IR & Mass Spectrometry

Spectroscopy — Seeing Molecules Without Seeing Them

Part 1 of 7 — Introduction to Structure Determination

A practicing organic chemist almost never sees the molecule they have made. A flask might hold a colorless oil weighing a fraction of a gram, and the central question is always the same: what is it? Spectroscopy answers that question by measuring how a sample interacts with electromagnetic radiation. Different frequencies of light probe different features of a molecule, and from the pattern of absorption (or, in mass spectrometry, of masses) we reconstruct the structure.

The key idea is that energy is quantized. A molecule can only absorb a photon whose energy exactly matches the gap between two of its allowed energy states:

$\Delta E = h\nu = \frac{hc}{\lambda}$

Here $h$ is Planck's constant, $\nu$ is the frequency, and $\lambda$ is the wavelength. Because $\Delta E \propto \nu$, higher-frequency (shorter-wavelength) radiation carries more energy. This single relationship organizes the entire electromagnetic spectrum into a ladder of techniques, each tuned to a different molecular motion.

The Four Workhorse Techniques

Modern structure elucidation rests on four complementary methods. Each interrogates a different physical process, so each returns a different kind of structural information.

Notice the energy ordering. Mass spectrometry uses a high-energy electron beam (~70 eV) that is violent enough to knock out an electron and shatter the molecule. IR uses mid-energy photons that merely make bonds stretch and bend. NMR uses very low-energy radio waves that only reorient nuclear spins. As you move from MS toward NMR you move down the energy ladder and gain ever finer detail about connectivity.

Why combine them? No single technique is sufficient. MS might tell you a compound has the formula $\text{C}_4\text{H}_8\text{O}$, but dozens of structures share that formula. IR can then reveal a carbonyl. NMR pins down exactly how the atoms connect. The power is in the overlap.

Checkpoint — Energy and the Spectrum

The Logic of an Unknown Problem

Throughout this suite we will repeatedly attack the same kind of problem: given spectra, deduce the structure. It helps to have a fixed mental workflow. Experienced chemists almost always proceed in this order:

1. Get the molecular formula. The mass spectrum gives the molecular mass (from the molecular ion, $\text{M}^{+\bullet}$); isotope peaks and the nitrogen rule refine the formula.
2. Compute the degrees of unsaturation. This number — derived purely from the formula — tells you how many rings plus $\pi$ bonds the molecule must contain before you look at any other data.
3. Identify functional groups with IR. A strong band near $1700\ \text{cm}^{-1}$ screams carbonyl; a broad band near $3300\ \text{cm}^{-1}$ suggests O–H or N–H.
4. Assemble connectivity with NMR (covered in a later course).

Step 2 is so useful that we introduce it now. For a compound of formula $\text{C}_c\text{H}_h\text{N}_n\text{O}_o\text{X}_x$ (where X is a halogen), the degrees of unsaturation (also called the index of hydrogen deficiency, IHD) are:

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

Oxygen is ignored because it does not change the hydrogen count of a saturated formula.

Worked example. Benzaldehyde has the formula $\text{C}_7\text{H}_6\text{O}$. Then

$\text{DoU} = \frac{2(7) + 2 + 0 - 6 - 0}{2} = \frac{14 + 2 - 6}{2} = \frac{10}{2} = 5$

Five degrees of unsaturation: four come from the benzene ring (three $\pi$ bonds + one ring) and the fifth from the aldehyde $\text{C}=\text{O}$. Before recording a single spectrum we already suspect an aromatic ring bearing a carbonyl — exactly what benzaldehyde is.

Checkpoint — Degrees of Unsaturation

What to Carry Forward

• Spectroscopy = light meets matter. Quantized energy gaps mean each technique uses a specific slice of the electromagnetic spectrum, governed by $\Delta E = h\nu = hc/\lambda$.
• Four complementary tools. MS (mass/formula), IR (functional groups), and $^1$H / $^{13}$C NMR (connectivity). Their overlap, not any one alone, solves structures.
• A fixed workflow. Formula $\rightarrow$ degrees of unsaturation $\rightarrow$ functional groups $\rightarrow$ connectivity.
• Degrees of unsaturation, $\text{DoU} = (2c + 2 + n - h - x)/2$, convert a bare formula into a count of rings plus $\pi$ bonds before any spectrum is interpreted.

