MCAT Biology : Organic Analyses and Lab Techniques

Study concepts, example questions & explanations for MCAT Biology

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Example Questions

Example Question #2 : How To Identify Compounds

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

Possible Answers:

5

10

6

2

4

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 #2 : How To Identify Compounds

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

Possible Answers:

Propanone

2-butanol

1,1-dibromoethane

Acetaldehyde

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 #31 : Organic Analyses And Lab Techniques

Q2

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

Possible Answers:

Four

Ten

Two

Six

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.  

Example Question #1 : Mass Spectroscopy

Determining the molecular ion peak (parent peak) in mass spectroscopy allows you to determine what characteristic of a mystery molecule?

Possible Answers:

Molecular charge

Nuclear charge

Molecular weight

Functional groups

Correct answer:

Molecular weight

Explanation:

The molecular ion peak is determined using mass spectrometry. The parent peak is formed when a mystery molecule does not fragment, and simply loses an electron. This means that the mass to charge ratio of this peak will allow us to determine the molecular weight of the compound.

Example Question #81 : Organic Chemistry, Biochemistry, And Metabolism

Which of the following statements is true concerning infrared spectroscopy?

Possible Answers:

IR spectroscopy is useful in determining the size and shape of a compound's carbon skeleton.

Functional groups can be identified by looking in the fingerprint region of the spectrum.

When the infrared light frequency matches the frequency of bond vibration in a molecule, a peak is recorded on the spectrum. 

An IR spectrometer shines infrared light on a compound and records the positions where the light is blocked by the compound. This results in the spectrum's peaks.

Correct answer:

When the infrared light frequency matches the frequency of bond vibration in a molecule, a peak is recorded on the spectrum. 

Explanation:

IR spectroscopy allows you to identify what functional groups are present in a compound. The IR spectrum is created by recording the frequencies at which a polar bond's vibration frequency is equal to the infrared light's frequency.

The fingerprint region is separate from the function group region, and generally corresponds to carbon-carbon or carbon-hydrogen interactions. While the spectrum can show what groups are present in a compound, it cannot be used to find the position of these groups or provide a carbon skeleton.

Example Question #1 : Ir Spectroscopy

Which of the following statements is true concerning infrared (IR) spectroscopy?

Possible Answers:

The fingerprint region is most easily used to determine the functional groups in the molecule

IR spectroscopy is used to determine the shape of the carbon backbone

IR spectroscopy can be used to easily determine molecular mass

IR spectroscopy is used to determine the frequency of vibrations between atoms

Correct answer:

IR spectroscopy is used to determine the frequency of vibrations between atoms

Explanation:

IR spectroscopy is most commonly used to determine the functional groups found in the molecule being observed. This is done by observing the vibration frequencies between atoms in the molecule. It does not easily reveal the size or shape of the molecule's carbon skeleton. Although the fingerprint region is unique for every molecule, it is very difficult to read when attempting to determine the molecule's functional groups. Most functional group peaks are observed in the functional group region adjacent to the fingerprint region.

Example Question #32 : Organic Analyses And Lab Techniques

In IR spectroscopy, the vibration between atoms is caused by which of the following?

Possible Answers:

The number of protons in a nucleus

The movement of electrons to higher energy levels

Dipole moments between atoms

The overall molecular weight of the molecule

Correct answer:

Dipole moments between atoms

Explanation:

Infrared (IR) spectroscopy takes advantage of the electrical difference between atoms in a polar bond. These dipole moments, when exposed to infrared radiation, stretch and contract in what appears to be a vibrating motion between the atoms. The different vibrational frequencies in the molecule allow for the compound to be "read" using IR spectroscopy.

Example Question #11 : Compound Identification

Ultraviolet spectroscopy is used to detect conjugate double bonds in a compound. The longer the chain of conjugated double bonds in a compound, the longer the absorbed wavelength of UV light.

Which of the following compounds would result in the longest absorbed wavelength?

Possible Answers:

Butadiene

Ethene

1,3-dimethylhexatriene

1,4-pentadiene

Correct answer:

1,3-dimethylhexatriene

Explanation:

When predicting the absorbed wavelength, a general rule of thumb is that butadiene will absorb around  217nm. For each additional conjugated double bond, you add 30-40nm. In addition, an alkyl group will add approximately 5nm.

Since 1,3-dimethylhexatriene has three consecutive conjugated double bonds, as well as two alkyl groups attached to the conjugate system, it will have the longest absorbed wavelength.

Example Question #1 : Ir Spectroscopy

Which of the following functional groups exhibits the highest frequency in an infrared (IR) spectrum?

Possible Answers:

Ketone

Ester

Aldehyde

Nitrile

Alcohol

Correct answer:

Alcohol

Explanation:

An alcohol (-ROH) exhibits a strong, broad absorbance peak at about 3500cm-1. A nitrile's (-RCN) characteristic absorbance peak is at about 2200cm-1. Carbonyl groups have strong, sharp peaks from 1700cm-1 to 1750cm-1, depending on the type of carbonyl group. For instance, an ester (-RCO2R'-) has an absorbance at about 1750cm-1, while a ketone (-ROR'-) has an absorbance at around 1710cm-1.

Example Question #61 : Analytical Chemistry

An alcohol group in a compound would result in a broad dip around what part of the infrared (IR) spectrum?

Possible Answers:

1200cm-1

1700cm-1

2800cm-1

3400cm-1

Correct answer:

3400cm-1

Explanation:

There are a couple of key functional group spectra that you must memorize. A carbonyl group will cause a sharp dip at about 1700cm-1, and an alcohol group will cause a broad dip around 3400cm-1.

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