MCAT Biology : Compound Identification

Study concepts, example questions & explanations for MCAT Biology

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Example Questions

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Example Question #1 : Compound Identification

Imagine the H-NMR spectroscopy of a propane molecule. 

How many peaks represent the 2-carbon?

Possible Answers:

7

3

6

2

Correct answer:

7

Explanation:

In order to determine how many peaks will be associated with the hydrogens of this carbon, you need to determine how many neighboring hydrogens surround the central carbon. Both of the terminal carbons have three hydrogens, so there are six total hydrogens neighboring the central carbon.

Since the number of peaks is given by the number of neighboring hydrogens plus one, there will be seven peaks on the spectrum for the 2-carbon. This is known as a septet.

Example Question #2 : Compound Identification

Synthetic testosterone is typically synthesized in yams and then used by athletes to boost their physical performance across various sports. This practice has been deemed illegal by most major sports authorities. Testing for synthetic testosterone use is accomplished by comparing the chemical composition of synthetic testosterone to natural testosterone.

Which method would be the most useful in identifying the chemical composition differences between natural and synthetic testosterone? 

Possible Answers:

Magnetic resonance imaging 

Mass spectrometry 

Chromatography 

Radio isotope transferance labeling

Correct answer:

Mass spectrometry 

Explanation:

Mass spectrometry is used to identify the chemical composition of samples and, therefore, is the best choice to look at the differences between natural and synthetic testosterone.

Example Question #3 : Compound Identification

All of the following molecules would exhibit two distinct singlets in a 1H-NMR spectrum except __________.

Possible Answers:

1,2,4,5-tetramethylbenzene

1,3,5-trimethylbenzene

1,4-dimethylbenzene

2,4-hexadiyne

methyl-tert-butyl ether

Correct answer:

2,4-hexadiyne

Explanation:

2,4-hexadiyne has only one 1H-NMR signal, as the two terminal methyl groups are identical and will have the same chemical shift.

1,2,4,5-tetramethylbenzene has two singlets: one for the four methyl groups and one for the two aromatic protons. Likewise, 1,3,5-trimethylbenzene will have two singlets: one for the three methyl groups (nine hydrogens total) and one for the three aromatic protons, which are all identical.

Methyl tert-butyl ether also has two singlets, one corresponding to the tert-butyl methyl protons, and one corresponding to the methoxy protons.

Finally, 1,4-dimethylbenzene has two singlets, one for the methyl groups, and one for the four aromatic protons, which are all identical.

Example Question #1 : Compound Identification

Which of the following molecules will result in a single signal with the most splits?

Possible Answers:

Ethyl acetate

2-bromopropane

1-bromopropane

Methane

Correct answer:

2-bromopropane

Explanation:

The number of splits that a peak will experience is dependent on the number of neighboring hydrogens that are not chemically equal to the hydrogen in question. In 2-bromopropane, the hydrogen on the middle carbon is attached to two methyl groups, meaning that there are six neighboring hydrogens. This results in a peak that is split into seven peaks.

Methane only has one peak, and does not split. 1-bromopropane has a peak that is split into six peaks, and ethylacetate has a peak that is split into four peaks.

Example Question #5 : Compound Identification

Which of the following most likely represents the H-NMR spectrum of the molecule shown below?

Mcat7

Possible Answers:

One quartet, one triplet, one doublet, and one singlet

One quartet, one triplet, and three doublets

One quartet, two triplets, one doublet, and one singlet

One quartet, one triplet, two doublets, and one singlet

Two triplets, two doublets, and one singlet

Correct answer:

One quartet, one triplet, two doublets, and one singlet

Explanation:

There are four total aromatic protons, consistent with two sets of identical pairs. This would result in two distinct aromatic signals, each having a doublet and each integrating two protons.

The methyl protons next to the ketone would be deshielded by the electron withdrawing ketone group, resulting in a downfield shift. The signal would be a singlet, since there are no neighboring protons to the methyl group.

Finally, the ethyl group would have two signals, one for the two protons next to the aromatic ring (shifted downfield because of the aromatic ring), and one highly shielded peak corresponding to the terminal protons. The protons next to the aromatic ring will result in a quartet from the three neighboring hydrogens, while the terminal peak will be a triplet from the two neighboring hydrogens.

The final result is one quartet (ethyl), one triplet (ethyl-terminal), two doublets (aromatic), and one singlet (methyl).

Example Question #6 : Compound Identification

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Which of the following observations would most likely be seen when performing an H-NMR on 1-ethyl ethanoate (above)?

