Spectroscopy

In a nutshell

NMR spectroscopy shows you how the carbon and hydrogen atoms in a molecule are arranged, so it is the technique that pins down an exact structure.

In this subtopic you read ¹³C and ¹H (proton) NMR spectra, predict them, and then combine NMR with elemental analysis, mass spectra and IR spectra to deduce a structure. Every chemical shift you need is given on the Data Sheet: the skill is reading spectra, not memorising numbers.

Assumed knowledge: Analytical techniques (IR and mass spectra), Amount of substance (empirical and molecular formulae).

Core content

NMR spectroscopy: chemical environments

Everything in NMR rests on one idea: atoms in the same chemical environment give one signal (peak).

• Number of peaks = number of different environments (of carbon in ¹³C NMR, of hydrogen in ¹H NMR).
• Atoms are in the same environment (they are equivalent) if they are bonded to exactly the same arrangement of atoms in the molecule.
• Symmetry makes atoms equivalent, so a symmetrical molecule gives fewer peaks than it has atoms.

Still don't get it? · equivalent atoms

Imagine taking a photo of a molecule, secretly swapping two of its hydrogen atoms, and taking a second photo. If nobody could tell the two photos apart, those two hydrogens are "the same kind" of hydrogen.

Now build it up one step at a time:

1. Look at one atom and list what it is bonded to, then what those atoms are bonded to.
2. Do the same for a second atom of the same element.
3. If the two lists are identical (including where they sit in the molecule), the atoms are equivalent.

In propan-2-ol, CH₃CH(OH)CH₃, both CH₃ carbons are bonded to the same CH(OH) carbon in the middle. Swap them and the molecule is unchanged, so they share one peak. The middle carbon is different (it carries the OH), so it gets its own peak: 2 carbon environments.

The exam version: count the different environments, not the atoms, and check for symmetry first. One environment gives one peak.

NMR spectroscopy: chemical shift

The horizontal axis of every NMR spectrum is the chemical shift, δ, measured in parts per million (ppm).

• δ is the position of a signal relative to the signal of TMS, which is set at δ = 0 (more on TMS below).
• The δ value depends on the type of environment. Atoms near electronegative atoms (O, N, Cl, Br) or a C=O group appear at higher δ (further left).
• You match each δ value to an environment using the shift chart given on the Data Sheet.

NMR spectroscopy: carbon-13 NMR

A ¹³C spectrum tells you three things, in this order:

1. Number of peaks = number of carbon environments.
2. δ of each peak = the type of carbon environment (from the Data Sheet).
3. Put 1 and 2 together to suggest possible structures.

Every ¹³C spectrum in an OCR exam is proton decoupled: each carbon environment shows up as a single line, never split.

Approximate ¹³C shifts (given on the Data Sheet):

Ethyl ethanoate, CH₃COOCH₂CH₃, has four carbons, all in different environments, so it gives four peaks:

Reading it against the Data Sheet:

• 171 ppm: in the C=O range (160 to 220).
• 60 ppm: in the C-O range (50 to 90), the carbon bonded to the ester oxygen.
• 21 and 14 ppm: both in the C-C range (0 to 50). You cannot always tell which alkyl carbon is which when two sit in the same range, and you do not need to.

Peak height in ¹³C NMR does not count carbons. Use only the number of peaks and their δ values.

NMR spectroscopy: proton NMR, shifts and peak areas

A high resolution ¹H spectrum gives you everything a ¹³C spectrum does, plus two extra layers:

1. Number of peaks = number of proton environments.
2. δ of each peak = the type of proton environment (from the Data Sheet).
3. Relative peak areas = the relative numbers of protons in each environment.
4. Splitting pattern = the number of non-equivalent protons on the adjacent carbon (next section).
5. Put it together to suggest possible structures.

Approximate ¹H shifts (given on the Data Sheet):

The area under each peak is proportional to the number of protons in that environment. Exams show this as:

• an integration trace: a stepped line over each peak, where the height of each step is proportional to the number of protons; or
• ratio numbers printed by each peak.

Either way you get a ratio, not absolute numbers. Scale it to fit the molecular formula: a 1 : 1 ratio in a compound with 6 H means 3 : 3.

Aromatic protons: identify them from their shift (6.2 to 8.0 ppm). You are not expected to analyse their splitting patterns.

