negative and one positive peak and a blank DEPT-90, while 3-methylbutan-2-one shows an all-positive DEPT-135 with three peaks and a single DEPT-90 peak. As a limiting-case sanity check, both isomers must obey $\sum(\text{H count} \times \text{degeneracy}) = 10$ : for pentan-3-one, $2\times(2\,\text{CH}_2) + 2\times(3\,\text{CH}_3) = 4+6=10$ ✓; for 3-methylbutan-2-one, $1\times(1\,\text{CH}) + 1\times(3\,\text{CH}_3) + 2\times(3\,\text{CH}_3) = 1+3+6=10$ ✓.

Example 3 (exam/competition level) — A trisubstituted cyclohexenone with an ambiguous quaternary olefinic carbon. Consider 4,4-dimethylcyclohex-2-en-1-one, a substrate common in total-synthesis problem sets, with the ring numbered C1 (carbonyl), C2–C3 (the enone double bond), C4 (the quaternary carbon bearing two methyl groups), C5, C6 (the remaining ring CH₂'s), and two equivalent gem-dimethyl carbons. The broadband ¹³C spectrum shows seven distinct shifts: C1 (carbonyl, ~199.5 ppm), C3 (β-carbon of the enone, ~160.0 ppm — deshielded by conjugation with the carbonyl, pushed further downfield than a typical alkene carbon), C2 (α-carbon, ~127.5 ppm), C4 (quaternary sp³, ~34.0 ppm), C6 (CH₂ adjacent to carbonyl, ~37.8 ppm), C5 (CH₂ adjacent to the quaternary carbon, ~32.5 ppm), and the gem-dimethyl carbons (~27.9 ppm, double intensity from the two equivalent methyls). DEPT-90 shows a single peak at 127.5 ppm. This is the critical diagnostic step: it proves that C2, the α,β-unsaturated carbon closer to the carbonyl, is a CH, while C3 at 160.0 ppm — despite being part of the same conjugated alkene and despite looking, at first glance, like it "should" also carry a proton by analogy to simple cyclohexenone — is not the methine here. In unsubstituted cyclohex-2-enone both C2 and C3 are CH; the specific substitution pattern of 4,4-dimethylcyclohex-2-enone leaves C3 still protonated as well in the actual compound, so a careful worked check is needed: is C3 also CH, and does DEPT-90 miss it or confirm it? Re-examine the structure: in 4,4-dimethylcyclohex-2-en-1-one, C3 (the β-carbon, conjugated, at 160.0 ppm) is protonated — it bears one vinylic hydrogen — while C4, bearing the two methyl substituents, is the quaternary sp³ carbon with zero attached protons. So the correct DEPT-90 spectrum shows two peaks, not one: at 127.5 ppm (C2) and at 160.0 ppm (C3), both methine, both vinylic. This is the trap the example is built to expose — a student who assumes "the carbon closer to the carbonyl in a push-pull enone is the only CH" is factually wrong, and only the trigonometric fact $I_1(90°)=1,\ I_0(90°)=0$ resolves it by direct evidence rather than assumption: both alkene carbons that actually bear a proton light up in DEPT-90, and no reasoning about conjugation shortcuts that measurement. Continuing the analysis: DEPT-135 shows positive peaks at 127.5 and 160.0 ppm (the two vinylic CH's, confirmed from DEPT-90), a positive peak at 27.9 ppm (the gem-dimethyl CH₃'s, double intensity), and two negative peaks at 37.8 and 32.5 ppm (C6 and C5, the two distinguishable ring CH₂'s — distinguishable because the molecule's substitution pattern breaks any symmetry that might otherwise have equated them). The broadband spectrum has two peaks with no DEPT-135 counterpart: 199.5 ppm (the carbonyl, obviously quaternary) and 34.0 ppm (C4, the quaternary sp³ ring carbon bearing the two methyls) — both correctly inferred as quaternary purely by the subtraction logic of Mechanism 2, Step 4, with no need to guess. As the limiting-case arithmetic check: total attached protons predicted by DEPT $= 1(\text{C2}) + 1(\text{C3}) + 2(\text{C6}) + 2(\text{C5}) + 2\times3(\text{gem-dimethyl}) = 1+1+2+2+6 = 12$ , matching the molecular formula $\text{C}_8\text{H}_{12}\text{O}$ exactly (12 hydrogens), with the two quaternary carbons (C1 and C4) contributing zero — a full internal consistency check that the DEPT-based assignment is self-consistent with the known formula.

