The oxidative cleavage of compounds Glucose (A) and Fructose (B) with $H I O_{4}$ leads to the formation of acids and aldehydes. The ratios of formic acid to formaldehyde for (A) and (B) respectively are

$\frac{5}{1}, \frac{3}{2}$

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$\frac{5}{2}, \frac{3}{1}$

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$\frac{3}{2}, \frac{5}{2}$

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$\frac{4}{2}, \frac{5}{1}$

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Solution

The correct option is A
$\frac{5}{1}, \frac{3}{2}$

$G l u c o s e 5 H I O_{4} - --- \to 5 H C O O H + H C H O; F r u c t o s e 5 H I O_{4} - --- \to 3 H C O O H + 2 H C H O + C O_{2}$

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	List I		List II
A	Hexokinase	1	Conversion of fructose-6-phosphate to fructose-1-6-diphosphate
B	Triose phosphate dehydrogenase	2	Conversion of glucose to glucose-6-phospahte
C	Phosphoglyceromutase	3	Conversion of 1, 3-diphosphoglyceraldehyde to 1, 3-diphosphoglyceric acid
D	Phosphofructokinase	4	Conversion of 3-phosphoglyceric acid to 2-phosphoglyceric acid

Match the above lists and choose the correct option.

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The oxidative cleavage of compounds Glucose (A) and Fructose (B) with HIO4 leads to the formation of acids and aldehydes. The ratios of formic acid to formaldehyde for (A) and (B) respectively are

The correct option is A 51,32 Glucose5HIO4−−−−→5HCOOH+HCHO;Fructose5HIO4−−−−→3HCOOH+2HCHO+CO2

The oxidative cleavage of compounds Glucose (A) and Fructose (B) with $H I O_{4}$ leads to the formation of acids and aldehydes. The ratios of formic acid to formaldehyde for (A) and (B) respectively are

The correct option is A
$\frac{5}{1}, \frac{3}{2}$

$G l u c o s e 5 H I O_{4} - --- \to 5 H C O O H + H C H O; F r u c t o s e 5 H I O_{4} - --- \to 3 H C O O H + 2 H C H O + C O_{2}$