h nmr problem solving examples

The OChem Whisperer

Guide to Solving NMR Questions

How to solve any nmr question.

Solving NMR questions is easier than you think. All you need is a step-by-step process to help guide you through each question. And here it is…

Most NMR questions on an exam involve determining a specific structure rather than memorizing and repeating various NMR values. Typically, you will be given an NMR spectra and a molecular formula (sometimes an IR spectra will be provided). I have put together a few ideas that might make this process a bit easier.  I am in the process of putting together a more concise document than this as a study aid. This post is meant to walk you through the thought process of how to tackle this type of problem. The description is a bit long (….so hold on!), but once you get it, you can just use the algorithm to solve your NMR problems. Here are some reference values and a couple of proton NMR spectra:

Proton NMR Reference Values

(cem.msu.edu)

(process-nmr.com)

(Our example 1H NMR spectra for this post; unknown source)

Start with an algorithm to get you on track

When staring an NMR question, you can use the following algorithm to help guide you through the thought process:

(above should say C2H5Cl = C2H6)

Calculate Degrees of Freedom

We notice the first thing says calculate degrees of unsaturation… what is that ? This allows us to determine if there are any double bonds or rings (cyclic structures) in the compound. Let’s look at an example; the formula is C 4 H 8 O 2 . Now, we need to compare this formula with the formula of a completely saturated hydrocarbon (all single bonds…no double bonds):

This formula tells us how many hydrogens we need to have a carbon compound with NO double bonds or rings. Let’s look at our example…

If I have a compound with 4 carbons, this 4 refers to n. Now, if we plug this in the formula, we get

C 4 H 2(4)+2 = C 4 H 10

This says that if we have a compound with only 4 carbons, we need 10 hydrogens to have a compound with no double bonds or rings. If we look at our example, we have C 4 H 8 O 2 . For now, all we need to look at is the C 4 H 8 when dealing with degrees of unsaturation (we will discuss what to do with heteroatoms a bit later). What we now want to do is subtract the hydrogens in our example ( C 4 H 8 ) from the saturated formula ( C 4 H 10 ):

C 4 H 10 – C 4 H 8 = 2 hydrogens

This leaves us with a value of 2. Now, the last things we do to get our degrees of unsaturation is divide this number by 2:

How to use the degrees of unsaturation to get the answer

This leaves us with 1; therefore, we have 1° of unsaturation. So, what does this mean? Each degree of unsaturation equates to a double bond or ring. Here are a few examples to further clarify:

1° of unsaturation = 1 double bond or 1 cyclic structure

2° of unsaturation = 2 double bonds; 1 alkyne; 1 double bond and 1 cyclic structure; 2 cyclic structures

3° of unsaturation = 3 double bonds; 2 double bonds and 1 cyclic structure; etc…

When you see 4° of unsaturation, think benzene; 3° of unsaturation for the 3 double bonds and 1° of unsaturation for the ring.

In our example, it means we have one double bond or one cyclic structure in our compound. Let’s look at few ring systems:

The larger the ring, the more stable the ring (with this series). Three and four-membered rings are rare. Usually, you will hear more about 5- and 6-membered rings. Since this is the case, we more than likely have a double bond (but never rule out a ring until you have looked at the NMR spectra).

Next, if we look at the algorithm, we need to consider the other atoms (other than carbon and hydrogen) in our formula…oxygen. Since we have degrees of freedom…think carbonyl. That is all we can do for now with our algorithm. Let’s now look at our spectra and see if we can start to determine the structure from the peaks:

Interpreting the spectra – splitting patterns

We have three peaks: a quartet around 4 ppm; a singlet around 2 ppm; and a triplet around 1 ppm. At this point, let’s review singlet, doublet, triplet, quartet, and multiplet:

What do these peaks refer to and how to we get the specific peak pattern? Well, we use the n+1 rule to figure out the pattern:

Putting the fragments together

Once you have your fragments, it is a matter of figuring out how to put them together. By looking at the spectra and where the peaks show up (ppm), you can figure out how the fragments go together.

Click  NMR pictures to see the images as a PDF.

Don’t forget IR Spectroscopy!

