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*,…
Compound A is a D-aldopentose that can be oxidized to an optically inactive aldaric acid B. On Kiliani-Fischer chain extension, A is
converted into C and D; C can be oxidized to an optically active aldaric acid E, but D is oxidized to an optically inactive aldaric acid
F.
What is the structure of compound F?
• Use the wedge/hash bond tools to indicate stereochemistry where it exists.
You do not have to explicitly draw H atoms.
If a group is achiral, do not use wedged or hashed bonds on it.
• Show stereochemistry in a meso compound.
• Do not include lone pairs in your answer. They will not be considered in the grading.
All 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.
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_forwardRibose, 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.arrow_forwardHyaluronic 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_forward
- 1. Trehalose is a disaccharide that can be obtained from fungi, sea urchins and insects. Acid hydrolysis of trehalose yields only D-glucose. Trehalose is hydrolysed by α-glucosidase and not by β-glucosidase enzymes. Methylation of trehalose followed by hydrolysis yields two molar equivalents of 2,3,4,6-tetra-O-methyl-D-glucopyranose.From the following experimental data, deduce the structure of trehalose.What will be the effect of trehalose on Fehling’s solution? 2.Suggest a test you will use to show that a given food substance contains proteinarrow_forwardThe anticoagulant heparin is a polysaccharide that contains alternating residues of -D- glucuronic acid-6- sulfate and N-sulfo-D-glucosamine-6sulfate connected by (1 B 4)- glycosidic linkages. Draw a part of heparin that shows one each of the two residues.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_forward
- A 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_forwardRibose, 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_forward(b) For the following reactions, draw structures (including any stereoisomers) for the products A-C. 1. LIAIH4 2. H30arrow_forward
- Trehalose, 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_forward5. Provide suitable responses for questions (a) – (). 6 CH2OH 4 OH OH 3 OH (a) What is the relative configuration of the above monosaccharide? (b) Which labeled carbon is the anomeric carbon? (c) Trace and identify the acetal in the above monosaccharide. (d) Draw the hemiacetal that results from above acetal. (e) Draw the open chain equivalent of the sugar in part (d). (f) Classify the monosaccharide below as a D-sugar or an L-sugar. H. OH O. OH CH,OH OH OHarrow_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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