Skip to main content
Ch. 23 - Carbohydrates and Nucleic Acids
Wade - Organic Chemistry 9th Edition
Wade9th EditionOrganic ChemistryISBN: 9780135213728당신이 사용하는 게 아니라요?교과서 변경
모든 교과서Wade 9th EditionCh. 23 - Carbohydrates and Nucleic Acids문제 46
23장, 문제 46

Glucose is the most abundant monosaccharide. From memory, draw glucose in
(a) the Fischer projection of the open chain.
(b) the most stable chair conformation of the most stable pyranose anomer.
(c) the Haworth projection of the most stable pyranose anomer.

검증된 단계별 안내
1
Understand the structure of glucose: Glucose is an aldohexose, meaning it contains six carbons and an aldehyde group. It has the molecular formula C₆H₁₂O₆ and contains four chiral centers.
For part (a), draw the Fischer projection of open-chain glucose: Start by placing the aldehyde group (-CHO) at the top. Then, arrange the hydroxyl (-OH) groups on the chiral carbons in the following order: on C2, the -OH is on the right; on C3, the -OH is on the left; on C4, the -OH is on the right; and on C5, the -OH is on the right. The terminal carbon (C6) is a CH₂OH group at the bottom.
For part (b), identify the most stable chair conformation of the most stable pyranose anomer: Glucose cyclizes to form a six-membered ring (pyranose form). The most stable anomer is β-D-glucopyranose, where the -OH group on the anomeric carbon (C1) is equatorial. Draw the chair conformation with all bulky groups (e.g., -OH and CH₂OH) in equatorial positions to minimize steric hindrance.
For part (c), draw the Haworth projection of the most stable pyranose anomer: Start with a six-membered ring. Place the oxygen atom at the top right of the ring. For β-D-glucopyranose, the -OH group on the anomeric carbon (C1) is drawn above the plane of the ring. The remaining -OH groups are positioned based on their orientation in the Fischer projection: C2 -OH is below, C3 -OH is above, C4 -OH is below, and C6 -CH₂OH is above.
Double-check your drawings for consistency: Ensure that the Fischer projection, chair conformation, and Haworth projection are consistent with each other and represent the correct stereochemistry of glucose. Verify that the β-anomer has the -OH group on the anomeric carbon in the equatorial position (chair) or above the plane (Haworth).

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.

주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Fischer Projection

The Fischer projection is a two-dimensional representation of a molecule, particularly useful for depicting the stereochemistry of carbohydrates. In this format, the carbon chain is arranged vertically, with the most oxidized carbon at the top. The horizontal lines represent bonds that project out of the plane towards the viewer, while vertical lines represent bonds that extend behind the plane. Understanding this projection is essential for accurately drawing the open-chain form of glucose.
추천 영상:
09:56
Monosaccharides - Drawing Fischer Projections

Chair Conformation

The chair conformation is a three-dimensional arrangement of cyclohexane rings that minimizes steric strain and torsional strain, making it the most stable form of cyclohexane derivatives. For glucose in its pyranose form, the chair conformation allows for optimal positioning of substituents, which affects the molecule's stability and reactivity. Recognizing how to draw and interpret this conformation is crucial for understanding the stability of glucose's cyclic forms.
추천 영상:
03:29
Understanding what a conformer is.

Haworth Projection

The Haworth projection is a way to represent cyclic sugars in a two-dimensional format that emphasizes the cyclic structure and stereochemistry. In this representation, the ring is depicted as a planar structure, with substituents drawn above or below the plane to indicate their orientation. This projection is particularly useful for visualizing the anomeric carbon and distinguishing between different anomers of sugars, such as alpha and beta forms of glucose.
추천 영상:
04:03
Monosaccharides - Haworth Projections
관련 실천
교과서 질문

An aliphatic aminoglycoside is relatively stable to base, but it is quickly hydrolyzed by dilute acid. Propose a mechanism for the acid-catalyzed hydrolysis.

1365
views
교과서 질문

Ribonucleosides are not so easily hydrolyzed, requiring relatively strong acid. Using your mechanism for part (a), show why cytidine and adenosine (for example) are not so readily hydrolyzed. Explain why this stability is important for living organisms.

842
views
교과서 질문

Without referring to the chapter, draw the chair conformations of

(b) α-D-allopyranose (the C3 epimer of glucose).

850
views
교과서 질문

Without referring to the chapter, draw the chair conformations of

(c) β-D-galactopyranose (the C4 epimer of glucose).

882
views
교과서 질문

Without referring to the chapter, draw the chair conformations of

(a) β-D-mannopyranose (the C2 epimer of glucose).

905
views
교과서 질문

All of the rings of the four heterocyclic bases are aromatic. This is more apparent when the polar resonance forms of the amide groups are drawn, as is done for thymine here. Redraw the hydrogen-bonded guanine-cytosine and adenine-thymine pairs shown in Figure 23-24, using the polar resonance forms of the amides. Show how these forms help to explain why the hydrogen bonds involved in these pairings are particularly strong. Remember that a hydrogen bond arises between an electron-deficient hydrogen atom and an electron-rich pair of nonbonding electrons.

627
views