Here is a simple guide showing you the most common IR values

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When you look at an NMR spectrum, do you see only a bunch of disordered lines or peaks? Then you have come to the right place. This site was established to provide people interested in NMR with a library of NMR spectroscopy problems. Interpretation of spectra is a skill that requires pattern recognition and/or practice to master the chaos. This site provides one dimensional spectra of different nuclei, COSY, HSQC, HMBC and some less common spectra of various compounds for you to interpret, together with worked solutions. Hopefully, these problems will provide a useful resource to help you better understand NMR spectral interpretation.

A series of about 50 problems is available in printable form. The print version will not be developed further. The PDF documents are still available (german only). Get them here:

The step-by-step approach works nearly perfectly on the desktop and most tablets. On cellular phones, PowerPoint documents are sometimes displayed incorrectly. There is a workaround, but it is unfortunately much too complicated to be practical. As an alternative, all documents can be saved locally. Feel free to follow the download links. A free PowerPoint viewer is available in both the iOS and Android AppStores. As a bonus, there would be no need for a permanent internet connection. Of course, this offline method also works with desktop computers.

The author is always grateful for criticism, suggestions, references to errors and information about the design via email:

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Chemistry Steps

Organic Chemistry

Nuclear magnetic resonance (nmr) spectroscopy.

At times solving an NMR problem leads to two or more plausible structures satisfying the given data. We have seen that 13 C NMR is usually decoupled and therefore there is no splitting of signals which limits the information we can get as to how many hydrogens are connected to a carbon atom.

And even combining 1 H and 13 C NMR spectra may not give definite evidence for choosing only one structure.

This is where a technique called DPET ( distortionless enhancement by polarization transfer ) becomes very useful. All it does (and that’s a lot and very handy) is it differentiates the carbons based on the number of hydrogens it is bonded to.

So, instead of simply saying hey this is a carbon, and this is another one, it tells us if it is a C, CH, CH 2 , or a CH 3 . Isn’t that nice?

Let’s see how we get this information in DEPT.

Depending on the carbon type, the signal in DEPT can be pointing up or down while being at the same ppm values as in the regular 13 C NMR. Another possibility in DEPT is the lack of a given signal . Here is the summary of DEPT signals :

DEPT-90 and DEPT-135 are different types of DEPT experiments and we won’t go over the mechanisms here but rather use this data as it is. The aim of this article is to explain the application of DEPT in solving NMR spectra .

As an example , let’s see this (stimulated) 13 C NMR combined with the DEPT experiments:

Notice how the ppm values are retained but depending on the signals in DEPT we can tell if the carbon is a C, CH, CH 2 or a CH 3 group .

In case you needed, here are the chemical shift values for 13 C NMR:

Show Me a Good Example of DEPT NMR Problem

Let’s discuss a specific NMR problem where the final structure is only determined using the DEPT data.

The proton and carbon NMR spectra of a compound with the formula C 5 H 9 Br are shown below. The DEPT experimental results are also provided in the table.

Purpose a plausible structure based on the data provided.

I went over the steps for solving NMR problems with lots of examples which you can find here but for now let’s quickly apply those and see what we get.

First, determine the hydrogen deficiency index . Replacing the Br with an H we get C 4 H 10 which corresponds to one degree of unsaturation . Therefore, the compound has a double bond or a ring .

Now, looking at signal at about 4.7 ppm in the proton, and the ones above 100 on the carbon, we know that it must be a double bond rather than a ring.

Next, look at the signal splitting in 1 H NMR; two triplets indicate a -CH 2 -CH 2 – fragment which is connected to Br on one end since it is downfield (3.3 ppm) . And the other CH 2 must be connected to the double bond since the signal is still more downfield than if it was a regular alkyl group.

I’ll put this table for 1 H NMR shifts for a reference:

So, let’s put down the groups we have so far:

Two of these X groups must be hydrogens because of the integration of the signal at ~4.7 ppm.

The other X group is a methyl group which we can deduce from the integration.