Limiting cases, scope & common traps

Situation	What happens	Why
Quaternary carbon (any substitution)	Absent from DEPT-90 and DEPT-135, always	$I_0(\theta)\equiv0$ for all θ — no proton magnetization to transfer, by construction (Mechanism 1, Step 1)
θ mis-set slightly off 90°	DEPT-90 shows faint CH₂/CH₃ "leakage" peaks	$\cos\theta \neq 0$ exactly if pulse calibration drifts; leakage scales as $\cos(90°+\epsilon)\approx-\sin\epsilon$ , small but nonzero
Unusual $^1J_{CH}$ (e.g., alkynyl C–H, ~250 Hz, vs. typical sp³ ~125–145 Hz)	Attenuated or phase-distorted signal if $\tau$ is set for the "average" $J$	The fixed $\tau=1/2J$ delay assumes a single characteristic $J$ ; a mismatched $J$ desynchronizes the antiphase buildup in Mechanism 1, Step 2
Two accidentally overlapping shifts, e.g., a CH₂ and a CH₃ at nearly the same ppm	Apparent partial cancellation (up peak nearly cancelling a down peak) in DEPT-135	Phases genuinely oppose; resolve by re-running at higher field or using DEPT-90 to separately confirm any CH content in that region
Exchangeable or quadrupolar-adjacent carbons (e.g., C–OH, C–NH)	Ordinary CH/CH₂/CH₃ rules still apply — DEPT only "sees" the carbon-bound protons, not O–H/N–H	The $^1J_{CH}$ transfer pathway is specific to carbon-attached protons; heteroatom-bound protons play no role in the sequence
Symmetric molecule with degenerate carbons	Single DEPT peak represents multiple magnetically equivalent carbons at full combined intensity	Signal-counting logic (equivalence by symmetry) applies identically to DEPT as to the underlying ¹³C spectrum

Common trapMisconception: "DEPT-135 shows quaternary carbons faintly."

A very common mistake is to assume that quaternary carbons show up in DEPT-135 as small, weak peaks that are just harder to see than protonated carbons. This is not correct even in principle: the derivation in Mechanism 1 shows that a quaternary carbon's intensity is exactly zero for every value of θ, because the entire DEPT pulse sequence is a relay race that starts on protons and only reaches carbon by that relay. A carbon with no attached proton is never entered into the relay at all — there is no small residual signal to look for, weak or otherwise. Any peak that appears at a "quaternary" shift in a real DEPT-135 spectrum is either genuine noise, an artifact of imperfect pulse calibration bleeding through from a nearby strongly coupled CH₃, or (most often) an assignment error where that shift actually belongs to a nearby protonated carbon.

Common trapMisconception: "A CH₂ pointing down and a CH₃ pointing up is a fixed rule regardless of experiment parameters."

The up/down convention (CH, CH₃ positive; CH₂ negative) is specific to θ = 135°. If an instrument is run at a different tip angle — some laboratories use 90° combined with a separate "DEPT-45" experiment (θ=45°, where all of CH, CH₂, and CH₃ appear positive, useful for a clean "all protonated carbons" overview) — the sign convention changes entirely, since $I_2(45°) = \sin90°=+1$ , positive, not negative. Always check which θ was used before reading peak phase as a multiplicity label; the physics is in the $\sin\theta\cos^{n-1}\theta$ formula, not in a fixed memorized color code.

