Organic Chemistry
12th Edition
ISBN: 9781118875766
Author: T. W. Graham Solomons, Craig B. Fryhle, Scott A. Snyder
Publisher: WILEY
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Chapter 22, Problem 24P
Interpretation Introduction
Interpretation:
The understanding of the specificity of bacterial oxidation for commercial synthesis of the compound vitamin C is to be explained.
Concept Introduction:
▸ The chemical name for vitamin C is L-ascorbic acid. It is obtained commercially by converting the D-sorbitol into the L-ascorbic acid with the formation of an
▸ The process involves two major steps which are the formation of
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One step in the gluconeogenesis pathway for the biosynthesis of glucose is the partial reduction of 3-phosphoglycerate to give glyceraldehyde 3-phosphate. The process occurs by phosphorylation with ATP to give 1,3-bisphosphoglycerate,
reaction with a thiol group on the enzyme to give an enzyme-bound thioester, and reduction with NADH.
-OPO3²- Enz-SH
H-C-OH
ATP
CH₂OPO3²-
3-phosphoglycerate
O
0-0--0
O
ADP
CH₂CH3
substitute for
1,3-bisphosphoglycerate
C
H-C-OH
CH₂OPO3²-
1,3-bisphosphoglycerate
O=C
CH3-SH
substitute for
Enz-SH
H
H-C-OH
|
CH₂OPO3²-
PO4³-
O.
S-Enz
H-C-OH
glyceraldehyde 3-phosphate
Propose a structure for the first intermediates in the reaction of 1,3-bisphosphoglycerate with a thiol group on the enzyme to form an enzyme-bound thioester. Assume a basic group on the enzyme catalyzes the formation of this
intermediate.
To simplify the drawing process, substitute the structures below for the 1,3-bisphosphoglycerate and Enz-SH.
CH₂OPO3²-
(Enzyme-bound
thioester)
NADH/H*
NAD*,…
Prostaglandins are a class of eicosanoids, fatty acid derivatives involved in a variety of important phenomena, including fever, inflammation, and pain. The first step in
prostaglandin synthesis involves the conversion of the 20-carbon fatty acid arachidonic acid to PGG2 by the enzyme prostaglandin endoperoxide synthase, or cyclooxy:
Ibuprofen is one of several inhibitors of cyclooxogenase used therapeutically.
Arachidonic acid
20₂
0.5
1.0
1.5
2.5
3.5
cyclooxrygenase
PGG₂
JOH
محمد
The following kinetic data were obtained for the enzyme cyclooxygenase in the absence of any inhibitor (1), and in the presence of ibuprofen (2) at a concentration of 48
[arachidonic acid] (uM)
(1) v(uM/min) (2) v(uM/min)
23.5
32.2
36.9
41.8
44.0
16.67
25.25
30.49
37.04
38.91
Ibuprofen
A. Plot the data in standard Michaelis-Menten form as well as in double-reciprocal form.
B. Determine Vmax and Km for the enzyme.
C. Based on this limited data set, what type of inhibition does ibuprofen likely exhibit?…
Ribose, 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.
When D-ribose is treated with methanol in the presence of an acid catalyst, two cyclic
acetals, A and B, are formed, both with molecular formula C,H,0, These are separated,
and each is treated with sodium periodate (Section 10.8C) followed by dilute aqueous
acid. Both A and B yield the same three products in the same ratios.
он о
CHO
СНО
H+ CH,OH A +B
ÕH
1. NalO,
2. H,0*
НО
CHO + CHOH + CH,OH
ÕH
CH,OH
Isomeric cyclic
acetals with molecular
formula CH12O,
D-Ribose
(C;H1605)
From this information, deduce whether the cyclic hemiacetal formed by D-ribose is four-
membered, five-membered, or six-membered.
Chapter 22 Solutions
Organic Chemistry
Ch. 22 - Prob. 1PPCh. 22 - Prob. 2PPCh. 22 - Prob. 3PPCh. 22 - Prob. 4PPCh. 22 - Prob. 5PPCh. 22 - Prob. 6PPCh. 22 - Prob. 7PPCh. 22 - Prob. 8PPCh. 22 - Practice Problem 22.9 What products would you...Ch. 22 - Prob. 10PP
Ch. 22 - Prob. 11PPCh. 22 - Prob. 12PPCh. 22 - Prob. 13PPCh. 22 - Prob. 14PPCh. 22 - Prob. 15PPCh. 22 - Prob. 16PPCh. 22 - Prob. 17PPCh. 22 - Prob. 18PPCh. 22 - Prob. 19PPCh. 22 - Prob. 20PCh. 22 - Prob. 21PCh. 22 - Prob. 22PCh. 22 - Prob. 23PCh. 22 - Prob. 24PCh. 22 - Prob. 25PCh. 22 - Prob. 26PCh. 22 - Prob. 27PCh. 22 - Prob. 28PCh. 22 - Prob. 29PCh. 22 - Prob. 30PCh. 22 - Prob. 31PCh. 22 - Prob. 32PCh. 22 - Prob. 33PCh. 22 - Prob. 34PCh. 22 - Prob. 35PCh. 22 - Prob. 36PCh. 22 - Prob. 37PCh. 22 - Prob. 38PCh. 22 - Arbutin, a compound that can be isolated from the...Ch. 22 - Prob. 40PCh. 22 - Prob. 41PCh. 22 - Prob. 42PCh. 22 - Prob. 43PCh. 22 - 22.44 The following reaction sequence represents...Ch. 22 - 22.45
The NMR data for the two anomers...Ch. 22 - Shikimic acid is a key biosynthetic intermediate...