And now the i nteresting part realted to DEPT . There are three combinations of putting two hydrogens and a methyl group on the double bond:

All of these would be good candidates based on the data from the proton and carbon NMR. But only the last structure matches the data from the DEPT experiments which indicate the presence of three CH 2 groups (three negative signals in DEPT-135):

I do want to mention that the structure of a double bond can be analyzed using the J coupling values and a powerful NMR spectrometer will give a resolution good enough to exclude the other candidates based on the coupling. However, DEPT makes things easier without the need for a lot of complicated analysis.

• NMR spectroscopy – An Easy Introduction
• NMR Chemical Shift
• NMR Chemical Shift Range and Value Table
• NMR Number of Signals and Equivalent Protons
• Homotopic Enantiotopic Diastereotopic and Heterotopic
• Homotopic Enantiotopic Diastereotopic Practice Problems
• Integration in NMR Spectroscopy
• Splitting and Multiplicity (N+1 rule) in NMR Spectroscopy
• NMR Signal Splitting N+1 Rule Multiplicity Practice Problems
• 13 C NMR NMR
• DEPT NMR: Signals and Problem Solving
• NMR Spectroscopy-Carbon-Dept-IR Practice Problems

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H NMR Table Practice Problems

Enumerate the differences between the IR and  1 H NMR spectra of butyramide and those of  N , N -diethylbutyramide. 

The methyl-H highlighted in the isopentane structure appears at 0.87 ppm, while that of 2-methylbut-2-ene appears at 1.56 ppm. Rationalize this observation. 

Even though alkene hydrogens usually show up at similar chemical shifts between 5 and 6 ppm, the chemical shifts of the labeled alkene hydrogens below are different (H a = 6.009 ppm; H b = 7.034 ppm). Rationalize the difference.

Predict the chemical shifts of the labeled hydrogens. 

Provide a good estimate of the chemical shift for each labeled hydrogen in the compound below.

For the hydrogen indicated in red, sketch the expected 1 H NMR signal showing its estimated chemical shift.

For the hydrogen indicated in red, draw the expected 1 H NMR signal showing its estimated chemical shift.

Fill out the values in the 1 H NMR table given below for the indicated hydrogens in the following compound.

Fill out the 1 H NMR data in the following table for the molecule given below.

Fill out the following 1 H NMR table for the indicated hydrogens in the molecule given below.

Draw the expected 1 H NMR signal for the indicated hydrogen. Show the integration and chemical shift for it.

Identify the labeled proton (or a set of protons) with the higher chemical shift.

Identify the labeled proton (or set of protons) with the higher chemical shift.

Which statement explains why the methine proton has a higher chemical shift than the methyl proton?

Compound X can be oxidized by chromic acid to form a benzenedicarboxylic acid. Draw the structure of compound X from the given 1 H NMR spectrum.

Identify the labeled set of protons whose 1 H NMR signal occurs at a higher chemical shift.

Without looking at tables of 1 H NMR values, rank the protons in the given compound from highest to lowest chemical shifts. Use 1 for proton with the highest chemical shift, 2 for the second highest, and so on.

Describe the chemical shifts of the protons in the benzene ring in phenol.

Use the letters a, b, c, and so on to designate each set of chemically equivalent protons, starting with the lowest chemical shift in the 1H NMR spectrum. Include the multiplicity of each signal.

Explain how the chemical shifts of chloromethane, dichloromethane, and trichloromethane at δ 3.06, δ 5.47, and δ 7.26 correlate with pK a in 1 H NMR.

A) The hydrogen in trichloromethane is the least acidic and has the lowest chemical shift due to the three electron-withdrawing chlorine atoms, followed by dichloromethane with two electron-withdrawing chlorine atoms, and lastly chloromethane with one electron-withdrawing chlorine atom. 

B) The hydrogen in trichloromethane is the most acidic and has the highest chemical shift due to the three electron-withdrawing chlorine atoms, followed by dichloromethane with two electron-withdrawing chlorine atoms, and lastly chloromethane with one electron-withdrawing chlorine atom.

C) The hydrogen in trichloromethane is the most acidic and has the lowest chemical shift due to the three electron-donating chlorine atoms, followed by dichloromethane with two electron-donating chlorine atoms, and lastly chloromethane with one electron-donating chlorine atom.

D) The hydrogen in chloromethane is the most acidic and has the lowest chemical shift due to one electron-withdrawing chlorine atom, followed by dichloromethane with two electron-withdrawing chlorine atoms, and lastly trichloromethane with three electron-withdrawing chlorine atoms.

