Spectroscopy in Structure Determination Questions and Answers

Answer: C

The compound is CH₃-CH₂-CH(CH₃)-OH, butan-2-ol.

Read the spectrum: the triplet at 0.9 (3H) is a CH₃ next to a CH₂; the doublet at 1.1 (3H) is a CH₃ next to a single CH; the multiplet at 1.5-1.6 (2H) is the CH₂; the broad singlet at 3.6 (1H) exchanges with D₂O, so it is the OH; and the sextet at 3.9 (1H) is a CH-O proton coupled to five neighbouring protons (CH₃ plus CH₂).

Only butan-2-ol has a CH₃CH₂ group, a CH₃CH group and a CH(OH) proton with five neighbours. 2-Methylpropan-1-ol would show a six-proton doublet, butan-1-ol no methyl doublet, and tert-butyl alcohol a single nine-proton singlet.

Answer: B

The correct match is (i-c) (ii-d) (iii-b) (iv-a).

1. p-Xylene shows two singlets in ¹H-NMR, the four ring H and the six methyl H (c).
2. 4-Methylcyclohexanone, a saturated ketone, has the weak n→π* band near 280 nm with ε about 30 (d).
3. 1-Butyne has a C≡C stretch in the IR, which the key places at 2210 cm⁻¹ (b).
4. 2,2-Dimethylpropane has twelve equivalent H, one ¹H signal, and two kinds of carbon, two ¹³C signals (a).

Option A gives p-xylene a single proton signal, which ignores the aromatic H, and options C and D give the UV band of a ketone to an alkyne.

Answer: C

The structure is (C), ethyl 4-bromobutanoate, Br-CH₂CH₂CH₂-CO-OCH₂CH₃.

The quartet at 4.1 (2H) and triplet at 1.25 (3H) with the same J show an ethyl ester, OCH₂CH₃. The remaining 2H triplet and the 4H multiplet need a chain of three CH₂ groups: the triplet is the CH₂ at one end of the chain (the key assigns it to CH₂Br), and the multiplet is the two inner CH₂ groups.

(A) and (B) contain a CH bearing Br and would show a methine signal and, for (A), a methyl doublet. (D) has no ethyl ester: it would show no quartet and its CH₃ would be part of a propyl group.

Answer: C

The compound shows 06 signals.

The lactone is 4-methyldihydrofuran-2(3H)-one, with a stereocentre at C4 (the carbon carrying the methyl). Count the proton environments:

1. the CH₃ group (a doublet);
2. the C4 methine H;
3. the two H atoms of C3, next to the carbonyl;
4. the two H atoms of C5, next to the ring oxygen.

Because C4 is a stereocentre, the two protons of each neighbouring CH₂ group are diastereotopic: one is cis to the methyl and one trans, so they are chemically different and give separate signals. That gives 1 + 1 + 2 + 2 = 6 signals.

03 counts only the kinds of carbon bearing H, and 05 treats one CH₂ as a single environment; both overlook diastereotopicity. 08 counts each of the eight protons separately, but the three methyl protons are equivalent and give one signal.

Answer: C

The CD₃ carbon gives a septet with intensities 1 : 3 : 6 : 7 : 6 : 3 : 1.

In acetone-d₆ each methyl carbon carries three deuterium atoms, and deuterium has spin I = 1. The number of lines is 2nI + 1:

\[ 2nI + 1 = 2(3)(1) + 1 = 7 \]

The intensities are not binomial, because each deuterium has three equally likely orientations (+1, 0, −1), not two. One D gives 1:1:1; adding a second gives 1:2:3:2:1; adding the third gives 1:3:6:7:6:3:1, the familiar pattern of the CDCl₃ triplet extended.

1:6:15:20:15:6:1 is the binomial septet from six spin-1/2 nuclei, the pattern of a proton next to two CH₃ groups. The quintet is the pattern for two deuteriums (a CD₂ group), and a sextet does not arise from three spin-1 nuclei.

Answer: C

The compound is (C), 5,5,6-trimethyl-5,6-dihydro-2H-pyran-2-one.

C₈H₁₂O₂ has three degrees of unsaturation: a ring, a C=C and a C=O. The IR band at 1720 cm⁻¹ fits an α,β-unsaturated lactone carbonyl.

