(2) Read each statement carefully and pick the one that is incorrect in its information.

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1 Organic Chemistry - Problem Drill 17: IR and Mass Spectra No. 1 of Which statement about infrared spectroscopy is incorrect? (A) IR spectroscopy is a method of structure determination based on the amount of infrared light absorbed by a compound. (B) The energy of the photon needed to excite the molecule is based upon the masses of the bonded atoms and the bond strength. (C) Similar functional groups have similar atoms and so they will absorb photons of similar energy. (D) The frequencies of the infrared light correspond to stretching and bending between atoms. (E) If absorption of energy is to take place, the vibration must not change the dipole moment of the bond. One can determine the structure of a compound based on the amount of infrared light that is absorbed by the compound. Go back and review the theory behind infrared spectroscopy. The bond strength and the masses of the bonded atoms will determine the energy needed to cause an excitation. Go back and review the theory behind infrared spectroscopy. Since similar functional groups contain similar atoms, they do absorb photons of similar energy. Go back and review the theory behind infrared spectroscopy. The frequency of the photons of infrared light absorbed to correspond to stretching and bending between atoms. Go back and review the theory behind infrared spectroscopy. E. Correct! If absorption of energy is to take place, the vibration must result in a change of the dipole moment of the bond. (1) Recall the fundamentals of IR spectroscopy. IR spectroscopy is a method of structure determination based on the amount of infrared light absorbed by a compound. The energy of the photon needed to excite the molecule is based upon the masses of the bonded atoms and the bond strength. Also, absorption of energy is to take place, the vibration must change the dipole moment of the bond. If the vibration does not change the dipole moment of the bond, then the absorption will not be seen. The frequencies of absorbed radiation correspond to stretching and bending between the atoms. Similar functional groups have similar atoms and so they will absorb photons of similar energy. (2) Read each statement carefully and pick the one that is incorrect in its information. Therefore, the correct answer is (E).

2 No. 2 of Which molecular vibration is depicted below? (A) Asymmetric in-plane bend (B) Symmetric in-plane bend (C) Asymmetric stretch (D) Symmetric stretch (E) Symmetric out-of-plane bend A. Correct! This is an example of an asymmetric in-plane bend. This is not an example of a symmetric in-plane bend. Go back and review the different molecular vibrations seen in IR spectroscopy. This is not an example of an asymmetric stretch. Go back and review the different molecular vibrations seen in IR spectroscopy. This is not an example of a symmetric stretch. Go back and review the different molecular vibrations seen in IR spectroscopy. This is not an example of a symmetric out-of-plane bend.. Go back and review the different molecular vibrations seen in IR spectroscopy. (1) Recall the different vibrations of molecules. Six vibrations were listed but we only had to choose from four of them: Symmetric in-plane bend: Asymmetric stretch: Symmetric stretch: Asymmetric in-plane bend: Symmetric out-of-plane bend: Therefore the correct answer is (A).

3 No. 3 of Calculate the index of hydrogen deficiency for the formula C 10 H 12 O 2. (A) 3 (B) 4 (C) 5 (D) 6 (E) & This is not the correct index of hydrogen deficiency. Go back and review how the index is calculated. This is not the correct index of hydrogen deficiency. Go back and review how the index is calculated. C. Correct! The index of hydrogen deficiency for this formula is 5. This is not the correct index of hydrogen deficiency. Go back and review how the index is calculated. This is not the correct index of hydrogen deficiency. Go back and review how the index is calculated. (1) Recall the equation for calculating the index of hydrogen deficiency: I = C + 1-1/2H + 1/2N (2) Determine C, H, and N: C = number of carbons is molecular formula H = number of hydrogens and halogens N = number of nitrogens and phosphorus In the above example, we were given the formula C 10 H 12 O 2, so: C = 10 H = 12 (3) Calculate the index of hydrogen deficiency: I = ½(12) + ½(0) I = 11-6 I = 5 This means that the molecule contains 5 pi bonds, rings or combination of the two. Therefore, the correct answer is (C).

4 No. 4 of Where would you find the absorption due to the stretch of the nitrogen-hydrogen bond of an amine? (A) cm -1 (B) cm -1 (C) cm -1 (D) cm -1 (E) cm -1 The region from cm -1 is called the fingerprint region and contains many absorptions but not the N-H stretch. Go back and review the IR interpretation trends. The region from cm -1 is where one finds alkenes and aromatic stretches but not the N-H stretch. Go back and review the IR interpretation trends. The region from cm -1 is where one finds carbonyl stretches and aromatic overtones but not the N-H stretch. Go back and review the IR interpretation trends. The region from cm -1 is where one finds alkane, alkene, and C-H stretches, the aldehyde C-H stretch and the O-H stretch from carboxylic acids but not the N-H stretch. Go back and review the IR interpretation trends. E. Correct! The absorption due to the stretching of the N-H bonds of amines is found in the region from cm -1. (1) Recall the absorptions of functional groups from the table given in the tutorial. Once again, you will need to memorize this information as you will be expected to know it on an exam. The nitrogen-hydrogen bond absorbs in the cm -1 range. By looking at this region, you can tell the substitution on the nitrogen of the amine. If there is one hydrogen on the nitrogen, you will see one peak in this region. If there are two hydrogens bound to the nitrogen, you will see two peaks in this region. Therefore, the correct answer is (E).

