# Identifying organic compounds using data

### Identifying organic compounds using data

#### Lessons

In this lesson, we will learn:

• How to combine evidence from multiple analytical methods to suggest organic structures.

Notes:

• Identifying organic compounds based on their analytical data is very important for chemists. Remember that we don’t ‘see’ organic molecules, groups or bonds.
We have lots of ways to analyse compounds (NMR, IR spectroscopy and many others) which give us evidence that certain groups or bonds are present. For example:
• When you look at an IhotR spectrum and see a stretch at around 1750cm-1, you aren’t ‘seeing’ a C=O bond, you’re just seeing that IR radiation has been absorbed by the substance in the sample you just analysed.
Chemists know that this absorption is a very common, clear and consistent feature of compounds when they are known to contain C=O bonds. So this absorption becomes very strong evidence that you have a C=O bond in your molecule.
It’s like looking at an unknown animal footprint and matching it to footprints of known animals.

Chemists piece different analytical data together and conclude what they think the molecule is. Some analytical methods are much more powerful and conclusive than others, but this is what chemists would do if given a totally unidentified molecule.
This lesson contains a few worked examples and some exercises to practice this.

• Exam questions to identify molecules based on analytical data can give you a lot of different information but it will almost always use most of the following:
• A molecular or empirical formula.
• Mass spectrometry, giving the molecular ion peak.
• 1H NMR containing the chemical shifts ($\delta$) in ppm, the splitting pattern and possibly the integration (the number of protons that make the signal).
• Look carefully if you are given two 1H NMR spectra, one with D2O solvent and one without. -OH peaks will not show up in D2O but they will without it. This is done because -OH signals are inconsistent in the solvent – the proton can exchange with it.
Learn how to interpret 1H NMR because this should form the backbone of your analysis to find the structure.
• 13C NMR containing the chemical shifts.
The questions might also give the IR spectrum or some information about test-tube reactions for any functional groups in the molecule.

• Identifying a structure based on analytical data: Worked example 1 .
A chemist isolates compound X and found the following test results of X after analysis.

• Molecular formula of C9H12.

• Mass spectrum: Molecular ion peak found at m/z = 120.

• 1H NMR with the following signals:
• 6.7 ppm, singlet.
• 2.2 ppm, singlet.

• 13C NMR with the following signals:
• 138 ppm.
• 128 ppm.
• 20 ppm.

Suggest a structure for compound X giving reasons using the data provided.

Before we start answering, remember you are being asked for a structure, which is all the atoms and the bonds that connect them. You just need a way to get to that. The molecular formula or empirical formula is very useful for the atoms, while NMR is very good for groups of atoms – how they’re connected.

The molecular formula of C9H12 is unlike that of an alkane or alkene – the C:H ratio is too even. There are examples in Introduction to NMR: Carbon-13 (13C) NMR and Proton (1H) NMR showing that using general formulae as a guide, the molecular formula gives us big clues on what type of molecule we have. The only major hydrocarbon group that has close to a 1:1 C:H ratio is benzene, C6H6. This molecule then almost certainly has a benzene ring, and with three more carbons to add, these are likely saturated alkanes (CnH2n+2) which would bring more H into the formula than C.
Leaving this for now, we almost definitely have a benzene ring, and it probably has three alkyl carbons attached to it in some way.

The mass spectrum molecular ion peak doesn’t really give us much clues in this example – we have the formula already, which matches the molecular ion peak as the molecular mass.

The 1H NMR spectrum gives us a big clue as to what the structure might be – a symmetrical substituted benzene ring. We definitely have a benzene ring, then, but we have only two proton signals for twelve protons .
To look in more detail, we have:
• A singlet at 2.2 ppm, which is just slightly out of the range of alkyl protons. These are the protons of the alkyl attachments to the ring, which would be slightly higher ppm than typical alkyl protons due to the aromatic ring, and would be singlets as there are no protons on the adjacent carbon atom that is part of the aromatic ring.
• A singlet at 6.7 ppm, right in the middle of the typical range for aromatic protons. A singlet suggests this environment has no adjacent protons – the substituent(s) must be attached there and since there are two adjacent carbons there must be substituents on both. Otherwise, it would be a doublet signal due to an adjacent C-H!

These are all the signals we have – one aromatic signal and one substituent signal. From the formula C9H12 we have assigned a benzene ring (C6H6) with three carbon and six hydrogen atoms left unassigned. How do the three carbon atoms fit on the ring?
• If there was one substituent on the ring, there would be at least three aromatic signals. We can rule this out already – there’s no point explaining further.
• If there were two substituents on the ring, there would be at least two different aromatic environments no matter how you arrange them around the ring. This is a no go.
• If there were three substituents on the ring, there could be only one aromatic proton environment if they were in the 1,3,5 positions relative to each other and each substituent was identical. We have three carbon atoms left to assign to the molecule and by taking three hydrogen atoms off the ring, we have C3H9 left to assign. That’s three -CH3 methyl groups on there! This also matches the 1H NMR spectrum, as it would give one aromatic signal (the C-H protons on the ring where it is not substituted) and one alkyl signal (the C-CH3 methyl protons).

Moving forward with this – 1,3,5-trimethylbenzene – we can check the 13C NMR spectrum for confirmation. We would expect three signals, one for the methyl carbons, one for the substituted aromatic carbons and one for the unsubstituted aromatic carbons.
That’s exactly what we get. The 13C NMR spectrum contains:
• A signal at 128 ppm, probably the unsubstituted aromatic carbon.
• A signal at 138 ppm, probably the substituted aromatic carbon.
• A signal at 20 ppm, which is the methyl carbon environment.

