Question

Which of the following is the correct evidence for cyclic glucose not having free $-CHO$ group?

• A. It does not react with Schiff’s reagent
• B. It does not react with hydroxylamine
• C. Both (a) and (b)
• D. None of the above

Answer 1

The correct option is C Both (a) and (b)

Despite having aldehyde group, ​glucose does not give Schiff’s test & ​it does not form the hydrogen sulphite (bisulphite) addition product with $NaHS{O}_3$.​ Schiff’s reagent is a dilute solution of ​rosaniline hydrochloride whose pink ​colour has been discharged by passing $S{O}_2$. Aldehyde restores pink colour​ when treated with Schiff’s reagent​ (Magenta solution in $H_{2}S{O}_3$).​ The pentaacetate of glucose does not ​react with hydroxylamine indicating​ the absence of free –CHO group. These behaviour couldn’t be​ explained by open chain structure. ​It was proposed that one of ​ –OH group may add to –CHO group ​to form a cyclic structure.​ Hence, the correct answer is option (c).