1559Tch24423-43810/20/054:54 AM Page428 EQA 428.Chapter 24 CARBOHYDRATES:POLYFUNCTIONAL COMPOUNDS IN NATURE HOCH, CH,OH OH CH2OH OH (d) OH OH OH HO OH OH CH,OH OH CH2OH o-Furanose B a-Pyranose B HOCH> CH2OH HO -H CH2OH OH CH,OH OH (e) OH HO OH HO { HO OH CH>OH IOH CH>OH a-Furanose a-Pyranose & 36.No.They are all hemiacetals and therefore are capable of readily interconverting their o and B anomers. HO CH,OH 00H 一0 37.(a) 一OH (b)HO OH OH OH HO ⊕ OH HOCH2 CH,OH ⊕ HO个 HO 07 OH (c) HO OH (d) ∠OH HO (d)is an unusual case where the CH,OH is forced to be axial to allow all four OH's to be equatorial. 38.Base-catalyzed enolization allows the interconversion to take place.But notice that the product is not an ordinary enol.It has hydroxy groups on both carbons of the double bond:It is an enediol.Therefore. when it tautomerizes it has the option of losing a proton from either of two hydroxy groups,giving either the original ketone or the isomeric aldehyde. CH2OH CHOH HC-OH HC-OH H H20 =0 C-Q: C-OH -OH Ketose Enolate Enediol HC-0 Hc-O HC- 6: H2O OH HC-OH- H←一 OH Aldose Other enolate
(d) (e) 36. No. They are all hemiacetals and therefore are capable of readily interconverting their and anomers. 37. (a) (b) (c) (d) (d) is an unusual case where the CH2OH is forced to be axial to allow all four OH’s to be equatorial. 38. Base-catalyzed enolization allows the interconversion to take place. But notice that the product is not an ordinary enol. It has hydroxy groups on both carbons of the double bond: It is an enediol. Therefore, when it tautomerizes it has the option of losing a proton from either of two hydroxy groups, giving either the original ketone or the isomeric aldehyde. Aldose Other enolate H2O OH H2O OH OH HC O HC OH HC O C OH HC O C O O CH2OH C Ketose Enolate Enediol H2O OH H2O OH CHOH C HC C OH OH HC OH O O C HO O HO OH OH CH2OH HOCH2 O HO HO OH OH CH2OH O HO HO OH OH O HO OH OH OH OH -Furanose -Pyranose OH O HO HO OH HO O CH2OH CH2OH OH CH2OH OH CH2OH HOCH2 HO H OH CH2OH OH OH OH HOCH2 CH2OH -Furanose O -Pyranose CH2OH HO OH OH OH O CH2OH OH CH2OH OH 428 • Chapter 24 CARBOHYDRATES: POLYFUNCTIONAL COMPOUNDS IN NATURE 1559T_ch24_423-438 10/20/05 4:54 AM Page 428
1559T_ch24_423-43811/8/0510:14Pa9e429 ⊕ EQA Solutions to Problems.429 (a)CH2OH H,C=0 C=0 →C02 that is.2 formaldehyde I CO, CH2OH H,C=0 (b)HC=O HCO2H HC-OH HCO,H HC 一OH HCO,H that is.4 formic acid 1 acetaldehyde HC -OH HC-OH HC=0 CH3 CH (e)CH2OH H,C=0 (HC -OH4一4HC0H that is,4 formic acid 2 formaldehyde H2C=0 40.(a)0 COOH COOH (iii) HO- -H HO- H HO -H H -OH H -OH H -OH COOH CH-OH D-Threonic acid D-Tartaric acid D-Threitol COOH CH=NNHCHs (b)(i) COOH OH H C-NNHC Hs o HO- -H HO- H H -OH H 一OH H —OH CH-OH COOH D-Xvlonic acid D-Xylaric acid
39. Each carbon atom in the starting sugar is alongside its product after HIO4 cleavage. The number of hydrogens on each carbon remains the same before and after cleavage. (a) (b) (c) 40. (a) (i) (ii) (iii) (iv) (b) (i) (ii) COOH COOH D-Xylaric acid OH HO H H OH H D-Xylonic acid CH2OH COOH OH HO H H OH H D-Threose phenylosazone* CH2OH NNHC6H5 CH NNHC6H5 H OH C D-Threitol CH2OH CH2OH HO H H OH D-Tartaric acid COOH COOH HO H H OH D-Threonic acid COOH CH2OH HO H H OH (HC OH)4 CH2OH H2C O H2C O that is, 4 formic acid 2 formaldehyde CH2OH 4 HCO2H OH HC O HC HC HC OH OH HC OH CH3 CH3 HCO2H HCO2H HCO2H HCO2H HC O that is, 4 formic acid 1 acetaldehyde O CH2OH CH2OH C H O 2C CO2 H O 2C that is, 2 formaldehyde 1 CO2 Solutions to Problems • 429 1559T_ch24_423-438 11/8/05 10:14 Page 429