finalreview07 spectroscopy

finalreview07 spectroscopy - Infrared (IR) and-nuclear...

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Unformatted text preview: Infrared (IR) and-nuclear magnetic resonance (NMR) spectroscopy, and mass spectrometry, are important tools for determining the structures of unknown compounds. These techniques are also essential for confirming that the products of chemical reactions actually have the structures anticipated. Each spectral method is independently useful with simple molecules; but when used together, these methods make it possible to determine the structures of very complicated molecules. Massspectrometry must be used when the molecular formula is not known, because neither IR nor NMR spectroscopy- dan provide this information. 7 _ _ ‘ Infrared radiation is absorbed by molecules, causing them to move to excited vibrational states. Different types of bonds, such as 'C=C, 0—H or C=O, have different vibrational frequencies and, therefore, absorb IR radiation at different, and characteristic, frequencies (or wavelengths). Thus an IR spectrum is useful for determining whether various functional groups are present or absent in a molecule. IR spectra are usually plots of % transmission against wavelength. (in micrometers, am) or wavenumberj ,(the reciprocal of wavelength, which is proportional to frequency, and is expressed in cm“). Wavenumber (1/9») and frequency (all) differ by a constant factor, the speed of light. Infrared Spectroscopy Remembering a few key absorption frequencies and the molecular vibrations responsible for them is very useful when deciphering IR spectra. Table 1 provides a concise summary of important absorption frequencies. Table 1. Approximate Infrared Absorption Frequencies -1 ‘ flitglggr Magnetic Resgngngg Spectroscopy ‘ Some nuclei have magnetic moments and, in a strong magnetic field, absorb electromagnetic radiation in the radio—frequency range, giving rise to an NMR spectrum. An IR spectrum of a compound gives a picture of what functional groups are present in a molecule, but gives very little information about the number and type of hydrogen and carbon atoms in the molecule. 1H and 13C NMR spectroscopy prdvides this information. For 1H NMR spectra, integration of the areas under signals in a spectrum gives the relative number of equivalent protons producing each signal, and the n+1 rule for spin—spin splitting of signals allows one to determine the number of protons on atoms adjacent to the proton producing the signal. In addition to being familiar with these two tools, knowing the approximate chemical shifts for six types of protons is extremely useful. Table 2. 1H NMR Chemical Shifts H attached to 5 S . 1 :Eiiiiiiiii ’ Spectroscopy “C NMR spectra are normally recorded as proton-decoupled spectra and the signals consist of a series of Single lines. Thus, the number of lines in the spectrum corresponds to the number of the different types of carbon atoms in the molecule. Several techniques are available to determine how many protons are attached to each type of carbon atom, greatly increasing the usefulness of 13C NMR spectra. As with 1H NMR, knowing the approximate chemical shifts for six types of carbon atoms is extremely Useful. Table 3. 