Asparagine and aspartic acid are two naturally occurring amino acids. Their structures and isoelectric points are shown in Table 8.1. Table 8.1 amino acid structure isoelectric point asparagine HOOCCH(NH2)CH2CONH2 5.41 aspartic acid HOOCCH(NH2)CH2COOH 2.77

1. (a)Define isoelectric point.[1]
2. (b)Draw the structures of asparagine and aspartic acid at pH2.[2]
3. (c)Asparagine and aspartic acid are treated separately with an excess of LiAlH4. Draw the structures of the organic products of these reactions.[2]
4. (d)(i)Propanedioic acid, HOOCCH2COOH, is treated with an excess of thionyl chloride, SOCl2. Propanedioyl chloride, ClOCCH2COCl, is formed. Write an equation for this reaction.[1]
5. (d)(ii)Propanedioyl chloride reacts with an excess of asparagine to form compound G with molecular formula C11H16N4O8. Each molecule of compound G has four amide groups. Draw the structure of compound G.[2]
6. (e)Asparagine is hydrolysed with an excess of hot NaOH(aq). Draw the structure of the organic product of this reaction.[2]
7. (f)A polymer can form from asparagine, HOOCCH(NH2)CH2CONH2, as the only monomer. Draw a length of the polymer chain containing three monomer residues. Clearly label the repeat unit of the polymer on your diagram.[3]
8. (g)(i)Aspartic acid exists in two optically active forms. Plane polarised light is passed through pure samples of these two optically active forms in solutions of the same concentration. Describe two similarities and one difference in their effect on the plane polarised light. similarities difference[2]
9. (g)(ii)Give the term used to describe a mixture of equal amounts of the two optically active forms.[1]

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