4.9 Spectroscopy and chromatography
Posted by: intechemistry on: September 23, 2010
Update: Sept 22nd. IR spectra of alkanes, alcohols, alkyl halides, alkyl bromide, ketones, aldehydes, carbox acids, esters, amines, amides, aromatics and alkenes great info from the University of Colorado (US), Department of Chemistry and Biochemistry: IR Spectroscopy Tutorial – all common functional groups. (.doc format, & only about 350 kB !! 
) I like their Table of Characteristic IR Absorptions very much.
Update: Sept 20th 2011. Here’s the VIDEOS:
Royal Society of Chemistry:Proton Nuclear Magnetic Resonance (NMR) [not shown in class]
James Mungall’s videos: 1. NMR spectroscopy – Introduction to proton nuclear magnetic resonance
2. NMR spectroscopy – Integration
3. NMR spectroscopy – Chemical shift and regions of the spectrum
4i. NMR spectroscopy – Coupling
4ii. NMR spectroscopy – Coupling
5i. NMR spectroscopy – Examples of NMR spectra
5ii. NMR spectroscopy – Examples of NMR spectra
HPLC and GC videos found here.
.
Update: 10 Sept 2011. Here’s the powerpoints:
Organic spectroscopy – infra-red
Organic spectroscopy – mass spec
& we will examine the spectra below in class..
.
As promised here’s more spectra to practice analysis on: Additional Spectra 1 Additional Spectra 2 Additional Spectra 3 Additional Spectra 4
NOTE: Additional Spectra 5 had the incorrect 1H NMR. Here is the corrected spectra: Additional Spectra 5 – IH NMR corrected
Suggested method:
1) Take note of any additional information.
2) Get a ‘feel’ for the molecule by looking at the IR for functional groups, N.B. 1H NMR may also reveal some types of functional group {Q: which types?}
3) Look for the molecular ion and try and get a molecular mass. {at A-level, you will almost always be given M.S. with M+1 (although beware: alcohols are well known for not displaying a (or giving a very small) molecular ion peak.
4) Look for base peak and other large peaks especially those magic numbers, 15, 29, 43(propyl and varients), 43 acyl, 77 phenyl C6H5. Lookd for 3:1 ratios of molecular ion peaks separated by 2 mass units (indicates one Cl atom), look for 1:1 ratios for molecular ion peaks separated by 2 mass units (indicated one Br atom). Look at mass differences between significant peaks (tells you what fragment broke off)
5) Try and make a proposed structure before doing the NMR (it will make life easier)
6) Tabulate NMR data (should actually be done for M.S. and I.R.) and look for magic number peaks, e.g. almost 1 ppm (methyl) 1.5 ppm (methylene CH2 group) 7-7 ppm complex peak set (aromatic) etc.
7) No of peaks. The number of peaks gives the number of H environments. Note: CH3-CH2(a)-CH2(b)-CH2(c)CH2Cl might have an NMR where CH2(a) and CH2(b) overlap giving an integration height of 4. Their environments are quite similiar so may overlap.
8) Use integration heights to count the number of H atoms in any particular environment. Methyls (v. common) will usually have heights of 3,6,9,12 etc. and this number can help with determining which NMR peak is a methyl. Remember however the integration height is a relative number of H, not necessarily the actual(absolute) number of H. A molecular formula will belp translate the relative integration height into actual numbers of H.
9) Remember peaks e.g. Methyl can ‘shift’ if next to electron withdrawing groups which cause deshielding and expose the proton nucleus to the applied magnetic field more, (causing an increase in the energy gap to flip to the high opposing field state)
10) Look at spin spin coupling. If say 3 peak appear and only 2 are split then the ones that split must be next to each other. Each peak/environment causes the other to split.
11) When you think you have got a structure, RE-CHECK to make sure ALL the spectroscopic data fits the structure. If it doesn’t then your proposed structure probably has an error.