Here is the ¹H spectrum of ethyl ethanoate (environments a, b, c as in the figure above):

NMR spectroscopy: spin-spin splitting and the n + 1 rule

In a high resolution ¹H spectrum, a peak is split into a cluster of lines by the protons on the adjacent carbon atom(s).

The n + 1 rule: a proton with n non-equivalent protons on the adjacent carbon(s) gives a signal split into n + 1 peaks.

Apply it to ethyl ethanoate:

• a (CH₃-C=O) is a singlet: the adjacent carbon (the C=O carbon) has 0 H.
• b (O-CH₂) is a quartet: the adjacent CH₃ carbon has 3 H (3 + 1 = 4).
• c (CH₃) is a triplet: the adjacent CH₂ carbon has 2 H (2 + 1 = 3).

A quartet plus a triplet (areas 2 and 3) is the signature of an ethyl group, -CH₂CH₃.

Three rules stop most errors:

• Count the H on the neighbouring carbon, never the H on the carbon itself. A CH₂ next to a CH₃ is a quartet, not a triplet.
• Equivalent protons do not split each other. The 3 H of one CH₃ never split themselves, and propanone's six equivalent H give one singlet.
• O-H and N-H protons usually appear as a singlet and are not usually seen as split peaks (Data Sheet note). They are often broad.

If a proton has neighbours on both sides, add up the H on all adjacent carbons to get n.

Still don't get it? · the n + 1 rule

Think about flipping coins. Flip one coin and there are 2 possible results: 0 heads or 1 head. Flip two coins and there are 3 results: 0, 1 or 2 heads. Flip three coins and there are 4 results: 0, 1, 2 or 3 heads.

So n coins always give n + 1 results. And the middle results happen more often: with two coins, "1 head" can happen two ways (HT or TH), so the results come in a 1 : 2 : 1 pattern.

Now the chemistry, one step at a time:

1. Each H on the next-door carbon acts like one coin. It nudges the proton you are looking at in one of two ways.
2. Each possible combination of nudges puts that proton's signal in a slightly different place, so one peak becomes a cluster of lines.
3. Two neighbouring H give three combinations (1 : 2 : 1), so you see a triplet. Three neighbouring H give four (1 : 3 : 3 : 1), a quartet.

The exam version: the signal is split into n + 1 peaks, where n is the number of non-equivalent protons on the adjacent carbon. Always say which group the peak belongs to and how many H are on the adjacent carbon.

NMR spectroscopy: predicting a spectrum

To predict a spectrum for a given molecule, work through the molecule one environment at a time:

1. Draw the structure and label each environment (use symmetry).
2. ¹³C: one peak per carbon environment, each with a δ range from the Data Sheet.
3. ¹H: for each proton environment give the δ range, the relative number of protons, and the splitting from the n + 1 rule.

Quoting the Data Sheet range (for example "3.0 to 4.3 ppm, HC-O") is safer than inventing a single exact value. A full worked prediction is in the Worked examples.

NMR spectroscopy: TMS, deuterated solvents and D₂O

TMS, tetramethylsilane, Si(CH₃)₄, is the standard for chemical shift measurements. A little is added to the sample and its signal is defined as δ = 0, for both ¹³C and ¹H NMR.

Why TMS works well as the standard:

• All its H atoms are equivalent (and all its C atoms), so it gives a single, strong peak.
• That peak lies away from almost all other signals, so it does not overlap them.
• It is inert, so it does not react with the sample.
• It is volatile (low boiling point), so it is easily removed afterwards.

Deuterated solvents, e.g. CDCl₃, are needed for ¹H NMR. The sample is dissolved in a solvent, and the solvent is present in far greater amount than the sample.

• An ordinary solvent such as CHCl₃ contains ¹H atoms, so it would give a huge solvent peak that swamps the sample's peaks.
• In CDCl₃ the hydrogen is deuterium, D (²H), which does not give a signal in a ¹H NMR spectrum. So the solvent is invisible.

D₂O identifies O-H and N-H protons by proton exchange. O-H and N-H peaks are hard to assign because their shifts are so variable. So you run the spectrum, shake the sample with D₂O, and run it again:

$R{-}OH + D_2O \rightleftharpoons R{-}OD + HOD$

• The H on the O (or N) exchanges for D.
• D gives no ¹H signal, so the O-H or N-H peak disappears.
• C-H protons do not exchange, so every other peak stays.

