Corrected exercises: spectroscopy and structure determination (202-SNU-RE)

The mass $73$ is ODD, so the compound contains an ODD number of nitrogens: the simplest is ONE nitrogen. The $M+1$ peak comes almost entirely from the presence of one carbon-$13$ atom in the molecule; since each carbon has a $1.1$ percent chance of being a carbon-$13$, the ratio $\frac{M+1}{M}$ is about $1.1\,n_{C}$ percent. Solving $1.1\,n_{C}=4.4$ gives $n_{C}=4$ carbons. It remains to complete the formula: $4$ carbons weigh $48$ and one nitrogen $14$, which makes $62$ and leaves $11$ units. Eleven hydrogens fit exactly and give $C_{4}H_{11}N$, whose degree of unsaturation is $\frac{2\times 4+2+1-11}{2}=0$: a saturated, acyclic molecule, which is consistent. It is a $C_{4}$ amine, for example butan-$1$-amine or diethylamine. Note the power of the method: three numbers read off a spectrum gave the complete molecular formula.

The common atoms of organic chemistry, nitrogen excepted, all share one property: their most abundant mass and their valence have the same PARITY. Carbon weighs $12$ and is tetravalent, even and even; oxygen weighs $16$ and is divalent, even and even; hydrogen weighs $1$ and is monovalent, odd and odd; the halogens are monovalent with an odd mass. Nitrogen is the exception: it weighs $14$, an EVEN mass, but it is trivalent, an ODD valence. Each nitrogen therefore introduces a shift in parity. A molecule with no nitrogen, or with an even number of nitrogens, necessarily has an EVEN nominal molar mass; an odd number of nitrogens makes it ODD. The rule is therefore not an empirical recipe but a direct consequence of the fact that the skeleton must have all its bonds satisfied, and it is the first reflex to have when facing a mass spectrum.

The first compound contains one CHLORINE. The theoretical ratio is the ratio of the abundances, $\frac{24.2}{75.8}=0.319$, that is $31.9$ percent, in agreement with the $32$ percent observed. Remember the characteristic $3$ to $1$ cluster profile. The second contains one BROMINE. The theoretical ratio is $\frac{49.3}{50.7}=0.972$, that is $97.2$ percent: the two peaks are practically the same height, a twin-tower profile that cannot be mistaken for anything else. These two clusters are the easiest to spot in all of mass spectrometry, and they let you conclude that the halogen is present BEFORE you even know the formula. Note that two chlorines would give a three-peak cluster in the ratio $9$ to $6$ to $1$, and that iodine, being monoisotopic, gives no cluster at all but betrays itself by a loss of $127$.

The loss of $15$ corresponds to a METHYL radical $CH_{3}$, which indicates at least one methyl on the molecule. The loss of $29$ can correspond to $CHO$ or to $C_{2}H_{5}$; additional information is therefore needed to decide, typically infrared, which will or will not show a carbonyl. The loss of $43$ corresponds either to a propyl $C_{3}H_{7}$ or, above all, to the ACYLIUM ion $CH_{3}CO^{+}$, an extremely common fragment. The consistency of the whole is instructive: losing $15$ and losing $43$ from the same starting molecule strongly suggests a $CH_{3}CO-$ group, which can give up its methyl or leave as a whole. We therefore lean toward a methyl ketone or an ethanoate ester. This reasoning illustrates the golden rule of interpretation: an isolated fragment proves nothing, it is the COMBINATION of losses that identifies a structural unit.

The fragment at $m/z=91$ has the formula $C_{7}H_{7}^{+}$. It arises from a BENZYLIC cleavage, that is, from the breaking of the bond just beyond the carbon attached to the ring, which here expels a methyl radical from $C_{6}H_{5}CH_{2}-CH_{3}$. The benzyl cation formed is already stabilized by resonance over the ring, but it does better: it immediately rearranges to the TROPYLIUM ion, a ring of SEVEN carbons bearing seven hydrogens and one positive charge. This ion has six $\pi$ electrons delocalized over a planar, fully conjugated ring: it therefore satisfies Hückel's rule with $n=1$ and it is AROMATIC. This aromaticity gives it an exceptional stability, far beyond that of an ordinary carbocation, and a very stable fragment is a fragment that accumulates: hence a base peak that is often much more intense than the molecular ion peak itself. The peak at $91$ is therefore an almost certain signature of an aromatic ring bearing a side chain of at least two carbons, and one of the few fragments you must know by heart.