18 Responses to "4.9 Spectroscopy and chromatography"
October 11, 2010 at 11:05 pm
salam sir,
why functional group of aldehyde should be at the end of molecular structure?
October 12, 2010 at 12:07 am
To: tuz, October 11, 2010 at 11:05 pm
You must have H-C=O for aldehyde leaving just one bond left for C to bond to a other atoms, so you cant get a CHO group at a ‘branch point’ e.g. you cant get H3C-CHO-CH3.
Ok, you might get say 8 C’s in a row and a CHO group coming off at the third C in that 8 C sequence, so the CHO itself may may look like it’s branching off from the longest C chain i.e. the CHO may look like a side group on the octanne type hydrocarbon chain, but in this case, you would BEGIN counting from the CHO group, meaning the longest C chain would be 7 C’s now (includng the C in CHO) and the ethyl group sticking out would be a side group. The name of the molecule I have in mind is 2-ethylheptanal. (rather than naming it an octane molecule with an aldehyde sticking out at position 3).
October 12, 2010 at 12:09 am
To: Nurul Afiqah:), October 11, 2010 at 11:44 pm
Additional spectra 1: butan-2-one
Additional Spectra 2: cyclohexene
Additional Spectra 3: 2,2-dimethylpropanal
Additional Spectra 4: ethylethanoate
Additional Spectra 5: propylamine
October 12, 2010 at 12:38 am
ooooooo..that’s wonderful..
sir,about the exercise,no-2,,how can i know that there is cyclo- in the structure??is there any specific value of nmr for cyclo-?
then,,it is because of cyclo-,there is no CH3- value which is 15 in the mass spectro?
October 12, 2010 at 12:47 am
The molecular formula (calculated from the empirical formula) suggests 2 double bond equivalents (see the slides for this unit). The possibilities are
1) Two double bonds (straight chain molecule)
2) A ring and double bond
3) two rings.
Option 3 isn’t feasible due to the No. of C’s in the molecule, leaving us with option 1 or 2. The integration heights and the complicated splitting suggest the ring. Getting ‘ring’ info from the mass spec isn’t easy for young minds, so I won’t mention it.
(& sorry, I 4got to return your salam earlier)
October 12, 2010 at 10:46 pm
sir, can you explain more about the integration height, and how to relate it with the number of H and the arrangement of the environment?
October 12, 2010 at 11:18 pm
salam sir,
for question no.2, how to know that it is ketone, not aldehyde?
October 13, 2010 at 12:20 am
To: ridhwan, October 12, 2010 at 10:46 pm
Integration height for a peak can be a fraction or multiple of the actual number of H’s in the group responsible for generating that particular peak. This heights are relative numbers.
Say you have ethanol, CH3-CH2-OH
You have peaks at 3.7, 2.6 and 1.2 ppm
The computer (or chemistry lecturer!) might gives the respective integration heights of these peaks to be 8 : 4 : 12 (those numbers might have been obtained by a person measuring the height in cm of the integration trace drawn on the spectrum by the NMR computer).
A different person might report the numbers as having integration height of 3.1 : 1.6 : 4.7 (perhaps they were measuring the height in inches). It doesn’t matter what unit was used as the relative sizes are the same.
The power of integration height is that it can really help in clarifying which peak corresponds to a certain environment; Because groups can shift, overlap, cross over one another, etc, the tables of peak position (i.e. ‘chemical shift’ measured in ppm) may not always be very helpful in identifying exactly which H environment gives which peak.
If you get a molecular formula then life becomes so much more simple because If we know molec formula, you can turn the relative integration heights into actual, absolute, numbers of H’s in that peak.
Say for example the heights (given above) were for the C2H6O then we can add up the total integration height and in this case, the height corresponds to the total number of H in the compound, i.e. 6 H’s. So the height for one H can be determined and used to calculate the absolute (real) number of H in that peak.
