1559T_ch24_423-43810/20/054:54 AM Page433 ⊕ EQA Solutions to Problems.433 CHO H -OH You now kno these sugars B.A.a H -OH D-Rihose Next.(iv)tells you that CHO COOH COOH CHOH CHoH←Tet 、H0- —H H -OH H-C-OH H OH H OH H-C-OH H OH H-C-OH H OH CHOH COOH COOH The stereocenters ir CH-OH =0 HO- H H- H- H -OH CH2OH D-Sedoheptulose CHO COOH Ho. HO- 一H HO- HO_ -H HNO, HO- -H HO- HO- OH OH CH-OH COOH Active Next, (iv) tells you that From this information, you now can work backward toward the unknown: The stereocenters in D-sedoheptulose must be 3S, 4R, 5R, and 6R. 47. Two aldoheptoses are formed. After HNO3 treatment, one gives an optically active diacid, the other an inactive meso compound. HNO3 COOH COOH Active CH2OH CHO HO H HO H HO H HO H H OH HO H HO H HO H HO H H OH D-Sedoheptulose O CH2OH CH2OH OH HO H H OH H OH H CHO CH2OH Aldohexose B CHOH C C OH HNO3, H2O,  H H C OH H OH COOH COOH This is said to be optically active Otherwise the product would be meso This carbon must be S CHOH C H C OH H C OH H OH COOH COOH HO H H OH H OH H OH D-Ribose You now know these carbons are R in sugars B, A, and D-sedoheptulose as well. CH2OH CHO OH H OH H OH H Solutions to Problems • 433 1559T_ch24_423-438 10/20/05 4:54 AM Page 433