Spectroscopy - GRE Subject Test: Chemistry
Card 0 of 20
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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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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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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Observe the underlined/bold hydrogen. In HNMR, how many spectral lines will that bolded hydrogen be split into?
Observe the underlined/bold hydrogen. In HNMR, how many spectral lines will that bolded hydrogen be split into?
The answer is:
Within 
NMR spectroscopy there are a couple important factors to understand, including the ppm shift (delta) and the splitting pattern. Here we are focusing on the splitting pattern for individual hydrogens. This is important for it lays the groundwork for understanding the patterns of peaks seen on large compound NMR’s.
When determining the splitting of any hydrogen you must use the n+1 rule. Before going into that rule we must understand two things, 1. Only nonequivalent hydrogens (protons) couple and 2. Usually they only couple to other hydrogens (protons) attached to adjacent carbons. Nonequivalent means the protons occupy their own unique spatial environment with different atoms surrounding them. Typically two hydrogens attached to the same carbon are equivalent (though this isn’t always the case and you must think about where the hydrogens are located in space and see whether they are adjacent to different or similar chemical groups).
The n+1 rule is performed as follows. n stands for the number of equivalent protons that are not equivalent to the proton of interest (the one we are trying to determine the splitting for). We multiply together each group of protons that are nonequivalent to the proton of interest. For example, lets say there are two groups of protons that are nonequivalent to the proton of interest, in group A there are 2 protons, and in group B there are 3 protons. We would use the n+1 rule for group A and get (2+1) = 3 and for group B get (3+1) = 4. We would then multiply these two numbers together to get the splitting for the proton of interest, thus there are 12 lines that the proton of interest would be split into. Proper use of this rule should allow you to get all NMR splitting questions correct (and elevate your understanding of why a certain NMR printout looks the way it does).
The answer is:
Within
When determining the splitting of any hydrogen you must use the n+1 rule. Before going into that rule we must understand two things, 1. Only nonequivalent hydrogens (protons) couple and 2. Usually they only couple to other hydrogens (protons) attached to adjacent carbons. Nonequivalent means the protons occupy their own unique spatial environment with different atoms surrounding them. Typically two hydrogens attached to the same carbon are equivalent (though this isn’t always the case and you must think about where the hydrogens are located in space and see whether they are adjacent to different or similar chemical groups).
The n+1 rule is performed as follows. n stands for the number of equivalent protons that are not equivalent to the proton of interest (the one we are trying to determine the splitting for). We multiply together each group of protons that are nonequivalent to the proton of interest. For example, lets say there are two groups of protons that are nonequivalent to the proton of interest, in group A there are 2 protons, and in group B there are 3 protons. We would use the n+1 rule for group A and get (2+1) = 3 and for group B get (3+1) = 4. We would then multiply these two numbers together to get the splitting for the proton of interest, thus there are 12 lines that the proton of interest would be split into. Proper use of this rule should allow you to get all NMR splitting questions correct (and elevate your understanding of why a certain NMR printout looks the way it does).
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Observe the underlined/bold hydrogen. In HNMR, how many spectral lines will that bolded hydrogen be split into?
Observe the underlined/bold hydrogen. In HNMR, how many spectral lines will that bolded hydrogen be split into?
The answer is:
Within 
NMR spectroscopy there are a couple important factors to understand, including the ppm shift (delta) and the splitting pattern. Here we are focusing on the splitting pattern for individual hydrogens. This is important for it lays the groundwork for understanding the patterns of peaks seen on large compound NMR’s.
When determining the splitting of any hydrogen you must use the n+1 rule. Before going into that rule we must understand two things, 1. only nonequivalent hydrogens (protons) couple and 2. usually they only couple to other hydrogens (protons) attached to adjacent carbons. Nonequivalent means the protons occupy their own unique spatial environment with different atoms surrounding them. Typically two hydrogens attached to the same carbon are equivalent (though this isn’t always the case and you must think about where the hydrogens are located in space and see whether they are adjacent to different or similar chemical groups).
The n+1 rule is performed as follows. N stands for the number of equivalent protons that are not equivalent to the proton of interest (the one we are trying to determine the splitting for). We multiply together each group of protons that are nonequivalent to the proton of interest. For example, lets say there are two groups of protons that are nonequivalent to the proton of interest, in group A there are 2 protons, and in group B there are 3 protons. We would use the n+1 rule for group A and get (2+1) = 3 and for group B get (3+1) = 4. We would then multiply these two numbers together to get the splitting for the proton of interest, thus 3 x 4 = 12 lines that the proton of interest would be split into. Proper use of this rule should allow you to get all NMR splitting questions correct (and elevate your understanding of why a certain NMR printout looks the way it does).
The answer is:
Within
When determining the splitting of any hydrogen you must use the n+1 rule. Before going into that rule we must understand two things, 1. only nonequivalent hydrogens (protons) couple and 2. usually they only couple to other hydrogens (protons) attached to adjacent carbons. Nonequivalent means the protons occupy their own unique spatial environment with different atoms surrounding them. Typically two hydrogens attached to the same carbon are equivalent (though this isn’t always the case and you must think about where the hydrogens are located in space and see whether they are adjacent to different or similar chemical groups).
