IChO 2003 · Theory problem 32

NMR puzzle

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An organic compound A (\(\ce{C8H10}\)) gives the following chain of reactions:

Figure 1

Based on the \({}^{1}\mathrm{H}\)-NMR spectra given, draw the structures of compounds A, B, C, D, E and F, and match the groups of the hydrogen atoms of each compound to the corresponding \({}^{1}\mathrm{H}\)-NMR peaks, as shown in the example.

Figure 2
Figure 3

General remarks: NMR spectra were ecorded in \(\ce{CDCl3}\) on a 60 MHz Perkin Elmer Spectrometer. Under ordinary conditions (exposure to air, light and water vapour) acidic impurities may develop in \(\ce{CDCl3}\) solutions and catalyse rapid exchange of some particular protons.

Official English text of the problem.

Answers

The supplied problem has no numbered subparts or stated point values. The compound labels are retained below.

  • A — Ethylbenzene, \(\ce{C6H5CH2CH3}\), SMILES CCc1ccccc1. A1: five aromatic H; A2: benzylic \(\ce{CH2}\), 2 H; A3: terminal \(\ce{CH3}\), 3 H.

  • B — (1-Bromoethyl)benzene, \(\ce{C6H5CH(Br)CH3}\), expected as a racemate. Enantiomer SMILES: C[C@H](Br)c1ccccc1 and C[C@@H](Br)c1ccccc1. B1: five aromatic H; B2: \(\ce{CHBr}\), 1 H; B3: \(\ce{CH3}\), 3 H.

  • C — Styrene (ethenylbenzene), \(\ce{C6H5CH=CH2}\), SMILES C=Cc1ccccc1. C1: five aromatic H; C2: ring-adjacent vinylic \(\ce{CH}\), 1 H; C3: the two nonequivalent terminal vinylic H, 2 H total.

  • D — Benzaldehyde, \(\ce{C6H5CHO}\), SMILES O=Cc1ccccc1. D1: aldehydic H, 1 H; D2 (the aromatic bracket labelled “D” in the scan): five aromatic H.

  • E — 1-Phenylethanol, \(\ce{C6H5CH(OH)CH3}\), molecular formula \(\ce{C8H10O}\), expected as a racemate. Enantiomer SMILES: C[C@H](O)c1ccccc1 and C[C@@H](O)c1ccccc1. E1: five aromatic H; E2: \(\ce{CH(OH)}\), 1 H; E3: hydroxyl H, 1 H; E4: \(\ce{CH3}\), 3 H.

  • F — Acetophenone (1-phenylethan-1-one), \(\ce{C6H5COCH3}\), SMILES CC(=O)c1ccccc1. F1: five aromatic H; F2: carbonyl-adjacent \(\ce{CH3}\), 3 H.

Solution

Chemical shifts below are approximate readings from the supplied spectra, in ppm. Each five-proton aromatic envelope contains the ortho, meta and para protons of a monosubstituted benzene ring; assigning them one label does not imply that all five are chemically equivalent.

A — Identification of the starting hydrocarbon

For \(\ce{C8H10}\), the degree of unsaturation is

\[ \mathrm{DBE}=\frac{2(8)+2-10}{2}=4. \]

This is consistent with a benzene ring. The spectrum identifies the remaining two carbons as an ethyl group:

Label Approximate \(\delta\) / ppm Assignment Multiplicity
A1 7.2 Five aromatic H Aromatic envelope
A2 2.6 Benzylic \(\ce{CH2}\), 2 H Quartet
A3 1.3 Terminal \(\ce{CH3}\), 3 H Triplet

The methylene protons are split by three neighbouring methyl protons, while the methyl protons are split by two methylene protons. Thus A is ethylbenzene.

B — Benzylic bromination

The reagent drawn is N-bromosuccinimide, SMILES O=C1CCC(=O)N1Br. It replaces a benzylic hydrogen by bromine, giving (1-bromoethyl)benzene. The balanced reaction is

\[ \ce{C6H5CH2CH3 + C4H4BrNO2 -> C6H5CH(Br)CH3 + C4H5NO2}. \]

The other product is succinimide, SMILES O=C1CCC(=O)N1.

Label Approximate \(\delta\) / ppm Assignment Multiplicity
B1 7.2–7.6 Five aromatic H Multiplet
B2 5.2 \(\ce{CHBr}\), 1 H Quartet
B3 2.0 \(\ce{CH3}\), 3 H Doublet

Replacing a benzylic H with electronegative bromine shifts the remaining methine H substantially downfield. The methine is split by three methyl H, and the methyl group is split by one methine H.

A stereogenic centre is created. Under achiral radical bromination conditions, the product is expected to be racemic; the two enantiomers have identical spectra in the stated achiral solvent.

