Chemistry Reference
In-Depth Information
18.31
Hydrolysis occurs at all glycosidic, ester, and amide linkages. The products are as
follows:
H
H
H
O
OCH
O
N
H
3
H
N
O
H
OH
nicotinic acid
ammonia
ribose (2 mol)
N
H
2
N
N
H
3
PO
4
N
N
H
phosphoric acid
(2 mol)
adenine
18.32
The formula for uridine
mono
phosphate is shown in Sec. 18.10 of the text, and the
formula for
α
-D-glucose is shown in eq. 16.3. The structure of UDP-glucose is
O
H
H
N
O
H
OH
O
O
O
N
OH
O
P
O
P
O
H
H
O
-
-
OH
O
O
OH
OH
18.33
There are no N-H bonds in caffeine. Therefore (unlike adenine and guanine),
caffeine cannot form N-glycosides. Basic sites are indicated by the red in ESP maps.
Thus the non-methylated nitrogen in the 5-membered ring is most basic. Suppose
you did not have a color-coded ESP map. How could you determine which nitrogen
is most basic? First, remember that a base is an electron pair donor. Next, examine
the nitrogens. Two of the nitrogens are amides (see Sec. 11.7). One nitrogen has its
lone pair delocalized as part of an aromatic ring (see Sec. 13.6). Only one nitrogen
has an electron pair that is not involved in bonding. Thus, this will be the most basic
nitrogen.
O
C
H
3
C
H
3
lone pair part of
aromatic ring
N
N
amide
non-bonded lone pair in
sp -hybridized orbital
N
2
O
N
amide
C
H
3
18.34
a.
For theobromine and theophylline, the most basic nitrogen is the one with
the non-bonded lone pair available in an
sp
2
-hybridized orbital.
