CHE 232

Answers to Spectroscopy Problems

Problem 1

(a) We know from the M+2 peak that this compound has a Cl atom in it. (If it were a Br atom, the M+2 peak would be the same size as the M⁺ peak.) We know from the M+1 peak that there are six C atoms in this compound. That makes C₆Cl so far. 132 – 6 × 12 – 35 = 25, so we have 25 amu left to consider. We can't possibly have a compound with the formula C₆H₂₅Cl. However, we could have a compound with the formula C₆H₉ClO. Such a compound would have two degrees of unsaturation. Possibilities (there are others):

(b) These absorbances are characteristic of C(sp³)–H and either an unstrained ketone or an ester next to a double bond. We don't have any C(sp²)–H bonds (no peak at 3000–3100 cm^–1), according to the IR, nor do we have any double bonds next to carbonyls (no peak at 1600–1650 cm^–1), so we are most likely going to have an unstrained ketone. A ketone uses up one degree of unsaturation, so we must have another double bond or a ring. Considering the lack of C(sp²)–H stretches, the latter is more likely. Either the ring has to be six-membered or the ketone carbon can't be part of the ring, because cyclopentanones and smaller cycloalkanones stretch at higher frequency. Possibilities (there are others):

(c) The resonance at δ 200 confirms the ketone. We have five other C resonances, showing that the compound isn't symmetrical. None of the resonances are quartets, showing there are no Me groups. Four of the resonances are CH₂ groups, while one is a CH group. Also, the CH group is shifted considerably far downfield, suggesting it is bound to Cl. With four CH₂ groups, one CHCl group, and one C=O group, the most likely possibilities are:

Note that 4-chlorocyclohexanone is not a possibility, because it would show only four resonances in the ¹³C NMR spectrum. It has a plane of symmetry that makes C2 equivalent to C6, and C3 equivalent to C4.

(d) The resonance at 3.8 ppm is due to the CHCl group. With a multiplicity of dddd, it must have four different inequivalent neighbors that are two or three bonds away, i.e. on the same C or on the adjacent C's. In 2-chlorocyclohexanone, the H in the CHCl group has just two such neighbors, but in 3-chlorocyclohexanone it has four such neighbors. The compound is 3-chlorocyclohexanone.

(e) You bet! It's chiral. The compound has no plane of symmetry, and it has one stereocenter, at C3.

Problem 2

(a) The compound can have at most 13 C's and no Cl's or Br's. If it has 13 C's, its formula is C₁₃H₂, and it has 13 degrees of unsaturation. Not likely. Subtract one C, add 12 H's to get C₁₂H₁₄, with six degrees of unsaturation. This is more likely, especially if the compound has a phenyl ring, which uses up four degrees of unsaturation. Subtract one C, add 12 H's to get C₁₁H₂₆, an unreasonable structure because it has too many H's, but from this subtract CH₄, add O to get C₁₀H₂₂O (no degrees of unsaturation), or subtract CH₄, add O again to get C₉H₁₈O₂ (one degree of unsaturation). Or take C₁₀H₂₂O, subtract CO, add N₂ to get C₉H₂₂N₂ (no degrees of unsaturation). Possibilities:

(b) The compound has nine C's. This narrows the possibilities down to C₉H₁₈O₂ and C₉H₂₂N₂.

(c) The compound has no OH, no NH₂, and no C(sp²)–H. It has a carbonyl group of some sort. This rules out C₉H₂₂N₂. The carbonyl group might be an ester, or it might be a cyclopentanone. However, a cyclopentanone would have to have at least two degrees of unsaturation, and C₉H₁₈O₂ has only one. So we have a nine-carbon ester with no rings. Possibilities:

(d) There are nine C atoms, but only seven of them show in the ¹³C NMR spectrum. This suggests either that there are two pairs of equivalent C's, or that three C's are equivalent. The resonance at 168 ppm confirms the ester. The triplet at 70 ppm suggests C(=O)–O–CH₂. In addition to this there are two kinds of Me groups, two CH groups, and one more CH₂ group. If there were three equivalent CH₃ groups, they would all have to be attached to the same C, but since no singlet is seen, it is much more likely that there are two pairs of CH₃ groups. Each set of CH₃'s would be atatched to a CH group, so we have two inequivalent i-Pr groups. The only question is the location of the last CH₂ group. There are two possible structures: 2-methylpropyl 3-methylbutanoate, and 3-methylbutyl 2-methylpropanoate.

(e) The structure of the compound has been narrowed down to two possibilities. Now predict the ¹H NMR spectrum of the compound you drew for (d). Label each H atom H_a, H_b, etc., giving equivalent H's identical letters. Then predict the approximate chemical shift, multiplicity, and integration of each resonance that you would see in the ¹H NMR spectrum. If you can see the two possibilities, how might you distinguish them by ¹H NMR?

The biggest distinguishing factor in these two spectra is the multiplicity of the resonance at δ 4.1. In the first compound it is a doublet, while in the second it is a triplet.

(f) Nope! It's achiral. This is most easily seen by noting its lack of stereocenters.

Problem 3

(a) The molecular weight is odd. That means there is an odd number of N's. C₆N makes 86 amu, but C₅N makes 74 amu, leaving room for eleven H's. C₅H₁₁N has one degree of unsaturation, a perfectly reasonable structure. Subtract CH₄, add O to get C₄H₇NO, also a reasonable structure, with two degrees of unsaturation. Possibilities:

(b) The absorbance at 3340 cm^–1 is characteristic of N–H. There are no C(sp²)–H bonds, no CºN bonds, no C=O bonds. The most likely possibility is a monocyclic C₅H₁₁N with one or two H's attached to N. Possibilities:

(c) With five C atoms and only three resonances, the compound must have some symmetry. This rules out the third structure above. It likely has two pairs of two equivalent C's, with the fifth carbon being different again. Both cyclopentanamine or piperidine, the first two compounds above, are consistent with this information. You would be able to distinguish them by the multiplicities of the resonances, but this information is not given.