5 No. 5 of Which group usually has the strongest peak of the IR spectrum? (A) Alkyne (B) Carbonyl (C=O) (C) Carbon-carbon double bond (D) Carbon-nitrogen triple bond (E) C-H of an aldehyde Alkynes usually produce weak absorptions and therefore have small, weak peaks. If the alkyne is symmetrically substituted, it may not produce an absorption at all. Go back and review the IR interpretation trends. B. Correct! If present in the molecule, the carbonyl group typically has the strongest, most noticeable peak in the IR spectrum. The carbon-carbon double bond can produce a fairly strong absorption if appropriately substituted. However, it is not usually the strongest in the spectrum. Go back and review the IR interpretation trends. The carbon-nitrogen triple bond can produce a fairly strong absorption, however, it is not usually the strongest in the spectrum. Go back and review the IR interpretation trends. The carbon-hydrogen bond of aldehydes can produce a sharp but fairly week absorption. Go back and review the IR interpretation trends. (1) Recall the details of IR absorptions in a spectrum. Absorptions are described as either weak, medium, strong, or very strong. The stronger the absorption, the longer the peak appears in the spectrum. If a carbonyl is present, a very strong absorption occurs in the range of cm -1 which is almost in the center of the spectrum. Very little else absorbs in this area so when such an absorption is present, it is very hard to miss. The carbonyl peak is usually the strongest in the spectrum. Therefore, the correct answer is (B).

6 No. 6 of When approaching an IR spectrum for the first time, what is the first step you do in working to solve the structure that produces it? (A) Calculate the index of hydrogen deficiency. (B) Scan the spectrum and try to pick out obvious absorptions. (C) Think of possible structures from the formula you were given. (D) Consult a frequency table and try to match up absorptions with peaks in the spectrum. (E) None of the above. A. Correct! By calculating the index of hydrogen deficiency first, you get an idea of what to specifically look for in the IR spectrum. Scanning for obvious absorptions would not be the best way to approach an IR spectrum to determine the structure of a compound. Review the steps for solving IR spectra. Thinking of possible structures from the given formula would not be the best way to approach an IR spectrum. Review the steps for solving IR spectra. Consulting a frequency table and trying to match absorptions would not be the best way to approach an IR spectrum. Review the steps for solving IR spectra. The correct answer is listed in the above choices. (1) Recall the steps for approaching an IR spectrum to determine a molecule s structure. Step 1: Calculate the index of hydrogen deficiency (sometimes called the degrees of freedom in organic textbooks). Step 2: Using the formula and the index of hydrogen deficiency, predict possible functional groups that may be present in the molecule. Step 3: Look for the frequencies for those functional groups in the table of frequencies. Step 4: Confirm those functional groups are present in the IR spectrum, and then draw the structure. Anytime you are asked to determine the structure of a molecule, your first step should be to calculate the index of hydrogen deficiency. Therefore, the correct answer is (A).

7 No. 7 of What is mass spectrometry used for? (A) Determining the molecular formula and structure of an organic compound. (B) Determining the available electronic transitions of an organic compound. (C) Determining the electronegativities of the atoms comprising the molecule. (D) Determining the empirical formulas of compounds. (E) Determining the stereochemistry present in a compound. A. Correct! Mass spectrometry is used to determine the molecular formula and structure of an organic compound. Mass spectrometry is not used to determine the available electronic transitions of an organic compound. Go back and review the theory behind mass spectrometry. Mass spectrometry is not used to determine the electronegativities of the atoms comprising an organic compound. Go back and review the theory behind mass spectrometry. Mass spectrometry is not used to determine the empirical formula of an organic compound. Go back and review the theory behind mass spectrometry. Mass spectrometry does not typically give any information on the chirality of a compound. (1) Recall the theory behind mass spectrometry. Mass spectrometry is an analytical tool for identifying unknown compounds. It is useful for determining the molecular formula and the molecular structure of both large and small organic molecules. The unknown samples are bombarded with a stream of electrons which ionize the molecules. The ionized molecules are unstable and fragment. Based on the fragmentations, one can reconstruct the molecular structure of the molecules. (2) Read each statement carefully and choose the one that best describes the purpose of mass spectrometry. Therefore, the correct answer is (A).

