Introduction

Nuclear magnetic resonance (NMR) is a very powerful tool that gives information to help identify the structure of a compound. However, the NMR instrument does not simply spit out the structure of the sample. Instead, it gives data in the form of a spectrum from which the molecular structure can be deduced.

There are two main kinds of NMR: 1H NMR, which gives information about the hydrogen atoms in the molecule and 13C NMR, which gives valuable information about a particular carbon isotope (carbon-13). The most common NMR technique in Introductory Organic Chemistry is 1H NMR so that will be the topic of this chapter.

This chapter consists of 2 main parts:

  1. The Fundamentals of 1H NMR
  2. Solving 1H NMR problems and 1H NMR spectra

As a final note, this chapter is very different than any other in organic chemistry, so be prepared to think in new ways. Some students struggle with 1H NMR, but the tips and tricks presented in this chapter will vastly simplify these often complex problems.

 

 

The Fundamentals of 1H NMR

There are 4 fundamental concepts to understanding 1H NMR. A brief introduction to each is listed below followed by a figure that depicts each concept on an actual 1H NMR spectrum.

  1. The number of signals: The number of signals is equal to the number of distinct hydrogen atoms in the molecule, which can be found by analysis of the molecule (more on this later). Signals are indicated using the red boxes below.
  2. The multiplicity of the molecule: This refers to the number of “peaks” within a single signal. For example, there are 3 peaks within the signal located at 3.39 PPM in the spectrum below. If 3 peaks are in a signal, the signal is called a triplet, and it tells us valuable information about the molecule that we’ll talk more about later.
  3. Integration: This refers to the number of hydrogen atoms that a single signal represents. This is shown using the green boxes below.
  4. Location on the spectra: The location of the signal on the spectra gives us valuable information about both the functional groups present in the molecule and the environment surrounding the hydrogen atom. Information about location on the spectra is shown in blue below.

Picture1

Remember, this was just an introduction to each of these concepts. We’ll learn much more about each in the sections below.

 

  1. Number of Signals

One of the first key 1H NMR concepts is the number of signals a molecule will produce. The number of signals on the 1H NMR spectrum is equal to the number of distinct hydrogens in the molecule. If hydrogens are located in the exact same environment as one another, they are chemically equivalent and appear in the same signal in the 1H NMR spectrum.

If 2 hydrogen atoms are in the exact same environment, then these hydrogens are chemically equivalent and will appear in the same signal. If 2 hydrogens are not in the exact same environment, then the hydrogens are chemically distinct, and they will not appear in the same signal.

There are quite a few techniques to assess whether two hydrogens are chemically equivalent or not. Some of these involve complex mental rotations of the molecule. However, I believe the best, and simplest, way to decide this is by simply describing the location of the hydrogen using words. Let’s look at an example from the original spectrum above, comparing the 2 hydrogens explicitly shown:

Picture2

Red hydrogen: Directly attached to a carbon that is bonded to a –CH2Br group and a methyl group.

Blue hydrogen: Directly attached to a carbon that is bonded to a –CH2Br group and a methyl group.

These are clearly chemically equivalent hydrogens because they are in the same environment, so they will appear in the same 1H NMR signal.

 

Let’s do another example using the same molecule:

Picture3

Red hydrogen: Directly attached to a carbon that is bonded to a –CH2Br group and a methyl group.

Blue hydrogen: Directly bonded to a carbon that is bonded to a -Br and an ethyl group.

These two hydrogen are therefore distinct hydrogens, so they will appear in different signals in the 1H NMR.

 

Example

How many distinct signals will be present in the 1H NMR of the following molecule?

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Answer

First, let’s explicitly draw out the hydrogens and describe each’s environment.

Picture5

H1: Directly attached to a carbon atom on the ring. Between 2 fluorine atoms also coming off the ring.

H2: Directly attached to a carbon atom on the ring. Between a fluorine and a hydrogen on the ring.

H3: Directly attached to a carbon atom on the ring. Between 2 hydrogen atoms coming off the ring.

H4: Directly attached to a carbon atom on the ring. Between a fluorine and a hydrogen atom on the ring.

Note that the environments for H2 and H4 are exactly the same. Therefore, they are chemically equivalent hydrogens. All the other hydrogen atoms are in distinct environments in this molecule. Therefore, we have a total of 3 chemically distinct hydrogens, so there will be 3 signals in the 1H NMR spectrum (see color coding below for the signals).

Picture6

 

 2. Multiplicity

Multiplicity (or coupling) refers to the number of peaks within a signal. To determine the number of peaks a hydrogen will produce within its own signal, use the n+1 rule:

The n+1 rule tells us to count the number of neighboring hydrogens, and add 1. The result of this simple equation gives the number of peaks in the signal. For example, if you have 6 neighboring hydrogens, you will have 7 peaks in that signal.

 

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There are names given to a signal based on the number of peaks present:

# of neighboring H’s # of peaks in signal Coupling name
0 1 singlet  Picture8
1 2 doublet  Picture9
2 3 triplet  Picture10
3 4 quartet  Picture11
4 5 pentet/quintet  Picture12
5 6 sextet  Picture13
6 7 septet  Picture14

 

3. Integration

The integration, or the area under the curve, provides information about the number of hydrogens that are contained in a single signal. If a single signal has an integration of “2H,” this normally tells you the signal represents 2 chemically equivalent hydrogen atoms.