The n+1 rule is performed as follows. N stands for the number of equivalent protons that are not equivalent to the proton of interest (the one we are trying to determine the splitting for). We multiply together each group of protons that are nonequivalent to the proton of interest. For example, lets say there are two groups of protons that are nonequivalent to the proton of interest, in group A there are 2 protons, and in group B there are 3 protons. We would use the n+1 rule for group A and get (2+1) = 3 and for group B get (3+1) = 4. We would then multiply these two numbers together to get the splitting for the proton of interest, thus 3 x 4 = 12 lines that the proton of interest would be split into. Proper use of this rule should allow you to get all NMR splitting questions correct (and elevate your understanding of why a certain NMR printout looks the way it does).
Compare your answer with the correct one above
Observe the underlined/bold hydrogen. In HNMR, how many spectral lines will that bolded hydrogen be split into?
Observe the underlined/bold hydrogen. In HNMR, how many spectral lines will that bolded hydrogen be split into?
The answer is:
Within 
NMR spectroscopy there are a couple important factors to understand, including the ppm shift (Delta) and the splitting pattern. Here we are focusing on the splitting pattern for individual hydrogens. This is important for it lays the groundwork for understanding the patterns of peaks seen on large compound NMR’s.
When determining the splitting of any hydrogen you must use the n+1 rule. Before going into that rule we must understand two things, 1. only nonequivalent hydrogens (protons) couple and 2. usually they only couple to other hydrogens (protons) attached to adjacent carbons. Nonequivalent means the protons occupy their own unique spatial environment with different atoms surrounding them. Typically two hydrogens attached to the same carbon are equivalent (though this isn’t always the case and you must think about where the hydrogens are located in space and see whether they are adjacent to different or similar chemical groups).
The n+1 rule is performed as follows. N stands for the number of equivalent protons that are not equivalent to the proton of interest (the one we are trying to determine the splitting for). We multiply together each group of protons that are nonequivalent to the proton of interest. For example, lets say there are two groups of protons that are nonequivalent to the proton of interest, in group A there are 2 protons, and in group B there are 3 protons. We would use the n+1 rule for group A and get (2+1) = 3 and for group B get (3+1) = 4. We would then multiply these two numbers together to get the splitting for the proton of interest, thus 3 x 4 = 12 lines that the proton of interest would be split into. Proper use of this rule should allow you to get all NMR splitting questions correct (and elevate your understanding of why a certain NMR printout looks the way it does).
The answer is:
Within
When determining the splitting of any hydrogen you must use the n+1 rule. Before going into that rule we must understand two things, 1. only nonequivalent hydrogens (protons) couple and 2. usually they only couple to other hydrogens (protons) attached to adjacent carbons. Nonequivalent means the protons occupy their own unique spatial environment with different atoms surrounding them. Typically two hydrogens attached to the same carbon are equivalent (though this isn’t always the case and you must think about where the hydrogens are located in space and see whether they are adjacent to different or similar chemical groups).
The n+1 rule is performed as follows. N stands for the number of equivalent protons that are not equivalent to the proton of interest (the one we are trying to determine the splitting for). We multiply together each group of protons that are nonequivalent to the proton of interest. For example, lets say there are two groups of protons that are nonequivalent to the proton of interest, in group A there are 2 protons, and in group B there are 3 protons. We would use the n+1 rule for group A and get (2+1) = 3 and for group B get (3+1) = 4. We would then multiply these two numbers together to get the splitting for the proton of interest, thus 3 x 4 = 12 lines that the proton of interest would be split into. Proper use of this rule should allow you to get all NMR splitting questions correct (and elevate your understanding of why a certain NMR printout looks the way it does).
Compare your answer with the correct one above
Observe the underlined/bold hydrogen. In HNMR, how many spectral lines will that bolded hydrogen be split into?
Observe the underlined/bold hydrogen. In HNMR, how many spectral lines will that bolded hydrogen be split into?
The answer is:
Within 1H NMR spectroscopy there are a couple important factors to understand, including the ppm shift (delta) and the splitting pattern. Here we are focusing on the splitting pattern for individual hydrogens. This is important for it lays the groundwork for understanding the patterns of peaks seen on large compound NMR’s.
When determining the splitting of any hydrogen you must use the n+1 rule. Before going into that rule we must understand two things, 1. only nonequivalent hydrogens (protons) couple and 2. usually they only couple to other hydrogens (protons) attached to adjacent carbons. Nonequivalent means the protons occupy their own unique spatial environment with different atoms surrounding them. Typically two hydrogens attached to the same carbon are equivalent (though this isn’t always the case and you must think about where the hydrogens are located in space and see whether they are adjacent to different or similar chemical groups).
