1H NMR 11.2(d)2015, 11.2(e)2015 notes


An incredible and amazingly powerful tool for determining structure (and interaction!) in organic molecules. But the machines are expensive ūüė¶

Nuclei with a non zero magnetic spin can interact with an applied magnetic field, symbol, B0 (“B zero”) of an electromagnet. Hydrogen is such an atom. It has a nuclear spin of +1/2 and is the focus of NMR at CIE A-level chemistry. As it involves hydrogen (which only contain a single proton in its nucleus), this NMR is sometimes called 1H NMR, or “proton NMR”. Getting back to our applied magnetic field again, the magnetic H nuclei can either align with B0 or they can be aligned opposite to B0.

When aligned with B0, they are in a low energy configuration, i.e. a low energy state.

When opposed to B0 they are in a high energy configuration, i.e. a high energy state.

There is an energy gap between the two states. The value of this gap corresponds to the energy found in the radio wave part of the electromagnetic spectrum.

If radio waves  of an energy that exactly matched the energy gap are supplied to the molecule, then the hydrogen nuclei can be flipped from the low energy state to the high one. And this energy absorption can be absorbed.

This absorption is recorded and forms the basis of the NMR spectrum. {in practice it’s the re-release of this absorbed energy that’s actually detected)

A horizontal axis of “chemical shift“, which can essentially be thought of as an energy scale, has various peaks plotted along it.

The scale begins at a chemical shift value of 0 (zero) but is located on the right hand side of the horizontal axis. Usually the zero of a horizontal scale is located on the left. The symbol for chemical shift is the lower case Greek letter ‘delta’ (shown below)and has units of parts per million (ppm)delta symbolDelta (lower case), The symbol for chemical shift.



Four essential rules of NMR.

1) The number of peaks (energy absorptions) occurring reveals the number of different H environments.

2) The location of these peaks. The different environments mentioned in ‘rule 1’, arise from different levels of magnetic shielding by the electrons positioned between B0 and the spinning nucleus. The electrons essentially partially ‘cancel out’ some of the B0, which would mean the energy now needed to flip to an opposed alignment state, will be smaller compared to if no ‘cancellation’ of the B0 occurs.

If the shielding electrons are withdrawn (by an electron withdrawing atom or group nearby, e.g. carbonyl or fluorine etc) then the more the nucleus will be exposed to the ‘raw’ B0 and a greater energy gap exists to flip into an opposed state for the B0.

The result of this is that peaks form one environment can be found in a slightly different location from one molecule to another.

This effect is ‘short range’. For A-level, we only examine the effect of any electron withdrawing groups on the came carbon as hydrogen environment we are looking at or any electron withdrawing groups on the adjacent carbon to the environment we are looking at.

You are expected to use the 1H NMR chemical shift tables in your data book in exams to be able to identify the environments in an organic molecule.

3) The AREA UNDER THE PEAK gives the RELATIVE number of H’s in that environment compared to the other environment present. In butane, you would get two peaks as you have two environments. The methyl groups, CH3, are equivalent to each other, and one of the methylene groups, CH2, is equivalent to the other CH2 group. The area under these peaks will be 3:2. The ratio of this number is the same as the absolute number of H’s in the molecule, 6 H’s in the methyl groups, and 4 in the methylene groups. 3:2 = 6:4
Sometimes you get lucky, the numbers processed by the computer or examiner may happen to he the exact number of H’s in that particular environment. But don’t count on it. You may well have to scale up (or scale down!) the numbers for the final structure.

The older machines used to be an electronic integration of these peaks and draw a line going up which spanned the breadth of the peak. The line ‘going up’ represented the cumulative area under the peak. As the integration was done over the span of the peak, the area (and hence the distance this line went up) increased, So the area was translated into what was called the integration height. This ‘integration height’ terminology is common place, but I have not used it (I will try to only refer to it in terms of area under the peak) because there is another ‘height’ that is mentioned, and it comes up in the last ‘rule’ of NMR, which covers what’s known as “splitting”.

Here’s two spectra in which the integration height for the peaks in a 1H NMR spectrum for ethanol are shown. (you may need to click them to expand the image size)

NMR spectra of ethanol with integration height shown

The spectrum on the left has separate integral heights displayed (the kind of ‘reverse z’ shaped lines) for just the region of the peak only,¬† whereas the integration heights for the spectrum on the RHS is continuous across the whole spectrum. Either way gives you the same result.

Sources: http://www.chemistryexplained.com/Ne-Nu/Nuclear-Magnetic-Resonance.html
https://scilearn.sydney.edu.au/OrganicSpectroscopy/?type=NMR&page=Introduction respectively (quite a good site)

4) Splitting – High resolution spectra only. Splitting is sometimes called “spin-spin coupling”.

