ORGANIC CHEMISTRY TOPICAL: Molecular Spectroscopy Test 1 Time: 23 Minutes* Number of Questions: 18

* The timing restrictions for the science topical tests are optional. If you are using this test for the sole purpose of content reinforcement, you may want to disregard the time limit.

MCAT

DIRECTIONS: Most of the questions in the following test are organized into groups, with a descriptive passage preceding each group of questions. Study the passage, then select the single best answer to each question in the group. Some of the questions are not based on a descriptive passage; you must also select the best answer to these questions. If you are unsure of the best answer, eliminate the choices that you know are incorrect, then select an answer from the choices that remain. Indicate your selection by blackening the corresponding circle on your answer sheet. A periodic table is provided below for your use with the questions.

PERIODIC TABLE OF THE ELEMENTS 1 H 1.0

2 He 4.0

3 Li 6.9

4 Be 9.0

5 B 10.8

6 C 12.0

7 N 14.0

8 O 16.0

9 F 19.0

10 Ne 20.2

11 Na 23.0

12 Mg 24.3

13 Al 27.0

14 Si 28.1

15 P 31.0

16 S 32.1

17 Cl 35.5

18 Ar 39.9

19 K 39.1

20 Ca 40.1

21 Sc 45.0

22 Ti 47.9

23 V 50.9

24 Cr 52.0

25 Mn 54.9

26 Fe 55.8

27 Co 58.9

28 Ni 58.7

29 Cu 63.5

30 Zn 65.4

31 Ga 69.7

32 Ge 72.6

33 As 74.9

34 Se 79.0

35 Br 79.9

36 Kr 83.8

37 Rb 85.5

38 Sr 87.6

39 Y 88.9

40 Zr 91.2

41 Nb 92.9

42 Mo 95.9

43 Tc (98)

44 Ru 101.1

45 Rh 102.9

46 Pd 106.4

47 Ag 107.9

48 Cd 112.4

49 In 114.8

50 Sn 118.7

51 Sb 121.8

52 Te 127.6

53 I 126.9

54 Xe 131.3

55 Cs 132.9

56 Ba 137.3

57 La * 138.9

72 Hf 178.5

73 Ta 180.9

74 W 183.9

75 Re 186.2

76 Os 190.2

77 Ir 192.2

78 Pt 195.1

79 Au 197.0

80 Hg 200.6

81 Tl 204.4

82 Pb 207.2

83 Bi 209.0

84 Po (209)

85 At (210)

86 Rn (222)

87 Fr (223)

88 Ra 226.0

89 Ac † 227.0

104 Unq (261)

105 Unp (262)

106 Unh (263)

107 Uns (262)

108 Uno (265)

109 Une (267)

*

58 Ce 140.1

59 Pr 140.9

60 Nd 144.2

61 Pm (145)

62 Sm 150.4

63 Eu 152.0

64 Gd 157.3

65 Tb 158.9

66 Dy 162.5

67 Ho 164.9

68 Er 167.3

69 Tm 168.9

70 Yb 173.0

71 Lu 175.0

†

90 Th 232.0

91 Pa (231)

92 U 238.0

93 Np (237)

94 Pu (244)

95 Am (243)

96 Cm (247)

97 Bk (247)

98 Cf (251)

99 Es (252)

100 Fm (257)

101 Md (258)

102 No (259)

103 Lr (260)

GO ON TO THE NEXT PAGE.

2

as developed by

Molecular Spectroscopy Test 1 Passage I (Questions 1–6)

Table 1

A chemist carried out the reaction of tert-butyl chloride with sodium methoxide, resulting in the formation of two products:

IR Major peaks (cm–1)

Product A

Product B

3050 2900 1650

2900 1200

56

88

41

73

Reaction 1 After separation by distillation, Products A and B were analyzed using infrared spectroscopy (IR), proton nuclear magnetic resonance spectroscopy (NMR), mass spectroscopy (MS), and ultraviolet spectroscopy (UV). The results are as follows: Product A

MS Molecular weight (amu) Base peak (amu)

Product B

Product A was found to contain only carbon and hydrogen, while Product B contained carbon, hydrogen, and oxygen.

NMR Spectra 6H ∫

9H ∫

2H ∫

3H ∫

5

10

0

10

PPM (δ)

5

0

A. B. C. D.

PPM (δ)

UV Spectra

ε

200

nm

400

1 . Based on the mass spectra data given in Table 1, it can be determined that: both products contain a double bond. there is an ethyl group present in each product. both products contain at least one methyl group. no useful information can be obtained.