In Part 2 we zoom into infrared spectroscopy: what a bond vibration actually is, why we report frequencies as wavenumbers in $\text{cm}^{-1}$, and how the spectrum splits into a diagnostic region and a fingerprint region.

Exit Ticket — Part 1 Synthesis

Infrared Spectroscopy — Bonds That Behave Like Springs

Part 2 of 7 — How IR Works

When infrared light passes through a sample, the molecule absorbs photons whose energy matches the energy needed to make its bonds vibrate more vigorously. A covalent bond is not rigid; it behaves like two masses joined by a spring, constantly stretching, compressing, and bending. Infrared radiation (roughly $4000$ to $400\ \text{cm}^{-1}$) sits at exactly the right energy to drive these vibrations.

The two fundamental vibration types are:

• Stretching — the bond length oscillates (atoms move apart and together along the bond axis). These require more energy and appear at higher frequency.
• Bending — the bond angle oscillates (scissoring, rocking, wagging, twisting). These require less energy and appear at lower frequency.

A molecule with $N$ atoms has $3N - 6$ vibrational modes ($3N - 5$ if linear), so even a small molecule produces a rich spectrum. Crucially, only vibrations that change the molecule\u2019s dipole moment absorb IR light — a perfectly symmetric stretch (like $\text{N}\equiv\text{N}$ in $\text{N}_2$) is IR-inactive.

Why We Use Wavenumbers $(cm^{-1})$

IR spectra are plotted against wavenumber, symbol $\tilde{\nu}$, with units of $\text{cm}^{-1}$ (reciprocal centimeters). The wavenumber is simply the reciprocal of the wavelength:

$\tilde{\nu} = \frac{1}{\lambda} = \frac{\nu}{c}$

The appeal is that wavenumber is directly proportional to energy ($E = hc\tilde{\nu}$), so a band at $3000\ \text{cm}^{-1}$ corresponds to a vibration of exactly twice the energy of one at $1500\ \text{cm}^{-1}$. Higher $\tilde{\nu}$ means a stiffer or lighter oscillator.

The frequency of a stretch is governed by an idealized Hooke\u2019s law relationship:

$\tilde{\nu} = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}}$, where $\mu = \frac{m_1 m_2}{m_1 + m_2}$

where $k$ is the bond force constant (stiffness) and $\mu$ is the reduced mass of the two atoms. Two predictions follow directly, and they explain almost every trend in an IR table:

1. Stronger (stiffer) bonds vibrate faster. Because $\tilde{\nu} \propto \sqrt{k}$, a triple bond ($k$ large) absorbs at higher wavenumber than a double bond, which beats a single bond: $\text{C}\equiv\text{C}\ (~2150) > \text{C}=\text{C}\ (~1650) > \text{C}-\text{C}\ (~1000\ \text{cm}^{-1})$.
2. Lighter atoms vibrate faster. Because $\tilde{\nu} \propto 1/\sqrt{\mu}$, any bond to hydrogen (tiny $\mu$) absorbs high: C–H, O–H, and N–H all appear above $2850\ \text{cm}^{-1}$.

Checkpoint — The Spring Model

The Two Regions of an IR Spectrum

By long convention, chemists mentally divide every IR spectrum at roughly $1500\ \text{cm}^{-1}$:

The diagnostic (or functional group) region is where you do most of your interpretive work. The stretches of C=O, O–H, N–H, C–H, and the triple bonds all live here, and their positions are remarkably transferable from molecule to molecule. A band at $1715\ \text{cm}^{-1}$ means a carbonyl whether the rest of the molecule is a sterol or a solvent.

The fingerprint region arises from a tangle of single-bond stretches and bending modes coupled together. These vibrations are exquisitely sensitive to the whole molecular skeleton, so the pattern is essentially unique to each compound — like a fingerprint. You rarely assign individual fingerprint bands; instead you overlay the spectrum on a reference and check whether the two match peak-for-peak.

Practical rule. Use the diagnostic region to decide what functional groups are present; use the fingerprint region to decide which specific compound you have. Trying to assign every fingerprint band by hand is a common beginner trap and a waste of effort.

Checkpoint — Reading the Regions

Worked Example — Predicting Relative Band Positions

Without a table, order the C–H, O–H, and C=O stretches of acetic acid ($\text{CH}_3\text{COOH}$) from highest to lowest wavenumber, and justify each placement using the spring model.

Reasoning.