The n+1 rule is performed as follows. N stands for the number of equivalent protons that are not equivalent to the proton of interest (the one we are trying to determine the splitting for). We multiply together each group of protons that are nonequivalent to the proton of interest. For example, lets say there are two groups of protons that are nonequivalent to the proton of interest, in group A there are 2 protons, and in group B there are 3 protons. We would use the n+1 rule for group A and get (2+1) = 3 and for group B get (3+1) = 4. We would then multiply these two numbers together to get the splitting for the proton of interest, thus 3 x 4 = 12 lines that the proton of interest would be split into. Proper use of this rule should allow you to get all NMR splitting questions correct (and elevate your understanding of why a certain NMR printout looks the way it does).
The answer is:
Within 1H NMR spectroscopy there are a couple important factors to understand, including the ppm shift (delta) and the splitting pattern. Here we are focusing on the splitting pattern for individual hydrogens. This is important for it lays the groundwork for understanding the patterns of peaks seen on large compound NMR’s.
When determining the splitting of any hydrogen you must use the n+1 rule. Before going into that rule we must understand two things, 1. only nonequivalent hydrogens (protons) couple and 2. usually they only couple to other hydrogens (protons) attached to adjacent carbons. Nonequivalent means the protons occupy their own unique spatial environment with different atoms surrounding them. Typically two hydrogens attached to the same carbon are equivalent (though this isn’t always the case and you must think about where the hydrogens are located in space and see whether they are adjacent to different or similar chemical groups).
The n+1 rule is performed as follows. N stands for the number of equivalent protons that are not equivalent to the proton of interest (the one we are trying to determine the splitting for). We multiply together each group of protons that are nonequivalent to the proton of interest. For example, lets say there are two groups of protons that are nonequivalent to the proton of interest, in group A there are 2 protons, and in group B there are 3 protons. We would use the n+1 rule for group A and get (2+1) = 3 and for group B get (3+1) = 4. We would then multiply these two numbers together to get the splitting for the proton of interest, thus 3 x 4 = 12 lines that the proton of interest would be split into. Proper use of this rule should allow you to get all NMR splitting questions correct (and elevate your understanding of why a certain NMR printout looks the way it does).
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Which of the following spectroscopic techniques provides the most information about an organic molecule's framework/structure?
Which of the following spectroscopic techniques provides the most information about an organic molecule's framework/structure?
NMR spectroscopy is most useful for determining the type of nuclei (most commonly studied nuclei are 
and 
) present and their relative locations within a molecule. H-NMR is most commonly used because it is practically present in all organic compounds. This technique is useful for a complete determination of the structure of organic compounds. IR spectroscopy is best for determining the functional groups of a molecule, however, it does not give information of the electric environment like NMR. Mass spectrometry is a technique used to determine the amount and mass of substances present in a sample. UV-visible spectroscopy is used to determine the amount of analyte present in a given sample - this method is best for transition metals and/or highly conjugated compounds. Melting point analysis gives information about the purity of a sample, pure substances tend to have higher melting points and more narrow ranges than impure samples.
NMR spectroscopy is most useful for determining the type of nuclei (most commonly studied nuclei are
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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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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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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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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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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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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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In IR spectroscopy, the vibration between atoms is caused by which of the following?
In IR spectroscopy, the vibration between atoms is caused by which of the following?
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.
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.
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After taking an IR spectrum of a sample synthesized in the lab, you have 3 IR peaks. Peak 
has a 
transmittance, peak 
has a 
transmittance, and peak 
has a 
transmittance. Which peak has the greatest absorbance?
After taking an IR spectrum of a sample synthesized in the lab, you have 3 IR peaks. Peak
Transmittance (
) is the fraction of incident light transmitted through an analyte. Absorbance (
) is the amount incident light that is absorbed by the analyte. The equation that governs this relationship is:
Where 
is the power of the incident radiation and 
is the decreased power of the incident radiation due to the interactions between the absorbing analyte particles and the power of the incident radiation. So, as the percent transmittance increases the absorbance decreases. In this case, peak 
has the lowest transmittance, therefore it has the highest absorbance.
Transmittance (
Where
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What is the absorbance of an IR peak with a 25% transmittance?
What is the absorbance of an IR peak with a 25% transmittance?
There are two equations we can use to solve this question:
And
Let's show that each give us the same correct answer:
There are two equations we can use to solve this question:
And
Let's show that each give us the same correct answer:
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What is the absorbance of an IR peak with a 
transmittance?
What is the absorbance of an IR peak with a
Therefore,
Using the equation that relates absorbance to transmittance, we can convert 0.17 transmittance to absorbance:
Therefore the absorbance for this peak is 0.77.
Therefore,
Using the equation that relates absorbance to transmittance, we can convert 0.17 transmittance to absorbance:
Therefore the absorbance for this peak is 0.77.
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What is the absorbance of an IR peak with a 
transmittance?
What is the absorbance of an IR peak with a
Therefore,
Using the equation that relates absorbance to transmittance, we can convert 0.36 transmittance to absorbance:
Therefore the absorbance for this peak is 0.44.
Therefore,
Using the equation that relates absorbance to transmittance, we can convert 0.36 transmittance to absorbance:
Therefore the absorbance for this peak is 0.44.
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