2 . All of the NMR signals in Figure 1 are singlets because:

200

Figure 1

nm

400

A . the hydrogen atoms are bonded to carbon atoms that have no other hydrogen atoms on adjacent carbon atoms. B . the proton resonances occur at the same magnetic field. C . the carbon atoms within each molecule are identical. D . the hydrogen atoms in each molecule are identical.

GO ON TO THE NEXT PAGE.

KAPLAN

3

MCAT 3 . The NMR signal at 5.5 ppm for Product A most likely indicates the presence of: A. B. C. D.

vinyl protons. aromatic protons. an aldehyde proton. the proton of an alcohol group.

6 . Which two processes are likely to occur in the reaction of tert-butyl chloride with sodium methoxide? A. B. C. D.

Elimination and substitution Rearrangement and hydrolysis Photolysis and elimination [4+2] Cycloaddition and substitution

4 . Which of the following is responsible for the IR stretching peak at 1200 cm–l for Product B? A. B. C. D.

C–H C=O C–O C–C

5 . Which of the following correctly represents the structure of Products A and B, respectively? A.

O

CH3CH2 CH2

B.

CH3 CH2CH2CH2CH3

and CH 3 C CH2CH3

and

O

C . CH2 CH CH CH2 and CH 3CH2OCH2CH3

D.

CH3 CH3 C CH2

CH3 and CH3 C CH 3 OCH3

GO ON TO THE NEXT PAGE.

4

as developed by

Molecular Spectroscopy Test 1 Passage II (Questions 7–12)

7 . From the data given in Table 1, what compound did Student 1 have?

Three students were each given an unknown compound and asked to identify it by using only spectroscopic techniques. They were told that their compound was one of those listed below.

A. B. C. D.

Cyclohexyl ethyl ketone Propiophenone 3-Phenylpropionaldehyde 1-Phenyl-2-propen-1-ol

O C

C2H5

Cyclohexyl ethyl ketone

O C

8 . The peak at 57 amu in the mass spectrum of Student l’s compound can be attributed to loss of which of the following fragments?

C2H5

Propiophenone

A.

O CH 2CH2 C

C H

B.

3–Phenylpropionaldehyde

CH

CH

O

C.

CH2

C 2 H5

OH 1–Phenyl–2–propen–1–ol

D.

CH 3

The results of the students’ experiments are shown below.

IR peaks (cm–1) NMR (ppm)

MS peaks (amu)

Student 1 3050 2900 1710 1.0; 3H triplet 2.0; 2H quartet 7.0; 5H 134 (M+) 105 77 57

Student 2

Student 3 1700-1740

– 9.5; 1H triplet

–

–

–

9 . Which of the following correctly describes the signal at 2.0 ppm in the NMR spectrum of Student l’s compound? A . It is a quartet because there are 3 protons on adjacent carbon atom. B . It is a quartet because there are 4 protons on adjacent carbon atom. C . It is due to the aromatic protons in compound. D . It has an area twice as large as the area of signal at 7.0 ppm:

the the the the

GO ON TO THE NEXT PAGE.

KAPLAN

5

MCAT 1 0 . The signal at 2900 cm–l in Student l’s IR spectrum is the result of: A . protons absorbing radio waves as they change spin state. B . electrons changing to a higher energy antibonding orbital. C . a C–H bond absorbing energy as it goes to a higher energy vibrational state. D . several C–H bonds breaking as it absorbs infrared light.

1 1 . The NMR spectrum of Student 2’s compound revealed the presence of which of the following functionalities? A. B. C. D.

A ketone A phenyl group A vinyl group An aldehyde

1 2 . Which of the following could be Student 3’s compound? I. II. III. IV. A. B. C. D.

Cyclohexyl ethyl ketone Propiophenone 3-Phenylpropionaldehyde 1-Phenyl-2-propen-1-ol

I only III and IV only I, II, and III only I, II, and IV only

GO ON TO THE NEXT PAGE.

6

as developed by

Molecular Spectroscopy Test 1 Questions 13 through 18 are NOT based on a descriptive passage. 1 3 . Which of the following is NOT true? A . Ultraviolet spectroscopy involves electronic transitions. B . Mass spectroscopy involves the fragmentation of molecules. C . NMR spectroscopy utilizes radio waves. D . IR spectroscopy utilizes light of wavelength 200400 nm.

