Organic Chemistry
9th Edition
ISBN: 9781305080485
Author: John E. McMurry
Publisher: Cengage Learning
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Chapter 25.SE, Problem 34MP
Interpretation Introduction
Interpretation:
Description of a biological process.
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4. Glycolysis is a metabolic pathway that converts glucose into pyruvate while synthesizing
high-energy molecules. During one step of Glycolysis, glucose-6-phosphate (an aldose) is
converted to fructose-6-phosphate (a ketose) via an ene-diol intermediate.
OH OH O
OP OH OH
ОН
ОН
glucose-6-phosphate
OH OH OH
OP OH
fructose-6-phosphate
Draw arrow pushing mechanism to describe this isomerization. Use B: and HA as surrogates
for basic/acidic amino acid residues.
One 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-phosphate
In the biosynthesis of aromatic amino acids, erythrose-4-phosphate undergoes electrophilic addition to
phosphoenolpyruvate (PEP). Draw the products of this step, paying particular attention to regiochemistry.
ОН
OPOH
ОН
?
Enzyme
Erythrose 4-phosphate
Phosphoenolpyruvate
(PEP)
Chapter 25 Solutions
Organic Chemistry
Ch. 25.1 - Prob. 1PCh. 25.2 - Prob. 2PCh. 25.2 - Prob. 3PCh. 25.2 - Prob. 4PCh. 25.2 - Prob. 5PCh. 25.3 - Prob. 6PCh. 25.3 - Prob. 7PCh. 25.4 - Prob. 8PCh. 25.4 - Prob. 9PCh. 25.4 - Prob. 10P
Ch. 25.5 - Prob. 11PCh. 25.5 - Prob. 12PCh. 25.5 - Prob. 13PCh. 25.5 - Prob. 14PCh. 25.5 - Prob. 15PCh. 25.6 - Prob. 16PCh. 25.6 - Prob. 17PCh. 25.6 - Prob. 18PCh. 25.6 - Prob. 19PCh. 25.6 - Prob. 20PCh. 25.6 - Prob. 21PCh. 25.6 - Prob. 22PCh. 25.6 - Prob. 23PCh. 25.7 - Prob. 24PCh. 25.8 - Show the product you would obtain from the...Ch. 25.SE - Prob. 26VCCh. 25.SE - Prob. 27VCCh. 25.SE - Prob. 28VCCh. 25.SE - Prob. 29VCCh. 25.SE - Prob. 30MPCh. 25.SE - Prob. 31MPCh. 25.SE - Glucosamine, one of the eight essential...Ch. 25.SE - D-Glicose reacts with acetone in the presence of...Ch. 25.SE - Prob. 34MPCh. 25.SE - Prob. 35MPCh. 25.SE - Prob. 36APCh. 25.SE - Prob. 37APCh. 25.SE - Prob. 38APCh. 25.SE - Prob. 39APCh. 25.SE - Prob. 40APCh. 25.SE - Assign R or S configuration to each chirality...Ch. 25.SE - Prob. 42APCh. 25.SE - Prob. 43APCh. 25.SE - Prob. 44APCh. 25.SE - Prob. 45APCh. 25.SE - Prob. 46APCh. 25.SE - Prob. 47APCh. 25.SE - Prob. 48APCh. 25.SE - Prob. 49APCh. 25.SE - Prob. 50APCh. 25.SE - Prob. 51APCh. 25.SE - Prob. 52APCh. 25.SE - Prob. 53APCh. 25.SE - Prob. 54APCh. 25.SE - Prob. 55APCh. 25.SE - Prob. 56APCh. 25.SE - Prob. 57APCh. 25.SE - Prob. 58APCh. 25.SE - Prob. 59APCh. 25.SE - Prob. 60APCh. 25.SE - Prob. 61APCh. 25.SE - Prob. 62APCh. 25.SE - Prob. 63APCh. 25.SE - D-Mannose reacts with acetone to give a...Ch. 25.SE - Prob. 65APCh. 25.SE - Prob. 66APCh. 25.SE - Prob. 67APCh. 25.SE - Prob. 68APCh. 25.SE - Prob. 69APCh. 25.SE - Prob. 70APCh. 25.SE - Prob. 71AP
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- One of the steps in the pentose phosphate pathway for glucose catabolism is the reaction of sedoheptulose 7-phosphate with glyceraldehydes 3-phosphate in the presence of a transaldolase to yield erythrose 4-phosphate and fructose 6-phosphate. (a) The first part of the reaction is the formation of a protonated Schiff base of sedoheptulose 7-phosphate with a lysine residue in the enzyme followed by a retro-aldol cleavage to give an enamine plus erythrose 4-phosphate. Show the structure of the enamine and the mechanism by which it is formed. (b) The second part of the reaction is a nucleophilic addition of the enamine to glyceraldehyde 3-phosphate followed by hydrolysis of the Schiff base to give fructose 6-phosphate. Show the mechanism.arrow_forward5)A certain aerobic organism is able to metabolize the followingglycolipid CH,OH H H OH H HO OH A.Draw the 2 resulting structures that would occur upon initial hydrolysis of the O-glycosidic bond.arrow_forwardCholesterol-lowering drugs commonly target O HMG-COA synthase, the enzyme that catalyzes the synthesis of mevalonate O pyruvate decarboxylase complex, which catalyzes the synthesis of acetyl-CoA O mevalonate, an isoprene O HMG-COAreductase, the rate-limiting enzyme in cholesterol biosynthesis O HMG-CoA synthase, the enzyme that catalyzes the first step in cholesterol biosynthesisarrow_forward
- The aldonic acid of d-glucose forms a five-membered-ring lactone. Draw its structure.arrow_forwardIn an aqueous solution, d-glucose exists in equilibrium with two six-membered ring compounds. Draw the structures of these compounds.arrow_forwardThe hydrolysis of pyrophosphate to orthophosphate drives biosynthetic reactions such as DNA synthesis. In Escherichia coli, a pyrophosphatase catalyzes this hydrolytic reaction. The pyrophosphatase has a mass of 120 kDa and consists of six identical subunits. A unit of activity for this enzyme, U, is the amount of enzyme that hydrolyzes 10 umol of pyrophosphate in 15 minutes. The purified enzyme has a Vnax of 2800 U per milligram of enzyme. When (S] >> KM, how many micromoles of substrate can 1 mg of enzyme hydrolyze per second? Vnax = umol -s. mg- If cach enzyme subunit has one active site, how many micromoles of active sites, or (E]r, are there in 1 mg of enzyme? (Er = umol - mg-arrow_forward
- 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 Nearrow_forwardWrite down the reactions: D-Glucose + HNO3 →arrow_forward4. Identify the component monosaccharides of each of the following compounds and describe the type of glycosidic linkage in each. Но он Но OH HO он Но- Но- (a) OH (c) CH,OHO. (b) CH2OHO Lon OH HO H ČHOH H OH ÓH ОНarrow_forward
- Vanillin (4-hydroxy-3-methoxybenzaldehyde), the principal component of vanilla, occurs in vanilla beans and other natural sources as a b-d-glucopyranoside. Draw a structural formula for this glycoside, showing the d-glucose unit as a chair conformation.arrow_forwardDraw the product formed when CH3CH2CH2CH2CH=CH2 is treated with either(a) H2O, H2SO4; or (b) BH3 followed by H2O2, HO−.arrow_forwardIdentify 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) - mutarotationarrow_forward
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