In solution, glucose exists in

Question

BlackTom AI · Accepted Answer

Glucose exists predominantly as cyclic hemiacetals: the aldehyde at C-1 reacts intramolecularly with a hydroxyl (usually the C-5 OH) to form a six-membered pyranose ring (and in small amounts a five-membered furanose). This cyclization produces a hemiacetal center at C-1, giving the alpha and beta anomeric forms.

Relevant concepts:
- Hemiacetal vs acetal: a hemiacetal bears both an –OH and an –OR (or ring oxygen) on the same carbon; an acetal has two –OR groups and requires reaction of a hemiacetal with an alcohol (typically under acid catalysis) to form. In simple aqueous solution, free glucose does not convert to a full acetal unless an external alcohol/glycosylation reagent is present.
- Mutarotation: alpha and beta anomers interconvert via transient ring opening to the open-chain aldehyde; thus an equilibrium mixture includes both cyclic hemiacetals and a very small fraction of the open-chain form.

Step-by-step reasoning:
1. Start with D-glucose in water: the C-1 aldehyde can nucleophilically attack by the C-5 OH to form a cyclic hemiacetal (pyranose). This yields two stereochemical outcomes at C-1 (alpha and beta).
2. These cyclic forms can open back to the open-chain aldehyde; this reversible process establishes an equilibrium (mutarotation) among alpha-pyranose, beta-pyranose, minor furanoses, and the open-chain form.
3. Under normal aqueous conditions, the cyclic hemiacetal forms predominate (major species), while the open-chain aldehyde is present only in trace amounts; no full acetal forms unless an external alcohol reacts under acidic conditions.

Why E is correct:
Option E correctly describes the equilibrium between the open-chain aldehyde form and the cyclic hemiacetal forms (alpha and beta). This matches the known behavior and mechanism (mutarotation).

Why other options are incorrect:
- A and B mention acetals: these imply two OR groups (full acetal/glycoside), which are not the native forms in water unless glycosylation occurs. Thus incorrect.
- C and D state “only” one form exists; experimental equilibrium includes multiple cyclic hemiacetal anomers plus a tiny open-chain fraction, so “only” is false.

Conclusion: The correct choice is E) an equilibrium mixture of the open-chain form and cyclic hemiacetal forms, which matches the provided answer.

In solution, glucose exists in单项选择题

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Question textFill in the gap with ONE word from the text. Research can help to reduce the amount of Answer 1 Question 6[input] between fauna and humans.

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