In more powerful/newer NMR machines, a high resolution the spectrum can be obtained. What a old machine may have shown as one peak, is revealed to be being made up from a set smaller ‘constituent’ peaks. This is described as the peak is ‘split’.

What causes this splitting is the influence on the neighbouring hydrogen atoms, which like the electrons in ‘rule 2’ can change the TOTAL magnetic field felt by the H nucleus environment we are considering. This effect is also weak and we only need to consider only ADJACENT protons. (H’s on the same C will for A-level be classified as equivalent in this case)
While the effect is weak, it is extremely useful in providing clues about the molecules structure.

It turns out that the peak will be a singlet, doublet, triple, quartet, pentet, sextet or septet. A singlet means that the peak looks the same under high resolution as it did at low resolution, i.e. the peak hasn’t actually split. A doublet means the low resolution peak is split into two when viewed under high resolution. If the peak is split into three, it is called a triplet. If plit into 4, it’s called a quartet, etc.

Here we use the n+1 rule.
“A peak is split into¬† n+1 peaks, where n is the number of adjacent equivalent H.”

1,1-dichloroethane will give two peaks, there are two environments. The relative heights will be 3:1, which in this case happens to correspond to the actual number of H’s in that molecule… we got ‘lucky’! The peak with area 3 obviously refers to the CH3 group. It is split into a double because the number of adjacent H is one, n=1, so n+1=2.
The peak of area 1, is from the H on the CH(Cl)2 carbon. It is next to the CH3 group which has 3 equivalent H’s, so here n=3. n+1=4 so the CH group is split into a quartet.

When you are new to NMR, this can be a bit confusing, because we are looking at one type of H environment, but it can be split by the adjacent H.

1-bromoethane would have two peaks under low resolution spectrum. Each relative area may be quoted as 3:2. The peak of area 3 related to the CH3 environment. This peak will be split by the adjacent H’s. The adjacent group is CH2Br, i.e. there are 2 adjacent H, so the CH3 peak gets split into a triplet. The CH2 group (CH2Br group) is split into a quartet by the CH3 group.

When splits occur, e.g. a triplet, the peaks in the triplet are not all the same, but display different heights, in the triplet case we see a height ratio of 1:2:1.
The doublet has a 1:1 height ratio. A quartet has a 1:3:3:1 ratio. A pentet has a 1:4:6:4:1 ratio. This corresponds to values found in Pascals triangle. The “level of the triangle” (which is the same as the amount of numbers found in that level) happens to be the value of n+1. The tip has only one number, so n+1 = 1 and therefore n is equal to zero. the 1 in thr triangle means the peak is NOT split so there are no adjacent H’s to split it. “Level Two” of the pyramid has 1 and 1. i.e. two numbers, hence n+1=2 and so n=1, i.e. there is one adjacent H causing splitting and two peaks appear which is a doublet with height ratio’s of 1:1.
The third level of the triangle (or pyramid) has three numbers on it. i.e. it’s dealing with a triplet. The number in the triange at ‘level 3’ are 1 2 1, meaning the triplet seen on the spectrum have a height ratio of 1:2:1. This third level is the value of n+1, i.e. n+1=3 so n must =2, telling us this triplet was caused by two adjacent H atoms.

This splitting really helps tell us what is in the neighbouring group.

Note: H on OH and NH groups do not cause splitting. And splitting from the H in aldehyde’s is usually too small to be seen even under high resolution.

Perhaps the second most toughest bit of this is when you have an aryl group present in the molecule, e.g. ethyl benzene. The H atoms in the ring are essentially ‘linked’ to each other because of the delocalized electron system, hence you get a lot of interactions and the peak from the aryl environment shows lots of peaks (due to ‘spin-spin coupling’). These peaks usually appear around the 7.5 region on the chemical shift scale.

It although the peaks here are more complicated and ‘noisy’, we can still get information about the substitution pattern in the aryl ring.

In methyl benzene, i.e. mono-substituted benzene, we will see the aryl H’s contain three peaks, of area 2:2:1. The splitting if these peaks will respectively be be a doublet, a doublet of doublets, and a doublet.

NMR spectra of methylbenzene, ethylbenzene and propylbenzene:

methylbenzene, ethylbenzene and propylbenzene 1H NMR
Sources (#1 and #3): http://orgchem.colorado.edu/Spectroscopy/specttutor/arom.html
(#2): http://en.wikibooks.org/wiki/Organic_Chemistry/Spectroscopy

A 1,4-disubstitured benzene (with equivalent subsistent groups), will give a singlet (albeit a noisy one) in the aromatic region.