Propanoic acid before and after shaking with D₂O:

The peak at 11.7 ppm has disappeared, so it was the O-H proton of the COOH group.

Combined techniques: deducing a structure from all the data

Structure questions give you several kinds of data. Each one answers a different question:

A reliable order of attack:

1. Elemental analysis: % divided by A_r, then simplest whole number ratio (empirical formula).
2. Mass spectrum: molecular ion m/z divided by empirical formula mass gives the multiplier (molecular formula).
3. IR: look for C=O, O-H (alcohol or broad acid), N-H, C-O; note what is absent.
4. ¹³C NMR: count carbon environments; assign C=O, C-O, C-C.
5. ¹H NMR: shift, area and splitting for each peak; build fragments (ethyl groups, CH₃ singlets next to C=O or O).
6. Assemble and check: the structure must fit every piece of data, including the formula and the fragment peaks.

Two limits from the spec: questions are restricted to functional groups in the A level specification, and you will not be asked to interpret mass spectra of organic halogen compounds.

Worked examples

Example 1: choosing an isomer from a ¹³C spectrum

A compound C₃H₈O gives this proton decoupled ¹³C spectrum. Identify it.

1. Count the peaks: 2 peaks, so 2 carbon environments, but there are 3 carbons. Two carbons must be equivalent, so the molecule is symmetrical.
2. Check the isomers: propan-1-ol has 3 environments; methoxyethane has 3; propan-2-ol has 2 (its two CH₃ carbons are equivalent).
3. Check the shifts: 64 ppm is in the C-O range (50 to 90), the CH(OH) carbon. 25 ppm is in the C-C range (0 to 50), the two equivalent CH₃ carbons.

Answer: propan-2-ol, CH₃CH(OH)CH₃. Marks are lost by saying "2 peaks" without linking it to the equivalent CH₃ carbons.

Example 2: integration trace to structure (¹H NMR)

A compound C₄H₈O₂ shows a C=O absorption but no O-H absorption in its IR spectrum. Its ¹H spectrum is below. The integration trace step heights are 24 mm (3.7 ppm), 16 mm (2.3 ppm) and 24 mm (1.1 ppm). Deduce the structure.

Step 1: integration ratio. Divide every step height by the smallest (16 mm):

$\frac{24\ \text{mm}}{16\ \text{mm}} : \frac{16\ \text{mm}}{16\ \text{mm}} : \frac{24\ \text{mm}}{16\ \text{mm}} = 1.5 : 1 : 1.5$

$1.5 : 1 : 1.5 = 3 : 2 : 3 \quad (\times 2)$

Step 2: scale to the formula. The molecule has 8 H:

$3 + 2 + 3 = 8$

So the ratio is the actual number of protons: 3 H, 2 H and 3 H. (If the ratio had added up to 4, you would double every number.)

Step 3: each peak in turn.

• x, 3.7 ppm, 3H, singlet: HC-O range (3.0 to 4.3); a CH₃; the neighbouring atom has 0 H. So O-CH₃, attached through the O.
• y, 2.3 ppm, 2H, quartet: HC-C=O range (2.0 to 3.0); a CH₂; adjacent carbon has 3 H. So CH₂ next to C=O and next to a CH₃.
• z, 1.1 ppm, 3H, triplet: HC-R range (0.5 to 1.9); a CH₃; adjacent carbon has 2 H. So CH₃-CH₂.

Step 4: assemble. CH₃CH₂- joined to C=O, and -O-CH₃; C=O with no O-H fits an ester:

$CH_3CH_2COOCH_3$

Answer: methyl propanoate.

The trap: ethyl ethanoate, CH₃COOCH₂CH₃, is also C₄H₈O₂ with a singlet, a quartet and a triplet in a 3 : 2 : 3 ratio. Only the shifts tell them apart:

Example 3: predicting the spectra of ethanal

Predict the ¹³C and ¹H NMR spectra of ethanal, CH₃CHO.

Environments: 2 carbon environments (CH₃ and CHO) and 2 proton environments (the 3 H of CH₃, and the 1 H of CHO).