1 7 . A student recorded the NMR spectra of each of the following compounds. Disregarding chemical shifts, which one would possess a spectrum significantly different from the others? A . CH3CH2CH2OCH2CH2CH3 Cl B . CH3 CH2CCH 2CH3 Cl C . CH3CH2CH2Cl Cl D . CH3CH2CH2C Cl

1 4 . Which of the following compounds would give only one signal in its 1H NMR spectrum? A. B. C. D.

Ethanol Phenol 1,2-Dichloroethane tert-Butyl alcohol

Cl

1 8 . If a compound has an NMR spectrum with 3 signals in an area ratio of 1:3:4, which of the following compounds could it be? Cl

I.

CH 3 CH2C

CH 3

III. CH 3

CH 3

II. CH 3

1 5 . What type of spectroscopy would be the LEAST useful in distinguishing diethyl ether from chloroethane? A. B. C. D.

Infrared spectroscopy Nuclear magnetic spectroscopy Ultraviolet spectroscopy Mass spectroscopy

A. B. C. D.

C

CH3

Cl

Cl

Cl

Cl

C

C

H

H

CH 2

IV. CH 3

ClCH 2

C

Cl CH 3

H

III only IV only I and II only II and IV only

1 6 . Which of the following compounds is NOT infrared active? A. B. C. D.

CH3Cl CH3CH2OH CO Br2

KAPLAN

END OF TEST

7

MCAT ANSWER KEY: 1. C 6. A 2. A 7. B 3. A 8. B 4. C 9. A 5. D 10. C

8

11. 12. 13. 14. 15.

