Periodic acid splits glucose and fructose into formaldehyde and formic acid, Ratio of moles of formic acid in glucose and fructose is

Question

Periodic acid splits glucose and fructose into formaldehyde and formic acid, Ratio of moles of formic acid in glucose and fructose is (A) 1:2 (B) 5:3 (C) 1:1 (D) 2:3

Tardigrade Admin · Accepted Answer

CHO-CHOH- undersetg l u c o s eC H O H-CHOH-CHOH-CH2OH overset5 H I O4 arrow 5HCOOH+HCHO+5HIO3 textCH2 textOH - overset underset|. |. textO textC - underset textfructose textCHOH - textCHOH - textCHOH - textCH2 textOH overset5 textHIO4 arrow 3 textHCOOH + 2 textHCHO + 5 textHIO3 + textCO2 + textH2 textO So molar ratio is 5:3

Q. Periodic acid splits glucose and fructose into formaldehyde and formic acid, Ratio of moles of formic acid in glucose and fructose is

1:2

5:3

1:1

2:3

$C H O - C H O H - g l u cose C H O H - C H O H - C H O H - C H_{2} O H \to 5 H I O_{4} 5 H COO H + H C H O + 5 H I O_{3}$

$CH_{2} OH - C ∣ ∣ O - fructose CHOH - CHOH - CHOH - CH_{2} OH \to 5 HIO_{4} 3 HCOOH + 2 HCHO + 5 HIO_{3} + CO_{2} + H_{2} O$

So molar ratio is 5:3

Q. Periodic acid splits glucose and fructose into formaldehyde and formic acid, Ratio of moles of formic acid in glucose and fructose is

1:2

5:3

1:1

2:3

CHO−CHOH−glucoseCHOH​−CHOH−CHOH−CH2​OH→5HIO4​5HCOOH+HCHO+5HIO3​ CH2​OH−C∣∣O​−fructoseCHOH​−CHOH−CHOH−CH2​OH→5HIO4​3HCOOH+2HCHO+5HIO3​+CO2​+H2​O So molar ratio is 5:3

$C H O - C H O H - g l u cose C H O H - C H O H - C H O H - C H_{2} O H \to 5 H I O_{4} 5 H COO H + H C H O + 5 H I O_{3}$

$CH_{2} OH - C ∣ ∣ O - fructose CHOH - CHOH - CHOH - CH_{2} OH \to 5 HIO_{4} 3 HCOOH + 2 HCHO + 5 HIO_{3} + CO_{2} + H_{2} O$

So molar ratio is 5:3