¹³C prediction: 2 peaks.

• C=O carbon: 160 to 220 ppm.
• CH₃ carbon: 0 to 50 ppm (C-C).

¹H prediction: 2 peaks, areas 3 : 1.

Notice the pairing: the CH₃ is split by 1 neighbour (doublet), and the CHO is split by 3 neighbours (quartet). Each peak's splitting comes from the other group.

Example 4: combined techniques

An ester X contains 58.8% C, 9.8% H and 31.4% O by mass. Its spectra are shown below. Deduce the structure of X.

Step 1: empirical formula (per 100 g of X):

$n(\text{C}) = \frac{58.8\ \text{g}}{12.0\ \text{g mol}^{-1}} = 4.90\ \text{mol}$

$n(\text{H}) = \frac{9.8\ \text{g}}{1.0\ \text{g mol}^{-1}} = 9.80\ \text{mol}$

$n(\text{O}) = \frac{31.4\ \text{g}}{16.0\ \text{g mol}^{-1}} = 1.96\ \text{mol}$

Divide by the smallest (1.96 mol):

$\text{C} : \text{H} : \text{O} = 2.50 : 5.00 : 1$

2.50 is not a whole number, so multiply by 2 (do not round):

$\text{C} : \text{H} : \text{O} = 5 : 10 : 2 \quad \Rightarrow \quad \text{empirical formula } C_5H_{10}O_2$

Step 2: molecular formula from the mass spectrum.

$M_r(C_5H_{10}O_2) = (5 \times 12.0) + (10 \times 1.0) + (2 \times 16.0) = 102.0$

The molecular ion peak is at m/z 102, so the multiplier is 102 ÷ 102.0 = 1 and the molecular formula is C₅H₁₀O₂.

Step 3: IR spectrum.

• C=O at about 1740 cm^-1 (Data Sheet: 1630 to 1820).
• C-O at about 1180 cm^-1 (Data Sheet: 1000 to 1300).
• No broad O-H (2500 to 3300), so not a carboxylic acid. This fits an ester.

Step 4: ¹³C NMR.

5 peaks for 5 carbons, so no two carbons are equivalent: one C=O (174), one C-O (60) and three C-C (28, 14, 9).

Step 5: ¹H NMR.

Areas: 2 + 2 + 3 + 3 = 10, matching the 10 H. Two quartet and triplet pairs mean two ethyl groups: one on the C=O side (quartet at 2.3) and one on the O side (quartet at 4.1).

Step 6: assemble and check.

X is ethyl propanoate, CH₃CH₂COOCH₂CH₃. Check it against every piece of data:

• Formula C₅H₁₀O₂, M_r = 102.0: matches the molecular ion.
• Fragments: CH₃CH₂CO⁺ = 57, C₂H₅⁺ = 29, COOCH₂CH₃⁺ = 73 (and 102 minus 73 = 29, the mass of the lost C₂H₅).
• Rival isomers fail: propyl ethanoate would show a CH₃ singlet near 2.0; methyl butanoate an O-CH₃ singlet near 3.7; pentanoic acid a broad O-H in the IR and a COOH peak at 10 to 12 ppm.

Common exam mistakes

• Counting atoms, not environments. Propan-2-ol has 3 carbons but only 2 ¹³C peaks, because its two CH₃ carbons are equivalent. Check for symmetry first.
• Using the H on the carbon itself in the n + 1 rule. A CH₂ next to a CH₃ is a quartet (3 + 1), not a triplet (2 + 1). n is the number of H on the adjacent carbon.
• Vague splitting explanations. "It's a triplet because of the n + 1 rule" scores nothing. Say which group: "the CH₃ peak is a triplet because the adjacent carbon has 2 H".
• Treating integration as absolute numbers. It is a ratio. Scale it to the molecular formula: a 1 : 1 ratio in a compound with 6 H means 3 H and 3 H, and the numbers must add up to the total H.
• Splitting equivalent protons. Propanone gives one singlet; the three H on a single CH₃ never split each other.
• Expecting O-H or N-H peaks to be split, or to split their neighbours. They usually appear as a (often broad) singlet.
• Using ¹³C peak heights to count carbons. Only the number of peaks and their δ values carry information.
• Assigning a shift without the Data Sheet range. Quote the environment and range, e.g. "4.1 ppm is HC-O, 3.0 to 4.3". This is how you tell methyl propanoate from ethyl ethanoate.
• Saying CDCl₃ is used "because it dissolves the sample" or "because it is inert". The mark is for: the solvent has no ¹H atoms (D does not give a signal in ¹H NMR), so it gives no solvent peak.
• Getting the D₂O result wrong. The O-H or N-H peak disappears (it does not move or split), because the H exchanges with D. C-H peaks are unchanged.
• Mixing up the jobs of TMS and the solvent. TMS is the standard (δ = 0); CDCl₃ is the solvent; D₂O is added to identify O-H/N-H.
• Trying to analyse splitting of aromatic protons. Not required: identify them from their 6.2 to 8.0 ppm shift.
• Rounding an empirical formula ratio of 2.50 to 3 instead of multiplying everything by 2.
• Stopping before checking. In combined questions, show that your structure fits every piece of data: formula, molecular ion, IR, both NMR spectra.