13C. NMR Chemical Shifts C environment 5 ( ) Wear When molecules are ionized in the gas phase and the resulting ions passed through a magnetic or electric field, the ions are deflected by the field. Measuring the deflections yields a mass spectrum, in which the ions are arranged according to their mass-to-charge (m/z) ratio. Mass spectrometry deals with the generation and analysis of mass spectra. The usual way to ionize a molecule is to bombard it with a beam of electrons, which dislodge an electron, producing a molecular ion. The molecular ion has the same molecular mass as the molecule. If a high resolution mass spectrometer'is used, the exact molecular formula for the molecule can be obtained from the mass spectrum. Molecular ions often fragment into smaller ions and neutral components (radicals or small neutral molecules). The fragment ions may themselves undergo further fragmentation. Fragmentation patterns provide clues as to the structure of the molecule. Abundances of fragment ions reflect carbocation and radical stabilities. What effect does the conjugation of a carbonyl group with a carbon—carbon double bond have on the infrared absorption due to the C=O stretch? (A) It shifts to a lower frequency (longer wavelength). (B) It shifts to a higher frequency (shorter wavelength). (C) It has no effect. (D) The absorption due to the C=O stretch disappears. ............................... nun"..."nu-"nun"...nun-unu-nun-u.nun-un...u.-nu.-nun..."nun-u...nunu.u..."nu..."u..."nu..."-u.-u".u...u.-nun-nun"...-nun-uuunuun Knowledge Required: ( 1) Effect of resonance (in the nature of double bonds. (2) Relationship between frequency, or wavelength, and energy of electromagnetic radiation. (3) Recognition that the frequency of the absorbed infrared radiation corresponds to a vibrational frequency in the molecule. Thinking it Through: The resonance interaction between the C=C and the (3:0 groups introduces some single bond character into the carbonyl group as shown by the resonance structures below. 0) 0 _ + KW II | R2C=CH—C-R HRZC—CHr-C—R Less energy is required to stretch a carbon—oxygen single bond than a carbon—oxygen double bond. Consequently, the reduced double bond character of a conjugated carbonyl group will result in a lower stretching frequency than for a non-conjugated carbonyl group, and it will absorb lower frequency infrared radiation. Choice (A) is a statement of this fact and is the correct answer. Note that the stretching frequency of the carbon—carbon double bond would also be lowered. Etlttttéiltiiitltlltiiéf‘.Tit Spectroscopy Whichcompound is consistent with this IR spectrum? anglh Wave! 5 o mmnmntmmmm 4600 4000 3500 3000 2500 2000 1500 1000 800 600 ' Wavenumber. (cm-1) (A) 9H . '(B) ,9 cngcnzcngcncn3 CH3CH2CH2CCH3 (C) 9 ’ (n) (3 CH3CH2CH2CNH2 CH3CH2CHZCOH. ....................................................................................................................................................................................................................................... u Knowledge Required: (1) The approximate absorption frequencies for OH, NH2 and carbonyl groups. (2) Recognition that the absence of characteristic IR absorption bands for various functional'groups allows molecules containing those groups to be ruled out. Thinking it Through: The spectrum shows a strong absorption band at approximately 1720 cm”‘. Thus, the molecule contains a carbonyl group of some sort, and choice (A) may be ruled out. Choices (B), (C) and (D) all contain a carbonyl group, but they may be distinguished because choices (C) and (D) also contain other- functional groups that have other distinctive absorption bands. The strong and broad band covering the 2500—3300 cm“ region that is typical for the OH of a carboxylic acid is clearly absent. Therefore, choice (D) cannot be correct. The NH2 group of the amide in choice (C) would show two bands of medium strength near 3210001117", and in addition, the carbonyl band for amides normally comes below 1700 cm"1 (near 1650 cm“). The spectrum is clearly not that - of choice (C). Choice (B), a ketone, must be the correct answer. _ ' ' S P - 3 . Signals from how many sets of