The formula shows that the wavenumber increases with the force constant $k$, which measures the stiffness of the bond, and DECREASES with the reduced mass $\mu$, which measures the inertia of the two atoms. For the $C-H$ bond, the reduced mass is $\frac{12\times 1}{13}=0.92$: it is very small, because one of the two partners is the lightest atom there is. For the $C-O$ bond, it is $\frac{12\times 16}{28}=6.86$, seven times larger. At comparable force constants, this factor alone already divides the wavenumber by $\sqrt{7.4}\approx 2.7$. On top of this, $C-O$ is a single bond between two heavy atoms, a little less stiff than a $C-H$ bond. The result is indeed a factor close to three between $3000$ and $1100$ cm$^{-1}$. Remember the qualitative rule: bonds to HYDROGEN are always at the top of the spectrum, above $2500$, and everything below concerns bonds between heavy atoms. This is what explains the two-region structure of the infrared spectrum.

The force constant depends only on the electronic nature of the bond, and deuterium has exactly the same electrons as hydrogen: $k$ is therefore unchanged and only $\mu$ varies. We calculate $\mu(C-H)=\frac{12\times 1}{13}=0.923$ and $\mu(C-D)=\frac{12\times 2}{14}=1.714$. The ratio of the wavenumbers is $\sqrt{\frac{\mu(C-H)}{\mu(C-D)}}=\sqrt{\frac{0.923}{1.714}}=\sqrt{0.5385}=0.734$. The $C-D$ band therefore lies near $3000\times 0.734\approx 2200$ cm$^{-1}$, which experiment confirms. This is a region of the spectrum that is otherwise almost empty, which makes deuterium labelling very convenient for following one particular atom. Compare this calculation with the kinetic isotope effect met in the chapter on aromatic compounds: it is the same reduced mass that, there, slowed down the breaking of the bond.

The table gives, in decreasing frequency: acyl chloride $1800$, anhydride $1760$ and $1820$, ester $1735$, aldehyde $1725$, ketone $1715$, acid $1710$, amide $1650$. Setting aside the aldehyde and the ketone, which are not acid derivatives, we get chloride, anhydride, ester, acid, amide: word for word, the reactivity hierarchy established in the chapter on carboxylic acids. The coincidence is no coincidence, and the explanation fits in one sentence: the two rankings measure the same thing, the degree to which the heteroatom DONATES its lone pair to the carbonyl. The more this heteroatom donates, the more the resonance form with a single $C-O$ bond weighs, and the more the carbonyl bond has a WEAKENED double-bond character. A weakened bond has a smaller force constant $k$, hence a lower wavenumber; and a carbonyl whose positive charge has been neutralized by this donation is a less reactive electrophile. The nitrogen of the amide donates a lot, hence $1650$ and the lowest reactivity; chlorine donates almost nothing, hence $1800$ and the highest reactivity. An infrared spectrum therefore MEASURES, to within $10$ cm$^{-1}$, an electronic effect that the previous chapter had only reasoned out.

The anhydride contains TWO carbonyl groups linked by the same oxygen. These two oscillators do not vibrate independently: they are mechanically COUPLED through the atom that links them, and this interaction splits the vibration into two modes, one symmetric in which both carbonyls stretch in phase, the other antisymmetric in which one stretches while the other shortens. These two modes do not have the same energy, hence two distinct bands near $1760$ and $1820$ cm$^{-1}$. This pair is an unambiguous diagnostic: no other functional group gives two carbonyl bands so close together and so intense. As for conjugation, it lowers the frequency because a carbonyl conjugated with a double bond or an aromatic ring sees its $\pi$ electrons delocalize over the whole system. The $C=O$ bond loses some of its double-bond character, exactly as in the amide case, its force constant decreases, and the band drops by about $30$ cm$^{-1}$: an ordinary ketone at $1715$ thus falls to $1685$ if it is conjugated. Conversely, enclosing the carbonyl in a small strained ring forces the angle to close, which stiffens the bond and makes the band RISE, up to $1780$ cm$^{-1}$ for a cyclobutanone.

All have a degree of unsaturation of $\frac{2\times 3+2-6}{2}=1$. A has a carbonyl at $1725$ and above all the DOUBLET at $2720$ and $2820$ cm$^{-1}$, the exclusive signature of the $C-H$ bond of an aldehyde: A is PROPANAL. B has a carbonyl at $1715$ without this doublet: it is therefore a ketone, and the only possible one in $C_{3}H_{6}O$ is PROPANONE. C has no carbonyl but a broad band at $3350$, hence an alcohol; the band at $1645$ signals a $C=C$ double bond and the one at $3080$ hydrogens on $sp^{2}$ carbons: the unsaturation is therefore a $C=C$ and not a $C=O$. C is PROP-$2$-EN-$1$-OL. D has neither a carbonyl nor an $O-H$, and yet it must place its single unsaturation and its oxygen somewhere: the strong band at $1100$ indicates a single $C-O$ bond, hence an ether, and the unsaturation can then only be a RING. D is a cyclic ether, methyloxirane or its isomer oxetane. The reasoning to remember is the one applied to C and D: the degree of unsaturation must ALWAYS be placed, and it is often the absence of a band, more than its presence, that forces the conclusion.