(d) This is consistent with piperidine (the second compound). Cyclopentanamine would have a downfield peak that integrated to 1H and a broad resonance that integrated to 2H. -- The six H's at δ 1.5 consist of two inequivalent sets of four and two H's that happen to resonate in the same place. This happens a lot, especially with lower field (200 MHz) spectrometers!

Problem 4

(a) The compound has at most C₁₀. C₁₀H₂ is not a very reasonable formula, although a compound could be drawn for it. C₉H₁₄ has three degrees of unsaturation, which is reasonable. C₈H₁₀O has four degrees of unsaturation, which is also reasonable, especially if the compound has an aromatic ring. Other possibilities: C₇H₆O₂, C₇H₁₀N₂. Possibilities:

(b) These correspond to OH, C(sp²)–H, and C(sp³)–H. The compound has no carbonyl group. The most likely possibility is C₈H₁₀O or C₇H₆O₂, with an OH group and no C=O group. Some possibilities:

(c) The compound is chiral. This rules out all compounds that lack a stereocenter. The only three possibilities I can think of are:

(d) The 5H multiplet at δ 7.2 is characteristic of a phenyl ring. Only the first structure has one. The other resonances are consistent with the first structure. The other structures are not consistent with the ¹H NMR data. The compound is (+)-1-phenylethanol.

Problem 5

(a) C₁₄ is not a likely formula. C₁₃H₁₂, with 8 degrees of unsaturation, is possible if there are two Ph rings. C₁₂H₂₄ has only one degree of unsaturation. C₁₁H₂₀O has two degrees of unsaturation. Etc.

(b) The compound has a C(sp²)–H and C(sp³)–H bonds. If it has C=C bonds, they are more or less symmetrical. No C=O or O–H. (Note: C–O bonds are characterized by a strong absorbance in the fingerprint region at ca. 1100 cm^–1. If we had information about absorbances in the fingerprint region, we could rule ethers in or out at this point.)

(c) No C–O groups are present. No aromatic rings are present, or we would see more than one resonance in the C(sp²) region. The compound must have the formula C₁₂H₂₄. The compound is clearly highly symmetrical, so we must at least have identical groups attached to each side of the double bond. The multiplicities tell us that we have one kind of CH, one kind of CH₂, and one kind of CH₃ group. If the CH is attached to the double bond C, and two CH₂'s are attached to the CH, and one CH₃ is attached to each CH₂, as below on the left, then we have a structure consistent with all of the data. If, on the other hand, we attach the CH₂ to the double bond, then attach the CH to the CH₂, then attach two CH₃'s to each CH, as below on the right, we get a structure with only 10 C's, which is not consistent with the MS data. So the structure on the left is most consistent with the data. However, since we have a double bond with two different groups attached to each C, we might have either diastereomer, cis or trans.

(d) The cis isomer has two planes of symmetry. In the structure drawn below, one is vertical and perpendicular to the plane of the paper, and one is horizontal and perpendicular to the plae of the paper. The first plane relates the right left halves of the compound; the second plane relates the top and bottom halves. Each H on each CH₂ group is equivalent to three other H's, but the two H's on one CH₂ group are not related by any plane or axis, so they are diastereotopic. This is clearest if you replace one H_b with, say, Cl. You will generate two stereocenters. You will get a different diastereomer from if you replace H_c on the same atom with Cl.

Probably H_b and H_c will have nearly the same chemical shift. In this case, their signals will be superimposed, and a multiplet of 4H will be observed. If, however, their chemical shifts are sufficiently different, then a large coupling constant (ca. 15 Hz) will be observed between them.

The trans isomer has one plane of symmetry, relating the top and bottom halves of the compound. It also has a two-fold axis of symmetry. Imagine an axis in the plane of the paper, through the center of the double bond (the arrow in the picture below). If you imagine taking a copy of the compound and rotating it 180¡ about that axis, your copy is now exactly superimposed upon your original. All the atoms that are superimposed are chemically equivalent. Again, though, the two H's in each CH₂ group are related neither by an axis nor by a plane of symmetry, so they are diastereotopic. Nearly the same spectrum is expected for the trans isomer as for the cis isomer.

(e) The cis isomer is less stable than the trans isomer, so the compound in hand is the cis isomer.

Problem 6

(a) C₈H₆ is a possibility, with six degrees of unsaturation. C₇H₁₈ has too many H's, but C₆H₁₄O (no degrees of unsaturation) is a possibility, as are C₅H₁₀O₂ (one degree of unsaturation) and C₅H₁₄N₂ (no degrees of unsaturation).

(b) There are no C=O or C(sp²)–H groups, nor any NH₂ groups. The compound may be an acyclic alcohol, a cyclic diol, or a cyclic alcohol with an ether group.

(c) There is just one kind of C attached to O, and it has one H on it: C₂CHOH. This rules out a cyclic alcohol with an ether group, which should show more than one kind of C–O. The other two kinds of C are CH₂ groups. There are no CH₃ groups; this rules out an acyclic alcohol. We must have a cyclic diol. There are four possibilities:

(d) The CH attached to O is a dd. It must have two inequivalent neighbors. This rules out the first two structures, in which each CHOH has four neighbors. It is not possible to distinguish the latter two compounds unambiguously.

(e) This rules out the cis isomer, which has a plane of symmetry. The compound is trans-1,2-cyclopentanediol.

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