Key definitions

• Chemical environment: atoms are in the same environment (equivalent) if they are bonded to the same arrangement of atoms in the molecule; each environment gives one NMR signal.
• Chemical shift, δ: the position of an NMR signal relative to the signal of TMS (δ = 0), measured in ppm; it depends on the type of environment.
• Proton decoupled ¹³C spectrum: a ¹³C spectrum in which each carbon environment appears as a single, unsplit peak.
• Relative peak area (integration): the area under a ¹H peak, which is proportional to the number of protons in that environment; it gives the relative numbers of each type of proton.
• Spin-spin splitting: the splitting of a ¹H signal into several peaks by non-equivalent protons on the adjacent carbon atom(s).
• The n + 1 rule: a proton with n non-equivalent protons on the adjacent carbon(s) gives a signal split into n + 1 peaks.
• TMS, tetramethylsilane, Si(CH₃)₄: the standard for chemical shift measurements; its single peak is set at δ = 0.
• Deuterated solvent (e.g. CDCl₃): a solvent whose H atoms are replaced by deuterium, so it gives no peak in a ¹H NMR spectrum.
• Proton exchange with D₂O: H atoms in O-H and N-H bonds exchange with D from D₂O, so the O-H or N-H peak disappears from the ¹H spectrum.

Specification

• I can use a ¹³C NMR spectrum to find the number of carbon environments. (6.3.2 a i)
• I can identify the types of carbon environment from chemical shift values on the Data Sheet. (6.3.2 a ii)
• I can suggest possible structures from a ¹³C spectrum, knowing exam spectra are proton decoupled. (6.3.2 a iii)
• I can use a high resolution ¹H spectrum to find the number of proton environments. (6.3.2 b i)
• I can identify the types of proton environment from chemical shift values, including aromatic protons. (6.3.2 b ii)
• I can find the relative numbers of each type of proton from an integration trace or ratio numbers. (6.3.2 b iii)
• I can use the n + 1 rule to find the number of non-equivalent protons adjacent to a given proton. (6.3.2 b iv)
• I can suggest possible structures from a ¹H spectrum. (6.3.2 b v)
• I can predict the ¹³C or ¹H NMR spectrum of a given molecule. (6.3.2 c)
• I can explain the use of TMS as the standard for chemical shift measurements. (6.3.2 d i)
• I can explain why deuterated solvents such as CDCl₃ are needed. (6.3.2 d ii)
• I can explain how D₂O identifies O-H and N-H protons by proton exchange. (6.3.2 d iii)
• I can deduce a structure by combining elemental analysis, mass spectra, IR spectra and NMR spectra. (6.3.2 e)

Related notes

• Analytical techniques: IR spectroscopy, mass spectrometry and elemental analysis
• Amount of substance: empirical and molecular formula calculations
• Basic concepts of organic chemistry: structural, displayed and skeletal formulae, and isomers
• Aromatic compounds: the benzene ring behind the 6.2 to 8.0 ppm protons
• Carbonyl compounds: aldehydes and ketones, with C=O at 160 to 220 ppm
• Carboxylic acids and esters: the commonest NMR unknowns
• Chromatography and qualitative analysis: the other half of 6.3 Analysis
• Organic synthesis: confirming the product of a synthetic route