First piece of information, the NUMBER of signals: it gives the number of sets of EQUIVALENT hydrogens, that is, of distinct chemical environments. It is a measure of symmetry. Second, the chemical SHIFT $\delta$, expressed in parts per million: it measures the electronic environment of each hydrogen, hence which atoms surround it. Third, the INTEGRATION, that is, the area under each peak: it is proportional to the NUMBER of hydrogens the signal represents, and it must be stressed that it gives only RATIOS, never absolute numbers. Fourth, the MULTIPLICITY: the splitting of the signal into several lines tells us the number of hydrogens carried by the NEIGHBOURING atoms, according to the $n+1$ rule. These four pieces of information are independent and cross-check each other: it is precisely this cross-checking that makes the elucidation reliable.

A hydrogen placed in a magnetic field is not bare: the electrons around it circulate and create a small induced field that opposes the applied field. The hydrogen therefore feels a field slightly weaker than the nominal field: it is said to be SHIELDED, protected, by its electron cloud. Deshielding is the opposite situation: when a neighbouring electronegative atom pulls electron density toward itself, the cloud that surrounded the hydrogen thins out, the protection decreases, and the hydrogen feels a field closer to the nominal field. It then resonates at a higher frequency, which is recorded as a LARGER chemical shift, toward the left of the spectrum, the region called downfield. In chloroethane, the very electronegative chlorine strongly depletes the hydrogens of the neighbouring carbon, which come out near $3.5$ ppm, and slightly less those of the more distant methyl, near $1.5$ ppm; in ethane, no atom pulls on the electrons and the six hydrogens stay at $0.9$ ppm. The chemical shift is therefore a direct, local measure of the inductive effect, a notion met in the very first chapter of the course.

First reason, tetramethylsilane has TWELVE hydrogens, all rigorously equivalent, carried by four identical methyls around one silicon: it therefore gives a single, sharp, intense signal, even in very small amounts. Second reason, silicon is less electronegative than carbon, so it pushes electron density toward its methyls instead of withdrawing it: its hydrogens are therefore MORE shielded than those of practically any organic molecule. We can add that it is chemically inert, volatile and therefore easy to remove after the measurement, and soluble in the usual solvents. This second reason answers the next question: since almost every organic group contains an atom more electronegative than silicon, almost all the hydrogens of the organic world are less shielded than those of tetramethylsilane and so come out to its left, between $0$ and $12$ ppm. The scale was thus built so that the usual values are positive, which is a convenience and not a physical necessity: a few rare compounds, such as certain cyclophanes whose hydrogens sit above an aromatic ring, do indeed have NEGATIVE shifts.

The molecule has three distinct environments. The methyl of the acyl part, $CH_{3}CO-$, is next to a carbonyl but has no hydrogen on the adjacent atom: it gives a SINGLET integrating for $3$ hydrogens, near $2.05$ ppm. The methylene $-OCH_{2}-$ is carried by a carbon bonded to oxygen and has three neighbouring hydrogens: it gives a QUARTET integrating for $2$ hydrogens, near $4.12$ ppm. The terminal methyl $-CH_{3}$ has two neighbouring hydrogens: it gives a TRIPLET integrating for $3$ hydrogens, near $1.25$ ppm. We therefore expect three signals in the integration ratio $3:2:3$. This triplet and this quartet, in this $3:2$ ratio, make up the ETHYL pattern, the most common in all of organic NMR, which you must be able to recognize at a glance.

Both hydrogens are indeed next to the same ester group, but not to the same ATOM of that group, and that is the whole difference. The methyl at $2.05$ ppm is carried by the CARBONYL carbon: the atom it is directly bonded to is a carbon, admittedly depleted by the neighbouring oxygen, but a carbon all the same. The deshielding it undergoes is therefore SECOND-hand, transmitted through an intermediate atom, and the inductive effect fades sharply from one bond to the next, as the chlorobutanoic acids of the previous chapter had quantified. The methylene at $4.12$ ppm, on the contrary, is carried by a carbon bonded directly to the OXYGEN of the ester group. It therefore feels the inductive effect head-on, with no intermediary, from a very electronegative atom. The two-ppm gap measures exactly this difference between a direct deshielding and a transmitted one. This gives a general and very rewarding reading rule: a hydrogen beyond $3.5$ ppm on a saturated molecule is almost always carried by a carbon bonded directly to an oxygen or a halogen, and this observation alone places the heteroatom in the skeleton.