It’s important to note that the integrations are displayed in ratios of hydrogens. Therefore, if you have a symmetric molecule, you must correct for the symmetry by multiplying by the correct factor (see figure below).

Picture15

 

 

 4. Location on the spectra

The location of the signal on the spectrum (often called the chemical shift) also gives us valuable information about the atoms in the molecule. This can tell us (1) about the presence of certain functional groups on the molecule or (2) about a hydrogen’s position relative to an electronegative atom.

  1. Identifying Functional Groups Based on Location on the Spectrum

Functional groups often show up at highly predictable regions in the 1H NMR spectrum. For example, a hydrogen coming off a benzene ring normally shows up between 6.5 – 8 parts per million (PPM). Therefore, if you see a signal in that region of the 1H NMR spectrum, you have a strong indication that the molecule contains a benzene ring.

Below is a figure that depicts that location of common functional groups on the 1H NMR spectrum.

Picture16

 

2. General Location on the Spectrum

The terms upfield and downfield are used to refer to the general location of particular signals on an 1H NMR spectrum. Downfield is towards the left side of the NMR spectrum, towards higher PPM values, while upfield is more rightward on the spectrum, moving more towards lower PPM values.

All other factors being equal, a hydrogen that is closer to an electronegative atom, such as a halogen or an oxygen atom, appears more downfield on an 1H NMR spectrum. In the example below, note that the red hydrogens (signal boxed in red) are more downfield than the blue hydrogens (signal boxed in blue) because the red hydrogens are closer in proximity to the bromine.

Picture17

Keep it Simple

Hydrogens directly attached to oxygen atoms (like in alcohols or carboxylic acids) often, but not always, appear as broad peaks on the 1H NMR spectrum. They are generally very easy to spot because of this.

Picture18

That being said, alcohols and carboxylic acids don’t always appear this way, so don’t solely rely on a broad peak to identify one of these functional groups.

 

 

Solving 1H NMR Problems

Now that we know the core concepts of 1H NMR, we’ll learn how to actually solve 1H NMR problems. There are four steps:

  1. Find the number of degrees of unsaturation
  2. Note the position of the signals in the spectrum
  3. Find the patterns in the spectrum
  4. Piece the patterns together to form the molecule

 

  1. Find the number of degrees of unsaturation

A fully saturated molecule has the maximum number of carbon-hydrogen bonds possible. This occurs in molecules with all single bonds that do not contain a ring.

Maximum number of hydrogen bonds possible = (2)(#C) +2

If a molecule has a double bond, a triple bond, or a ring, it is not fully saturated with hydrogens. We can quantitate how unsaturated a molecule is (called the degrees of unsaturation) by plugging the information given to us in the molecular formula into the mathematical formula below:

Degrees of unsaturation=  (2C+2+#N-#X-H)/2

Where #X is the number of halogens (F, Cl, Br, or I) in the molecule and #N is the number of nitrogen atoms in the molecule.

We can also obtain the degrees of unsaturation simply by looking at the molecule. A degree of unsaturation occurs in a molecule with either a double bond, triple bond, or a ring. For example, if a molecule has a double bond and a ring, it would have 2 degrees of unsaturation. Similarly, a molecule with a triple bond would also have 2 degrees of unsaturation.

 

Picture19

 

    2. Note the position of the signals in the spectrum

Take a moment to look at the full spectrum to see if any of the peaks are located in the common functional group regions that were discussed in the “Location of the spectrum” section above. For example, if the molecule has 4 degrees of unsaturation and a signal at ~8 PPM on the 1H NMR, then that’s a very good indication that molecule contains a benzene ring.

 

   3. Find the patterns in the spectrum

1H NMR can be very difficult until there is full understanding of this point: 1H NMR in introductory organic chemistry is all about patterns.

The figure below depicts the spin-splitting patterns of molecular fragments that regularly show up in an introductory organic chemistry class. Full knowledge of these fragments will help you be able to solve 1H NMR very efficiently. Be very, very familiar with these patterns.

Picture20

Note that these patterns alone typically won’t help you solve the entire problem, but it often helps to solve a large portion of the molecule.

 

  4. Piece the patterns together to form the molecule

You’ve already found the fragments of the molecule. Now, simply piece them together to form the entire molecule.

 

Example

Picture21

Answer
  1. Find the number of degrees of unsaturation

The molecular formula of the molecule for this 1H NMR is C4H8O. Plugging this information into the equation

This tells us there is either a double bond or a ring in the molecular structure.

    2. Note the position of the signals in the spectrum

Doing a quick scan of the location of the peaks in the spectrum, none are located in any area that is indicative of a particular functional group.

   3. Find the patterns in the spectrum

Picture22

Using the figure of common 1H NMR patterns, we were able to identify an ethyl and a methyl group.

  4. Piece the patterns together to form the molecule

We now must piece the molecule together. Since we know we have an ethyl group and a methyl group in the molecule, we are left with 1 carbon, 1 oxygen, and 0 hydrogen atoms to piece together. Also note that the signals from the methyl group and the ethyl group are entirely separate from one another because there is no coupling between these groups.

There is not a peak that could potentially belong to an –OH in our spectrum, so we can rule out an alcohol functional group. That means our oxygen must occur in the molecule via either a carbonyl or an ether functional group. Since there is also one degree of unsaturation in our molecule, we have a strong indication the functional group present is a carbonyl group.

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