The chemist behind it

DEPT was introduced in 1982 by Ovide W. Sørensen, working with Ray Freeman's group at Oxford, as a refinement of an earlier related method (INEPT, Insensitive Nuclei Enhanced by Polarization Transfer, developed slightly earlier by Sørensen, Freeman, and coworkers) that suffered from intensity distortions when multiple different $^1J_{CH}$ couplings were present in the same sample. INEPT's problem was that its final-readout intensities depended sensitively on how precisely the fixed delay matched the actual $J$ for each carbon, so peaks would appear distorted, sometimes even inverted, in ways that depended on details of molecular structure rather than purely on proton count. DEPT's refinement built in an extra layer of coherence transfer and averaging that made the relative editing (CH vs. CH₂ vs. CH₃) far more robust to the exact value of $J_{CH}$ across a typical range of organic carbons, which is why DEPT rather than INEPT became the default routine multiplicity-editing method in essentially every organic chemistry teaching lab and research facility from the 1980s onward.

Origins · 1 minOve Wolf Sørensen and how this was discovered

Ove Wolf Sørensen

b. 1954 · Danish–British NMR methodology, Oxford & ETH Zürich, early 1980s

Sørensen worked within Ray Freeman's Oxford physical chemistry group during the period when pulsed multidimensional NMR was rapidly maturing from a physics curiosity into the everyday workhorse of structural organic and biological chemistry. His contributions, alongside Freeman, Geoffrey Bodenhausen, and others, were principally in the product-operator formalism itself — the systematic bookkeeping (used throughout the two mechanisms above) that let spectroscopists predict, from first principles, exactly what an arbitrary sequence of pulses and delays would do to a coupled spin system, rather than relying on classical vector pictures that broke down for coupled multi-spin systems. DEPT is one of the clearest illustrations of that formalism's power: a single trigonometric identity, $\sin\theta\cos^{n-1}\theta$ , replaces what would otherwise be an intractable spin-dynamics problem for the practicing chemist.

History

- 1976 — Overhauser-effect-based polarization transfer ideas (SPI, selective population inversion) show that ¹H magnetization can be relayed to insensitive nuclei.

- 1979 — INEPT introduced by Morris and Freeman, giving large ¹³C sensitivity gains via coherence transfer from ¹H.

- 1982 — Doddrell, Pegg, and Bendall (and independently Sørensen with Freeman) develop DEPT, correcting INEPT's

J

-mismatch sensitivity and enabling clean multiplicity editing.

- Mid-1980s — DEPT-90/DEPT-135 pairs become a standard default acquisition alongside routine broadband ¹³C spectra in organic structure-elucidation laboratories worldwide.

- 1990s–present — DEPT is largely complemented (not replaced) by 2D HSQC/HMBC methods for complex molecules, but remains the fastest one-shot method for simple multiplicity editing on small to medium organic structures.

Why does DEPT use a proton pulse angle θ instead of just always using 90°?

Because θ = 90° only reveals CH carbons — at that angle $\cos\theta = 0$ , which kills the $\cos^{n-1}\theta$ factor for every CH₂ ( $n=2$ ) and CH₃ ( $n=3$ ) signal, leaving only the $n=1$ CH term. To see CH₂ and CH₃ at all, θ must be moved away from 90° so that $\cos\theta \neq 0$ . The angle 135° is chosen because it gives all three types (CH, CH₂, CH₃) simultaneously nonzero and, crucially, gives CH₂ the opposite sign from CH and CH₃ — a phase-based label built directly into the physics rather than needing a separate experiment for each multiplicity. Running the full trio (DEPT-45, 90, 135) gives complete redundancy, but 90 and 135 alone are normally sufficient for unambiguous assignment.

Can DEPT tell me anything about a quaternary carbon's environment, even though it never shows a peak for it?

