Describe, using curly arrows, how limonene is biosynthesized from geranyl pyrophosphate. ham OH OH Geranyl pyrophosphate Limonene Synthase Water -$ Limonene 0=d²-5 0=0-5 OH OH
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- The following molecule has a glycosidic bond that is best described as HỌ – CH2 НО HỌ – CH2 HO OH O ОН alpha-1,4-glycosidic bond beta-1,6-glycosidic band alpha-1,6-glycosidic bond beta-1,4-glycosidic bond ОН OH OH.H2OOne pathway for the metabolism of D-glucose 6-phosphate is its enzyme-catalyzed conversion to D-fructose 6-phosphate. Show that this transformation can be accom- plished as two enzyme-catalyzed keto-enol tautomerisms. СНО CH,OH C=0 enzyme catalysis Но Но H OH H H. O- H- -HO- ČH,OPO, ČH,OPO, D-Glucose 6-phosphate D-Fructose 6-phosphateRaloxifene is a drug used in the treatment of osteoporosis in postmenopausal women and those on glucocorticoids. In a study conducted by Izgelov et al. (2018), the two metabolites derived from raloxifene are shown and their intestinal metabolism are discussed. он Raloxifene 6-Glucuronide HO Raloxifene onge HQ. OH OH OH Raloxifene 4'-Glucuronide Determine the isomer configuration of the labelled double bond (shown in red) from the structure of raloxifene by writing the letter of the configuration in the provided box below. STRICTLY write the letter only.
- The dehydration of citrate to yield cis-aconitate, a step in the citric acid cycle, involves the pro-R “arm’’ of citrate rather than the pro-S arm. Which of the following two products is formed?Eleostearic acid, C18H30O2, is a rare fatty acid found in the tung oil used for finishing furniture. On ozonolysis followed by treatment with zinc, eleostearic acid furnishes one part pentanal, two parts glyoxal (OHC-CHO), and one part 9-oxononanoic acid [OHC(CH2)7CO2H]. What is the structure of eleostearic acid?The following molecule has a glycosidic bond that is best described as HỌ— CH2 НО HO-CH2 НО OH O OH alpha-1,4-glycosidic bond beta-1,6-glycosidic bond alpha-1,6-glycosidic bond beta-1,4-glycosidic bond OH OH OH. H2O
- Give an acceptable name for the major organic product formed by the following reaction: (CH₂CH₂CH₂CH₂2) N + H₂SO4Draw and label a strcture of: Cervonic acid 22:6 Δ4,7,10,13,16,19. This is a fatty acid found in cold fish.Cellulose, the major structural material of wood and plants, consists of a chain of B-D-glucose molecules connected by B-1,4-glycosidic bonds. In comparison, the amylose component of starch consists of a-D-glucose molecules connected by a-1,4-glycosidic bonds. Identify the following structures, denoted A and B, as the structures of cellulose or amylose. A В 0 Н CH2OH H . н Н OH H Н CH₂OH 0 Н ОН Н ОН Н Н ОН О О Н Н CH₂OH -0 Н OH H Н CH₂OH 0 Н ОН Н ОН Н Н ОН OB is the structure of cellulose, while A is the structure of amylose. O A and B both represent the structure of amylose. O A is the structure of cellulose, while B is the structure of amylose. O A and B both represent the structure of cellulose. Н 0 0 H Н CH₂OH до Н ОН Н Н H OH CH₂OH 0 Н ОН Н Н Н ОН Н О Н H CH₂OH -о Н ОН Н Н ОН CH₂OH 0 H OH HA Н ОН Н H 0
- N-NHPH CHO CH2OH H- C=N-NHPH Но- -H- 3 equiv Но- -H- 3 equiv Но- PHNHNH2 PHNHNH, H- H- OH H- -O- H- H- -HO- H- -HO- ČH2OH ČH2OH ČH2OH D-glucose osazone D-fructose + NH3 + PHNH2 + 2 H20 The final step in the formation of the osazone from glucose is the reaction of the keto imine with 2 equivalents of phenylhydrazine to yield the osazone plus ammonia. This reaction involves the following intermediate steps: 1. Addition of phenylhydrazine to the imine and proton transfer to yield intermediate 1; 2. Elimination of ammonia to yield phenylhydrazone 2; 3. Addition of phenylhydrazine to the ketone to yield tetrahedral intermediate 3; 4. Proton transfer yields carbinolamine 4; 5. Elimination of water yields the final product osazone. Write out the mechanism on a separate sh of paper and then draw the structure of tetrahedral intermediate 3. NH НО- -H H- -ОН H- -ОН ОН Glucose keto imine Previous NeIdentify the organic functional group and reaction type for the following reaction. The reactant is a(n) - carboxylic acid hexose - Aldohexose - aldotetrose -deoxyhexose -carboxylic acid tetrose - ketohexose The product is a(n) - carboxylic acid tetrose - aldotetrose -alcohol hexose -aldohexose -carboxylic acid hexose - alcohol tetrose The reaction type is - hemiacetal formation -hydrolysis -oxidation( Benedict’s) -acetal formation -reduction( hydrogenation) - mutarotationRibose, a carbohydrate with the formula shown, forms a cyclic hemiacetal, which, in principle, could contain either a four-membered, five-membered, or six-membered ring. To determine which ring is formed, ribose is treated with methanol in the presence of an acid catalyst. The products are then isolated and treated with NalO, then with H30*. он MEOH 1. NalO. 2. H,O* A & B MEOH + products HO H* isomeric cyclic acetals with formula CgH1205 ÕH Ribose, CSH1005