The smallest aldose which is able to form cyclic hemiacetal is/are:

D-Glyceraldehyde

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D-Erythrose

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D-Therose

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D-Ribose

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Solution

The correct options are
B D-Erythrose
C D-Therose
Minimum 4 carbons are required to form a 5 membered cyclic hemiacetal hence D-Erythrose and D-Therose are the smallest aldose which is able to form cyclic hemiacetal.

Options B and C are correct.

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Similar questions

Pentose sugars formed in Calvin’s cycle
A. Glyceraldehyde phosphate B. Xylulose C. Erythrose D. Ribulose

Which of the following sugars howe the same number of carbon as present in glucose ?

(a) Fructore (b) Erythrose (c) Ribulose (d) Ribose

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