Using the cm example: 8 : 4 : 12, total height = 24. Total H from molec. formula was 6, so 24/6 ( = 4) is the height for 1 H, so the peak with height 8 must contain 2 H’s, the peak with height 4 must contain 1 H, and the peak with height 12 must contain 3 H’s.
This helps us assign the peak with 2H (the one occurring at 3.7 ppm), as very likely containing a CH2 group (note: it MUST also fall inside the chemical shift range for CH2 groups, and looking on the table you should see that it does fall indeed fall in the ‘zone’ for CH2′s). The peak at 2.6 ppm contained 1 H and so is probably an OH, and the peak at 1.2 ppm was calculated to have 3 H and so is probably a methyl group.
Does that make it any better?
October 13, 2010 at 12:21 am
To: syabila, October 12, 2010 at 11:18 pm
In the book, the Q says ‘ketone isomers’
October 13, 2010 at 6:20 am
Salam sir…
do we need to memorize the shapes of the peaks in the IR spec?
October 13, 2010 at 10:21 pm
To Amirul, October 13, 2010 at 6:20 am
Salam.
Knowing a few I.R. ‘magic number’ peaks allows you to rapidly ‘see’ what molecule (functional group) you have. It’s quite ‘do-able’ and saves a lot of time ‘cos you won’t have to look up the peaks in the data book.
But you don’t have to have to memorise them.
To Amirul, October 13, 2010 at 4:33 pm
The size of the (M+1)+. peak relative to the M+. peak tells you how many carbons there are in the molecule. If the (M+1)+. peak is 8.8% the height of the M+. peak then you have 8 C’s present. This ‘works’ because the natural abundance of carbon-12 to carbon-13 is 1.1%
If the (M+1)+. peak is 17.6% the height of the M+. peak, then there are 12 carbons in the molecule. A molecule with 16 has a 16 x 1.1% chance of any particular carbon atom being a C-13 isotope.
C-13 is one mass unit higher than C-12 which is where the ’1′ in (M+1)+. comes from.
October 31, 2010 at 1:00 pm
Sir,
for Q19b)(iii), how do we actually come up with the ratio 2:3:3?
I dont really understand it because the height of the peak doesnt really correspond to the ratio given in the answer.
Besides, what is the significant of knowing the ratio to determine the number of H in a particular H-environment?
One more question, how can we apply Pascal triangle to determine a substance in a given H-NMR?
thank you again sir!
October 31, 2010 at 7:59 pm
– for organisational purposes, I’ve moved your comment here. I can’t remember the original time you posted the comment. Below is my reply–
We get the 2:3:3 ratio from the fact the molecule is given to us and we are supposed to deduce from the structure what the integration heights should be.
The use of the word peak can be confusing.
I want you to visualise a mountain in your mind….
In NMR, the term peak means the ‘whole mountain’ {or mountain range if the peak is split due to adjacent H’s}. Peak height does NOT mean ‘mountain height’ but the whole area under the whole mountain from the time it begins sloping upwards on one side until the sloping ends on the other side.
The integration of that mountain (i.e. the total area underneath it) is called the peak height. It’s not where the highest part where the line or ‘spike’ reaches.
If the peak is split then the relative heights of the spikes is discussed in terms but in terms of ratio’s
Pascal’s triangle shows the n+1 rule. The quantity of numbers in a particular row of the triangle is n+1. If you saw a peak was split into 5, then n+1 = 5 and n must therefore be 4. The ‘n’ is the number of adjacent protons to the group under investigation. Pascal’s triangle also gives the relative heights of the spikes within a split peak. Consider the same split peak as before, its split into 5. They will have height ratios of 1:3:6:3:1. The ratio’s are what is found on the 4th row of the table and as before can also be used to determine number of adjacent H’s. One can use the ratio’s if observable peaks if the total number of splits is difficult to see clearly or overlaps slightly with a different peak, but the ratios are not used much at A-level.
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October 6, 2010 at 10:35 pm
thank you once again sir. it helps a lot