Q. In aqueous solution glucose remains as
A) Only in open chain form
B) Only in pyranose form
C) Only in furanose forms
D) In all three forms in equilibrium
Explanation
(d)
$\beta - D -$glucose D-glucose $a - D -$glucose
$( \approx 64\%)\ (open\ chain\ \approx 0.02\%)\ ( \approx 34\%)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.