Each neighbouring hydrogen has a spin that, in the applied field, can take only TWO orientations, parallel or antiparallel, in practically equal proportions. The small magnetic field it creates therefore adds to the field felt by the observed hydrogen in half of the molecules, and subtracts from it in the other half. The observed hydrogen consequently resonates at two slightly different frequencies: its signal is SPLIT in two. With a second neighbour, each of these two lines splits in turn, as the figure shows on the lower level, giving four lines in all. But two of them overlap exactly in the centre, since the combination in which the first spin adds and the second subtracts gives the same field as the reverse combination. We therefore observe THREE lines, the central one being twice as tall as the others because it collects two of the four combinations: hence the ratio $1:2:1$. The intensities of a multiplet thus follow Pascal's triangle, $1:1$ for a doublet, $1:2:1$ for a triplet, $1:3:3:1$ for a quartet, and one can check that the sum of the lines is always $2^{n}$, that is, the number of spin combinations.

For $1$-bromopropane $CH_{3}CH_{2}CH_{2}Br$: the $CH_{2}$ carried by the bromine comes out near $3.4$ ppm, it has two neighbours so it is a TRIPLET integrating for $2$; the central $CH_{2}$ comes out near $1.9$ ppm, it has $2+3=5$ neighbours so it is a SEXTET integrating for $2$; the methyl comes out near $1.0$ ppm, it has two neighbours so it is a TRIPLET integrating for $3$. That makes three signals, $2:2:3$. For $2$-bromopropane $\left(CH_{3}\right)_{2}CHBr$: the six hydrogens of the two methyls are EQUIVALENT by symmetry and have only one neighbour, so a DOUBLET integrating for $6$ near $1.7$ ppm; the central hydrogen has six neighbours, so a SEPTET integrating for $1$ near $4.3$ ppm. That makes only two signals, $6:1$. It is $2$-bromopropane that NMR identifies faster: two signals instead of three, an integration of $6:1$ that cannot be mistaken, and a septet whose sight alone is enough to conclude. This is the ISOPROPYL pattern, the second major pattern to recognize at once after the ethyl pattern.

First condition, the coupling hydrogens must be NONEQUIVALENT. Equivalent hydrogens do couple with each other physically, but this coupling has no observable effect on the spectrum: this is why the six hydrogens of benzene, or the four of $1,2$-dichloroethane, give a simple singlet whereas the rule applied naively would predict a multiplet. Second condition, the coupling constants with the different neighbours must be EQUAL so that the lines overlap cleanly. It holds well between hydrogens carried by neighbouring $sp^{3}$ carbons, where $J$ is about $7$ Hz everywhere, but it fails on an alkene or a substituted aromatic ring, where the cis, trans and geminal couplings have very different values. One then obtains not a symmetric multiplet but a doublet of doublets, with four lines of equal intensity. This is why a vinylic hydrogen with two neighbours almost never gives a clean triplet.

The hydrogen carried by the oxygen of an alcohol is EXCHANGEABLE: it constantly hops from one molecule to another by proton transfer, catalyzed by the traces of acid or water that are always present. This hopping is much faster than the time scale of the measurement, so the instrument sees not a hydrogen attached to a specific neighbour but an average: the coupling information is erased and the signal appears as a singlet, often broadened by the exchange process itself. Its shift is moreover very variable, from $1$ to $5$ ppm depending on concentration, solvent and temperature, which rules out using it as a landmark. The confirmation test is the HEAVY WATER test: a few drops of $D_{2}O$ are added to the sample and it is shaken. The exchangeable hydrogen is replaced by a deuterium, which does not resonate in the proton window. The signal therefore DISAPPEARS from the spectrum, which proves in hindsight that it corresponded to a hydrogen carried by an oxygen or a nitrogen. It is a thirty-second manipulation that removes the most frequent ambiguity of an elucidation.

The constant of $16$ Hz corresponds to the TRANS isomer, the one of $10$ Hz to the CIS isomer. The reason is geometric: the strength of the coupling between two hydrogens carried by the two carbons of a double bond depends on the dihedral angle between them, according to a relation known as the Karplus relation. This angle is $180$ degrees in the trans arrangement and $0$ degrees in the cis arrangement, and the orbital overlap that transmits the spin information is maximal at $180$ degrees. Remember the ranges: $J_{trans}$ from $12$ to $18$ Hz, $J_{cis}$ from $6$ to $12$ Hz, which leaves an overlap zone near $12$ Hz but allows a decision in the great majority of cases. No other technique in the chapter can do this, and it is worth seeing why. Mass spectrometry weighs the molecule, but two stereoisomers have exactly the same mass and give nearly identical fragmentations. Infrared sees the functional groups, but the two isomers have the same ones; it does give an indication through the out-of-plane bending band, near $970$ cm$^{-1}$ for the trans, but it is less reliable. Carbon-$13$ NMR gives different shifts but with no sure rule. Only COUPLING, because it depends explicitly on GEOMETRY and not on composition, provides the stereochemical information, and this is the deep reason why one must measure the $J$ constants and not just count the lines.