Possible Answers:

One singlet and one quartet

One singlet and two triplets

One doublet and one triplet

A singlet, a doublet, and a triplet

A singlet, a triplet, and a quartet

Correct answer:

A singlet, a triplet, and a quartet

Explanation:

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Looking at the structure above, we can see that the molecule only contains three carbons bonded to protons. These carbons are labeled 1, 2 and 3.

An important concept in NMR questions is determining if two carbons on the same compound will have protons split identically, and thus indistinguishable in an NMR (i.e. will those two carbons represent two individual peaks or one large peak?). In this case, C1 and C3 are clearly distinguishable from C2, since C1 and C3 are bonded to 3 hydrogens, while C2 is only bonded to two. Because C2 is adjacent to a three proton carbon, we know that the splitting pattern will display at least one quartet. This will narrow our answer choices down to two options.

Because C1 and C3 contain the same number of protons, we need to determine if they will represent one large peak, or two separate peaks. Looking at the compound, we can see that C3 is adjacent to a two-proton carbon in C2, while C1 is not adjacent to any proton-bonded carbons; therefore, we can expect that C1 will not be split by any protons, and will display a singlet, and C3 will be split by 2 protons, and will display a triplet.

As a final result, we would expect to see one singlet, one triplet and one quartet.

Example Question #7 : Compound Identification

Which of the following statements is false about H-NMR spectroscopy?

Possible Answers:

Electron donating groups cause better shielding of the protons

Aldehydes and carboxylic acids will result in large peak shifts to the left of the spectrum

Peak splitting is caused by chemically equivalent hydrogens

A molecule of 3-pentanol will have four peaks

Electron withdrawing groups near chemically equivalent hydrogens will shift their peaks toward the left on the spectrum

Correct answer:

Peak splitting is caused by chemically equivalent hydrogens

Explanation:

Peak splitting is not caused by equivalent hydrogens, but rather neighboring hydrogens that are not chemically equivalent. In order to determine the number of peaks, we simply add one to the number of neighboring, nonequivalent hydrogens.

Peak shifts are caused by electron withdrawing groups, which will deshield the nucleus and shift the peak to the left. Electron donating groups stabilize the position of a peak by shielding the nucleus. 3-penanol will have four peaks due to its symmetry: one peak for the terminal methyl groups, one peak for the intermediate -CH2 groups, one peak for the -CH on the third carbon, and one peak for the hydroxy hydrogen.

Example Question #3 : How To Identify Compounds

How many hydrogen peaks appear in the H-NMR spectrum of 3-pentanone?

Possible Answers:

2

4

5

6

10

Correct answer:

2

Explanation:

3-pentanone contains ten hydrogens in total; however, 3-pentanone is a symmetric compound. The four hydrogens on the carbons next to the ketone have the same spin, and the six hydrogens on the methyl carbons have the same spin. The correct answer is two hydrogen peaks.

Example Question #9 : Compound Identification

According to HNMR spectroscopy, which of the following molecules would result in a peak at 9.5ppm?

Possible Answers:

1,1-dibromoethane

Acetaldehyde

Propanone

2-butanol

Correct answer:

Acetaldehyde

Explanation:

When dealing with peaks in NMR spectroscopy, remember that withdrawing groups on a molecule will push the proton signal farther to the left, or more downfield. Aldehydes have a distinctive peak at 9.5 ppm due to the effect of the oxygen atom in close proximity to the hydrogen.

Example Question #5 : Organic Chemistry

Q2

How many unique peaks would one expect to see on an 1H-NMR reading of the compound shown above?

Possible Answers:

Six

Two

Ten

Four

Correct answer:

Two

Explanation:

The molecule shown is completely symmetrical. This means that the hydrogens adjacent to the two carbons on the left of the ketone and the hydrogens adjacent to the carbons on the right of the ketone will have identical splitting patterns.  

Let's focus on the right side. The farthest carbon has three hydrogens that will be split by two adjacent hydrogens. The carbon between the terminal methyl and the ketone has two hydrogens, split by three. On each side we will have two 3-hydrogen triplets and two 2-hydrogen quartets, totaling two unique and distinctive peaks composed of six and four hydrogens, respectively.  

As an aside, in NMR readings, if the number of protons of each peak has a common denominator, it can likely be simplified. For example, a reading of this NMR might be reduced from a 6-H peak and 4-H peak, to a 3-H and 2-H peak, respectively. Do not get confused if the number of hydrogens in the reading does not match up to the number of hydrogens in the molecule; it just means it was most likely simplified.  

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