JEE Main 2017 · 2 Apr · Shift 1 · Chemistry Q57

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ChemistrySingle correct+4 / -1Question 57 of 90

Which of the following compounds will behave as a reducing sugar in an aqueous KOH solution ?

A furanoside with OCH₃ at the anomeric carbon

A sugar drawn in Haworth form: a five-membered ring with HOH₂C at C5, CH₂OH and OCH₃ on C2, HO on C3 and OH on C4.

A pyranose whose carbon next to the ring O carries CH₂OCH₃

A sugar drawn in Haworth form: a six-membered ring with HOH₂C on one carbon next to O, CH₂OCH₃ on the other, and three OH groups.

A furanose with OCOCH₃ at the anomeric carbon

A sugar drawn in Haworth form: a five-membered ring with HOH₂C at C5, CH₂OH and OCOCH₃ on C2, HO on C3 and OH on C4.

A furanose with only CH₂OH (and H) at C2

A sugar drawn in Haworth form: a five-membered ring with HOH₂C at C5, CH₂OH on C2, HO on C3 and OH on C4.

Answer and worked solution

Official answer: C

A reducing sugar needs a free hemiacetal (anomeric OH). The acetate ester in (3) is hydrolysed by aqueous KOH, freeing the anomeric hydroxyl, so it reduces.

(1) is a methyl glycoside, stable to base.
(2) and (4) have no anomeric oxygen substituent (C-linked or H).

Topics:Carbohydrates

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JEE Main 2017 · 2 Apr · Shift 1, Chemistry question 57