MCAT Organic Chemistry Review

Spectroscopy

11.3 Nuclear Magnetic Resonance Spectroscopy

Nuclear magnetic resonance (NMR) spectroscopy is the most important spectroscopic technique to understand for the MCAT. NMR spectroscopy is based on the fact that certain atomic nuclei have magnetic moments that are oriented at random. When such nuclei are placed in a magnetic field, their magnetic moments tend to align either with or against the direction of this applied field. Nuclei with magnetic moments that are aligned with the field are said to be in the α-state (lower energy). The nuclei can then be irradiated with radiofrequency pulses that match the energy gap between the two states, which will excite some lower-energy nuclei into the β-state (higher energy). The absorption of this radiation leads to excitation at different frequencies, depending on an atom’s magnetic environment. In addition, the nuclear magnetic moments of atoms are affected by nearby atoms that also possess magnetic moments.

Magnetic resonance imaging (MRI) is a noninvasive diagnostic tool that uses proton NMR, as shown in Figure 11.3. Multiple cross-sectional scans of the patient’s body are taken, and the various chemical shifts of absorbing protons are translated into specific shades of grey. This produces a picture that shows the relative density of specific types of protons; for instance, a dark area on a T1-weighted MRI tends to correspond to water, whereas a light area indicates fattier tissue. Comparison with normal MRI then allows the diagnostician to detect abnormalities in the scanned region. We mention this to explain the relevance of NMR spectroscopy to medicine; the MCAT will not test you on the details of how MRI works.

Figure 11.3. Magnetic Resonance Imaging (MRI)

A typical NMR spectrum is a plot of frequency vs. absorption of energy. Because different NMR spectrometers operate at different magnetic field strengths, a standardized method of plotting the NMR spectrum as been adopted. This standardized method, which is the only one seen on the MCAT, uses an arbitrary variable called chemical shift (δ), with units of parts per million (ppm) of spectrometer frequency. The chemical shift is plotted on the x-axis, and it increases toward the left (referred to as downfield). To make sure that we know just how far downfield compounds are, we use tetramethylsilane (TMS) as the calibration standard to mark 0 ppm; when counting peaks, make sure to skip the TMS peak.

KEY CONCEPT

TMS provides a reference peak. The signal for its 1H atoms is assigned δ = 0.

Nuclear magnetic resonance is most commonly used to study 1H nuclei (protons), although any atom possessing a nuclear spin (with an odd atomic number, odd mass number, or both) can be studied, such as 13C, 19F, 17O, 31P, and 59Co. The MCAT, however, only tests knowledge of 1H–NMR.

BRIDGE

Nuclei with odd mass numbers, odd atomic numbers, or both, will have a magnetic moment when placed in a magnetic field. Not all nuclei have magnetic moments (12C, for example). Atomic numbers and mass numbers are discussed in more detail in Chapter 1 of MCAT General Chemistry Review.

PROTON NMR (1H–NMR)

Most hydrogen (1H) nuclei come into resonance 0 to 10 ppm downfield from TMS. Each distinct set of nuclei gives rise to a separate peak. This means that if multiple protons are chemically equivalent, having the same magnetic environment, they will lead to the same peak. For example, Figure 11.4 depicts the 1H–NMR of dichloromethyl methyl ether, which has two distinct sets of 1H nuclei. The single proton attached to the dichloromethyl group (Ha) is in a different magnetic environment from the three protons on the methyl group (Hb), so the two classes will resonate at different frequencies. The three protons on the methyl group are chemically equivalent and resonate at the same frequency because this group rotates freely, and on average, each proton sees an identical environment.

Figure 11.4. 1H–NMR Spectrum of Dichloromethyl Methyl Ether Peak a: Dichloromethyl proton; Peak b: Methyl protons.

MCAT EXPERTISE

To determine how many peaks will be in the spectrum, see if you can describe protons differently using words. In the dichloromethyl methyl ether shown in Figure 11.4, one could call Ha the hydrogen on the carbon with two chlorides and all three Hb hydrogens the ones in the methyl group. It would not be possible to describe each of these three hydrogens as distinct from each other because they rotate freely in space.

The peak on the left (a) is from the single dichloromethyl proton, and the taller middle peak is from the three methyl protons (b). The height of each peak is proportional to the number of protons it contains. Specifically, if we were to analyze the area under the peaks, called the integration, we would find that the ratio of (a) to (b) is 1:3, corresponding exactly to the ratio of protons that produced each peak.

Now that we know which peak is which, let’s talk about their respective positions on the spectrum. We can see that the peak for the single proton (a) is fairly far downfield compared with the other protons. This is because it is attached to a carbon with two electronegative chorine atoms and an oxygen atom. These atoms pull electron density away from the surrounding atoms, thus deshielding the proton from the magnetic field. The more the proton’s electron density is pulled away, the less it can shield itself from the applied magnetic field, resulting in a reading further downfield. With this same reasoning, we know that if we had an electron-donating group, such as the silicon atom in TMS, it would help shield the 1H nuclei and give it a position further upfield. This is why tetramethylsilane is used as the reference or calibration peak; everything else in proton NMR will be more deshielded than it.

KEY CONCEPT

·        Each peak or group of peaks that are part of a multiplet represents a single group of equivalent protons

·        The relative area of each peak reflects the ratio of the protons producing each peak

·        The position of the peak (upfield or downfield) is due to shielding or deshielding effects, which reflect the chemical environment of the protons

Now, let’s make it a little more interesting. Consider a compound containing protons that are within three bonds of each other: in other words, a compound in which there are hydrogens on two adjacent atoms. When we have two protons in such close proximity to each other that are not magnetically identical, spin–spin coupling (splitting) occurs. Let’s use the molecule in Figure 11.5 to demonstrate this concept.

