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Solved Examples on Biomolecules

Question: 1,
a) Supply structures for H through K. Given:

b) Explain the last step (c). What is net structural change (d) Name this overall method. (e) Discuss the possibility of epimer formation.


a) H is an oxime HOCH2(CHOH)4CH = NOH; I is the completely acetylated oxime, AcOCH2(CHOAc)4CH = NOAc that loses 1 mole of HOAc to form J, AcOCH2(CHOAc)4 CºN; K is an aldopentose, HOCH2(CHOH)3CHO.

b) The acetates undergo transesterification to give methyl acetate freeing all the sugar OH’s. This is followed by reversal of HCN addition.

c) There is loss of one C from the carbon chain.

d) Wohl degradation

e) The a-CHOH becomes the –CH = O without any configurational changes of the other chiralcarbons. Thus no epimers are formed.

Get Chapter Test for this Topic

Question: 2, a) Show how an aldohexose can be used to synthesize 2-ketohexose. (b) Since glucose is converted to fructose by this method, what can you say about the configurations of C3, C4 and C5 in the sugars.



Here aldohexose reacts with one molecule of phenylhyrazine which condenses with the aldehyde group to give phenylhydrazone. When warmed with excess of phenyl hydrazine, the secondary alcoholic group adjacent to the aldehyde group is oxidised by another molecule of phenylhydrazine, to a ketonic group. With this ketonic group, the third molecule of phenylhydrazine condenses to give osazone. The phenylhydrazinyl group is transferred from osazone to C6H5CHO giving C6H5CH = N×NHC6H5 and a dicarbonyl compound called an osone. The more reactive aldehyde group of the osone is reduced, not the less reactive keto group and it gives the 2-ketohexose.


b) The configurations of these carbons which are unchanged in the reactions, must be identical in order to get the same osazone

Question:3,  Why should isoelectric point for Aspartic acid (2.98) be so much lower than that of leucine?

Solution: This may be explained by considering following ion equilibria


It is apparent  that ions (A) and (B) are neutral, while (C) is a cation and (D) is dianion. In species (D), the anion is derived from the second —COOH group present in aspartic acid and is not possible in leucine. At neutral pH a significant concentration of (D), will be present in aqueous solution. It will therefore, be necessary to decrease the pH of such a solution if the formation of (D) is to be suppressed to a stage where anions and cations are present in equal concentration
(the isoelectric point).


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