The two events being independent, the probability that two given carbons are both carbon-$13$ is $0.011\times 0.011=1.21\times 10^{-4}$, about one molecule in $8300$. In other words, in nearly all the molecules that contribute to the signal, a carbon-$13$ has only carbon-$12$ neighbours, which have a ZERO nuclear spin and therefore cannot couple with anything. Carbon-carbon coupling does exist in that one molecule in $8300$, but it produces satellite lines so weak that they are lost in the noise. This is a rare case where a low isotopic abundance is an ADVANTAGE: it radically simplifies the spectrum. Note in passing that this same low abundance is also the weakness of the technique, since it makes the signal about six thousand times less intense than that of the proton and requires much longer acquisition times.

Decoupling the proton means continuously irradiating, during the measurement, all the resonance frequencies of the hydrogens. These then flip so quickly between their two spin orientations that the observed carbon feels only a zero AVERAGE: the carbon-hydrogen coupling, which would otherwise be very large since the hydrogens are directly bonded, disappears entirely. What is gained is considerable: each nonequivalent carbon gives a single SINGLET, so that counting the lines is enough to count the nonequivalent carbons, and the intensity, concentrated in one line instead of spread over a multiplet, becomes usable. What is lost is the multiplicity information, hence the number of hydrogens carried by each carbon. It is recovered by a complementary experiment called DEPT, which distinguishes $CH_{3}$, $CH_{2}$, $CH$ and quaternary carbons by the SIGN or the presence of their peak. One last point is often forgotten: in decoupled carbon-$13$, the integration is NOT proportional to the number of carbons, unlike in proton NMR, because of the nuclear Overhauser effect and the different relaxation times.

For ORTHO-dimethylbenzene, the axis of symmetry passing between the two substituted carbons makes carbons $1$ and $2$ equivalent, then $3$ and $6$, then $4$ and $5$: that gives three signals for the ring, plus one for the two equivalent methyls, so FOUR signals. For the META isomer, carbons $1$ and $3$ are equivalent, $4$ and $6$ are, but $2$ and $5$ are each alone of their kind: that gives four signals for the ring, plus one for the methyls, so FIVE signals. For the PARA isomer, the symmetry is much higher: carbons $1$ and $4$ are equivalent, and the four carbons $2$, $3$, $5$ and $6$ are all equivalent to one another; that gives two signals for the ring, plus one for the methyls, so THREE signals. The three isomers therefore give three different counts, $4$, $5$ and $3$, and the conclusion is clear: merely counting the lines of the carbon-$13$ spectrum, without even looking at the shifts, unambiguously identifies which of the three isomers we hold. No other technique in the chapter does this as fast, which illustrates the particular purpose of the method: measuring SYMMETRY.

The compound has a degree of unsaturation of $1$ and four carbons all NONEQUIVALENT, since there are four signals for four carbons: the molecule therefore has no symmetry. The signal at $171$ ppm falls in the range of esters, acids and amides; since the formula contains no nitrogen and this shift is too low for a ketone, it is an ESTER or an acid carbon. The one at $60.4$ ppm is in the range of carbons bonded to an oxygen: it is an $-OCH_{2}-$, which decides in favour of the ester. The signals at $21.0$ and $14.2$ ppm are two purely aliphatic carbons. Putting it together: an ester carbonyl, a carbon bonded to oxygen, and two methyls or similar carbons. The compound is ETHYL ETHANOATE $CH_{3}COOCH_{2}CH_{3}$, with $171$ for the carbonyl, $60.4$ for the $OCH_{2}$, $21.0$ for the methyl of the acyl group and $14.2$ for the terminal methyl. We check the consistency with the proton NMR of exercise $3$, which indeed gave three hydrogen signals for four carbons, the carbonyl carrying none.

The degree of unsaturation is $\frac{2\times 5+2-10}{2}=1$, and the signal at $206.5$ ppm falls squarely in the range of KETONES and aldehydes, well beyond the $185$ ppm that caps esters and acids. The single unsaturation is therefore this carbonyl, and no oxygen is left for another functional group. The decisive point is the count: the molecule has FIVE carbons but gives only THREE signals, which requires two pairs of carbons to be equivalent in pairs. A symmetry element exchanging the two halves of the molecule is therefore needed, and the only way to achieve this with a carbonyl is to place it exactly in the MIDDLE of the chain. The compound is PENTAN-$3$-ONE $CH_{3}CH_{2}COCH_{2}CH_{3}$: its axis of symmetry passes through the carbonyl, making the two methylenes equivalent on one hand and the two methyls on the other. We then assign $206.5$ ppm to the carbonyl, $36.4$ to the two $CH_{2}$ in the alpha position, and $7.9$ to the two terminal $CH_{3}$. Its straight-chain isomer carrying the carbonyl at position $2$, pentan-$2$-one $CH_{3}COCH_{2}CH_{2}CH_{3}$, has no symmetry at all: its five carbons are all in different environments and it would give FIVE distinct signals. The two isomers are therefore told apart at first glance, even before any assignment, and this example sums up the particular purpose of the technique: in carbon-$13$ NMR, always begin by COUNTING the lines and comparing them with the number of carbons in the molecular formula, because the gap between the two numbers is a direct measure of symmetry.