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- Treatment of -D-glucose with methanol in the presence of an acid catalyst converts it into a mixture of two compounds called methyl glucosides (Section 25.3A). In these representations, the six-membered rings are drawn as planar hexagons. (a) Propose a mechanism for this conversion and account for the fact that only the OH on carbon 1 is transformed into an OCH3 group. (b) Draw the more stable chair conformation for each product. (c) Which of the two products has the chair conformation of greater stability? Explain.arrow_forwardb) Disaccharide E is a reducing sugar. It is hydrolyzed by an α-glycosidase enzyme, which means it contains an α- glycoside link. Treatment of E with Ag2O and excess Mel gives an octamethyl derivative F. Hydrolysis of F in dilute aqueous acid gives the pair of molecules shown below. Write the structures of E and F. (If the stereochemistry at a particular carbon is not determined by the above data, indicate this with a wavy line as shown below.) HO OMe OMe Is is MeO MeO MOH OMe mOH OMe OMearrow_forwardAll the glucose units in dextran have six-membered rings. When a sample of dextran is treated with methyl iodide and Ag2O and the product ishydrolyzed under acidic conditions, the final products are 2,3,4,6-tetra-O-methyl-d-glucose, 2,4,6-tri-O-methyl-d-glucose, 2,3,4-tri-O-methyl-d-glucose, and 2,4-di-O-methyl-d-glucose. Draw a short segment of dextran.arrow_forward
- Hyaluronic acid, a component of connective tissue, is the fluid that lubricates joints. It is a polymer of alternating N-acetyl-D-glucosamine and D-glucuronic acid subunits joined by B-1,3'-glycosidic linkages. Draw a short segment of hyaluronic acid.arrow_forwardrivatives: Nucleophilic Acyl Substitution Reactions - EOC O C-O [References] One step in the gluconeogenesis pathway for the biosynthesis of glucose is the partial reduction of 3-phosphoglycerate to give glyceraldehyde 3-phosphate. The process occurs by phosphorylation with ATP to give 1,3-bisphosphoglycerate, reaction with a thiol group on the enzyme to give an enzyme-bound thioester, and reduction with NADH. H-C-OH ATP ADP CH₂OPO32- 3-phosphoglycerate 0= OPO32- C H-C-OH CH₂OPO32 1,3-bisphosphoglycerate Enz-SH PO43- O= C S-Enz H-C-OH CH₂OPO3² (Enzyme-bound thioester) NADH/H+ O=C NAD*, Enz-SH H H-C-OH CH₂OPO3²- glyceraldehyde 3-phosphate Propose a structure for the first intermediate in the reaction of the enzyme-bound thioester with NADH to form glyceraldehyde 3-phosphate. To simplify the drawing process, substitute the structure below for the enzyme-bound thioester. S-CH3 substitute for the CH2CH3 enzyme-bound thioester You do not have to consider stereochemistry. • You do not have…arrow_forwardIsoerythrogenic acid, C18H26O2, is a acetylic fatty acid that turns vivid vle on exposure to UV light. On Catalytic hydrogenation over a palladium catalyst, five molar equivalents of hydrogen are absorbed, and stearic acidarrow_forward
- Ribose, 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 NaIO, then with H₂O*. HO OH OH OH Ribose, C5H10O5 H 9.81 ** MeOH H* A & B isomeric cyclic acetals with formula C6H12O5 Assuming that ribose formed a five-membered ring cyclic hemiacetal, draw the structure of the sodium periodate digestion products. ▼ • Use the wedge/hash bond tools to indicate stereochemistry where it exists. • Draw one structure per sketcher. Add additional sketchers using the drop-down menu in the bottom right corner. 1. NalO4, 2. H₂O* MeOH products 4 SIF Previous Nextarrow_forwardTrehalose, C12H22O11, is a nonreducing sugar that is only 45% as sweet as sugar. When hydrolyzed by aqueous acid or the enzyme maltase, it formsonly d-glucose. When it is treated with excess methyl iodide in the presence of Ag2O and then hydrolyzed with water under acidic conditions, only2,3,4,6-tetra-O-methyl-d-glucose is formed. Draw the structure of trehalose.arrow_forwardA D-aldopentose A is oxidized to an optically inactive aldaric acid with HNO3. A is formed by the Kiliani–Fischer synthesis of a D-aldotetrose B, which is also oxidized to an optically inactive aldaric acid with HNO3. What are the structures of A and B?arrow_forward
- Ribose, 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 NaIO4 then with H30*. OH Меон 1. NalO4 2. H30* A & B MEOH + products HO H. H* isomeric cyclic acetals with formula CgH1205 ÕH ÕH Ribose, C5H1005 Assuming that ribose formed a six-membered ring cyclic hemiacetal, draw the structure of compounds A and B.arrow_forward4) Structure: the appropriate box. CHO -OH H- HO- -H HO -H Write the requested cyclized form of the given carbohydrates in B-pyranose a-furanose CH₂OHarrow_forwardA chemist synthesized compound X as a racemic mixture. When the ketone group in X was enzymatically reduced to the corresponding alcohol, a 100% yield was obtained of the product shown below. Choose the statement that best describes this result. ОН enzyme C;H1 `OCH,CH; pH 4.0 C3H1 `OCH,CH3 ОН ÕH X (racemic) (100% yield) One enantiomer of compound X reacts quickly with the enzyme. The other enantiomer of compound X is unreactive, but rapidly equilibrates with the reactive enantiomer under the reaction conditions. Since compound X was racemic, it makes sense that only a single product was obtained. O The product is a meso compound, so either enantiomer of compound X gives the same product. One enantiomer of compound X reacts quickly with the enzyme, while the other enantiomer of compound X remains unchanged.arrow_forward
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