D 16. C 17. D 18. C C

D B B

as developed by

Molecular Spectroscopy Test 1 MOLECULAR SPECTROSCOPY TEST 1 EXPLANATIONS Passage I (Questions 1–6) 1. The correct answer for question 1 is choice C . Mass spectroscopy involves the removal of an electron from the molecule that is to be analyzed, creating a positively charged molecular ion. This molecular ion is identical to the original neutral molecule except that it is missing one electron, making it positively charged. The molecular ion is usually unstable and will itself fragment to give other positively charged fragments. These positively charged fragments are separated on the basis of their charge to mass ratios. When these fragments are accelerated by an electric field and passed through a magnetic field, they will be deflected by different amounts. Ions with different masses strike the electric multiplier at different times; their appearance are recorded on a chart. The instrument can be calibrated to give the molecular weight and approximate number of fragments. Since a methyl group is 15 atomic mass units, the loss of a 15 amu fragment from the molecular ion generally indicates that a methyl group must have been present in the molecule. Therefore, choice C is correct. An ethyl group (choice B) would have been indicated by a peak at 29 amu less than the molecular weight. Choice A is incorrect since the mass spectrum indicates only the masses of atoms and groups and not bonding relationships. Answer D is false because the fragmentation of molecules does provide much useful information regarding the structure of the molecule. Again, the correct answer is choice C. 2. The correct answer for question 2 is choice A. The multiplicity, or shape, of the signal is a result of the protons on adjacent atoms. The signal from protons on a particular carbon is split by protons on adjacent carbons due to spin-spin coupling. The information that is provided by the splitting of the proton NMR signals, along with the chemical shift and peak area data, accounts for most of the usefulness of proton NMR to identify compounds. In this question, all signals in the NMR spectrum of both compounds are singlets. This implies that none of the hydrogen atoms in the molecules of either product A or B have any other hydrogen atoms on adjacent atoms. For choice B to be correct, there would be only one type of hydrogen atom in each of products A and B and there would be only one peak in the entire NMR spectrum. This is not the case as both product A and B have two separate but unsplit proton signals. Choice C is also incorrect. Because this is proton NMR, no information about carbon atoms are obtained directly and so choice C is incorrect. Choice D is incorrect because it is simply another way of phrasing choice B. Again, choice A is correct. 3. The correct answer for question 3 is choice A . To answer this question, you have to remember the location on the NMR spectrum of the more common types of protons. These are called the chemical shifts of protons. The shifts are commonly expressed on the delta scale as parts per million or ppm values. You should have some of the more common chemical shifts committed to memory; a list of them can be found in your organic chemistry home study book. You should certainly know that aldehyde and carboxylic acid protons appear far downfield (downfield is to the left in the spectrum): the carboxylic acid proton appears between 10.5 and 12 and the aldehyde proton appears between 9 and 10. The MCAT has included these chemical shifts more than once. You should also know that if electronegative elements are close to the proton of interest, the proton will be deshielded, meaning that its signal will appear further downfield. A useful mnemonic is: deshield = downfield = left. Anyway, vinyl protons appear at around 5-6 ppm and so answer A is correct. Aromatic protons usually give a signal between 6 and 8.5 ppm (choice B), an aldehyde proton comes at 9-10 ppm (choice C), and those which are part of an alcohol group come between 3 and 5 ppm (choice D). Answers B, C, and D are wrong, and again, the correct answer is A. Again, if you had trouble with this question, consult your organic home study book. 4. The correct choice for question 4 is C . In order to answer this question, you need to remember the IR absorbance values of some major functional groups. The functional groups that are most readily identified by IR are the carbonyl group, with a sharp absorption between 1700 to 1750 cm–1, the hydroxyl group, showing a strong broad absorption band between 2500 and 3600 cm –1, the exact region depends upon what type of O–H linkage is in question: alcohol, carboxylic acid, or phenol. Keep in mind that all you have to know is that a hydroxyl group will show up as a strong broad band in this region. Another absorption that you should be familiar with is that of a C–O single bond stretch. This appears as a strong sharp absorption in the region of 1000 to 1300 cm –1. This type of absorption is seen for alcohols, ethers, and esters. With this in mind, an absorption at 1200 cm –1 will most likely correspond to the stretching of a carbon/oxygen single bond. This makes choice C the correct response. Choices A and B are incorrect as the absorbance values are incorrect for these functional groups. Choice D is also wrong: carbon-carbon single bonds show only very weak IR absorbances if they show up at all. Again, choice C is the correct answer. 5. The correct answer for question number 5 is choice D . The UV spectrum for product A indicates that there is some form of double bond present, so you could have narrowed the answer choices down to C and D. The fact that it peaks below 200 nm suggests that there is an isolated double bond, rather than a conjugated system. Don't worry if you didn't know this though; you can identify the products without the UV data. In fact, the NMR spectrum is probably the most useful in identifying the products. You can see that product A has two types of hydrogen atoms because there are two signals. The ratio of these two peak areas is given as 1 to 3. It is very important to understand that the ratio of peak areas only reveals the ratio of the number of hydrogens in the compound, not the actual numbers of hydrogens in the compound. The ratio provided by the NMR data for product A matches only the first compound shown in choice D, which has two types of hydrogen in a 2 to 6 or 1 to 3 ratio. The first compound in choice A would have 4 types of hydrogens, so this answer choice can be eliminated. The first compound in choice B would have 3 types of hydrogen and so is also incorrect. The first compound in choice C would seem to be a possibility because it has 2 types of hydrogen atom but in an incorrect ratio: 4 to 2 or 2 to 1. Here we do not even have to analyze the second compound in each answer choice to get the correct answer KAPLAN