Compound A, with a molecular formula C 3 H 4 , shows two singlets in its 1 H NMR spectrum (with a ratio 3 : 1). Compound B is formed when compound A undergoes an acid-catalyzed hydration. Compound B also gives a positive iodoform test, and shows a singlet in its 1 H NMR spectrum. Determine the structures A and B.

Match the right chemical shifts of the C-2 hydrogen in the 1 H NMR spectra recorded at 3.75 ppm, 6.62 ppm, and 7.38 ppm for pyrrole, furan, and tetrahydrofuran.

An NMR spectrum of an unknown compound obtained from a 400-MHz NMR spectrometer shows protons at a position 840 Hz downfield from the TMS. (a) Determine the chemical shift of these protons. (b) Determine their chemical shift in an 80-MHz NMR spectrometer. (c) By how many Hz their chemical shift should be downfield from TMS using an 80-MHz NMR instrument?

The NMR spectrum of ethylbenzene shows a broad peak around δ 7.2 in the aromatic region. Determine how many different types of protons are present in ethylbenzene and explain the presence of the broad peak in the aromatic regions.

• Each peak in the spectrum corresponds to a fragment of the molecule.
• Each peak contains THREE different kind of information about the molecular fragment: Integration, Chemical Shift, and Multiplicity (or splitting) .  You CANNOT solve the spectrum without using all three pieces of information.
• Once you have connected all the fragments, it does NOT mean you have successfully solved the structure.  You must make sure that each connection is consistent with the information in the spectrum.  CHECK IT!  (The NMR Mosaic is excellent at helping you make sure the pieces truly match.)
• Use the molecular formula (if given) to figure out the degree of unsaturation (the number of rings and/or multiple bonds).
• Use the Infrared spectrum (if given) to figure out the functional groups present.
• Look for any recognizable groups in the NMR such as aromatic rings, aldehydes, carboxyllic acids, or alcohols.
• Use the integration to determine the number of hydrogens contributing to each peak.
• OFTEN (but not always) the smallest integration is 1 hydrogen.
• OFTEN (but not always) peaks around 0.7-1.1 d are methyl groups (3 hydrogens).
• If you know the molecular formula, the integral per hydrogen equals the total integration of all peaks divided by the total number of hydrogens.
• Then make the molecular fragment  each peak corresponds to (remember, each peak corresponds to one fragment of the molecule, often a but now always a CH x ) :
• Use the integration to calculate the number of hydrogens contributing to each peak (a CH, CH 2 , CH 3 , etc.).
• Use the chemical shift to determine which (if any) functional group(s) is attached to that fragment.
• Use the multiplicity to determine how many hydrogens are adjacent to the fragment .  
• Put the fragments together.  You have two kinds of information telling you how to connect the fragments: 
• The chemical shifts tells you which fragments are connected to which functional groups.
• The splitting tells you how many hydrogens are on immediately adjacent carbons.
• If the pieces don't all match , something is not right.  Find the piece(s) that fit the worst and check them first; they are frequently the ones interpreted incorrectly.  Is the splitting correct?  Can the chemical shifts be for another functional group?  Can a CH really be an OH?  
• Check the interpretation of the splitting.  For instance a quartet almost always means the peak is next to a CH3 group.  But once in a while it may be next to a CH2 and a CH.  Either can be right, and often it is impossible to tell which is correct until you start putting the fragments together.  It just depends on the specifi spectrum you are solving.
•     It can help you recognize when you have made a mistake or interpreted a peak incorrectly for that particular spectrum: all the pieces won't fit.  
•     It can help you interpret complex splitting that my students call "messtets."

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12.10.2 MS, IR and NMR Problems

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Solving molecular structures

Solving spectroscopy problems

Worked example

Putting it all together

Practice Problems

Using spectroscopy to determine structure

Really good practice

A workbook of unknowns

Spectroscopy worksheet

NMR/IR/MS practice problems

Spectroscopy practice problems

Practice problems IR/MS/NMR

Huge set of practice problems

Spectroscopy Practice exam and answers

** IR, MS and NMR practice exams