The degree of unsaturation is $\frac{2\times 4+2-8}{2}=\frac{10-8}{2}=1$, the two oxygens being ignored. All the structures considered therefore have EXACTLY one unsaturation, and only one: either a double bond, $C=O$ or $C=C$, or a ring. This constraint is the most useful in the exercise, because it forbids, for example, imagining both a carbonyl and a carbon-carbon double bond, or a carbonyl and a ring. It also forces us to PLACE the unsaturation in every case, even when the spectrum shows no carbonyl, which is precisely the trap of compound S.

P: the extremely broad band from $2500$ to $3300$ cm$^{-1}$ is the exclusive signature of the $O-H$ of a CARBOXYLIC ACID, its breadth coming from the dimer with two hydrogen bonds studied in the previous chapter; the carbonyl at $1710$ confirms it. P is a carboxylic acid. Q: a carbonyl at $1735$ and a strong $C-O$ band near $1240$, with no $O-H$ at all: it is an ESTER, and the position $1735$ is exactly the one expected. R: a broad band at $3400$ signals an alcohol, and $1715$ a KETONE; the absence of the doublet at $2720$ and $2820$ confirms that it is not an aldehyde. R therefore carries both groups, alcohol and ketone. S: an alcohol at $3350$, but NO carbonyl. Since the unsaturation must be placed somewhere and it is neither a $C=O$ nor a $C=C$, for lack of a band near $1650$ and near $3080$, the unsaturation can only be a RING; the strong band at $1100$ further indicates a single $C-O$ bond, hence a cyclic ether. S is a cyclic ether bearing an alcohol. T: the doublet at $2720$ and $2820$ is the exclusive signature of the ALDEHYDE, confirmed by the carbonyl at $1725$; the band at $3400$ adds an alcohol. T is a hydroxy aldehyde.

For P, a carboxylic acid of formula $C_{4}H_{8}O_{2}$: two structures fit, butanoic acid and $2$-methylpropanoic acid. For Q, an ester of formula $C_{4}H_{8}O_{2}$: four structures fit, ethyl ethanoate, methyl propanoate, propyl methanoate and $1$-methylethyl methanoate. What prevents infrared from deciding lies in the very nature of the technique: it detects BONDS, not skeletons. Now the four esters have exactly the same bonds, an ester carbonyl, $C-O$ bonds, $C-H$ bonds and $C-C$ bonds; only their ARRANGEMENT differs, and the arrangement has almost no effect on the vibration frequencies of the functional groups. The spectra differ only in the region known as the fingerprint region, below $1500$ cm$^{-1}$, where the vibrations involve the whole molecule: this region is indeed characteristic of each compound and allows identification by COMPARISON with a reference spectrum, but it is far too complex to be interpreted line by line. Infrared can therefore say “it is an ester” and can say “it is THIS ester, because its fingerprint matches the one in the catalogue”, but it cannot build an unknown structure.

First unanswered question: what is the MOLAR MASS, hence the molecular formula? Infrared says nothing about it, and it is mass spectrometry that answers, through the molecular ion peak and the isotope cluster. Second question: how many hydrogens are there in each environment, and who is next to whom? PROTON NMR answers, through integration and multiplicity. Third question: how many nonequivalent carbons are there, in other words what is the symmetry of the molecule? CARBON-$13$ NMR answers, by simply counting the lines. A fourth question can be added, that of cis or trans stereochemistry, which only the coupling constant of proton NMR answers reliably. This division of roles is the reason why a serious elucidation always chains the four techniques in the order mass, infrared, proton, carbon.