9

because only choice D is possible based on the NMR spectrum of that compound. In any case we will look at how the second answer choices are possible. The NMR spectrum of product B has two signals, implying two types of hydrogen atoms, and the ratio of these hydrogens is 1 to 3. The second compound of choice A would have three types of hydrogen and is thus incorrect. The second compound in choice B would have two types of hydrogen but they are in a 4 to 4 ratio and this can therefore be eliminated. The second compound in choice C has two types of hydrogen but the ratio is 4 to 6 and is incorrect. The second compound in choice D has two types of hydrogen in the correct ratio, 3 to 9 or 1 to 3. Thus choices A, B, and C could not be correct and choice D is the right answer. 6. The correct answer for question 6 is choice A. Since product A has a double bond, it is the product of an elimination reaction because the original compound has no double bonds. You would also expect an elimination product because the reaction in the passage is between a tertiary alkyl halide and a strong base. In addition, you are told that product B contains oxygen and no chlorine, so it must have lost the chlorine and gained an oxygen containing group. Because chlorine is a good leaving group, product B is most likely the result of a substitution reaction. This would be logical since elimination is almost always accompanied by the formation of some substitution product. You could have also derived the answer from the experimental data in the passage, but you really should be know that the reaction presented in this passage is a model reaction for elimination and nucleophilic substitution. Sodium ethoxide is a very strong base and would cause elimination to produce the alkene. The ethoxide ion can also act as a nucleophile, replacing the chloride. The other answers are wrong for a variety of reasons, of which I will mention the most obvious. Choice B involves hydrolysis, or reaction with water. Water could not be present with sodium ethoxide, which would react with any water present. Choice C mentions photolysis, or reaction with light, which is not given as a reaction condition, so this is also wrong. Choice D is also wrong because there are no dienes in the starting materials which could undergo a [4 + 2] cycloaddition. A cycloaddition reaction is one in which another 2 unsaturated molecules combine to form a cyclic molecule. Cycloaddition reactions are further classified according to the number of pi electrons involved in the transition state. A [4+2] reaction means that 6 pi electrons are involved in the transition state: 4 pi electrons from one reactant and 2 from another. Again, A is the correct response. Passage II (Questions 7–12) 7. The correct answer to question 7 is choice B. Student 1's compound showed a peak at about 7 ppm in the NMR, which indicates it is aromatic. We can therefore eliminate choice A, since cyclohexyl ethyl ketone contains no aromatic ring. It can also be rejected because its molecular weight would be 140 atomic mass units, not 134 amu as indicated by the MS for student 1's compound. An ethyl group is indicated by the presence of a triplet and a quartet in the NMR, which student 1 has. The molecular weight is indeed 134 amu with fragments peaks as expected from such a compound. The fragments are the phenyl ring at 77 amu, the ethyl carbonyl fragment at 57 amu, and the phenyl carbonyl fragment at 105 amu. The IR peak of student 1's compound at 1710 cm–1 also indicates the presence of a C=O group, so this also rules out choice D is this does not possess a carbonyl group. Choice C can be eliminated because it would have shown a peak at 9.5 ppm in the NMR due to the aldehyde hydrogen, and the compound in choice D would have shown a peak at 3400 cm –1 in the IR spectrum. So the correct choice for this question is choice B. 8. The correct answer for question 8 is choice B. The molecular weight of the compound is 134 amu-indicated by the parent peak. Subtracting the weight of the fragment recorded (57 amu) from the total weight gives a difference of 77 amu. Of the groups listed, only the phenyl group C 6H 5 - has a weight of 77 amu. This gives choice B as the correct answer. The other groups have weights as follows: carbonyl - 28 amu, ethyl - 29 amu, methyl - 15 amu. Again the correct answer is choice B. 9. The correct answer to question 9 is choice A . The splitting of a proton signal is due to spin-spin coupling of the hydrogen with hydrogens on an adjacent carbon. A single signal is split into n+1 peaks, where n is the number of adjacent hydrogens. Because the signal is split into a quartet or four signals, the correct answer must be A. Choice B is incorrect because the signal that would be produced by four adjacent hydrogens in question would be a quintet or 5 peaks. Aromatic protons would absorb in the 6 to 8.5 ppm region, so choice C is incorrect. Lastly, the area of the signal is irrelevant to this question. In addition, the signal at 2.0 ppm would be 2.5 times smaller than the aromatic signal so choice D is not right. Again, the correct answer is A. 10. The correct answer for question 10 is choice C. Answer A is wrong because this process does not occur in IR spectroscopy but is the basis for NMR spectroscopy. The process of promoting electrons from a molecular orbital to an antibonding orbital mentioned in B is the basis for UV spectroscopy. IR spectroscopy, in contrast is a result of polar covalent bonds absorbing IR light and vibrating with more energy. The area around 2900 cm –1 is characteristic of the C–H bond and so answer C is correct. IR light usually does not have enough energy to break covalent bonds, so choice D is incorrect. Again, answer choice C is the correct response. 11. The correct answer for question 11 is choice D. Aldehyde protons absorb in the area around 9 to 10 ppm. The only aldehyde in the list is 3-phenylpropionaldehyde, so this must be the compound that student 2 had. Since it has 2 alpha hydrogen atoms adjacent, the aldehyde hydrogen would be a triplet. All of the other answer choices do not absorb in the region 9 to 10 ppm, so again, choice D is the correct answer. 12. The correct answer here is choice C . The dominant absorbance at 1700-1740 cm –1 in the IR spectrum is characteristic of a carbonyl group. Since compounds I, II and III all have carbonyl groups, being either ketones or aldehydes, student 3 could have had any one of them. Only compound IV does not contain a carbonyl group. You might think that this enol could convert to a tautomer containing the keto functionality, but note that the double bond is not adjacent to the alcohol-bearing carbon and so no interconversion occurs. The correct answer is C.

