Which one of the following statements is not correct?
A. D (-) fructose exists in furanose structure
B. D (+) glucose exists in pyranose structure
C. In sucrose the two monosaccharides are held together by peptide linkage
D. Maltose is a reducing sugar
E. Pentaacetate of glucose does not react with hydroxylamine
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → C only
The question asks for the statement that is not correct. Each statement is judged:
(A) D (−) fructose exists in furanose structure — True. Fructose forms a five-membered ring (C2 to C5 through oxygen), like furan.
(B) D (+) glucose exists in pyranose structure — True. The cyclic forms of glucose have a six-membered ring, like pyran.
(C) In sucrose the two monosaccharides are held together by peptide linkage — False. Glucose and fructose units in sucrose are joined by a glycosidic linkage between C1 of α-D-glucose and C2 of β-D-fructose. A peptide linkage (–CO–NH–) joins amino acids in proteins.
(D) Maltose is a reducing sugar — True. A free aldehyde group can be produced at C1 of the second glucose unit.
(E) Pentaacetate of glucose does not react with hydroxylamine — True. In the pentaacetate the cyclic hemiacetal OH is acetylated, so no free –CHO group is formed; this is evidence for the cyclic structure of glucose.
NCERT (2026-27 edition):
Structure of Fructose – “… It also exists in two cyclic forms which are obtained by the addition of –OH at C5 to the (>C=O) group. The ring, thus formed is a five membered ring and is named as furanose with analogy to the compound furan.”
Cyclic Structure of Glucose – “… 2. The pentaacetate of glucose does not react with hydroxylamine indicating the absence of free —CHO group. … The six membered cyclic structure of glucose is called pyranose structure (α– or β–), in analogy with pyran.”
Disaccharides – …
Disaccharides – “… (ii) Maltose: Another disaccharide, maltose is composed of two α-D-glucose units in which C1 of one glucose (I) is linked to C4 of another glucose unit (II). The free aldehyde group can be produced at C1 of second glucose in solution and it shows reducing properties so it is a reducing sugar.”