For $1755$ cm$^{-1}$, two explanations are consistent with the table. First: an ordinary ESTER, whose reference value is $1735$, slightly raised by a neighbouring electron-withdrawing substituent. Second: a KETONE enclosed in a strained ring, the ring strain raising the frequency well above the usual $1715$, a cyclopentanone coming out near $1745$ and a cyclobutanone near $1780$. For $1690$ cm$^{-1}$, two explanations as well. First: a CONJUGATED ketone or aldehyde, conjugation lowering the value by about $30$ cm$^{-1}$ from $1715$ or $1725$. Second: an AMIDE, whose reference value is $1650$, raised by a substituent or by the absence of hydrogen bonding. The complementary measurement that separates them is immediate in both cases and does not even require changing technique. For $1755$, we look for a strong $C-O$ band near $1200$ to $1250$ cm$^{-1}$: if present, it is an ester; if absent, it is a cyclic ketone. For $1690$, we look at the region near $3300$ cm$^{-1}$: one or two $N-H$ bands signal the amide, and their absence points back to the conjugated carbonyl, which aromatic bands near $1600$ and $1500$ would confirm. The general lesson is that an isolated band is a HYPOTHESIS and not a conclusion: it is the consistency of the whole spectrum that identifies.

The total integration is $6+1=7$ hydrogens, in agreement with the formula. The SEPTET integrating for a single hydrogen requires six equivalent neighbours, and the doublet integrating for six requires that these six hydrogens have only one neighbour: this is the ISOPROPYL pattern, two identical methyls attached to the same carbon bearing a single hydrogen. The shift of $4.32$ ppm for this central hydrogen is clearly deshielded, which places the bromine atom on this carbon. U is therefore $2$-BROMOPROPANE $\left(CH_{3}\right)_{2}CHBr$. Its isomer, $1$-bromopropane, would on the contrary give THREE signals, a triplet at $3.4$ ppm integrating for $2$, a sextet near $1.9$ integrating for $2$, and a triplet near $1.0$ integrating for $3$. The two spectra look nothing alike, neither in the number of signals, nor in the integration, nor in the multiplicities: this is the ideal case where NMR instantly settles a question that infrared could not even approach.

The total integration is $3+2+5=10$ hydrogens, in agreement with $C_{8}H_{10}$. The degree of unsaturation is $\frac{2\times 8+2-10}{2}=4$, which strongly suggests an AROMATIC ring, which uses up exactly four. The triplet at $1.22$ integrating for $3$ and the quartet at $2.62$ integrating for $2$ make up the ETHYL pattern. That leaves $5$ aromatic hydrogens, which corresponds to a MONOsubstituted ring, since a benzene carries six and only one has been replaced. V is therefore ETHYLBENZENE. The shift of $7.20$ ppm is explained by a mechanism specific to aromatics: in the applied field, the six delocalized $\pi$ electrons start circulating around the ring and generate a RING CURRENT, whose induced field opposes the applied field at the centre of the ring but ADDS to it on the outside, precisely where the hydrogens are. These are therefore strongly deshielded, well beyond what a simple inductive effect would explain, and come out between $6.5$ and $8$ ppm. The signal is an unresolved multiplet because the ortho, meta and para positions differ slightly and couple with each other with unequal constants.

The total integration is $6+1+1=8$ hydrogens, in agreement with the formula. We again recognize the isopropyl pattern, a doublet for $6$ hydrogens and a septet for $1$. The signal at $11.8$ ppm, a broad singlet integrating for a single hydrogen, is characteristic of the $O-H$ of a CARBOXYLIC ACID: no other organic hydrogen comes out so far downfield. Two causes add up for this extreme shift, the inductive effect of the neighbouring carbonyl and the hydrogen bonding of the dimer, which stretches the $O-H$ bond and uncovers the proton further. This uses up the functional group and the single unsaturation. W is therefore $2$-METHYLPROPANOIC ACID $\left(CH_{3}\right)_{2}CHCOOH$, the septet at $2.55$ ppm corresponding to the hydrogen carried by the carbon alpha to the carbonyl. The breadth of the signal and the absence of coupling prove that this hydrogen is EXCHANGEABLE. The confirming manipulation is the HEAVY WATER test: a few drops of $D_{2}O$, a shake, and the signal at $11.8$ ppm DISAPPEARS from the spectrum, the hydrogen having been replaced by a deuterium that is invisible in this window.

V contains the ETHYL pattern, recognizable by its triplet-quartet pair in the integration ratio $3:2$, with the triplet always further upfield than the quartet. W contains the ISOPROPYL pattern, recognizable by its doublet-septet pair in the ratio $6:1$. What makes them immediately identifiable is that they combine TWO redundant pieces of information: a characteristic multiplicity and a characteristic integration ratio, and the two must agree. A doublet integrating for $6$ can hardly be anything other than two equivalent methyls, and a septet is so rare that it almost suffices on its own. Spotting these patterns before any other analysis saves most of the time of an elucidation: the atoms they use up are removed from the count, and the reasoning is carried out on the rest.

