**Question 30: The shown structure is a cyclic D-monosaccharide. Which of the following statements is true?****Molecular Structure Explanation:**The diagram presents a cyclic D-monosaccharide. It features a five-membered ring with an oxygen atom. Attached to the ring are various hydroxyl (OH) groups and hydrogen (H) atoms. Specifically, the carbon atoms in this cyclic structure are bonded as follows:- The first carbon is bonded to a hydroxymethyl group (HOCH2).- The second carbon is bonded to an oxygen atom and a hydrogen atom.- The third and fourth carbons both have hydroxyl groups connected to hydrogen atoms.- The fifth carbon has a hydroxyl group and a hydrogen atom, along with a terminal methanol group (CH2OH).**Statements to Evaluate:**A. The chemical can react with methanol to form a hemiacetal in solution without forming an open chain structure.B. The chemical needs to form an open chain structure to be able to react with methanol to produce a glycosidic linkage in solution.C. Both A and B.D. Neither A nor B.**Explanation:**In the context of chemical reactions, particularly those involving carbohydrates and methanol, consider the following:- **Hemiacetal Formation (Statement A):** Hemiacetals generally form when an alcohol reacts with an aldehyde or ketone; in the case of this cyclic monosaccharide, it can already behave as a hemiacetal due to its structure.- **Glycosidic Linkage Formation (Statement B):** For glycosidic linkages, it typically involves an open-chain form undergoing a reaction with an alcohol to form an acetal, highlighting the necessity for an open-chain structure.Hence, after analyzing the given chemical structure and the nature of its reactions:- It can be seen that option **B** is correct.**Correct Answer:**B. The chemical needs to form an open chain structure to be able to react with methanol to produce a glycosidic linkage in solution.

Aldehyde and ketone are both functional groups bearing the carbonyl carbon (-C=O)Chemically,…

Answered: 30. The shown structure is a cyclic…

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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Which linear monosaccharide below corresponds to this cyclic one? A) A B) B C) C D) D E) None of them
3). reactions of the following alkyl iodide. Danny uses potassium tert-butoxide to promote the reaction while Claire uses water and high temperatures. One of their reactions has provided a single elimination isomer while the other has provided a mixture of elimination isomers. Organic chemistry students Danny and Claire are both performing elimination Me H20, heat Danny's Reaction Claire's Reaction H. Me a) Which student's reaction (Danny or Claire) has provided a single elimination isomer? b) Draw all the elimination isomers obtained from both students' reactions. c) Which of the dienes drawn (by you) in question 3b is the most stable? Circle that diene above.
The structure of the dipeptide Gly-Asn is given by . The structures of the amino acids Gly and Asn are given below. NH2 H. ČH2 H3N-C-CO H. H3N-C-CO H. Asn Gly NH2 CH2 H H H2 C COO 2. H3N-C -C H. 1. H&N C C-NH C COO NH, NH2 CH2 H. H2 C C NH2 3. H3N-C C- -C COO 4. 0OC-C NH3" Click Save and Submit to save and submit. Click Save All Ansuwers to save all answers. MacBook Air esc 吕0 DII DD F1 F2 F3 F4 F5 F6 F7 F8 F9 @ 2$ & 1 2 4 5 7 8 Q W E R Y | A S D F G K C V N M < CO # 3 HICH
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Ramipril is an ACE inhibitor that contains an ester, amide, and other functional groups. Part: 0/2 Part 1 of 2 Hydrolysis of the ester in ramipril forms ramiprilat, which is the active ACE inhibitor. Modify the structure of ramipril below to show the structure of ramipriat. ОН H x □: ✓
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