Indirectly, yes — by subtraction. Compare the peak list of the ordinary broadband ¹³C spectrum against the peak list of DEPT-135 (or the union of DEPT-90 and a full-multiplicity DEPT-135). Any shift present in the broadband spectrum but absent from every DEPT spectrum must belong to a carbon with zero attached protons. This does not tell you what kind of quaternary carbon it is directly, but combined with the chemical shift itself (a carbonyl near 165–215 ppm, a fully substituted aromatic carbon near 130–150 ppm, a quaternary sp³ carbon near 30–45 ppm) the shift plus the "missing from DEPT" observation together usually pin down the carbon type unambiguously.

Does DEPT work for carbons attached to two chemically different but accidentally equivalent protons, like a diastereotopic CH₂?

Yes, with a caveat. The derivation in Mechanism 1 assumes the two attached protons are magnetically equivalent, meaning they experience the same $J_{CH}$ and evolve identically during the delay. Diastereotopic CH₂ protons (common next to a stereocenter) are chemically inequivalent and can, in principle, have slightly different one-bond couplings and different proton chemical shifts, but their $^1J_{CH}$ values to the same carbon are normally close enough that the $\sin2\theta$ editing behavior is essentially unaffected. The practical result is that diastereotopic CH₂ carbons still show up cleanly as negative DEPT-135 peaks; what changes is only that their attached protons may appear as an AB pattern (not a simple integrated peak) in the corresponding ¹H spectrum, a separate issue from the carbon-level multiplicity editing DEPT performs.

Why is DEPT so much more sensitive than a plain proton-decoupled ¹³C pulse-acquire experiment?

Because DEPT's signal originates from proton magnetization, not carbon magnetization. At thermal equilibrium, proton nuclei have roughly four times the gyromagnetic ratio of carbon-13, and since equilibrium polarization scales with $\gamma^3$ in signal-to-noise terms (one factor from the population difference, one from the induced EMF, one from the natural linewidth/coherent averaging benefits), the enhancement is substantial — commonly cited as a theoretical maximum around $(\gamma_H/\gamma_C)\approx4$ in simple population-transfer terms, with additional gains from faster proton $T_1$ relaxation allowing shorter recycle delays between scans. In practice, DEPT spectra of protonated carbons can be acquired in a fraction of the time needed for an equivalent signal-to-noise plain ¹³C spectrum, which is part of why it is run routinely rather than being reserved for special cases.

If a molecule has perfect symmetry, like para-xylene, how does DEPT handle carbons that are chemically equivalent by symmetry?

Exactly as it handles any other equivalent set — 13C NMR: Sensitivity, Shifts, and Signal Counting already establishes that magnetically equivalent carbons give a single peak with intensity proportional to how many carbons contribute. DEPT does not change this equivalence logic at all; it simply overlays the CH/CH₂/CH₃ phase and shape information onto whatever peaks the symmetry analysis predicts. For para-xylene, the two methyl carbons are equivalent and give one CH₃ peak (positive in DEPT-135), and the aromatic CH carbons (four total, related in equivalent pairs) give one shift, appearing in DEPT-90 as well as DEPT-135, while the two quaternary aromatic carbons attached to the methyl groups are absent from both DEPT spectra, exactly as expected.

Is it possible for a real CH, CH₂, or CH₃ carbon to accidentally give zero DEPT-135 signal and be mistaken for quaternary?

In principle, yes, though it is uncommon in a well-tuned experiment. This happens if the carbon's actual one-bond coupling $J_{CH}$ deviates substantially from the value the instrument's fixed delay $\tau = 1/2J$ was calibrated for — for example, an unusually strained or electronically unusual C–H bond with a $J_{CH}$ far from the typical 125–145 Hz range assumed for ordinary sp³ carbons. Because the antiphase coherence buildup in the spin-echo period depends sensitively on matching $\tau$ to the true $J$ , a badly mismatched coupling can leave a genuinely protonated carbon with anomalously weak or even near-zero DEPT intensity. This is one reason experienced spectroscopists always cross-check a "missing" DEPT peak against the expected molecular formula and, where in doubt, confirm with an independent method such as HSQC before concluding a carbon is quaternary.