• The O–H stretch involves a bond to hydrogen, so $\mu$ is tiny and the band is pushed very high — around $2500$–$3300\ \text{cm}^{-1}$ (broad, because of hydrogen bonding in the carboxylic acid).
• The C–H stretches also involve hydrogen, so they too sit high, near $2850$–$3000\ \text{cm}^{-1}$, just below the O–H envelope.
• The C=O stretch is between two relatively heavy atoms (larger $\mu$) but the bond is stiff (double bond, large $k$). The stiffness wins partially but the heavier masses keep it well below the X–H bands, landing near $1710\ \text{cm}^{-1}$.

Answer: O–H $\approx$ C–H $>$ C=O, i.e. roughly $3000\ \text{cm}^{-1}$ (O–H/C–H) versus $1710\ \text{cm}^{-1}$ (C=O). The key insight is that any bond to hydrogen outruns even a stiff double bond between heavy atoms, because the reduced-mass term dominates.

Exit Ticket — Part 2 Synthesis

Key IR Absorptions — The Diagnostic Bands

Part 3 of 7 — A Working Table of Functional Groups

This part is the practical heart of IR. A handful of bands in the diagnostic region ($4000$–$1500\ \text{cm}^{-1}$) account for the vast majority of functional-group identifications. Memorize their positions, but more importantly learn to read their shape and intensity, because two bands at the same wavenumber can mean different things depending on whether they are broad or sharp, strong or weak.

Reading Shape and Position, Not Just Wavenumber

The O–H story. A free O–H (no hydrogen bonding, as in a dilute gas-phase sample) is a sharp spike near $3600\ \text{cm}^{-1}$. In a normal liquid sample the molecules hydrogen-bond to one another, and because each O–H now experiences a slightly different environment the band smears into a wide, rounded hump from about $3200$ to $3550\ \text{cm}^{-1}$. The carboxylic acid O–H is the most dramatic case: extensive hydrogen-bonded dimers produce an enormous band sprawling from $2500$ all the way to $3300\ \text{cm}^{-1}$, often swamping the C–H stretches. Breadth itself is the clue.

The C–H dividing line at $3000\ \text{cm}^{-1}$. This is one of the most useful tricks in IR. Carbon hybridization controls C–H stretch position:

• $sp^3$ C–H (more s-poor, longer/weaker bond) appears just below $3000\ \text{cm}^{-1}$.
• $sp^2$ and $sp$ C–H (more s-rich, shorter/stiffer bond) appear just above $3000\ \text{cm}^{-1}$.

So a band peeking above $3000\ \text{cm}^{-1}$ flags unsaturation (alkene or aromatic C–H), while bands purely below it indicate a saturated, alkyl-only skeleton.

The carbonyl is the loudest band in the spectrum. Because the C=O bond has a large dipole and a large change in dipole upon stretching, its absorption near $1700\ \text{cm}^{-1}$ is usually the strongest, sharpest peak present. Its exact position then subdivides the carbonyl family (covered next).

Checkpoint — Shape Tells the Story

Worked Example — Deduce the Functional Group from IR Alone

An unknown liquid gives these prominent IR bands:

• a very broad band from $2500$ to $3300\ \text{cm}^{-1}$,
• a strong, sharp band at $1710\ \text{cm}^{-1}$,
• bands just below $3000\ \text{cm}^{-1}$.

Which functional group is present?

Step 1 — Interpret the high-wavenumber region. A band that is very broad and stretches from $2500$ up to $3300\ \text{cm}^{-1}$ is far too wide to be an ordinary alcohol O–H or an amine N–H. This breadth, overlapping the C–H region, is the hallmark of a carboxylic acid O–H (hydrogen-bonded dimer).

Step 2 — Interpret the $1710\ \text{cm}^{-1}$ band. A strong, sharp absorption here is a carbonyl, C=O.

Step 3 — Combine. A broad acidic O–H and a carbonyl, occurring together, point unambiguously to a carboxylic acid ($-\text{COOH}$): the O–H and the C=O are the two halves of the same group. The $sp^3$ C–H bands below $3000\ \text{cm}^{-1}$ tell us the rest of the molecule is an alkyl chain.

Conclusion: the compound is a carboxylic acid such as propanoic acid, $\text{CH}_3\text{CH}_2\text{COOH}$. The lesson: a lone C=O could be many things, but C=O plus a very broad O–H = carboxylic acid. Had the broad O–H been absent, we would instead suspect an aldehyde or ketone.

