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Solutions for Carbohydrates and Nucleic Acids
39 This problem requires logic, inference, and working backwards, just as it did for Emil Fischer.
HHO
OHH
COOH
HHO
COOH
This is the only optically 
active 5-carbon aldaric acid.
came
from
HHO
OHH
CH2OH
HHO
CHO
HHO
OHH
CHO
HHO
CH2OH
OR
 This could 
be arabinose.
 This could 
be arabinose.
(a) Glucose and mannose degrade by one carbon to give arabinose. Parts (a) and (b) determine the structure 
of arabinose.
From the nitric acid oxidation, 
we do not know which end is 
the alcohol and which end is 
the aldehyde.
(b) Once we get to the 4-carbon sugars, then we can tell which 5-carbon sugar must be arabinose.
HHO
OHH
CH2OH
HHO
CHO
Ruff
HHO
CHO
OHH
CH2OH
This possible arabinose degrades to this 4-carbon sugar, oxidation of which gives an optically 
active aldaric acid. Conclusion: this starting material is not arabinose.
HNO3
HHO
COOH
OHH
COOH
This is optically active; it 
has two asymmetric 
carbons and no plane of 
symmetry.
OHH
OHH
CH2OH
HHO
CHO
Ruff
OHH
CHO
OHH
CH2OH
This possible arabinose degrades to this 4-carbon sugar, oxidation of which gives an optically 
inactive aldaric acid. Conclusion: this starting material is arabinose, and the 4-carbon sugar is 
erythrose.
HNO3
OHH
COOH
OHH
COOH
This is optically inactive; 
it has two asymmetric 
carbons and a plane of 
symmetry. This is meso-
tartaric acid.
(This structure is rotated 180° in the 
second part of part (b) to put the 
aldehyde at the top, as is customary.)
OHH
OHH
CH2OH
HHO
CHO
OHH
CHO
OHH
CH2OH
erythrosearabinose
CONCLUSIONS from (a) and (b):
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