protons would be observed in the 1H NMR spectrum for this compound? ...................................................................................................................................................................................................................................... .. Knowledge Required: (1) That chemically equivalent protons-are generally magnetically equivalent. (2) That symmetry causes sets of protons to be chemically equivalent._ Thinking it Through: To give rise to different NMR signals, sets of protons must be magnetically (chemically) n'onequivalent. The simplest test for chemical nonequivalence is to substitute some other atom for a proton in each set. If the substitution results in the same compound, the sets are chemically equivalent. Otherwise they are chemically nonequivalent. The two CH3 groups and the two CH2 groups in the two ethyl groups are chemically nonequivalent, and each one gives rise to a distinct signal (for a total of four signals). The Symmetry in the molecule makes the protons that are ortho to the ethoicy group chemically equivalent, but nonequivalent to the likewise chemically equivalent protons ortho to the ethyl group. Thus, the protons on the benzene ring give rise to two distinct signals, and these two signals plus the four from the two ethyl groups gives six distinct signals for the NMR spectrum. Therefore, choice (D) is the correct answer. Choice (C) is often selected because of the incorrect assumption that the two CH3 group will be equivalent because they are both attached to a carbon atom. Likewise, choice B would be selected if one erroneousl assumed that both eth l rou s were e uivalent. Spectroscopy Which structure is consistent with this 1H NMR spectrum? (A) CHgCHZOCH2CHC12 (B) 9 ClCHZCH2CCH2CH3 (C) ('3' (D) (I)I CICHZCHZCOCHQCH3 ClCHgCHZOCCHQCHg Knowledge Required: (1) Spin—spin splitting patterns. (2) Factors that affect chemical shifts. (3) Relative chemical shifts. Thinking it Through: The best way to approach this question is to compare the expected 1H NMR spectrum for each choice with the given spectrum. The given spectrum shows a quartet, two triplets of equal intensity, and a more intense triplet in that order from lowest to highest field. Only the spectra of choices (B), (C) and (D) would show the same type and number of splitting patterns (mutiplets) as are seen in the given spectrum. The spectrum of choice (A) would show two triplets of different intensity, a doublet. and a quartet. It is clearly incorrect. Although the spectra of choices (A), (B), and (D) have the same number and type of multiplets, the order of the signals from lowest to highest field will differ. Only in choice (C) is the methylene of the ethyl group attached to a heteroatom so that the quartet would come at lowest field. The spectra of choices (B) and (D) would both have the two equally intense triplets at lowest field followed by the quartet and then the more intense triplet. Therefore, choice (C) appears to be the correct answer. Indeed, the spectrum of choice (C) would show the correct order of the u uartet, two tri lets of eual intensit , and then the more intense trilet from lowest to hi hest field. S P - 5. How many signals would you expect in the proton- decoupled l3C NMR spectrum for this compound? (A) (B) (C) Knowledge Required: (1) Ability to discern to whether or not various carbon atoms are chemically equivalent. (2) That symmetry causes carbon atoms to be chemically equivalent. Thinking it Through: In a proton-decoupled 13C NMR spectrum, each different type of carbon atom will give rise to a singlet signal. As a result, the number of signals is equal to the number of different types of carbon atoms in the molecule. A plane of symmetry passing through the two substituted carbon atoms in the benzene ring causes the two carbon atoms ortho to the acetyl group to be equivalent. This is also true for the two carbon atoms ortho to the methyl group. There are a total of seven different types of carbon atoms in the molecule, and choice C) is the correct answer. 