We have found a structure that is consistent with and backed up by all the data. It is drawn out below:

• Identifying a structure based on analytical data: Worked example 2 .
A chemist isolates compound Y and found the following test results of Y after analysis.

• Empirical formula of C2H4O.

• Mass spectrum: Molecular ion peak found at m/z = 88.

• 1H NMR with the following signals:
• 1.2 ppm, triplet.
• 2.0 ppm, singlet.
• 4.1 ppm, quartet.

• 13C NMR with the following signals:
• 15 ppm.
• 20 ppm.
• 60 ppm.
• 175 ppm.

Suggest a structure for compound Y giving reasons using the data provided.

The difference from the first and second examples here should be noticed – we have an empirical formula . That means that the formula has been divided down into smallest whole numbers. We need the molecular formula first, and for that we should match it to the molecular ion peak from the mass spectrum.

C2H4O would have a relative molecular mass of 44. Multiplying this by two would give us a molecular mass of 88, matching the molecular ion peak in the mass spectrum.

Multiplying the empirical formula by two then will give us our molecular formula: C4H8O2. This formula is typical of a hydrocarbon chain that contains one carbonyl (C=O) bond with the rest of the molecule saturated. This could be an ester or a carboxylic acid at this stage. Remember that -COOH signals in carboxylic acids have an obvious peak at 11-12 ppm in the 1H NMR while esters do not. This is the next logical step to check.

The 1H NMR spectrum has three signals for eight protons:
• 1.2 ppm, triplet.
• 2.0 ppm, singlet.
• 4.1 ppm, quartet.

Most importantly, there is no -COOH peak. From the last paragraph, we can effectively rule out a carboxylic acid, while an ester is still in the running.

If we have an ester, and we have four carbons in the molecule (as the formula C4H8O2 suggests) then one of the carbon atoms won’t have any protons attached to it (C-COOR) so the other three are responsible for these signals. Take the -COO- ester bonds out of the C4H8O2 molecular formula and we have C3H8 left. So we have three different hydrocarbon environments to find on what is almost definitely an ester.

All three of these signals can reasonably be alkyl signals; the 4.1 ppm quartet signal is noticeably higher but alkyl signals are pushed downfield (higher ppm) when next to electronegative atoms like O. This 4.1 ppm quartet could be the alkyl environment next to the singly-bonded O atom (COOR). The fact it is a quartet means 3 protons, surely a -CH3 group, is probably attached to it on the other side.

The 2.0 ppm singlet is very high in the expected range for alkyl protons (0-2 ppm) so it is probably the protons of the carbon attached to the C=O carbon of the ester group. This carbon atom won’t have any protons attached to it, that’s why the 2.0 ppm signal is a singlet. It must be a -CH3 chain end next to the -COOR part of the molecule.

The 1.2 ppm triplet is a standard alkyl environment and with a triplet splitting suggests a -CH2- is next to it. This must be what is directly attached to the -COO- atom of the ester group. That signal itself is a quartet so we must have a -CH3 chain end for the 1.2 ppm triplet.

This leaves us with a suggested molecule of CH3COOCH2CH3. Let’s look at the 13C NMR to confirm: we are expecting an ester environment around 170-180 ppm and three different signals all in the alkyl region!

The 13C NMR has exactly this:
• 15 ppm.
• 20 ppm.
• 60 ppm.
• 175 ppm.

The 175 ppm signal is undoubtedly the -COO- ester carbon. The only signal somewhat unclear is the 60 ppm. It is slightly out of the alkyl range but like in 1H NMR, being adjacent to electronegative atoms can push 13C NMR signals downfield so 60 ppm is probably expected for the CH3COOCH2CH3 carbon atom. The 15 ppm and 20 ppm signals are the chain end -CH3 carbons; the CH3COO- carbon more than likely 20 due to its proximity to the ester carbon.

Again, we have suggested the structure of the compound that is consistent with all the data we have been given. Ethyl ethanoate is drawn out below:
• Introduction
Identifying compounds by analytical data
a)
Using analytical methods - exam questions.

b)
Worked example 1.

c)
Worked example 2.

• 1.
Identify the structure of the compounds given analytical data.
Identify compounds A, B and C below:
a)
Compound A was isolated and had the following data:

• Molecular formula of C6H12O.

• Mass spectrum: Molecular ion peak found at m/z = 100.

• 1H NMR with the following signals:
• 1.0 ppm, doublet.
• 1.9 ppm, multiplet.
• 2.1 ppm, singlet.
• 2.3 ppm, doublet.

• 13C NMR with the following signals:
• 20 ppm.
• 30 ppm.
• 40 ppm.
• 45 ppm.
• 215 ppm.

Suggest a structure for compound A.

b)
Compound B was isolated and had the following data

• Molecular formula of C5H10O2.

• Mass spectrum: Molecular ion peak found at m/z = 102.

• 1H NMR with the following signals:
• 1.2 ppm, doublet.
• 2.5 ppm, multiplet.
• 3.7 ppm, singlet.

• 13C NMR with the following signals:
• 19 ppm.
• 34 ppm.
• 50 ppm.
• 175 ppm.

Suggest a structure for compound B.

c)
Compound C was isolated and had the following data

• Molecular formula of C4H5.

• Mass spectrum: Molecular ion peak found at m/z = 106.

• 1H NMR with the following signals:
• 2.3 ppm, singlet.
• 7.0 ppm, doublet.

• 13C NMR with the following signals:
• 21 ppm.
• 129 ppm.
• 135 ppm.

Suggest a structure for compound C.