Checkpoint — Triple Bonds and the Triple-Bond Region

Lower-Region Clues Worth Knowing

Although the fingerprint region is mostly used for compound matching, a few of its bands are reliable enough to be diagnostic:

• C–O stretch of alcohols, ethers, and esters: a strong band around $1050$–$1300\ \text{cm}^{-1}$. In an ester you often see two C–O stretches.
• N–H bend of amines: a band near $1600\ \text{cm}^{-1}$ that can be mistaken for C=C.
• Aromatic overtone/combination bands and out-of-plane C–H bends ($690$–$900\ \text{cm}^{-1}$) that hint at the substitution pattern of a benzene ring.

A productive habit is to interpret the diagnostic region first to build a hypothesis, then check whether the fingerprint region is consistent with it (for instance, confirming a strong C–O stretch when you suspect an ester), rather than trying to assign the fingerprint region cold.

Summary heuristic. Scan an IR spectrum top-down: (1) Is there anything above $3000$? (O–H, N–H, $sp^2$/$sp$ C–H.) (2) Is there a triple-bond band near $2100$–$2260$? (3) Is there a strong band near $1700$? (carbonyl). (4) Then use position/shape to refine. Three questions resolve most unknowns.

Exit Ticket — Part 3 Synthesis

Mass Spectrometry — Weighing Molecules and Their Pieces

Part 4 of 7 — The Molecular Ion and Isotope Patterns

Mass spectrometry (MS) does something the other techniques cannot: it measures the mass of a molecule directly. In the most common method, electron-impact (EI) ionization, the gaseous sample is bombarded by a high-energy electron beam (~70 eV). A collision knocks one electron out of the molecule, producing a positively charged radical cation called the molecular ion:

$\text{M} + e^- \rightarrow \text{M}^{+\bullet} + 2e^-$

The instrument then accelerates these ions through electric and magnetic fields that sort them by their mass-to-charge ratio, $m/z$. Because almost every ion carries a single positive charge ($z = 1$), the $x$-axis of a mass spectrum reads essentially as mass. Peaks are plotted as vertical lines whose height is relative abundance, with the tallest peak (the base peak) set to 100%.

The single most valuable peak is the molecular ion $\text{M}^{+\bullet}$, because its $m/z$ value is the molecular mass of the compound. From that mass we begin reconstructing the molecular formula.

Finding (and Trusting) the Molecular Ion

The molecular ion is normally the highest-mass peak in the spectrum — but with caveats. Some molecules fragment so readily that $\text{M}^{+\bullet}$ is tiny or absent (branched alkanes and alcohols are notorious). A few rules keep you honest:

• The molecular ion must correspond to a chemically sensible loss from any higher fragments — you should never see a loss of, say, $4$ or $14$ mass units between the top peaks, because no stable neutral of that mass exists in a simple cleavage.
• Watch for the nitrogen rule (developed fully in Part 5): a molecule with an odd molecular mass contains an odd number of nitrogen atoms. An even nominal mass means zero or an even number of nitrogens. This is a powerful first filter on the formula.

Worked example — degrees of unsaturation from $\text{M}^{+\bullet}$. Suppose $\text{M}^{+\bullet}$ appears at $m/z = 78$ and other data establish the formula as $\text{C}_6\text{H}_6$. The degrees of unsaturation are

$\text{DoU} = \frac{2(6) + 2 - 6}{2} = \frac{14 - 6}{2} = 4$

Four degrees of unsaturation in a $\text{C}_6\text{H}_6$ framework is the classic signature of benzene (three $\pi$ bonds plus one ring). The molecular ion alone, combined with a formula, immediately constrains the skeleton.

Checkpoint — The Molecular Ion

Isotope Peaks: M+1 and M+2

Real elements are mixtures of isotopes, and MS resolves them. This produces small satellite peaks just above the molecular ion that are surprisingly informative.

The M+1 peak and carbon counting. Carbon is $98.9\%$ $^{12}\text{C}$ and $1.1\%$ $^{13}\text{C}$. A molecule containing $n$ carbons therefore has roughly an $n \times 1.1\%$ chance of containing one $^{13}\text{C}$, producing a peak one mass unit higher, $\text{M}+1$. Reading this backward gives a carbon count:

$n_\text{C} \approx \frac{\text{relative height of M}+1}{1.1\%}$

For example, an M+1 peak about $6.6\%$ as tall as $\text{M}^{+\bullet}$ implies roughly $6.6/1.1 \approx 6$ carbons.

