Organic Chemistry 202-SNU-RE • CEGEP, Montreal
Structure determination is the one chapter where almost nothing is calculated: you read. That is exactly why it goes wrong, because a piece of data read too fast, a carbonyl band at taken for , steers the whole conclusion.
This sheet leaves the tables in your course notes and gathers the order of reading: the molecular formula first, the functional groups next, the skeleton last, and a systematic cross-check before concluding.
The thread of the chapter
No single technique gives the structure: mass gives the formula, infrared gives the functional groups, NMR gives the skeleton, and it is the CROSS-CHECKING of the four that concludes.
This number is calculated before opening a single spectrum. It immediately steers the reading of the infrared, and it eliminates most of the structures you would be tempted to propose.
Two fragments to know: is the tropylium cation, the signature of a benzylic group; is the acylium , the signature of a methyl ketone.
The aldehyde is also signalled by a DOUBLET of bands at and cm, absent from every ketone: it is the only infrared way to separate them with certainty.
The abundance of carbon- is only : two carbon- atoms are practically never neighbours, which is why they do not couple with each other.
These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.
What not to write
“: I subtract the two oxygens as well.”
What to write
“Oxygen is IGNORED: .”
Why: A divalent oxygen slips into a chain without using up a hydrogen. This is also why nitrogen, being trivalent, adds one unit to the numerator.
What not to write
“There is an peak, so the molecule contains a chlorine.”
What to write
“Its HEIGHT must be read: of for a chlorine, almost for a bromine.”
Why: The two isotopes of bromine are almost equally abundant, those of chlorine in a ratio of three to one. The relative height is therefore a measurement, not a detail.
What not to write
“The molar mass is even, so there is a nitrogen.”
What to write
“An ODD mass signals an ODD number of nitrogens: an even mass signals zero or two nitrogens.”
Why: Nitrogen is the only common element whose valence and mass have different parities. It is this anomaly that makes the rule usable.
What not to write
“There is a broad band near , so it is an alcohol.”
What to write
“The acid band is MUCH broader, from to , and it overlaps the bands: it is its breadth that tells it apart.”
Why: A carboxylic acid forms dimers through very strong hydrogen bonds, which broadens the band enormously. The final check remains the carbonyl near .
What not to write
“There is a band near , so it is a ketone.”
What to write
“ points to an ester, to a ketone, to an amide: twenty wavenumbers are enough to change the functional group.”
Why: This is the richest piece of data in the whole chapter. A plain “there is a carbonyl” earns almost nothing, whereas the precise value names the derivative.
What not to write
“The carbonyl is at , so it is impossible to decide between aldehyde and ketone.”
What to write
“We look near and : this DOUBLET of bands exists only in the aldehyde.”
Why: The two functional groups have carbonyls too close together, against , to decide on this criterion alone. NMR, with its signal near ppm, gives the same answer.
What not to write
“The carries three hydrogens, so it comes out as a quartet.”
What to write
“We count the NEIGHBOURS: the of an ethyl sees the two hydrogens of the , so , a TRIPLET.”
Why: The hydrogens of the same carbon are equivalent and do not couple with each other. The ethyl pattern, triplet plus quartet, is the most frequent in exam spectra.
What not to write
“The curve rises by cm, so there are two hydrogens.”
What to write
“The integration gives PROPORTIONS: we scale to the smallest signal, then adjust to the total number of hydrogens in the molecular formula.”
Why: There is no absolute unit in NMR. The ratio can describe eight hydrogens as well as sixteen: it is the molecular formula that decides.
What not to write
“The neighbouring carbon should couple and give a doublet.”
What to write
“With proton decoupling, each nonequivalent carbon gives ONE SINGLET: the abundance of carbon- is too low for two of them to be neighbours.”
Why: An abundance of means about one chance in that two neighbouring carbons are both carbon-. The spectrum therefore simply counts the distinct carbons.
We always follow the same order, from the most general to the most precise.
If we have the mass spectrum → establish the molar mass, apply the nitrogen rule and read the isotope cluster
Example: at announces a chlorine
If we have the molecular formula → calculate the degree of unsaturation BEFORE opening the infrared
Example: a degree of zero rules out every carbonyl
If we read the infrared → look first for the carbonyl and its exact position, then the and bands
Example: points to an ester
If we read the proton NMR → count the signals, read the integration, then the multiplicities
Example: a triplet and a quartet sign an ethyl
If we have the carbon NMR → count the signals to measure the symmetry
Example: two equivalent carbons give a single peak
A structure is proposed only after cross-checking at least two techniques, and it is VERIFIED on the way back down: every piece of data in the problem must be explained by the structure retained.
We read the value in ppm and work back to the environment.
If between and ppm → hydrogen of an alkane, far from any withdrawing group
Example: the terminal methyl of a chain
If between and ppm → hydrogen alpha to a carbonyl
Example: the methyl of an acetate
If between and ppm → hydrogen carried by a carbon bonded to an OXYGEN
Example: the methylene of an ethyl ester
If between and ppm → aromatic hydrogen
Example: the five hydrogens of a phenyl
If beyond ppm → aldehyde near , carboxylic acid from to and very broad
Example: a singlet near ppm signs the aldehyde
The shift measures DESHIELDING: the richer the environment in electronegative atoms or circulating electrons, the further the signal moves to the left of the spectrum.
