Organic Analyses and Lab Techniques - MCAT Biology
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An IR spectrum reading is taken before and after treating acetone with the reducing agent 
. What IR peak readings would be seen for the reactant acetone and for the predicted product?
An IR spectrum reading is taken before and after treating acetone with the reducing agent
Treating acetone, a secondary carbonyl, with a reducing agent, such as sodium borohydride (NaBH4), will yield a secondary alcohol as the product.
When using IR spectroscopy, carbonyl (C=O) groups display characteristic peaks at approximately 1700cm-1, while alcohol groups (O-H) display characteristic peaks around 3300cm-1. The acetone would, therefore, initially have a characteristic peak at roughly 1700cm-1. After the reduction reaction is complete, the resulting 2-propanol would display a characteristic peak roughly at 3300cm-1.
Treating acetone, a secondary carbonyl, with a reducing agent, such as sodium borohydride (NaBH4), will yield a secondary alcohol as the product.
When using IR spectroscopy, carbonyl (C=O) groups display characteristic peaks at approximately 1700cm-1, while alcohol groups (O-H) display characteristic peaks around 3300cm-1. The acetone would, therefore, initially have a characteristic peak at roughly 1700cm-1. After the reduction reaction is complete, the resulting 2-propanol would display a characteristic peak roughly at 3300cm-1.
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Which of the following observations would most likely be seen when performing an H-NMR on 1-ethyl ethanoate (above)?
Which of the following observations would most likely be seen when performing an H-NMR on 1-ethyl ethanoate (above)?
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.
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.
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Imagine the H-NMR spectroscopy of a propane molecule.
How many peaks represent the 2-carbon?
Imagine the H-NMR spectroscopy of a propane molecule.
How many peaks represent the 2-carbon?
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.
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.
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Which of the following statements is false about H-NMR spectroscopy?
Which of the following statements is false about H-NMR spectroscopy?
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.
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.
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Approximately where would a carbonyl peak be found on an IR spectrum?
Approximately where would a carbonyl peak be found on an IR spectrum?
It is important to memorize a couple key functional groups, and where they are located on an IR spectrum. If you see a sharp peak near 1700cm-1, you can assume it is made by a carbonyl group.
Similarly, a wide peak around 3000cm-1 will be made by a hydroxyl group.
It is important to memorize a couple key functional groups, and where they are located on an IR spectrum. If you see a sharp peak near 1700cm-1, you can assume it is made by a carbonyl group.
Similarly, a wide peak around 3000cm-1 will be made by a hydroxyl group.
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Which of the following statements is true concerning infrared spectroscopy?
Which of the following statements is true concerning infrared spectroscopy?
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.
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.
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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?
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?
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.
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.
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All of the following molecules would exhibit two distinct singlets in a 1H-NMR spectrum except __________.
All of the following molecules would exhibit two distinct singlets in a 1H-NMR spectrum except __________.
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.
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.
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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?
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?
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.
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.
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How many hydrogen peaks appear in the H-NMR spectrum of 3-pentanone?
How many hydrogen peaks appear in the H-NMR spectrum of 3-pentanone?
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.
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.
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How many unique peaks would one expect to see on an 1H-NMR reading of the compound shown above?
How many unique peaks would one expect to see on an 1H-NMR reading of the compound shown above?
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.
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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Which of the following functional groups exhibits the highest frequency in an infrared (IR) spectrum?
Which of the following functional groups exhibits the highest frequency in an infrared (IR) spectrum?
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.
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.
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According to HNMR spectroscopy, which of the following molecules would result in a peak at 9.5ppm?
According to HNMR spectroscopy, which of the following molecules would result in a peak at 9.5ppm?
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.
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.
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Which of the following molecules will result in a single signal with the most splits?
Which of the following molecules will result in a single signal with the most splits?
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.
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.
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Which of the following statements is true concerning infrared (IR) spectroscopy?
Which of the following statements is true concerning infrared (IR) spectroscopy?
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.
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.
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The visible spectrum is typically between 390-700nm. Shorter wavelengths are responsible for the purple end of the color spectrum, while red is perceived in the eye from the longer wavelengths. 
-carotene is able to absorb light at a maximum value of 497nm.
Based on this information, what color does the eye perceive 
-carotene to be?
The visible spectrum is typically between 390-700nm. Shorter wavelengths are responsible for the purple end of the color spectrum, while red is perceived in the eye from the longer wavelengths.
Based on this information, what color does the eye perceive
The wavelength of 497nm corresponds to a blue-green color, however 
-carotene absorbs this wavelength. This means it reflects the complementary color on the opposite end of the color spectrum. This gives 
-carotene a red-orange appearance, as only the reflected wavelengths will be returned to the eye.
The wavelength of 497nm corresponds to a blue-green color, however
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An alcohol group in a compound would result in a broad dip around what part of the infrared (IR) spectrum?
An alcohol group in a compound would result in a broad dip around what part of the infrared (IR) spectrum?
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.
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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Determining the molecular ion peak (parent peak) in mass spectroscopy allows you to determine what characteristic of a mystery molecule?
Determining the molecular ion peak (parent peak) in mass spectroscopy allows you to determine what characteristic of a mystery molecule?
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.
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.
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Which of the following most likely represents the H-NMR spectrum of the molecule shown below?
Which of the following most likely represents the H-NMR spectrum of the molecule shown below?
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).
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).
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An unknown compound is analyzed using infrared spectroscopy. A strong, sharp peak is observed at a frequency of 1750cm-1. What functional group is present?
An unknown compound is analyzed using infrared spectroscopy. A strong, sharp peak is observed at a frequency of 1750cm-1. What functional group is present?
An ester has a characteristic IR absorption at about 1750cm-1. A saturated ketone has an absorption at about 1710cm-1, while an unsaturated ketone has an absorption between 1650cm-1 and 1700cm-1. A nitrile has an IR frequency of about 2200cm-1, while an alcohol has a strong, broad peak at about 3400cm-1.
Carbonyl compounds all have peaks between roughly 1650cm-1 and 1750cm-1. Ketone peaks are generally observed at the lower end of this range, while aldehydes and esters are toward the higher end of the range.
An ester has a characteristic IR absorption at about 1750cm-1. A saturated ketone has an absorption at about 1710cm-1, while an unsaturated ketone has an absorption between 1650cm-1 and 1700cm-1. A nitrile has an IR frequency of about 2200cm-1, while an alcohol has a strong, broad peak at about 3400cm-1.
Carbonyl compounds all have peaks between roughly 1650cm-1 and 1750cm-1. Ketone peaks are generally observed at the lower end of this range, while aldehydes and esters are toward the higher end of the range.
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