The total integration is $6+4=10$, and the formula $C_{8}H_{10}$ gives four unsaturations, hence an aromatic ring. A singlet integrating for $6$ at $2.30$ ppm corresponds to TWO equivalent methyls carried by the ring, the shift being typical of a benzylic position. Four aromatic hydrogens remain, so the ring is DISUBSTITUTED. The fact that the two methyls are equivalent and that the four aromatic hydrogens are too requires a high symmetry: it is the PARA isomer, $1,4$-dimethylbenzene. The aromatic signal is a singlet not because these hydrogens have no neighbours, but because they are all EQUIVALENT to one another. Now the first condition of the $n+1$ rule, recalled in exercise $4$, is that coupling is observable only between NONEQUIVALENT hydrogens: equivalent nuclei do couple physically, but this coupling has no effect on the appearance of the spectrum. It is the same phenomenon that reduces benzene itself, or $1,2$-dichloroethane, to a single singlet. Note, incidentally, that the ortho isomer would also give two signals, but with an aromatic multiplet and not a singlet, since its four hydrogens split into two nonequivalent pairs.

The $M+1$ peak is $8.8$ percent, so the number of carbons is $\frac{8.8}{1.1}=8$. The mass $120$ is EVEN, so the number of nitrogens is zero or even; the absence of any $N-H$ band in the infrared and the fact that eight carbons already weigh $96$ lead us to rule out nitrogen. This leaves $120-96=24$ units to distribute between hydrogen and oxygen. One oxygen uses $16$ and leaves $8$ hydrogens: the formula $C_{8}H_{8}O$ fits and does weigh $120$. The competing hypothesis with no oxygen would require $24$ hydrogens for eight carbons, which exceeds the limit of $2n+2=18$: it is impossible. The formula is therefore $C_{8}H_{8}O$, and the degree of unsaturation is $\frac{2\times 8+2-8}{2}=\frac{18-8}{2}=5$. The absence of an $M+2$ cluster also rules out chlorine and bromine.

The doublet at $2720$ and $2820$ cm$^{-1}$ is the exclusive signature of the $C-H$ bond of an ALDEHYDE, and the strong band at $1700$ confirms a carbonyl. The bands at $1605$ and $1575$ belong to the $1450$-$1600$ range of skeletal vibrations of an AROMATIC ring, and the one at $3030$ corresponds to $C-H$ bonds on $sp^{2}$ carbons. The absence of any band above $3100$ rules out alcohol, acid and amine. We therefore have an aromatic aldehyde, which uses four unsaturations for the ring and one for the carbonyl: exactly the five calculated in a), and nothing is left to place. The carbonyl comes out at $1700$ and not at $1725$ because it is CONJUGATED with the aromatic ring: its $\pi$ electrons delocalize over the whole system, the $C=O$ bond loses some of its double-bond character, its force constant decreases and the band drops by about $25$ cm$^{-1}$, in line with the rule of exercise $2$. This lowering is therefore not an anomaly to be excused, it is an independent CONFIRMATION of the conjugation.

The total of the integrations is $3+2+2+1=8$ hydrogens, exactly the formula. The singlet at $9.95$ integrating for $1$ is the hydrogen of the ALDEHYDE, which always comes out between $9.5$ and $10$ ppm and has no neighbour. The singlet at $2.44$ integrating for $3$ is a METHYL in a benzylic position, with no neighbour either. Four aromatic hydrogens remain, arranged as two doublets of $2$ hydrogens each. This configuration is very restrictive: four aromatic hydrogens mean a DISUBSTITUTED ring, and the fact that they form two EQUIVALENT pairs, each coupled only to the other pair, imposes PARA substitution. In the ortho or meta isomer, the four hydrogens would all be nonequivalent and would give a set of complex multiplets, never two clean doublets of two hydrogens. This pair of symmetric doublets is moreover so characteristic that it is recognized at once: it is the visual signature of the para-disubstituted benzene.

Let us count the signals expected for a benzaldehyde carrying a methyl in the para position. The para-disubstituted ring with two DIFFERENT groups has four nonequivalent carbons: the carbon bearing the carbonyl group, the carbon bearing the methyl, and the two pairs of intermediate carbons, equivalent in pairs. Added to these are the carbonyl carbon and the methyl carbon, making SIX signals in all. The spectrum shows exactly six, which confirms it. The assignment is immediate: $192.0$ ppm for the aldehyde carbon, a typical value for the $190$ to $220$ range; $145.5$ for the aromatic carbon bearing the methyl; $134.2$ for the one bearing the carbonyl; $129.8$ and $129.7$ for the two pairs of aromatic carbons bearing hydrogens; $21.8$ for the methyl. The structure is therefore $4$-METHYLBENZALDEHYDE, $CH_{3}-C_{6}H_{4}-CHO$, also known as para-tolualdehyde. The four techniques agree with no contradiction, which is the criterion of a successful elucidation.