A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.
When to use it: The problem gives a molecular formula and several spectra.
Concluding sentence
“The degree of unsaturation is , the carbonyl is at cm and the spectrum shows an ethyl pattern: the compound is ethyl ethanoate.”
The trap: Proposing a structure from the first compatible piece of data, without checking the others: the marking scheme counts the verification as much as the proposal.
Marking: 1 mark the degree of unsaturation, 1 mark per technique used, 1 mark the structure, 1 mark the final verification.
When to use it: The problem asks you to assign each peak to a group of hydrogens.
Concluding sentence
“The triplet at ppm integrating for H has two neighbours: it is the methyl of an ethyl group.”
The trap: Forgetting to check that the sum of the integrations matches the total number of hydrogens: it is the check that catches forgotten signals.
Marking: 0.5 mark per correctly assigned signal, 1 mark the complete table, 0.5 mark the verified sum.
Five minutes of checking recover more marks than one more problem started in a hurry.
The sum of the integrations
Add up the assigned hydrogens and compare with the number in the molecular formula.
A total of for a formula with signals a forgotten signal, often a broad .
Is the degree of unsaturation used up?
Check that the proposed structure carries exactly the number of multiple bonds and rings announced by the calculation.
A degree of and a structure with neither carbonyl nor ring: the proposal is incomplete.
Is every piece of data in the problem explained?
Take the retained structure and go back down over every spectrum supplied, including the bands that served no purpose.
An unexplained band at signals a forgotten functional group.
Consistency between techniques
Cross-check at least two techniques before concluding: infrared gives the functional group, NMR must confirm it.
A carbonyl at with no hydrogen near ppm contradicts the hypothesis of an ethyl ester.
A compound of formula shows in the infrared an intense band at cm and no broad band near .
In proton NMR: a triplet at ppm integrating for H, a singlet at ppm for H, a quartet at ppm for H. Determine the structure.
Step 1
Degree of unsaturation: , the oxygens being ignored.
Why
A single degree: there will be one double bond OR one ring, not both. The infrared will say which.
Step 2
Band at cm: it is an ESTER, and not a ketone at or an acid at .
Why
The exact value names the derivative. The absence of a broad band near confirms that it is not a carboxylic acid.
Step 3
Triplet at ppm for H and quartet at ppm for H: it is an ETHYL group.
Why
The triplet has two neighbours, the quartet has three: they see each other. The shift of places the methylene on the oxygen of the ester.
Step 4
Singlet at ppm for H: a methyl with NO neighbour, alpha to the carbonyl.
Why
The absence of coupling requires a neighbouring carbon with no hydrogen, here the carbonyl carbon. The shift confirms the alpha position.
Step 5
Sum of the integrations: hydrogens, consistent with .
Why
The count closes the reading: no signal has been forgotten, which in particular rules out a broad acidic hydrogen.
Step 6
Structure: ethyl ethanoate, .
Why
It combines the ester, the isolated methyl alpha to it and the ethyl carried by the oxygen. All four carbons and both oxygens are placed.
Step 7
Verification: every piece of data is explained, and carbon NMR would show four signals, all the carbons being nonequivalent.
Why
We go back down over the data one by one. An extra prediction, here the count of carbon signals, secures the conclusion.
The conclusion, written out
“The compound is ethyl ethanoate: the carbonyl at cm points to an ester, the triplet and the quartet make up the ethyl carried by the oxygen, and the singlet at ppm is the methyl alpha to the carbonyl.”
The classic mistake on this problem: Proposing butanoic acid: it would have a very broad band near and a signal beyond ppm, both absent.
Because an oxygen is divalent: it slips into a chain between two atoms without costing the molecule a single hydrogen. Nitrogen, being trivalent, adds one unit to the numerator of the formula instead, and halogens are counted like hydrogens since they occupy a single bond.
By the height of the peak two units above the molecular ion peak. Chlorine has two isotopes in a ratio of three to one, which gives a peak at about thirty-two percent. Bromine has two almost equally abundant isotopes, which gives two peaks of practically equal height.
It names the derivative. An acyl chloride absorbs near eighteen hundred, an ester near seventeen thirty-five, an aldehyde near seventeen twenty-five, a ketone near seventeen fifteen and an amide near sixteen fifty. Conjugation lowers the value by about thirty units, ring strain raises it.
You count the hydrogens carried by the NEIGHBOURING carbons, and the signal splits into that number plus one. The hydrogens of the same carbon are equivalent and do not couple with each other. A methyl next to a methylene therefore gives a triplet, and that methylene gives a quartet: this is the ethyl pattern.
Because coupling with the hydrogens is removed by decoupling, and because carbon thirteen makes up only one percent of carbons. The probability that two neighbouring carbons are both carbon thirteen is therefore tiny. The spectrum then reads as a simple count of the nonequivalent carbons.
© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-snu-spectroscopy. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).