Molecular Spectroscopy Test 1 Discrete Questions 13. The correct answer is choice D . All are correct statements except choice D. The UV range NOT the IR range is from 200-400 nm wavelength and corresponds to the energy required for electronic transitions between the bonding or nonbonding molecular orbitals and antibonding molecular orbitals. Note that these frequencies are shown on the UV spectrum in passage I. These transitions make UV spectroscopy most useful for studying compounds that contain double bonds, especially conjugated double bonds. Therefore, choice A can be eliminated. MS, or mass spectroscopy studies compounds through the fragmentation of molecules, though the parent peak which is unfragmented also provides very useful information. So, choice B is also a correct statement. Choice C is also right as NMR, or nuclear magnetic resonance spectroscopy involves putting a sample containing the compound in a high intensity magnetic field and scanning through the radio-frequency range of the electromagnetic spectrum for particular absorptions. This leaves choice D. Infrared radiation is useful in the region of 3000 - 30,000 nm corresponding to 4000 - 600 cm –1, usually called wavenumbers. In this range the molecular vibrations of molecules are active and functional groups of molecules can be identified by their characteristic absorbance frequencies. Again the correct answer is D. 14. The correct answer to question 14 is choice C . In this question, you must understand how the signals from the hydrogens are generated. The position that a hydrogen absorbs is determined by the chemical environment it is in. If the chemical environments of two hydrogens are identical, they will give only one signal. Another way to say this would be that any of the equivalent hydrogens could be replaced by another group to yield the same compound. The difficulty therefore arises in determining whether hydrogens are in identical environments. To give only one absorbance in the NMR spectrum, a compound must have only one type of hydrogen atom in the entire molecule. This is the case for 1,2dichloroethane and so choice C is the correct answer. The other choices all have chemically different hydrogens. In these cases, choice A would give 3 signals, B would give 2, and D would give 2. Again, the correct answer is choice C. 15. The correct answer for question 15 is choice C . Mass spectroscopy, abbreviated MS, provides the molecular weight, a range of molecular fragments, and a fingerprint pattern that is unique to a compound. From all this, some information regarding the structure of the compound can be derived, so choice D is out. IR determines whether the compound contains certain functional groups and also gives a useful fingerprint pattern unique to a compound whereas proton NMR examines the molecular environment of the various kinds of hydrogen atoms and is related to where the signal comes on the spectrum, the relative numbers of each kind of hydrogen atom given by the area under each signal, and the number of hydrogen atoms on adjacent atoms. So, both of these techniques will provide some useful information and you can eliminate A and B. UV spectroscopy is useful for identifying compounds that contain double bonds, especially conjugated double bonds. As neither are present in the compounds in question, it obviously won't be a very good analytical technique in order to distinguish between them. Therefore UV will be the least useful, making choice C the correct response. 16. The correct answer for question 16 is choice D . In order to give rise to an IR signal, there should be some form of dipole moment induced in the covalently bonded molecule when it stretches or bends. Choices A and B are organic compounds with covalent bonds. They are also attached to electronegative groups and so a dipole will be induced here giving IR signals. The same thing applies to carbon monoxide in choice C. Here, the electronegative oxygen induces a dipole, so this too will give an IR spectrum. However, when bromine stretches or bends, a dipole is not induced and so it will be IR inactive making choice D the correct response. 17. The correct answer for 17 is choice B. The compounds in choices A, C, and D all have propyl groups with electronegative atoms nearby, so their spectra would look very similar, having three signals split in an identical manner. Granted, the ratio of hydrogens would be twice as big in the case of choice A, but the splitting pattern would be much the same as answer choices C and D. The chemical shifts of some groups also might differ slightly, but the question tells you to ignore this when considering which spectrum would differ the most. Only compound B has ethyl groups which would give a splitting pattern consisting of only two signals, different from the propyl groups. Again, choice B is the correct response. 18. The correct answer for question 18 is choice B. Compound I would give three signals but the area ratio, proportional to the ratio of the number of hydrogens in the compound, would be 3 to 2 to 3 and thus choice C can be eliminated. Compound II would give only two signals and therefore choice D can also be eliminated. Compound III would give only one signal as there is only one type of proton. Finally you can see that only compound IV has three types of hydrogen atoms in a 1:3:4 ratio. Therefore, choice B is correct.

KAPLAN

11