The M+2 peak and the halogens. Chlorine and bromine each have two abundant isotopes separated by two mass units, and the ratios are unmistakable:

So a molecular-ion cluster with a 3:1 M:M+2 pattern announces one chlorine; a near 1:1 M:M+2 pattern announces one bromine. (Sulfur contributes a weaker M+2 of about $4\%$.) These patterns are among the fastest "at a glance" diagnoses in all of spectroscopy.

Checkpoint — Isotope Patterns

Nominal Mass vs High-Resolution Mass

Everything above uses nominal (integer) masses. But many different formulas share the same nominal mass — for instance, $\text{N}_2$, $\text{CO}$, $\text{C}_2\text{H}_4$, and $\text{CH}_2\text{N}$ all have nominal mass $28$. High-resolution mass spectrometry (HRMS) resolves this ambiguity by measuring mass to four decimal places, exploiting the fact that exact isotopic masses are not whole numbers:

Because each formula has a unique exact mass, HRMS can pin down the molecular formula directly. For example $\text{CO}$ ($27.9949$) and $\text{N}_2$ ($28.0062$) differ by about $0.011\ \text{u}$ — invisible at unit resolution but trivial for HRMS. In routine work, nominal masses plus isotope patterns get you most of the way; HRMS is the arbiter when two formulas survive.

Key takeaway. The molecular ion gives mass; isotope satellites (M+1 for carbon, M+2 for Cl/Br) refine the formula; and high-resolution mass, when available, locks the formula in. Next we turn to what happens after ionization — fragmentation.

Exit Ticket — Part 4 Synthesis

Fragmentation — Reading the Wreckage

Part 5 of 7 — Neutral Losses, McLafferty, and the Nitrogen Rule

The high-energy electron beam in EI-MS does more than ionize: it deposits enough excess energy that the molecular ion often shatters. The molecular ion $\text{M}^{+\bullet}$ breaks apart into a smaller cation (which the instrument detects) plus a neutral fragment (which is invisible, because it carries no charge). Each cleavage shows up as a new peak at lower $m/z$, and the difference between peaks tells you the mass of the neutral that was lost.

$\text{M}^{+\bullet} \rightarrow [\text{fragment}]^+ + \text{neutral}$

The single most important interpretive move in MS is therefore subtraction: take the molecular ion mass, subtract the mass of a prominent fragment, and identify the neutral that accounts for the gap. A loss of $15$ is a methyl ($\text{CH}_3$); a loss of $18$ is water; a loss of $28$ is $\text{CO}$ or ethylene; a loss of $29$ is $\text{CHO}$ or $\text{C}_2\text{H}_5$.

Fragmentation is not random. Bonds break to give the most stable cation and the most stable neutral, so the dominant peaks reveal the weak points of the skeleton — which is exactly the structural information we want.

A Catalog of Common Neutral Losses

These recurring mass differences are worth committing to memory. When you see one of them between the molecular ion and a strong fragment, you can usually name the group that left.

Stable cations to look for. Some fragment ions are so stabilized that they dominate the spectrum:

• $m/z = 43$: the acylium ion $\text{CH}_3\text{CO}^+$, diagnostic of a methyl ketone, and also the propyl/isopropyl cation $\text{C}_3\text{H}_7^+$.
• $m/z = 77$: the phenyl cation $\text{C}_6\text{H}_5^+$, a giveaway for a monosubstituted benzene ring.
• $m/z = 91$: the tropylium ion $\text{C}_7\text{H}_7^+$, a beautifully stable aromatic seven-membered ring formed from a benzyl ($\text{C}_6\text{H}_5\text{CH}_2$–) group. A strong peak at $91$ almost always means a benzylic carbon.

Checkpoint — Naming the Neutral Loss

The McLafferty Rearrangement

Most fragmentations are simple bond cleavages, but one rearrangement is so common and so diagnostic that it has a name. The McLafferty rearrangement occurs in carbonyl compounds (aldehydes, ketones, acids, esters) that possess a hydrogen on the $\gamma$-carbon — the third carbon counting from the carbonyl.

