1559T_ch26_454-46810/20/0515:53Pa9e457 EQA Solutionsto Problems457 COOH 900 (g)HN-H NH2 (1)HN-H NH2 (7) (CH)NHCNH> (CH2)3NHCNHz C00 C00 HaN--H NH2 (12) H2N-H NH (14) (CH2)NHCNHz 900 H (1)H. H (7)H2 H(9.5)H, H(12) 12 CH OH OH 0- 29.(a)Arg.Lys:(b)Ala.Ser.Tyr.His,Cys:(e)Asp single and a single roup.Their average is the pl. ()(9.0+10.5/2=9.7 (d(9.2+6.1)/2=7.6 (e(8.2+2.0/2=5.1 (0(6.7+1.92=2.8 (g(12.5+9.0M2=10.8 h)(9.1+2.2)/2=5.7 31.(a)Because the R group is secondary.alkylation routes should be avoided.Use the Strecker synthesis. NH3 CH(CO,CH,CH) (g) (h) 29. (a) Arg, Lys; (b) Ala, Ser, Tyr, His, Cys; (c) Asp 30. First identify the net charge-neutral structure, which is always the one with a single
and a single charge. Choose the two pKa values that bracket that structure. They are, repectively, the pKa for deprotona￾tion of the most acidic group in that structure and the pKa for protonation of the most basic group. Their average is the pI. (c) (9.0
10.5)/2 9.7 (d) (9.2
6.1)/2 7.6 (e) (8.2
2.0)/2 5.1 (f) (3.7
1.9)/2 2.8 (g) (12.5
9.0)/2 10.8 (h) (9.1
2.2)/2 5.7 31. (a) Because the R group is secondary, alkylation routes should be avoided. Use the Strecker synthesis. (b) The R group is primary; now you have a choice. Either the Strecker synthesis starting with (CH3)2CHCH2CHO or a Gabriel-based method will do. 1. NaOCH2CH3, CH3CH2OH 2. BrCH2CH(CH3)2 3. H
, H2O, 
NH3 N O O CH(CO2CH2CH3)2 (CH3)2CHCH2CHCOO H
, H2O 1. NH3 2. HCN (CH3)2CHCHO (CH3)2CHCHCN NH2 (CH3)2CHCHCOO
NH3 (1) COO COO COO O H3N
H COOH CH2 OH H3N (7)
H CH2 OH H2N H (9.5) CH2 OH H2N H (12) CH2 H3N NH2 (1)

H COOH (CH2)3NHCNH2 H3N (7)
H COO H2N H (12) COO H2N H (14) COO NH2
(CH2)3NHCNH2 NH2
(CH2)3NHCNH2 NH (CH2)3NHCNH2 Solutions to Problems • 457 1559T_ch26_454-468 10/20/05 15:53 Page 457