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6 - ANSWERS to written answer questions 21 a 10 ‘ ‘b...

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Unformatted text preview: ANSWERS to written- answer questions 21. a. 10 ‘ ‘b. 4“““c; 2 d. 20 e. 1 f. 9. ' g. 10 11.3 . x 22. AG = - n F AB 11 = number of electrons transferred - F = Faraday constant 23. Thr__e_e. The glycosidic linkage must be between the two anomeric -OH groups, since the sugar is non-reducing. The anomeric —OH of each ghlcoSe can be either alpha or beta, so the glyCosidic linkages of the three isomers are: a1pha>a1pha; beta->beta; and alpha1->beta2. (Note that beta1->a1pha2 IS the Same molecule as a1pha1—>beta2, since both monosacchande components are gluCOSe) 24. II I H3C—(CH2)14—C—O—(l3H2 "C—o4CH 4 lo", I 7‘ CH3 II] _ I I 1+ 0 .- CHz—o— fi—O—CHz-CHz—N—CH3 ' , o " ’- ' CH3 4' 1011. OH... __ 4‘ I ' 'O—fi—O—CHZ - 2' ...
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