Figure 11.5. 1,1-Dibromo-2,2-dichloroethane

MNEMONIC

When dealing with 1H–NMR on the MCAT, think of a proton as being surrounded by a shield of electrons. As we add electronegative atoms or have resonance structures that pull electrons away from the proton, we Deshield and move Downfield.

Notice the two protons, Ha and Hb, on 1,1-dibromo-2,2-dichloroethane. Because of their proximity, the magnetic environment of Ha can be affected by Hb, and vice-versa. Thus, at any given time, Ha can experience two different magnetic environments because Hb can be in either the α- or the β-state. The different states of Hb influence the nucleus of Ha, causing slight upfield and downfield shifts. There is approximately a 50% chance that Hb will be in either of the two states, so the resulting absorption is a doublet: two peaks of identical intensity, equally spaced around the true chemical shift of Ha. Ha and Hb will both appear as doublets because each one is coupled with one other hydrogen. To determine the number of peaks present (as doublets, triplets, and so on), we use the n + 1 rule: if a proton has n protons that are three bonds away, it will be split into n+ 1 peaks. (One caveat: do not include protons attached to oxygen or nitrogen.) The magnitude of this splitting, measured in hertz, is called the coupling constantJ.

KEY CONCEPT

The splitting of the peak represents the number of adjacent hydrogens. A peak will be split into n + 1 subpeaks, where n is the number of adjacent hydrogens.

Let’s try a molecule that has even more coupled protons. In 1,1-dibromo-2-chloroethane, shown in Figure 11.6, the Ha nucleus is affected by two nearby Hb nuclei, which together can be in one of four different states: αααββα, or ββ.

Figure 11.6. 1,1-Dibromo-2-chloroethane

Although there are technically four different states, αβ has the same effect as βα, so both of these resonances occur at the same frequency. This means we will have three unique frequencies, αααβ or βα, and ββ. Ha will thus appear as three peaks (a triplet) centered on the true chemical shift, with an area ratio of 1:2:1.

KEY CONCEPT

Proton NMR is good for:

·        Determining the relative number of protons and their relative chemical environments

·        Showing how many adjacent protons there are by splitting patterns

·        Inferring certain functional groups

Now let’s move on to Hb. Because both hydrogens are attached to the same carbon, which can freely rotate, they will be magnetically identical. These hydrogens are three bonds away from one other hydrogen, Ha. This means that they will appear as a doublet. Because there are two of them, the integration for the doublet representing Hb will be larger than the triplet for Ha.

Table 11.2 shows the ratios for up to seven adjacent hydrogens, but it isn’t necessary to memorize this table for the MCAT. Just remember to follow the n + 1 rule for the proton of interest to determine the number of peaks. In addition, peaks that have more than four shifts will generally be referred to generically as a multiplet.

Number of Adjacent Hydrogens

Total Number of Peaks

Area Ratios

0

1

1

1

2

1:1

2

3

1:2:1

3

4

1:3:3:1

4

5

1:4:6:4:1

5

6

1:5:10:10:5:1

6

7

1:6:15:20:15:6:1

7

8

1:7:21:35:35:21:7:1

Table 11.2. Area Ratios for Peaks Split by Adjacent Hydrogens

Table 11.3 indicates the chemical shift ranges of several different types of protons. It is unnecessary to memorize this table, as it is fairly low-yield information. The values that are useful to memorize are the outliers like the deshielded aldehyde at 9 to 10 ppm, and the even more deshielded carboxylic acid between 10.5 and 12 ppm. Another popular peak on the MCAT is the hydrogen of an aromatic ring, which lies between 6.0 and 8.5 ppm. It is also worthwhile to know the general ranges for hydrogens on sp3-hybridized carbons (0.0 to 3.0 ppm; higher if electron-withdrawing groups are present), sp2-hybridized carbons (4.6 to 6.0 ppm), and sp-hybridized carbons (2.0 to 3.0 ppm). When electronegative groups are present, they pull electron density away from the protons. The more electron density that is pulled away from the proton, the more deshielded it will be and the further downfield the proton will appear.

Type of Proton

Approximate Chemical Shift δ (ppm) Downfield from TMS

RCH3

0.9

RCH2

1.25

R3CH

1.5

RC=CH

4.6–6.0

RC≡CH

2.0–3.0

Ar–H

6.0–8.5

RCHX

2.0–4.5

RCHOH/RCHOR

3.4–4.0

RCHO

9.0–10.0

RCOCH3

2.0–2.5

RCHCOOH/RCHCOOR

2.0–2.6

ROH

1.0–5.5

ArOH

4.0–12.0

RCOOH

10.5–12.0

RNH2

1.0–5.0

Table 11.3. Proton Chemical Shift Ranges

MCAT EXPERTISE

On Test Day, just counting the number of peaks and unique hydrogens may be enough to get you the correct answer. (Remember not to count the peak for TMS, though!) If you need to consider shifts, the main ones for Test Day are:

·        Alkyl groups: 0 to 3 ppm

·        Alkynes: 2 to 3 ppm

·        Alkenes: 4.6 to 6 ppm

·        Aromatics: 6 to 8.5 ppm

·        Aldehydes: 9 to 10 ppm

·        Carboxylic acids: 10.5 to 12 ppm

MCAT Concept Check 11.3:

Before you move on, assess your understanding of the material with these questions.

1.    What does nuclear magnetic resonance (NMR) spectroscopy measure? What is NMR spectroscopy generally used for?

2.    What are the units for chemical shift on a standardized NMR spectrum?

3.    What does it mean for a proton to be deshielded? How does this affect its peak in NMR spectroscopy?

4.    What is spin–spin coupling?