14.4: Proton Equivalence

Understanding the basics of NMR theory gets us ready to move on to the most important and practical part in this section, which is how to understand the ¹H NMR spectrum and elucidate the structure of a compound from ¹H NMR spectrum information. Let’s first take a look at an actual ¹H NMR spectrum.

Figure \(\PageIndex{1}\): The 1H NMR spectrum of methyl acetate

Generally, the information about the structure of a molecule can be obtained from four aspects of a typical ¹H NMR spectrum:

• Chemical equivalent and non-equivalent protons (total number of signals)
• Chemical shift
• Integration
• Signal splitting

Chemical Equivalent and Non-Equivalent Protons

In the above ¹H NMR spectrum of methyl acetate (Figure \(\PageIndex{1}\): ), we can see that there are three signals. The peak at the far right is for the standard reference compound tetramethylsilane (TMS, more discussions in chemical shift section), not for the compound. So the compound methyl acetate shows two signals in ¹H NMR spectrum. Why only two signals for a compound containing a total of six hydrogens?

This is because of chemical equivalence. The six total hydrogens can be divided into two groups: the three Ha protons in the methyl group bonded to the C=O are in the same chemical environment and are therefore chemically equivalent. All chemical equivalent hydrogens have the same resonance frequency when applied to an external magnetic field, so they show only one signal in a 1H NMR spectrum. The three H_b protons in the methyl group bonded with O atom are chemical equivalent as well and show the other signal. That is why there are two signals in total for compound methyl acetate.

 

The ability to recognize chemically equivalent and non-equivalent protons in a molecule is very important in understanding the NMR spectrum. For the compound with the given structure, we can predict how many signals there are in the 1H NMR spectrum. On the other side, if the ¹H NMR spectrum is available for an unknown compound, counting the number of signals in the spectrum tells us the number of different sets of protons in the molecule, and that is very important information to determine the structure of the compound.

Homotopic, Enantiotopic, or Diasterotopic CH₂ Protons

For relatively small molecules, a quick look at the structure is often enough to decide how many kinds of protons are present and thus how many NMR absorptions might appear. If in doubt, though, the equivalence or nonequivalence of two protons can be determined by comparing the structures that would be formed if each hydrogen were replaced by an X group. There are four possibilities.

• One possibility is that the protons are chemically unrelated and thus nonequivalent. If so, the products formed on substitution of H by X would be different constitutional isomers. In butane, for instance, the –CH₃ protons are different from the –CH₂– protons. They therefore give different products on substitution by X than the –CH₂ protons and would likely show different NMR absorptions.

 

• A second possibility is that the protons are chemically identical and thus electronically equivalent. If so, the same product would be formed regardless of which H is substituted by X. In butane, for instance, the six –CH₃ hydrogens on C1 and C4 are identical, would give the identical structure on substitution by X, and would show an identical NMR absorption. Such protons are said to be homotopic.

 

• The third possibility is a bit more subtle. Although they might at first seem homotopic, the two –CH₂– hydrogens on C2 in butane (and the two –CH₂– hydrogens on C3) are in fact not identical. Substitution by X of a hydrogen at C2 (or C3) would form a new chirality center, so different enantiomers (Section 5.1) would result, depending on whether the pro-R or pro-S hydrogen had been substituted (Section 5.11). Such hydrogens, whose substitution by X would lead to different enantiomers, are said to be enantiotopic. Enantiotopic hydrogens, even though not identical, are nevertheless electronically equivalent and thus have the same NMR absorption.

• The fourth possibility arises in chiral molecules, such as (R)-2-butanol. The two –CH₂– hydrogens at C3 are neither homotopic nor enantiotopic. Because substitution of a hydrogen at C3 would form a second chirality center, different diastereomers (Section 5.6) would result, depending on whether the pro-R or pro-S hydrogen had been substituted. Such hydrogens, whose substitution by X leads to different diastereomers, are said to be diastereotopic. Diastereotopic hydrogens are neither chemically nor electronically equivalent. They are completely different and would likely show different NMR absorptions.

Note

As you probably already realized, chemical equivalence or non-equivalence in NMR is closely related to symmetry. The protons that are symmetric to each other by a certain plane of symmetry are chemically equivalent.

Here, we will go through several examples for the first situation: predicting the number of signals in a 1H NMR spectrum given the structure of a compound. To do that, we need to count the number of distinct sets of protons in the molecule.

For each of the following molecules, the chemically equivalent protons are labeled in the same color to facilitate understanding.

• Benzene: all six protons are chemically equivalent (have the same bonding and are in the same chemical environment) to each other and have the same resonance frequency in a 1H NMR experiment, therefore show only one signal.
• Acetone: both methyl groups (two CH₃) are bonded with the C=O bond, so they are in the same chemical environment, and as a result, all six protons are chemically equivalent, showing only one signal.

The molecules in the next figure contain more sets of chemically equivalent protons.

• Acetaldehyde: The three Ha protons in the methyl group are chemically equivalent, and they are all bonded to an sp³-hybridized carbon; but they are different from the Hb proton that is bonded to an sp²–sp²-hybridized carbonyl carbon. Two signals total in ¹H NMR spectrum.
• 1,4-dimethylbenzene: all four aromatic protons are chemically equivalent because of the symmetry. The two methyl groups are equivalent to each other as well. Two signals total in ¹H NMR spectrum.
• 1,2-dimethylbenzene: both H_a protons are adjacent to a methyl substituent, while both H_c protons are two carbons away. So the four aromatic protons are divided to two sets. Both methyl groups are in the same bonding and symmetric to each other; they are equivalent. Three signals total in ¹H NMR spectrum.
• 1,3-dimethylbenzene: H_b is situated between two methyl groups, the two H_c protons are one carbon away from a methyl group, and H_d is two carbons away from a methyl group. Therefore, the four aromatic protons can be divided into three sets. The two methyl groups are equivalent. Four signals total in ¹H NMR spectrum.

Exercises

Exercise \(\PageIndex{1}\)

How many ¹H NMR signals would you predict for each of the following molecules?

Answer

Exercise \(\PageIndex{2}\)

Identify the indicated sets of protons as unrelated, homotopic, enantiotopic, or diastereotopic:

a.          b.    c.  

d.      e.     f. 

Answer

a. Enantiotopic. b. Diastereotopic. c. Diastereotopic. d. Diastereotopic. e. Diastereotopic. f. Homotopic

Exercise \(\PageIndex{3}\)

How many kinds of electronically nonequivalent protons are present in each of the following compounds, and thus how many NMR absorptions might you expect in each?

a. CH₃CH₂Br  b. CH₃OCH₂CH(CH₃)₂  c. CH₃CH₂CH₂NO₂  d. Methylbenzene  e. 2-Methyl-1-butene  f. cis-3-Hexene

Answer

a. 2. b. 4 c. 3. d. 4. e. 5. f. 3.

Exercise \(\PageIndex{4}\)

How many absorptions would you expect (S)-malate, an intermediate in carbohydrate metabolism, to have in its ¹H NMR spectrum? Explain.

Answer

4