In the rearrangement, the molecular ion adopts a six-membered cyclic transition state. The carbonyl oxygen plucks the $\gamma$-hydrogen, and simultaneously the bond between the $\alpha$ and $\beta$ carbons breaks. The result is the loss of a neutral alkene and the formation of a stabilized enol radical cation:

$\text{R–CO–CH}_2\text{CH}_2\text{CH}_2\text{R}' \rightarrow [\text{enol}]^{+\bullet} + \text{alkene}$

Why it matters. The McLafferty peak appears at a predictable, even-mass value (when no nitrogen is present) and signals both a carbonyl and a chain at least three carbons long bearing a $\gamma$-hydrogen. For example, 2-hexanone ($\text{M}^{+\bullet} = 100$) undergoes McLafferty to give a characteristic fragment at $m/z = 58$ (the enol of acetone), losing propene (mass 42). Recognizing this even-electron, even-mass fragment alongside an acylium $m/z = 43$ is strong evidence for a methyl ketone with a propyl tail.

Two clues, one structure. A methyl ketone like 2-hexanone shows BOTH the $\alpha$-cleavage acylium at $m/z = 43$ AND a McLafferty fragment at $m/z = 58$. Seeing the pair together is far more convincing than either peak alone.

Checkpoint — McLafferty

The Nitrogen Rule

One of the quickest deductions in all of mass spectrometry comes from the parity of the molecular mass.

The nitrogen rule: a neutral organic molecule containing C, H, O, S, and the halogens has an even nominal molecular mass if it contains zero or an even number of nitrogen atoms, and an odd nominal molecular mass if it contains an odd number of nitrogen atoms.

Why? Nitrogen is the oddity among common organic elements: it has an even atomic mass (14) but an odd valence (3). Carbon, oxygen, sulfur all pair even mass with even valence; hydrogen and the halogens pair odd mass with odd valence. Working through the bookkeeping, each nitrogen flips the expected parity of the molecular mass.

Worked example. An unknown has $\text{M}^{+\bullet}$ at $m/z = 59$, an odd number. The nitrogen rule immediately says the molecule contains an odd number of nitrogens — most simply, one nitrogen. A reasonable formula is $\text{C}_3\text{H}_9\text{N}$ (mass $59$), an amine such as propylamine or trimethylamine. Without computing anything further, the odd molecular ion has already told us a nitrogen is present.

A corollary for fragments. When an odd-mass molecular ion (one nitrogen) fragments by simple cleavage, the nitrogen-containing daughter ion often appears at an even $m/z$. So in nitrogen-containing molecules, prominent even-mass fragment ions frequently carry the nitrogen — a useful cross-check.

Exit Ticket — Part 5 Synthesis

Problem-Solving Workshop — IR and MS Together

Part 6 of 7 — Solving Real Unknowns

You now have two independent windows on a molecule: mass spectrometry (mass, formula, degrees of unsaturation, fragmentation) and infrared (functional groups). Neither alone is decisive, but combined they are formidable. This part is pure practice: a repeatable procedure followed by fully worked unknowns.

The combined workflow.

1. From MS, get the molecular ion $\text{M}^{+\bullet}$ and apply the nitrogen rule (odd mass $\rightarrow$ odd N).
2. Read isotope peaks: M+2 at $\sim 3:1$ means Cl, at $\sim 1:1$ means Br; M+1 height $\div\, 1.1\%$ estimates carbon count.
3. Propose a formula and compute degrees of unsaturation, $\text{DoU} = (2c + 2 + n - h - x)/2$.
4. From IR, identify functional groups: scan above $3000$ (O–H, N–H, unsaturated C–H), the triple-bond window ($2100$–$2260$), and the carbonyl region ($\sim 1700$).
5. Reconcile: the functional groups must consume the right number of degrees of unsaturation, and MS fragments must match plausible cleavages.

The art is in step 5 — making the two data sets tell one consistent story.

Worked Unknown 1 — An Oxygen Compound

Data. $\text{M}^{+\bullet} = 60$ (even). IR: a very broad band $2500$–$3300\ \text{cm}^{-1}$ and a strong band at $1715\ \text{cm}^{-1}$; C–H bands below $3000$. MS fragments at $m/z = 45$ and $m/z = 43$.

Step 1 — Nitrogen rule. Even mass $60$ $\rightarrow$ zero (or even) nitrogens. Assume no N.

Step 2 — Formula. A small even mass with a carbonyl and a broad acidic O–H suggests a carboxylic acid. Try $\text{C}_2\text{H}_4\text{O}_2$: mass $= 24 + 4 + 32 = 60$.

