question_answer

Which of the following compounds will behave as a reducing sugar in an aqueous KOH solution?                  JEE Main Solved Paper-2017

A)           

B)  

C)        

D)  

Correct Answer: A

Solution :

 [a] Ester in presence of Aqueous KOH solution give SNAE reaction so following reaction takes place [b] In above compound in presence of Aq. KOH \[\text{(S}{{\text{N}}^{\text{AE}}}\text{)}\] reaction takes place \[\propto -\]Hydroxy carbonyl compound is formed which give \[\oplus \text{ve}\]Tollen's test So this compound behave as reducing sugar in an aqueous KOH solution.