Monosaccharides are carbohydrates which cannot be hydrolysed into simpler molecules. Those containing an aldehyde group are called aldoses while those containing a keto group are called ketoses. The aldehyde group is present at C1 while the keto group is usually present at C2. All monosaccharides reduce Tollens’ reagent as well as Fehling’s solution and hence are called reducing sugars. Glucose has a six-membered cyclic structure called the pyranose structure, while fructose in sucrose has a five-membered cyclic structure called the furanose structure. During ring formation, C1 of glucose becomes chiral, and hence glucose exists in two cyclic forms called the α-anomer and the β-anomer. When two molecules of the same or different monosaccharides combine through a glycosidic linkage, a disaccharide is formed.
Consider the following statements about the structure of glucose.
(A) The α-form of glucose is obtained by crystallisation from a concentrated solution of glucose at 303 K.
(B) The two cyclic forms of glucose exist in equilibrium with the open chain structure.
(C) The β-form of glucose (m.p. 419 K) is obtained by crystallisation from hot and saturated aqueous solution at 371 K.
(D) Glucose forms a six-membered ring in which –OH at C-4 is involved in ring formation.
Which of the statements given above are correct? Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (B) only
The open chain structure of glucose could not explain why glucose exists in two crystalline forms. One –OH group adds to the –CHO group and forms a cyclic hemiacetal, which exists in two forms, α and β.
(A) The α-form of glucose is obtained by crystallisation from a concentrated solution of glucose at 303 K. — True. The α-form (m.p. 419 K) crystallises at 303 K; the β-form (m.p. 423 K) crystallises from a hot and saturated solution at 371 K.
(B) The two cyclic forms of glucose exist in equilibrium with the open chain structure. — True. The α and β rings and the open chain form are in equilibrium with one another.
(C) The β-form of glucose (m.p. 419 K) is obtained by crystallisation from hot and saturated aqueous solution at 371 K. — False. The β-form does crystallise at 371 K, but it melts at 423 K; 419 K is the melting point of the α-form.
(D) Glucose forms a six-membered ring in which –OH at C-4 is involved in ring formation. — False. It is the –OH at C-5.
C1 to C5 and the ring oxygen make the six-membered ring; the –OH at C-4 would give a five-membered ring.
NCERT (2026-27 edition):
10.1.2.1 Glucose – “…The α-form of glucose (m.p. 419 K) is obtained by crystallisation from concentrated solution of glucose at 303 K while the β-form (m.p. 423 K) is obtained by crystallisation from hot and saturated aqueous solution at 371 K.”
“…It was found that glucose forms a six-membered ring in which –OH at C-5 is involved in ring formation. This explains the absence of –CHO group and also existence of glucose in two forms … These two cyclic forms exist in equilibrium with open chain structure.”