144 —_____—_._._—_SI£MBL SP-6. Which structure is consistent with this 1H NMR spectrum? (A) 9 (B) 9 9 CH30CCH2CH3 I CH3CH20CCH2COCH2CH3 (c) (,3, 9 (D) 9 CH3CCH2CO CH2CH3 CH3CCH2CH3 Knowledge Required: (1) Spin—spin splitting patterns. (2) Factors that affect chemical shifts. (3) Relative chemical shifts. Thinking it Through: As was done previously, compare the expected spectrum for each choice with the given spectrum. The given spectrum shows three different signals: a quartet, a singlet, and a triplet in that order from lowest to highest field. The spectra of choices (A), (B), and (D) will all show the same type and number of multiplets as; the given spectrum. However, the spectrum of choice (C) would show four different signals: a quartet, two singlets, and a triplet in that order from lowest to highest field. Consequently, choice (C)may be eliminated. The spectra for choices (B) and (B) would show the same ordering of the multiplets as the given spectrum, but the spectrum for choice (A) would have the singlet (CH3 attached to an oxygen atom) at lower field than the quartet. Thus, choice (A) cannot be correct. Although the spectra of choices (B) and (D) exhibit the same ordering of the signals,-the actual chemical shifts for the quartets are significantly different; The methylenes of the ethyl groups in choice (B) are attached to an electronegative oxygen atom, which causes the quartet for the methylene protons toappear near 5 4.0. But, the methylene of the ethyl group in choice (D) is attached to a carbonyl carbon atom, which causes the quartet for the methylene protons to appear near 6 2.0. The given spectrum is in better agreement with choice (B) than with choice (D), and choice (B) is the correct answer. Note that the singlets for choices (B) and (D) would also come at different chemical shifts. The singlet for the methylene proton between the'two carbonyl groups in choice (B) would be expected to appear near 6 3.0, whereas the singlet for the meth l rou attached to the carbon 1 carbon atom would be ex ected to a n ear 'ust above the c uartet near 3 2.0. SP_-‘7 . Which compound would show a molecular ion at m/z 114 and a major fragment ion. at m/z 71'? (A) (9H (B) ' 9 CHgCHgCHgCHCHgCHZCHg CH3CH2CH2CCH2CH3 (C) O (D) Q II II ' .. CH3CH2CH2CH2CCHZCH3 CH3CH2CH2CCH2CH2CH3 Spectroscopy __________________-__—.—————— Knowledge Required: (1) That the mlz for the molecular ion is the same as the molecular mass. (2) Factors that affect carbocation and radical stabilities. (3) That carbocation and radical stabilities determine which fragmentation processes are most likely to occur. Thinking it Through: Choices (C) and (D) both have niolecular masses equal to 114, but choice (A) has a molecular mass equal to 116 and choice (B) has a molecular mass equal to 100. Thus choices (A) and (B) cannot give rise to a molecular ion at m/z 114 in their mass spectra, and they may be eliminated from consideration. Since choices (C) and (D) will both give a molecular ion at mlz 114, we must look at the fragment ions to distinguish between them. A major fragmentation of the molecular ion of ketones is for either alkyl group attached to the carbonyl carbon atom to be lost as an alkyl radical to generate the corresponding acyl cation. For choice (C), the loss of an ethyl radical (mass of 29) would produce an acyl cation at m/z 85 and