Question

Which of the following statement(s) is/are true about cyclic structure of glucose?

• A. No free $CHO$ group at both $\alpha$ and $\beta$ cyclic forms
• B. $\alpha$ and $\beta$ forms exist in equilibrium with an open chain structure.
  
• C. $\alpha$ and $\beta$ forms does not exist in equilibrium with an open chain structure.
  
• D. Both $\alpha$ and $\beta$ cyclic forms have hemiacetal linkage

Answer 1

Answer: (A), (B), (D)

Both $\alpha$ and $\beta$ cyclic forms have hemiacetal linkage

Open chain glucose form a six membered pyranose ring in which $-OH$ at C5 is involved in ring formation which leads to hemiacteal linkage.
Due to this arrangement, a new asymmetric carbon atom is created at the carbonyl carbon and it is referred to as anomeric carbon atom. Anomers are two different configurations that are possible for the anomeric carbon atom. The position of OH group differs at C1 in both the structures ( $\alpha$ and $\beta$ forms )
Since, carbonyl forms a hemiacetal linkage, there is no free CHO group in $\alpha$ and $\beta$ forms.
These two cyclic forms exist in equilibrium with an open chain structure.

Hence, statements (a), (b) and (d) are correct.