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CHAPTER 22 953 
 
aromatic ring. The integration of this signal is 5, which 
indicates that the ring is monosubstituted: 
 
 
 
The presence of a monosubstituted ring is confirmed by 
the four signals between 100 and 150 ppm in the 13C 
NMR spectrum (the region associated with sp2 
hybridized carbon atoms), just as expected for a 
monosubstituted aromatic ring: 
 
 
 
In the 1H NMR spectrum, the pair of triplets (each with 
an integration of 2) indicates a pair of neighboring 
methylene groups: 
 
 
 
These methylene groups account for the two upfield 
signals in the 13C NMR spectrum. 
Thus far, we have accounted for all of the atoms in the 
molecular formula, except for one nitrogen atom and two 
hydrogen atoms, suggesting an amino group. This would 
indeed explain the singlet in the 1H NMR spectrum with 
an integration of 2. 
We have now analyzed all of the signals in both spectra, 
and we have uncovered the following three fragments, 
which can be connected to each other in only one way: 
 
 
 
 
22.82. The product has four C-N bonds, each of which 
can be prepared via a reductive amination process. As 
such, the product can be made from the following 
starting materials (two equivalents of ammonia and two 
equivalents of a dialdehyde): 
 
 
The necessary dialdehyde has only three carbon atoms, 
but the starting material (benzene) has six carbon atoms. 
This suggests that we must somehow break apart the 
aromatic ring into two fragments. This might seem 
impossible at first, as we have seen that aromatic rings 
are particularly stable. We did, however, cover a 
reaction that destroys aromaticity (a Birch reduction will 
convert benzene into 1,4-cyclohexadiene). If a Birch 
reduction is followed by ozonolysis, the resulting 
dialdehyde can then be treated with ammonia and 
sodium cyanoborohydride (with an acid catalyst) to give 
the product: 
 
3) DMS
H
O
H
O
HN NH
1) Na, NH3, CH3OH
2) O3
4) [ H+ ], NaBH3CN, NH3Na, NH3
CH3OH
2) DMS
1) O3
NH3
NaBH3CN
[ H+ ]
 
 
22.83. The starting material exhibits a five-membered 
ring, while the product exhibits a six-membered ring that 
contains a nitrogen atom. Since we have not learned a 
way to insert a nitrogen atom into an existing ring, we 
must consider opening the ring, and then closing it back 
up again (in a way that incorporates the nitrogen atom 
into the ring). There are certainly many acceptable 
synthetic routes. One such route derives from the 
following retrosynthetic analysis. An explanation of 
each of the steps (a-d) follows. 
 
 
 
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