With reference to D-glucose, which of the following cannot be explained by its open chain structure?
(A) D-Glucose does not give Schiff's test.
(B) The pentaacetate of D-glucose does not react with hydroxylamine.
(C) Glucose is found to exist in two different crystalline forms.
(D) D-glucose reacts with mild oxidizing agents like \( Br_2-H_2O \) and forms gluconic acid.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A), (B) and (C) only
The open chain structure of glucose has a free aldehyde group (–CHO) at C-1 and five –OH groups. In fact, the –OH at C-5 adds to the –CHO group to form a six-membered ring (a cyclic hemiacetal), and in this ring C-1 becomes a new chiral centre. Facts that need a free –CHO group in large amount, or a chiral C-1, can be explained only by this cyclic structure.
(A) D-Glucose does not give Schiff’s test. — Cannot be explained by the open chain. A free –CHO group should give Schiff’s test (and the hydrogensulphite addition product); in the cyclic form there is almost no free –CHO.
(B) The pentaacetate of D-glucose does not react with hydroxylamine. — Cannot be explained by the open chain. On acetylation the –OH at C-1 of the ring is also acetylated, so the ring can no longer open and no free –CHO is left to form an oxime.
(C) Glucose is found to exist in two different crystalline forms. — Cannot be explained by the open chain. The two forms, α and β, are the two configurations of the new chiral centre at C-1 of the ring (anomers).
(D) D-glucose reacts with mild oxidizing agents like \( \ce{Br2-H2O} \) and forms gluconic acid. — Explained by the open chain. Bromine water oxidises the –CHO group to –COOH, which shows that glucose has an aldehyde group:
\( \displaystyle \ce{CHO-(CHOH)4-CH2OH} \xrightarrow{\ce{Br2}\ \text{water}} \ce{COOH-(CHOH)4-CH2OH} \)
NCERT (2026-27 edition):
Glucose – …
