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One mole of each of glucose and fructose...

One mole of each of glucose and fructose require x and y moles of periodic acid respectively, for oxidation. Hence, x/y is .

A

`1/1`

B

`2/1`

C

`5/4`

D

`9/4`

Text Solution

Verified by Experts

The correct Answer is:
C
`CH_(2)OH- underset("(Glucose)")((CHOH)_(4))-CHO + 5HIO_(4) to 5HCOOH + HCHO + 5HIO_(3)`
`CH_(2)OH - underset("(Fructose)")overset(O)overset(||)(C)-(CHOH)_(3) - CH_(2)OH + 4HIO_(4) to HCHO + 4HCOOH + 4HIO_(3)`
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