Step 3 — Degrees of unsaturation. $\text{DoU} = (2\cdot2 + 2 - 4)/2 = (6 - 4)/2 = 1$. Exactly one — accounted for by the C=O.

Step 4 — IR. The very broad $2500$–$3300$ band is the carboxylic-acid O–H; $1715$ is its C=O. Consistent with $-\text{COOH}$.

Step 5 — Fragments. Loss of $\text{OH}$ (17) from 60 gives $m/z = 43$ (acylium $\text{CH}_3\text{CO}^+$); loss of $\text{CH}_3$ (15) gives $m/z = 45$ ($\text{COOH}^+$). Both fit.

Answer: acetic acid, $\text{CH}_3\text{COOH}$. Every piece of data converges: even mass (no N), one degree of unsaturation spent on C=O, the diagnostic broad acid O–H, and fragments at 45 ($\text{COOH}^+$) and 43 (acylium).

Checkpoint — Reconciling the Data

Worked Unknown 2 — A Halogen and a Ring

Data. Molecular-ion cluster at $m/z = 112$ and $m/z = 114$ in a $\sim 3:1$ ratio. IR: C–H bands BOTH above and below $3000\ \text{cm}^{-1}$; no carbonyl, no broad O–H. Strong fragment at $m/z = 77$.

Step 1 — Isotopes. The $3:1$ doublet at M / M+2 is the unmistakable signature of one chlorine.

Step 2 — Nitrogen rule. The lower cluster mass $112$ is even $\rightarrow$ zero nitrogens.

Step 3 — Formula. Subtract one Cl (35) from 112: $112 - 35 = 77$ for the rest. A $\text{C}_6\text{H}_5$ group is $72 + 5 = 77$. So the formula is $\text{C}_6\text{H}_5\text{Cl}$ (chlorobenzene), mass $= 77 + 35 = 112$.

Step 4 — DoU. $\text{DoU} = (2\cdot6 + 2 - 5 - 1)/2 = (14 - 6)/2 = 4$. Four degrees = an aromatic ring.

Step 5 — IR and fragments. C–H above $3000$ confirms aromatic ($sp^2$) C–H, consistent with a benzene ring; absence of carbonyl/O–H fits a simple aryl halide. The fragment at $m/z = 77$ is the phenyl cation $\text{C}_6\text{H}_5^+$, formed by loss of the chlorine (35) from 112.

Answer: chlorobenzene, $\text{C}_6\text{H}_5\text{Cl}$. The $3:1$ isotope cluster (Cl), four degrees of unsaturation (aromatic ring), aromatic C–H above $3000$, and the $m/z = 77$ phenyl cation all tell the same story.

Checkpoint — Putting Clues Together

Worked Unknown 3 — Nitrogen in Disguise

Data. $\text{M}^{+\bullet} = 59$ (odd). IR: medium bands near $3300$–$3400\ \text{cm}^{-1}$ (possibly two), no carbonyl, no triple-bond band. MS fragment at $m/z = 30$.

Step 1 — Nitrogen rule. Odd mass $59$ $\rightarrow$ odd number of nitrogens; take one N.

Step 2 — Formula. Subtract N (14): $59 - 14 = 45$ for $\text{C}_x\text{H}_y$. A formula $\text{C}_3\text{H}_9\text{N}$ has mass $36 + 9 + 14 = 59$.

Step 3 — DoU. $\text{DoU} = (2\cdot3 + 2 + 1 - 9)/2 = (8 + 1 - 9)/2 = 0$. Zero degrees — a saturated amine, no rings or $\pi$ bonds.

Step 4 — IR. Medium bands at $3300$–$3400$ (one or two, depending on primary vs secondary) are N–H stretches. No carbonyl and no O–H broad band is consistent with a simple amine.

Step 5 — Fragment. $m/z = 30$ is the iminium/aminomethyl cation $\text{CH}_2=\text{NH}_2^+$, a classic amine fragment (note its even mass, the nitrogen-containing daughter of an odd-mass parent).

Answer: a propylamine, e.g. $\text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2$ (or an isomer such as trimethylamine, distinguished by the N–H count). The odd molecular ion (nitrogen rule), zero degrees of unsaturation, N–H stretches, and the $m/z = 30$ nitrogen fragment all agree.