loss of a butyl radical (mass of 57) would produce an acyl cation at mlz 57. For choice (D), both alkyl groups are the same, propyl, and loss of a - propyl radical (mass of 43) would produce an acyl cation at m/z 71. Choice (D) is the correct answer. This type of fragmentation for ketones is often called Ot—cleavage, because the bond between the carbonyl carbon atom and the (it-carbon atom is the one that is broken. or—cleavage is a major fragmentation reaction for alcohols, amines and all t es of carbon 1 com ounds. Which structure is consistent with this 13C NMR spectrum? 200 180 160 I40 120 100 80 60 40 20 Sculpt") ' ‘ D ...................................................................................................................................................................................................................................... .- Knowledge Required: (1) Approximate chemical shifts of various types of carbon atoms. (2) The number of lines in a l3C NMR spectrum is equal to the number of different types of carbon atoms in the molecule. Thinking it Through: The spectrum shows signals for six different types of carbon atoms in the molecule. Inspection of the four choicesshows that choices (A) and (C) each contain six different types of carbon atoms, but choices (B) and (D) each contain seven different types of carbon atoms. Choices (B) and (D) may be immediately eliminated. These two choices could still be eliminated even if they had contained only six types of carbon atoms because the given spectrum does not have a signal in the 5 190—200 region, where the signals for the carbonyl carbon atoms in (B) and (D) would appear. Both choices (A) and (C) have only four types of spl- hybridized carbon atoms and would show four signals below 5 100. However, the low intensities of the signals at 5 133.5 and 5 159 indicate that they are probably for quaternary carbon atoms. Since choice (A) has two quaternary carbon atoms and choice (C) has only one, choice (A) is more likely the correct answer. That choice (A) is the correct answer is confirmed by considering the signals at 5 55 and 6 64. Signals for spa-hybridized carbon atoms must be attached to heteroatoms to come at these chemical shifts. This is the case for choice (A), but not for choice (C . The two 53—h bridized carbon atoms in choice C) would come above 5 30. S ectro'sco :} 5N a?» z; 1 . Which ketone will show a carbonyl absorption 3 . Which is the most reasonable structure for a at the lowest frequency (cm'l) in the infrared? compound with this IR spectrum? (A) 9 - ‘ % CCH3 I a n U s 1 (B) 9 g CCH3 E Wavenumber, (cm-1) 0 (A) 9 CH3(CH2)4—-C'='C—COCH3 I | CCH3 o CH3(CH2)3 — c a c — CHCH3 0 (C) 0 (D) II I l CCHQCHg H—C EC” (CH2)4CCH3 0/ (D) 0 ' I | CH3(CH2)4-CH= CH— CH l‘ 2 . Which compound is consistent with this IR spectrum? 4 . Which structure is consistent with this IR spectrum? Wavelength. (pm) - 4 5 65 100 I T I ; an n S 60 'l' t 40 t a 20 I1 0 e 2000 3500 3000 2500 200018001600140012001000 300 600 400 Wavenumber, (cm-1) Wavenumber, (cm-1) (A) 9 CH3CH2CH2CCH3 (A) (I? CH3(CH2)6CHZOCCH3 (B) 9 ,- \ CHgCHgCHgCHgCH (B) 9 CH3(CH2)10COH (C) ' 9H CH3CH2CH2CHCH3 _ (C) (RI CH3(CH2)3CH (D) 9 CH3CH2CH2CH2COCH3 (D) O | l HOCH2(CH2)7CCH3 Spectroscopy 5 . Which structure is most consistent with this IR spectrum? El file i _: III!!%IIIH IIIIiEIII" mnzm—f—gmaw-‘q 4000 3500 3000 2500 200018001600140012001000 300 600400 Wavenumber. (cnr‘) (A) UNflz @NHCHg 6 . Which compound corresponds to this IR spectrum? 4000 3500 3000 2500 200018001600140012001000 800 600 400 Wavenumber. (cm-1) (A) CHZZCH(CH2)5CH3 (B) HCEC(CH2)5CH3 (c) 9 HCEC(CH2)5COCH3 (D) 9 CH3(CH2)5CCH2CH3 .. 7 . Which compound would produce this IR spectrum? Wavelength, (pm I 4 Ill III!!!!