• δ 6.95 and 5.9, two doublets with the same J: the β and α vinyl protons of the enone, coupled only to each other. So the carbon beyond the β-CH has no hydrogen.
• δ 4.53, quartet, 1H, coupled to the 3H doublet at 1.41: an O-CH(CH₃) group whose other neighbour has no hydrogen.
• δ 1.20 and 1.15, two 3H singlets: a gem-dimethyl carbon, its methyls made different by the nearby stereocentre.

Only (C) puts a quaternary CMe₂ between the vinyl CH and the OCH(CH₃). (A) has a methyl on the double bond, leaving only one vinyl H. In (B) the vinyl H would also couple to the ring CH, and there is no OCH quartet because the oxygen-bearing carbon carries two methyls. (D) is a pyranone whose O-CH=CH protons and ring CH next to the ketone do not fit a quartet at 4.53.

Answer: A

The order is III \(\gt\) I \(\gt\) II.

I is a saturated six-membered lactone (δ-valerolactone), with C=O stretching near 1735-1750 cm⁻¹, like an ordinary ester.

In II the C=C is conjugated with the carbonyl (an α,β-unsaturated lactone). Conjugation lowers the C=O bond order and the frequency, to about 1720 cm⁻¹.

In III the C=C is attached to the ring oxygen (an enol lactone, like vinyl acetate). The oxygen's lone pair is drawn into the C=C instead of the carbonyl, so the C=O has more double-bond character and absorbs higher, about 1760-1770 cm⁻¹.

So III is highest and II lowest, which rules out every order that puts I or II first.

Answer: B

The correct match is (a-ii), (b-iii), (c-i), (d-iv).

• (a) 1,2-Dichloroethane, ClCH₂CH₂Cl: all four protons are equivalent, so the ¹H NMR spectrum is a single singlet, near 3.7 δ.
• (b) 1,4-Dibromobenzene, C₆H₄Br₂ (M = 234 with two ⁷⁹Br): bromine has two isotopes, ⁷⁹Br and ⁸¹Br, in nearly equal amounts, so a dibromide shows M, M+2 and M+4 peaks at 234, 236 and 238 in the ratio 1 : 2 : 1.
• (c) 2-Methyl-1,3-butadiene (isoprene) is a conjugated diene, with λ_max near 220 nm.

Those three pairings are found only in option (B), which leaves (d) cyclopropane with (iv). Every other option mismatches at least one of (a) to (c), for example by giving the dibromide the single NMR singlet.

Answer: B

The compound is ethyl methacrylate, CH₂=C(CH₃)COOCH₂CH₃.

Read the data piece by piece:

• 1.30 (t, 3H) and 4.19 (q, 2H), both J = 7 Hz: an ethyl group on oxygen, O-CH₂CH₃, as in an ethyl ester.
• 1.93 (s, 3H): a methyl on a carbon with no hydrogens, here the alkene carbon.
• 5.58 and 6.15 (each d, 1H, J = 3 Hz): two non-equivalent =CH₂ protons with a small geminal coupling.
• IR 1720 cm⁻¹: a conjugated ester C=O.

Option A is an enol acetate: its ethyl CH₂ sits on carbon, near 2.2 ppm, not on oxygen at 4.19, and it has an acetyl singlet near 2.1 rather than a vinyl methyl. Option C, isopropenyl propanoate, likewise has its CH₂ quartet near 2.4 ppm instead of an OCH₂ at 4.19, and enol esters absorb near 1760 cm⁻¹. Option D, an enol ether ketone, has two vinyl protons coupled trans with J of about 12 Hz, not two 3 Hz doublets.

Answer: A

The compound is 1-methoxy-2,4-dinitrobenzene (2,4-dinitroanisole).

The singlet at 3.79 (3H) is the OCH₃. The three aromatic protons show the pattern of a 1,2,4-trisubstituted ring:

• 9.11 (d, J = 2 Hz): H-3, between the two nitro groups, coupled only meta to H-5 and most deshielded.
• 8.47 (dd, J = 8 and 2 Hz): H-5, ortho to H-6 and meta to H-3.
• 7.29 (d, J = 8 Hz): H-6, ortho to the methoxy group and so the most shielded.

2,6-Dinitroanisole is symmetric and would show only two aromatic signals (a 2H doublet and a 1H triplet). In 3,4- and 2,5-dinitroanisole no proton lies between two nitro groups, so neither can give a signal as far downfield as 9.11; and in the 2,5-isomer the 8 Hz doublet belongs to H-3, ortho to a nitro group, which would appear far downfield rather than at 7.29.