You know from Chapter 3 that the position of a signal in an NMR spectrum tells us about its environment. In acetic acid the methyl group is next to the electron-withdrawing carbonyl group and so is slightly deshielded at about δ 2.0 ppm and the acidic proton itself, attached to O, is very deshielded at δ 11.2 ppm. The same factor that makes this proton acidic—the O–H bond is polarized towards oxygen—also makes it resonate at low field. So far things are much the same as in ¹³C NMR. Now for a difference. In ¹H NMR the size of the peaks is also important: the area under the peaks is exactly proportional to the number of protons. Proton spectra are normally integrated, that is, the area under the peaks is computed and recorded as a line with steps corresponding to the area, like this.

Simply measuring the height of the steps with a ruler gives you the ratio of the numbers of protons represented by each peak. In many spectra this will be measured for you and reported as a number at the bottom of the spectrum. Knowing the atomic composition (from the mass spectrum) we also know the distribution of protons of various kinds. Here the heights are 6 mm and 18 mm, a ratio of about 1:3. The compound is C₂H₄O₂ so, since there are four H atoms altogether, the peaks must contain 1 × H and 3 × H, respectively. In the spectrum of 1,4-dimethoxybenzene there are just two signals in the ratio of 3:2. This time the compound is C₈H₁₀O₂ so the true ratio must be 6:4. The positions of the two signals are exactly where you would expect them to be from our discussion of the regions of the NMR spectrum in Chapter 3: the 4H aromatic signal is in the left-hand half of the spectrum, between 5 and 10 ppm, where we expect to see protons attached to sp² C atoms, while the 6H signal is in the right-hand half of the spectrum, where we expect to see protons attached to sp³ C atoms.

In this next example it is easy to assign the spectrum simply by measuring the steps in the integral. There are two identical methyl groups (CMe₂) with six Hs, one methyl group by itself with three Hs, the OH proton (1 H), the CH₂ group next to the OH (two Hs), and finally the CH₂CH₂ group between the oxygen atoms in the ring (four Hs).

Before we go on, a note about the solvent peaks shown in brown in these spectra. Proton NMR spectra are generally recorded in solution in deuterochloroform (CDCl₃)—that is, chloro form (CHCl₃) with the 1H replaced by 2H (deuterium). The proportionality of the size of the peak to the number of protons tells you why: if you ran a spectrum in CHCl3, you would see a vast peak for all the solvent Hs because there would be much more solvent than the compound you wanted to look at. Using CDCl₃ cuts out all extraneous protons. 2H atoms have different nuclear properties and so don’t show up in the ¹H spectrum. Nonetheless, CDCl₃ is always unavoidably contaminated with a small amount of CHCl₃, giving rise to the small peak at 7.25 ppm. Spectra may equally well be recorded in other deuterated solvents such as water (D₂O), methanol (CD₃OD), or benzene (C₆D₆).

مواضيع ذات صلة

Protons on saturated carbon atoms Chemical shifts are related to the electronegativity of substituents

Regions of the proton NMR spectrum

The differences between carbon and proton NMR

Kinetic versus thermodynamic products

Catalysis in carbonyl substitution reactions

What does third-order kinetics mean

Rate equations

Catalysis

Reaction intermediates

Why some reactions are done at low temperature

The route from reactants to products: the transition state

How fast do reactions go? Activation energies