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Analytical Techniques: Question 4

Syllabus 37.3, 37.4

Structured A2 10 marks

Compound M has molecular formula C5H10O2\text{C}_5\text{H}_{10}\text{O}_2 (Mr=102M_r = 102) and is a colourless liquid with a fruity smell.

Its carbon-13 NMR spectrum shows five distinct peaks (five different carbon environments).

Its proton (1H^1\text{H}) NMR spectrum shows four signals, with the data below:

δ\delta / ppm Relative peak area (integration) Splitting pattern
0.95 3H triplet
1.65 2H multiplet
2.30 2H triplet
3.65 3H singlet

You may use the following typical chemical shift ranges:

Environment of proton δ\delta / ppm
alkane, −CH3-\text{CH}_3, −CH2−-\text{CH}_2- 0.9–1.7
alkyl next to C=O 2.2–3.0
alkyl next to an electronegative atom (e.g. O) 3.2–4.0

(a) A structural isomer of compound M, isopropyl ethanoate, CH3COOCH(CH3)2\text{CH}_3\text{COOCH}(\text{CH}_3)_2, shows only four carbon-13 environments rather than five. Explain why isopropyl ethanoate gives one fewer carbon-13 environment than compound M, and state what the five separate environments observed for M tell you about whether M's carbon skeleton is branched in the same way. [3]

(b) Using the chemical shift ranges, integrations and splitting patterns given, assign each of the four 1H^1\text{H} signals to a specific proton environment in compound M. Hence deduce the structural formula of compound M, and give its name. [5]

(c) Explain, using the n + 1 rule, why the signal at δ=1.65\delta = 1.65 appears as a multiplet rather than as a simple triplet or quartet. [2]

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Worked solution

Part (a): Why isopropyl ethanoate gives one fewer carbon-13 environment

The number of 13C^{13}\text{C} peaks tells you the number of chemically distinct carbon environments, which is not always the same as the total number of carbon atoms, carbons related by symmetry give a single combined signal.

In isopropyl ethanoate, CH3COOCH(CH3)2\text{CH}_3\text{COOCH}(\text{CH}_3)_2, the two methyl groups attached to the central −OCH−-\text{OCH}- carbon are in identical environments (related by a mirror plane through the molecule), so they are chemically equivalent and produce only one signal between them, rather than two. That reduces the total carbon count of 5 down to just 4 distinct environments.

Compound M shows the full five separate environments, so none of its carbons are related by this kind of symmetry (unlike the two equivalent methyls in isopropyl ethanoate). This rules out such symmetric structures; the full structure is then determined by combining this with the 1H^1\text{H} NMR data in part (b).

Part (b): Assigning the proton spectrum and deducing the structure

Working through each signal using the chemical shift ranges given:

  • δ=0.95\delta = 0.95, triplet, 3H: this shift is in the plain alkane range (0.90.9–1.71.7), so it is a CH3\text{CH}_3 group; its triplet splitting (n+1=3n+1=3) shows it is next to a CH2\text{CH}_2 group (2 protons).
  • δ=2.30\delta = 2.30, triplet, 2H: this shift is in the alkyl-next-to-C=O range (2.22.2–3.03.0), so it is a CH2\text{CH}_2 group directly bonded to the carbonyl carbon; its triplet splitting shows it is next to another CH2\text{CH}_2 group (2 protons).
  • δ=3.65\delta = 3.65, singlet, 3H: this shift is in the alkyl-next-to-an-electronegative-atom range (3.23.2–4.04.0), so it is a CH3\text{CH}_3 group bonded directly to oxygen (OCH3\text{OCH}_3); being a singlet, it has no protons on any adjacent carbon.
  • δ=1.65\delta = 1.65, multiplet, 2H: by elimination, this is the remaining central CH2\text{CH}_2 group, sitting between the two CH2\text{CH}_2/CH3\text{CH}_3 fragments identified above (explained further in part (c)).

Chaining these fragments together in the only order consistent with the splitting, the terminal CH3\text{CH}_3 (triplet) is coupled only to the central CH2\text{CH}_2; the CH2\text{CH}_2 next to C=O (triplet) is coupled only to the same central CH2\text{CH}_2; and the OCH3\text{OCH}_3 (singlet) is isolated from all of them by the ester oxygen, gives:

CH3−CH2−CH2−C(=O)−O−CH3\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{C}(=\text{O})-\text{O}-\text{CH}_3

This is methyl butanoate, CH3CH2CH2COOCH3\text{CH}_3\text{CH}_2\text{CH}_2\text{COOCH}_3, which indeed has the molecular formula C5H10O2\text{C}_5\text{H}_{10}\text{O}_2 and five distinct, unbranched carbon environments, matching part (a).

Part (c): Why the δ = 1.65 signal is a multiplet

The central CH2\text{CH}_2 group (at δ=1.65\delta = 1.65) has two different carbon neighbours:

  • the terminal CH3\text{CH}_3 group, contributing 3 equivalent protons, and
  • the CH2\text{CH}_2 group next to C=O, contributing 2 equivalent protons.

The simplified n + 1 rule used at this level assumes splitting by one set of equivalent neighbouring protons at a time. Here there are two different sets of neighbouring protons (3H and 2H) on either side, splitting the signal in two different ways simultaneously. The result is a more complex pattern than a simple triplet or quartet, so, rather than trying to name an exact number of lines, it is simply reported as a multiplet.

Final answers

  • (a) The two methyl groups in isopropyl ethanoate are equivalent by symmetry, giving 4 environments instead of 5; M’s five separate environments show it has no such symmetry-equivalent carbons.
  • (b) δ0.95\delta 0.95 triplet =CH3= \text{CH}_3; δ1.65\delta 1.65 multiplet == central CH2\text{CH}_2; δ2.30\delta 2.30 triplet =CH2= \text{CH}_2 next to C=O; δ3.65\delta 3.65 singlet =OCH3= \text{OCH}_3. Compound M is methyl butanoate, CH3CH2CH2COOCH3\text{CH}_3\text{CH}_2\text{CH}_2\text{COOCH}_3.
  • (c) The central CH2\text{CH}_2 has two non-equivalent neighbouring proton sets (3H and 2H), so the simple n + 1 rule cannot give a single multiplicity. It is reported as a multiplet.