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2.8: ¹H NMR Spectroscopy and Proton Equivalence

  • Page ID
    469364
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    Because each electronically distinct hydrogen in a molecule has its own unique absorption, one use of 1H NMR is to find out how many kinds of electronically nonequivalent hydrogens are present. In the 1H NMR spectrum of methyl acetate shown previously, for instance, there are two signals, corresponding to the two kinds of nonequivalent protons present, CH3C═O protons and –OCH3 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.
    Butane shows methylene and methane hydrogens are unrelated. Replacing one methylene or one methyl hydrogen with X creates constitutional isomers.
    • 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.
    Butane shows six methyl hydrogens homotopic with same N M R absorption. Replace one H with X to form only one substitution product.
    • 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.
    Butane shows dash-bonded and wedge-bonded hydrogens at C 2 labeled pro-S and pro-R, respectively. Replace either H with X to form substitution products that are enantiomers.
    • 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.
    (R)-2-butanol shows dash-bonded and wedge-bonded hydrogens at C 3 labeled pro-S and pro-R, respectively. Replace either H with X to form two substitution products that are diastereomers.
    Exercise \(\PageIndex{1}\)

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

    1. Chemical structure of 2-methyl-3-pentanone. Two arrows point toward the highlighted hydrogens at C 4.
    2. Chemical structure of cyclohexanol. Two arrows point toward the highlighted hydrogens at C 3. 
    3. Chemical structure of 3-methyl-2-butene. Two arrows point toward the highlighted hydrogens on methyl carbon and C 4. 
    4. Chemical structure of ethyl 2-bromopropionate. Two arrows point toward the highlighted hydrogens on ethyl carbon adjacent to oxygen. 
    5. Chemical structure of 2-isopropylcyclohexanone. Two arrows point toward the highlighted hydrogens on isopropyl terminals. 
    6. Chemical structure of cis-bicyclo[3.3.0]octane. Two arrows point toward the highlighted hydrogens on fusion carbons.
    Answer
    1. Enantiotopic
    2. Diastereotopic
    3. Diastereotopic
    4. Diastereotopic
    5. Diastereotopic
    6. Homotopic
    Exercise \(\PageIndex{2}\)

    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?

    1. CH3CH2Br
    2. CH3OCH2CH(CH3)2
    3. CH3CH2CH2NO2
    4. Methylbenzene
    5. 2-Methyl-1-butene
    6. cis-3-Hexene
    Answer
    1. 2
    2. 4
    3. 3
    4. 4
    5. 5
    6. 3
    Exercise \(\PageIndex{3}\)

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

    The ball-and-stick model of (S)-Malate shows a 4-carbon chain. C 1 and C 4 are carboxyl groups. C 2 is bonded to hydroxyl group.

    Answer

    4


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