§I" 000 3500 3000 2500 2000180016001400 12001000 800 600 400 Wavenumber. (cm-t) (A) 9 Gm (B) '9 0/(3011 (C) 0 6” (D) 9 O/COCH3 8 . Which is the correct order from lowest to highest field for the chemical shifts of the numbered sets of protons in the 1H NMR spectrum of this BrCI—IZOCHQCHa 1 2 3 compound? (A)3<2<1 (B)2<1<3 (C)1<2<3 (D)2<3<1 9 . Which of these compounds will show two triplets (among other signals) in their 1H NMR spectra? I ClCHZCHZCHCIZ 11 ClCHZCHZCH2C1 III CIZCHCHQCHCIE (A) I,II,III (C) I,III only (B) I only (D) II only S ectrosco 1 0 . Which 1H NMR CHBCHZSCH3 1 2 . Which structure is consistent with this IH spectrum is NMR spectrum? consistent with that of ethyl methyl sulfide? (A) (B) (B) 9' CH30QCOCH2CH3 a ' 2 1 o 6 (C) CH3CH2 9 \ (C) INQCH CH3CH2 (D) 9. CCH3 CH3CH2 (D) 1 3 . Which structure is consistent with this 1H NMR spectrum? 1 1 . How many signals would be observed in the proton-decoupled (A) (CH3)3COH 13C NMR spectrum of this compound? (3) CHacH2CH2CH20H (A) 5 (B) 6 (C) 7 (n) g (C) (EH3 7 CH3CHCH20H ' (D) OH | CH30H2CHCH3 Spectroscopy ____,____—._._.—_———-————-—-~— 1 4. Which structure is consistent with this 1H 1 6. Which structure is most consistent with this NMR spectrum? l3C NMR spectrum? B 7' 6 5 4 3 2 1 0 200 180 160 140 120 100 30 60 40 20 (1 5" (PPM) 50 (ppm) (A) 9 (A) (Ile g . (B) (EH3 IQ CH3CHCH2CCH3 (B) P. (C) 9 C0CH2CH3 CH3CH2CNHCH3 (D) (PH CH CH CHCH (C) 9 3 2 3 OCCH2CH3 1 7 . Which structure is in agreement with these 1H NMR and 13C NMR spectra? (D) 9 OCCH3 0/ (CDCIS) CH3 220 200 180 1.60 140 120 100 80 60 40 20 0 1 5 . The chemical 9 6° (ppm) . . | fflfts 1" the CH3—O—C—NHHCH2—CH3 3” C NMR spectrum for 1 2 3 4 3“ this compound 2H are 5 15.8, 40.1, 51.8 and 170.2. Which is the 2” correct set of chemical shifts for the carbon atoms? 3.0 2.5 2.0 1.5 1.0 0.5 Sufism) carbon atom O 1 2 3 4 (A) u CH3CH2CH2C0CH3 (A 51.8 40.1 170.2 15.8 ) (B) q (B) 170.2 15.8 51.8 40.1 CH3CH2CH25CH3 (C) 15.8 170.2 40.1 51.8 (C) 0 II (D) 51.8 170.2 40.1 15.8 CH3CH2CH20CCH3 (D) 9H CH3CH2CH2(|3CH3 CH3 d l 8 . Which structure is consistent with this 13C NMR spectrum? 200 200 130 160 140 120 100 so so 40 ficmpm) O (A) H CH3CH2~©CH2COH O (B) I I H2N~©—ccnzcn3 O | I CflgO-Q—CHZCH (D) CH3CH2 0 \ I I I CH3CH2 1 9 . Which compound would show major fragment ions at m/z 73 and 59‘? (A) 9H CH3CH2CHCH2CH3 (B) 9H CH3CH2CH2CHCH3 (C) 9H CH3CIIHCHCH3 CH3 (D) 9H CH3(IICH2CH3 CH3 i:i§§§l§i§§i§iifii§§§i§§ifi§§i§iiiiiiiéigf§§i§§i§§i§§i§§ifi§é§i§§§E§i§§§f§fi§§§§§§i§i§3§$€f§§§§ iiiégfi§§§§¥i§i§§i§§§§§§§§§§§l§ 7 2 0 . Which structure for a molecule with the molecular formula CllHMOz is in best agreement with this 13C NMR spectrum? 20 0 200 180 160 140 120 100 30 ED 40 8.; mm“) (A) O I I Q/cHZCHZCHZCOCH3 (B) 0 I I ©0HgCHZCOCH2CH3 (c) ' 9 CH2CH20CH CH3CH2 (D) 0 I I OCHZCHZCH CH30 2 1 . Which compound would show major fragment ions in its mass spectrum at m/z 85 and 57? (A) (I? _ CH3CH2CH2CHZCCH2CH3 (B) ' 9 CH3CH2CHZCCH2CH2CH3 (C) ('3' ' CH3CH2CH2CH2CH2CCH3 (D) 9 CH3CH2CH2C(IZHCH3 CH3 s? i Spectroscopy 22 . An unknown compound was found to be nmuwum mm Ilrluh IR spectrum. Which is a likely structure for the unreactive towards chromic acid and gave this unknown? 3500 3000 2500 200018001600140012001000 800 600 % Transmliiance Wavenumber, (cm-1) BCCADD .0123456 91111111 BCABDAAC e §§§;1.2 3 4 5‘6 7 3 ...
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This note was uploaded on 02/20/2012 for the course CHE 118B 118 taught by Professor Nasiri during the Winter '11 term at UC Davis.

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finalreview07 spectroscopy - Infrared (IR) and-nuclear...

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