JEE Main Chemistry question

Glucose reacts with acetyl chloride to form pentaacetyl glucose; it indicates the presence of:

(A)Five primary alcoholic groups
(B)Five secondary alcoholic groups
(C)Aldehyde as well as alcoholic group
(D)Five groupsCorrect

Answer: (D)

Solution

The formation of a pentaacetate shows that exactly five hydroxyl groups (a mix of secondary and primary –OH groups across the ring/chain) are available for acetylation in glucose. Option (b) alone is incomplete

See the full step-by-step solution

Create a free account to read the complete working, and practise questions like this in a timed test.

Create a free account

Question ID VK-150904