Inventory again
C
N
H
H
3
C H
122.4
3.71
1.17
7.60
50.40
18.89
N
C
H
H CH
3
119.8
1.02
8.15
4.50
17.47
44.62
C
N
H
H
3
C H
121.1
1.05
8.21
4.70
18.14
44.84
N
C
CH
3
H
124.3
1.38
4.00
53.97
16.63
CH
3
≈2.9
30.20
N
C
CH
3
CH
3
H
124.3
≈2.9
1.38
4.00
53.97
16.63
30.20
Finally
CH
C
O
H
N CH
C O
HN
H
C
C
O
N
H
CH
CO
N
Me
Me
Me
Me
Me
Finally we have to end with this molecule.
But what is the order of the amino acids?
No atom is missing anymore. We have eight open bonds
coming from the carbonyl groups, 8 open bonds at the amino
acid fragments and we still need one double bond equivalent.
The solution is easy: We have to place the carbonyl groups
between the amino acid fragments resulting in a ring. The ring
is the last missing double bond equivalent.
• four carbonyl groups (C
4
O
4
)
• four double bond equivalents
• for amino acid fragments (C
9
H
22
N
34
)
• five double bond equivalents
• molecular formula C
13
H
22
N
4
O
4
we have
we need
• all fragments together (C
13
H
22
N
4
O
4
)