C — Elimination to form styrene

Potassium tert-butoxide, SMILES CC(C)(C)[O-].[K+], removes a hydrogen from the methyl group while bromide leaves the adjacent carbon. This is an E2 dehydrobromination:

\[ \ce{C6H5CH(Br)CH3 + KOC(CH3)3 -> C6H5CH=CH2 + HOC(CH3)3 + KBr}. \]

The byproducts are tert-butanol, SMILES CC(C)(C)O, and potassium bromide, SMILES [K+].[Br-].

The arrow labelled “Reduction” in the scheme therefore represents an elimination. The product is styrene.

Label Approximate \(\delta\) / ppm Assignment Multiplicity
C1 7.2–7.5 Five aromatic H Multiplet
C2 6.7 Ring-adjacent vinylic \(\ce{CH}\), 1 H Approximately a doublet of doublets
C3 5.2 and 5.7 Terminal \(\ce{CH2}\), 1 H at each position Two approximately doublet-of-doublet patterns

This gives the stated integration:

\[ \boxed{\mathrm{C1:C2:C3}=5:1:2}. \]

The two terminal alkene hydrogens are nonequivalent: one is cis and the other trans to the ring-adjacent vinylic hydrogen. They consequently have different chemical shifts and different vicinal coupling constants. C2 couples separately to both.

D — Reductive ozonolysis

Cleavage of the side-chain double bond produces one carbonyl compound from each alkene carbon:

  • The ring-adjacent carbon gives benzaldehyde, D.
  • The terminal carbon gives methanal (formaldehyde), SMILES C=O.

A balanced net equation including the acidic zinc work-up is

\[ \ce{C6H5CH=CH2 + O3 + Zn + 2H3O+ -> C6H5CHO + HCHO + Zn^2+ + 3H2O}. \]

The reductive work-up preserves the aldehyde products.

Label Approximate \(\delta\) / ppm Assignment Multiplicity
D1 10 Aldehydic \(\ce{CHO}\), 1 H Approximately a singlet
D2, printed “D” 7.4–8.0 Five aromatic H Multiplets

The characteristic aldehydic resonance and the formula \(\ce{C7H6O}\) establish D as benzaldehyde.

E — Methyl addition to the aldehyde

The spectrum of E contains a methyl doublet, a methine quartet and an exchangeable hydroxyl proton. Together with five aromatic hydrogens, these identify 1-phenylethanol.

It is obtained from D by addition of methylmagnesium bromide, SMILES C[Mg]Br, followed by acidic work-up:

\[ \ce{C6H5CHO + CH3MgBr + H3O+ -> C6H5CH(OH)CH3 + Mg^2+ + Br- + H2O}. \]

The methyl group forms a bond to the aldehyde carbonyl carbon, and the resulting alkoxide oxygen is protonated during work-up. Addition to either face of the planar carbonyl group gives the two enantiomers in equal amounts under achiral conditions.

Label Approximate \(\delta\) / ppm Assignment Multiplicity
E1 7.3 Five aromatic H Aromatic envelope
E2 4.8 \(\ce{CH(OH)}\), 1 H Quartet
E3 2.2 Hydroxyl H, 1 H Exchangeable singlet
E4 1.5 \(\ce{CH3}\), 3 H Doublet

The methine proton couples to the three methyl protons, while the methyl protons couple to the methine proton. The acid-catalysed exchange mentioned in the problem explains why the hydroxyl proton does not produce resolved additional splitting. Its shift and linewidth depend on the exchange conditions.

F — Alcohol oxidation and closure of the reaction cycle

Pyridinium chlorochromate (PCC), SMILES c1cc[nH+]cc1.[O-][Cr](=O)(=O)Cl, oxidizes the secondary alcohol E to the ketone acetophenone, F.

Using \(\ce{[O]}\) to denote one oxygen-atom equivalent supplied by the oxidant, the balanced organic transformation is

\[ \ce{C6H5CH(OH)CH3 + [O] ->[PCC] C6H5COCH3 + H2O}. \]

Label Approximate \(\delta\) / ppm Assignment Multiplicity
F1 7.4–8.0 Five aromatic H Multiplets
F2 2.6 \(\ce{COCH3}\), 3 H Singlet

The methyl group has no adjacent proton-bearing carbon, so its signal is a singlet. Its downfield position is consistent with attachment to a carbonyl group. The methine and hydroxyl resonances of E disappear.

Finally, hydrazine, SMILES NN, with sodium hydroxide, SMILES [Na+].[OH-], effects a Wolff–Kishner reduction on heating:

\[ \ce{C6H5COCH3 + N2H4 -> C6H5CH2CH3 + N2 + H2O}. \]

The carbonyl group becomes a methylene group, regenerating ethylbenzene, A. The byproducts are dinitrogen, SMILES N#N, and water, SMILES O.

For completeness, the other species in the balanced work-up equations are ozone O=[O+][O-], zinc [Zn], hydronium [OH3+], zinc(II) ion [Zn+2], magnesium(II) ion [Mg+2], and bromide ion [Br-].

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.