Organic Chemistry
9th Edition
ISBN: 9781305080485
Author: John E. McMurry
Publisher: Cengage Learning
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Chapter 25.SE, Problem 28VC
Interpretation Introduction
Concept introduction:
We redraw the pyranose cyclic structure of the given L aldohexose as follows:
For the L sugar the CH2OH group is at the bottom of the ring.
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(d) Use the diagram below to complete the cyclic alpha form of structure V
(e) Circle the hemiacetal in cyclic alpha form of structure V.
(f) Redraw the cyclic alpha form of structure V but replace the OH group on the anomericcarbon with a methoxy group. Is this modified monosaccharide a reducing sugar or anonreducing sugar?
3a.
3b.
3c
3d.
3e.
CO₂
clavulanic acid
CH₂-OH
H
Answer the following questions about clavulanic acid.
Does clavulanic acid inhibit D-alanyl-D-alanine transpeptidase?
Does clavulanic acid contain a ß-lactam?
Does clavulanic acid contain a thiazolium ring?
What is the result of the treatment of penicillinase with clavulanic acid?
Does clavulanic acid form a covalent acyl-enzyme intermediate with
penicillinase?
can you please add an explanation.
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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- Trehalose, C12H22O11, is a nonreducing sugar that is only 45% as sweet as sugar. When hydrolyzed by aqueous acid or the enzyme maltase, it forms only d-glucose. When it is treated with excess methyl iodide in the presence of Ag2O and then hydrolyzed with water under acidic conditions, only 2,3,4,6-tetra-O-methyl-d-glucose is formed. Draw the structure of trehalosearrow_forwarda) The D-aldopentose A, C5H1005, reacts with HNO3 to yield an optically active aldaric acid B. Kiliani-Fischer chain extension of A produces a pair of D-aldohexoses C and D. C is converted by HNO3 to an optically active aldaric acid, but D is converted by HNO3 to an optically inactive aldaric acid. Write acyclic Fischer projections for A, B, C, D. b) 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 MeO MeO MOH OMe mOH OMe OMearrow_forwarda) The D-aldopentose A, C5H1005, reacts with HNO3 to yield an optically active aldaric acid B. Kiliani-Fischer chain extension of A produces a pair of D-aldohexoses C and D. C is converted by HNO3 to an optically active aldaric acid, but D is converted by HNO3 to an optically inactive aldaric acid. Write acyclic Fischer projections for A, B, C, D.arrow_forward
- The following observations are obtained after a D-hexose was made to react with several reagents: (1) The reactions of a D-hexose with (a) to (d) below yields an aldaric acid (a) NH₂OH, (b) (CH3CO)₂O, NaOCOCH 3, and, (c) NaOCH3, and then, (d) HNO3, H₂O (2) HNO3 oxidation of the same D-hexose gives an aldaric acid. Predict the structures of the three (3) possible hexoses that can undergo the above reactions?arrow_forward27. Which of the following statements about cholesterol is not correct? CH3 CH3 H. d но Cholesterol (a) Cholesterol is a steroid that contains a tetracyclic ring system. (b) Cholesterol is a steroid that contains 8 chiral carbons and can form 28 or 256 stereoisomers. (c) Each atom or group attached to a ring-junction carbon (i.e., carbons a -e) is in a trans or axial position. Because of this the tetracyclic ring system is mostly flat. (d) Cholesterol is used to synthesized vitamin D, bile acids, sex hormones, and adrenocorticoid hormones. (e) Cholesterol is not found in the cell membranes of animals.arrow_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_forward
- Kindly answer question f,g,harrow_forwardThe 1H NMR spectrum of d-glucose in D2O exhibits two high-frequency doublets. What is responsible for these doublets?arrow_forward10. For each of the following structures, decide whether the carbohydrate is: (a) an aldose or a ketose (b) a tetrose, pentose or hexosearrow_forward
- Kindly answer question e, f, garrow_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 NaIO4 then with H3O+. OH HO OH OH Ribose, C5H10O5 H MeOH H* A & B isomeric cyclic acetals with formula C6H₁2O5 1. NalO4 2. H₂O* MeOH products Assuming that ribose formed a six-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.arrow_forwardTreatment of a 258 mg sample of amylopectin by the methylation and hydrolysis procedure described yielded 12.4 mg of 2,3‑di‑O‑methylglucose. Determine what percentage of the glucose residues in amylopectin contained an (α1→6) branch. (Assume that the average molecular weight of a glucose residue in amylopectin is 162 g/mol and the molecular weight of 2,3‑di‑O‑methylglucose is 208 g/mol.) ( α1→6) branched glucose residues: %arrow_forward
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