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Solutions for Aromatic Compounds
N
C(CH3)3(CH3)3C
(CH3)3C (CH3)3C
(CH3)3C C(CH3)3
N
Analysis of structure 1 shows the three tert-butyl groups in unique environments in relation to the nitrogen. 
We would expect three different signals in the NMR, as is observed at –110° C. Why do signals coalesce as 
the temperature is increased? Two of the tert-butyl groups become equivalent—which two? Most likely, 
they are a and c that become equivalent as they are symmetric around the nitrogen. But they are not 
equivalent in structure 1—what is happening here? 
What must happen is an equilibration between structures 1 and 2, very slow at –110° C, but very fast at 
room temperature, faster than the NMR can differentiate. So the signal that has coalesced is an average of a 
and a' and c and c'. (This type of low-temperature NMR experiment is also used to differentiate axial and 
equatorial hydrogens on a cyclohexane.)
The NMR data prove that 1 is not aromatic, and that 1 and 2 are isomers, not resonance forms. If 1 were 
aromatic, then a and c would have identical NMR signals at all temperatures.
c'
b' a'
c
b a
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(d)
(c) Yes, the nitrogen should be basic. The pair of electrons on the nitrogen is in an sp2 orbital parallel to 
the ring and is perpendicular to the π system. It cannot be part of the π system as that would require two p 
orbitals occupying the same space.
(b) This molecule is electronically equivalent to cyclobutadiene. Cyclobutadiene is unstable and undergoes 
a Diels-Alder reaction with another molecule of itself. The tert-butyl groups prevent dimerization by 
blocking approach of any other molecule.
(a) Antiaromatic—only 4 π electrons.44
X
Z
Y
45
Mass spectrum: Molecular ion at 150; base peak at 135, M − 15, is loss of methyl.
Infrared spectrum: The broad peak at 3500 cm–1 is OH; thymol must be an alcohol. The peak at 1620 cm–1 
suggests an aromatic compound.
NMR spectrum: The singlet at δ 4.8 is OH; it disappears upon shaking with D2O. The 6H doublet at δ 1.2 
and the 1H multiplet at δ 3.2 are an isopropyl group, apparently on the benzene ring. A 3H singlet at δ 2.3 
is a methyl group, also on the benzene ring.
 Analysis of the aromatic protons suggests the substitution pattern. The three aromatic hydrogens 
confirm that there are three substituents. The singlet at δ 6.6 is a proton between two substituents (no 
neighboring Hs). The doublets at δ 6.75 and δ 7.1 are ortho hydrogens, splitting each other.
+ CH3 + OH + CH(CH3)2
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