8 No. 8 of What type of species does the mass spectrometer detect? (A) Carbanion (B) Neutral radical (C) Carbocation (D) Neutral compound (E) All of the above. Carbanions are not detected in mass spectrometry. Go back and review the details of mass spectrometry. Neutral radicals are not detected in mass spectrometry. Go back and review the details of mass spectrometry. C. Correct! Carbocations are detected in mass spectrometry. Neutral compounds are not detected in mass spectrometry. Go back and review the details of mass spectrometry. The mass spectrometer can detect only one kind of species. Go back and review the details of mass spectrometry. (1) Recall the theory behind mass spectrometry. When an electron stream is used to ionize a sample in mass spectrometry, an electron is kicked out from the original molecule forming a radical cation. This radical cation is called the molecular ion or parent peak usually begins to fragment into different species. In some cases, a compound can fragment into other cations, anions or radicals. However, the detector on the mass spectrometry only sees cationic species. Therefore, cations give rise to the peaks in a mass spectrum. Therefore, the correct answer is (C).

9 No. 9 of What is indicated when in a mass spectrum there is a M+2 peak of the same approximate size as the molecular ion peak? (A) There are two isotopes of chlorine present in the sample. (B) The sample is impure two compounds are present with a difference of two in their molecular weights. (C) There are two isotopes of bromine present in the sample. (D) The isotope 13 C is present in the molecule. (E) Nitrogen is present. If chlorine is present in a sample, the M+2 peak will be a third of the size of the molecular ion peak. Go back and review the information on interpreting mass spectra. An impure sample placed on a mass spectrometer would not yield the results described above. Go back and review the information on interpreting mass spectra. C. Correct! Bromine has two major isotopes found in nature in an almost 1:1 ration: 79 Br and 81 Br. If these are both present in a sample, it gives rise to a M+2 peak of similar intensity to the molecular ion peak. If 13 C is present in a sample, there is a M+1 peak not a M+2 peak. Go back and review the information on interpreting mass spectra. When nitrogen is present in a sample, its molecular weight will be an odd number. (1) Recall the theory behind mass spectrometry. The presence of isotopes is evident in a mass spectrum. If 13 C is present in a molecule, a M+1 peak will be seen in the spectrum. The relative abundance of the peak is low (i.e. very short peak0 because 13 C makes up about 1% of the total amount of carbon. The other two commonly seen isotopes in mass spectrums are chlorine and bromine. Chlorine s two most abundant isotopes are 35 Cl and 37 Cl. They are both present in nature with 35 Cl making up approximately 75% of the total amount of chlorine. When present in a mass spectrum, one sees the molecular ion (this molecule corresponds to one containing the 35 Cl isotope) and a M+2 peak which corresponds to the molecule that contains the 37 Cl. The molecular ion peak is approximately 3 times bigger (since 35 Cl s relative abundance is approximately 3 times more) than the M+2 peak. Bromine s two most abundant isotopes are 79 Br and 81 Br. They are both present in nature with 79 Br making up approximately 50.6% of the total amount of bromine. When present in a mass spectrum, one sees the molecular ion (this molecule corresponds to one containing the 79 Br isotope) and a M+2 peak which corresponds to the molecule that contains the 81 Br. The molecular ion peak and the M+2 peak are approximately the same size with the molecular ion peak being slightly higher than the M+2 peak. The presence of nitrogen only results in the molecular weight being an odd number. Therefore, the correct answer is (C).

10 No. 10 of What product(s) would you expect this ketone to yield after undergoing the McLafferty rearrangement? O (A) An alkene and an enol (B) An alkyl fragment and a carbonyl fragment (C) An alkyl fragment and a molecule of carbon monoxide (D) An alkyne and a ketone (E) An alcohol and a ketone A. Correct! This ketone will yield an alkene and an enol fragment after undergoing the McLafferty rearrangement. An alkyl fragment and a carbonyl fragment will not be obtained after a McLafferty rearrangement of a ketone. Go back and review the different types of fragmentations functional groups undergo. An alkyl fragment and a molecule of carbon dioxide will not be obtained after a McLafferty rearrangement of a ketone. Go back and review the different types of fragmentations functional groups undergo. An alkyne and a ketone will not be obtained after a McLafferty rearrangement of a ketone. Go back and review the different types of fragmentations functional groups undergo. An alcohol and a ketone will not be obtained after a McLafferty rearrangement of a ketone. Go back and review the different types of fragmentations functional groups undergo. (1) Recall the McLafferty rearrangement. Only carbonyl compounds can undergo the McLafferty rearrangement. The one above could be expected to rearrange in this manner since it is a ketone. The products of a McLafferty rearrangement are an alkene and an enol: H O OH And though drawn as the enol bearing the cationic radical, in reality either piece could be the cation radical. Therefore, the correct answer is (A).