Exit Ticket — Part 6 Synthesis

Synthesis & Review — The Complete Toolkit

Part 7 of 7 — Bringing It All Together

Across this suite we built a complete strategy for identifying an unknown organic compound from infrared and mass spectra. The central theme has been complementarity: mass spectrometry supplies the molecular mass, formula clues, and skeletal fragmentation, while infrared reveals the functional groups. Used together they constrain a structure far more tightly than either alone.

This final part consolidates everything into reference tables and a master strategy, then tests your integrated understanding with mixed problems.

The one-paragraph summary. Start with the mass spectrum: the molecular ion $\text{M}^{+\bullet}$ gives the molecular mass; the nitrogen rule (odd mass $\rightarrow$ odd N) and isotope peaks (M+1 for carbon count, M+2 at $3:1$ for Cl or $1:1$ for Br) refine the formula. Convert the formula to degrees of unsaturation, $\text{DoU} = (2c + 2 + n - h - x)/2$, to count rings and $\pi$ bonds. Turn to the infrared spectrum to name functional groups — broad O–H near $3300$, N–H near $3400$, C–H straddling $3000$, triple bonds at $2100$–$2260$, and the loud carbonyl near $1700$. Finally, reconcile: functional groups must match the degrees of unsaturation, and fragment masses must match sensible cleavages.

Master Reference — IR Diagnostic Bands

Carbonyl sub-positions (a finer cut of the $\sim 1700\ \text{cm}^{-1}$ band): anhydrides $\sim 1760 + 1820$ (two bands), esters $\sim 1735$, aldehydes $\sim 1725$, ketones $\sim 1715$, carboxylic acids $\sim 1710$, amides $\sim 1650$ (lowest). Conjugation with a C=C or aromatic ring lowers a carbonyl by $20$–$40\ \text{cm}^{-1}$.

Master Reference — MS Clues

Isotope patterns (M / M+2):

Common neutral losses (from $\text{M}^{+\bullet}$):

Diagnostic fragment ions:

Two rules to never forget: the nitrogen rule (odd molecular mass $\Rightarrow$ odd number of N atoms) and the McLafferty rearrangement ($\gamma$-H bearing carbonyls lose a neutral alkene to give an even-mass enol cation).

Mixed Review — Integrate the Techniques

The Decision Tree, in Words

When a fresh problem lands on your desk, run this top-down checklist. It rarely fails.

1. Is the molecular ion odd? If yes, suspect one nitrogen (nitrogen rule). Note the mass.
2. Is there an M+2? $3:1$ $\rightarrow$ Cl; $1:1$ $\rightarrow$ Br. Subtract the halogen mass before doing anything else.
3. How big is M+1? Divide by $1.1\%$ for a rough carbon count.
4. Write a trial formula; compute degrees of unsaturation. This is your budget of rings + $\pi$ bonds.
5. Scan IR above $3000$. Broad $\rightarrow$ O–H; medium (maybe split) $\rightarrow$ N–H; sharp at $3300$ $\rightarrow$ terminal alkyne; bands above $3000$ $\rightarrow$ unsaturated C–H.
6. Check $2100$–$2260$. Triple bond? Nitrile (sharp, $\sim 2250$) or alkyne (weak)?
7. Check $\sim 1700$. A strong band is a carbonyl; read its exact position to subtype it; remember conjugation lowers it.
8. Reconcile. Do the functional groups use up the degrees of unsaturation? Do the MS losses (15, 17, 18, 28, 29, 31, 45) and signature fragments (30, 43, 77, 91) match the proposed structure?

If all eight steps agree, you have your compound. If they conflict, the conflict itself tells you which assumption to revisit — usually a misread isotope ratio or an overlooked degree of unsaturation.

Final Checkpoint — Full Integration

Where This Leads

You can now take a mass spectrum and an IR spectrum of an unknown and extract its molecular mass, a plausible molecular formula, its degrees of unsaturation, its functional groups, and key skeletal features from fragmentation. That is the backbone of structure determination.

The remaining piece is connectivity — exactly which atom is bonded to which — and that is the province of NMR spectroscopy ($^1$H and $^{13}$C), the subject of the next course. With IR, MS, and NMR in hand, you will be able to take a colorless oil in a flask and write its full structure with confidence.

Carry these forward:

• Mass first ($\text{M}^{+\bullet}$), then formula (nitrogen rule, isotopes), then degrees of unsaturation.
• IR names functional groups; shape and exact position matter as much as the nominal wavenumber.
• Fragmentation reveals weak points; learn the common losses and signature ions.
• Always reconcile the two techniques into a single, self-consistent structure.