Source: http://orgchem.colorado.edu/Spectroscopy/specttutor/arom.html


A 1,4-disubstitured benzene (with non-equivalent substituent groups), will give a two peaks of area 1:1 and each appearing as a doublet.

1,4-non symmetrical benzene
Source: http://www.azom.com/article.aspx?ArticleID=11354

Some more spectra and a bit more discussion of substituted benzene patterns here: http://en.wikibooks.org/wiki/Organic_Chemistry/Spectroscopy

1,3-substitutions are harder to see but are possibly just on the edge of the syllabus learning requirements (and on the border of insanity!)

Disubstituted benzenes:
disub benzenes
Source: http://homepage.smc.edu/kline_peggy/Organic/NMR_disubst/o_m_p_HNMR.html

As is often the case at A-level, you must be able to use NMR skills/knowledge both ways.

The hardest thing is to be given a spectrum and asked to determine the structure. The easier skill is to be given a structure and predict it’s NMR structure. The examiners will often just as you you go from spectrum to structure, but when you are piecing together your structure, you can ‘go the other way’ to check to help make sure your ideas on the structure are correct.

PLEASE NOTE: ALWAYS, whenever possible, use accompanying data/clues/other spectra, like mass spectra, IN CONJUNCTION with info on an NMR spectrum. This makes piecing together info from NMR much easier.

ALSO….. ALWAYS TABULATE YOUR NMR and MASS SPECTRA data/analysis. It speeds up the process of structure determination.

An NMR table should have a row for each full peak. Column headings are “Peak number”, “Chemical shift”, “Relative area”, “Splitting pattern”, “Possible assignment”. You can add to those suggestions if you like, e.g. speculate on absolute number of H in a group.

P.S. The horizontal axis on an NMR spectrum is called Chemical Shift. (It is essentially a ratio of energies). The symbol for chemical shift is called “delta” and the units are parts per million (“ppm”) – which you can think of as being parts per million of some unit of energy. Remember these spectra start from the RIGHT HAND SIDE. That’s where the zero ppm is located, and you will often see a peak there from a standard reference, called tetramethysilane, Si(CH3)4, or “TMS” for short. It contains 16 H’s ALL in exact same environment. This compound helps with the analysis of NMR spectroscopy in ways I don’t think you need to know of.

P.S. If you have two molecules present, the area under the peaks of one molecule’s peaks, compared to the another molecules peaks, depends on the concentration of each compound. If TMS is in large excess, the areas of the peaks for the organic molecule under investigation will be small in comparison – but, the relative heights of the individual peaks within the organic molecule will be the same relative height to each other. It’s almost unknown that we measure the TMS peak relative to the organic molecule peak at A-level. We just normally focus on the peaks within the organic molecule itself.


The¬†deuterium oxide, D2O, heavy water, trick. Deuterium is symbolised by the 2(superscript)H, indicating it has an atomic mass of two. In fact, deuterium is even given it’s own elemental symbol “D”.

Deuterium is hydrogen with one proton and one neutron. The nucleus overall has a total magnetic spin of one (‘ordinary hydrogen has a nuclear spin of +1/2) Hence it’s interaction with the applied magnetic fields is of much different energy to that of 1H. It’s chemical properties are very similar to that of H (as it has the same electron configuration!). So LABILE groups, which means groups that can give off H+ ions, such as COOH, phenols, alcohols and amines (although the latter two are very poor acids) can give off H+ and accept D+.

D+ can be supplied from deuterium oxide, D2O, or “heavy water” as it’s sometimes called, which will over time, cause a labile H containing group to diminish in size on an NMR spectrum. (the spectrum us re-run after some time period and different spectra compared). This technique allows the detection of these labile groups and can be very useful.
OK, That’s all folks….
I’ll add some supporting spectra to this post when able, and post some practice spectra. Please note: spectroscopy is just a very small part of the syllabus. It’s easy to spend a lot of time on it. See the analysis of Paper 4 to gauge it’s likelyhood of appearing. IF they give you an NMR spectrum, it’s most likely going to be not too hard, as analysis of these spectra can be slightly time consuming and they have loads of other Paper 4 stuff to ask you about. So DON’T go into MASSIVE detail on NMR. Be sensible and, as always, look at the past years papers to see what level of knowledge is needed.
Challenge: predict the 1H NMR spectrum of 2-bromopropan-1-ol (bit tricky)

4 thoughts on “1H NMR 11.2(d)2015, 11.2(e)2015 notes

  1. I’m pretty sure “Carbon 13” NMR isn’t going to be covered, although (as is common with CIE !!!) unlike 11.2(d), 11.2(e) doesn’t specify which NMR students need to be familiar with!!! But I think it’s safe to say it’ll be 1H NMR only. That’s all I’ll cover in the class and on here anyway.


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