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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 41

  • question_answer
    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

    Correct Answer: B

    Solution :

    [b] \[CHO-CHOH-\underset{glu\cos e}{\mathop{CHOH}}\,-CHOH-CHOH-C{{H}_{2}}OH\]\[\xrightarrow{5HI{{O}_{4}}}HCOOH+HCHO+5HI{{O}_{3}}\] \[C{{H}_{2}}OH-\overset{O}{\mathop{\overset{||}{\mathop{C}}\,}}\,-\underset{fructose}{\mathop{CHOH}}\,-CHOH-CHOH-C{{H}_{2}}OH\]\[\xrightarrow{5HI{{O}_{4}}}3HCOOH+2HCHO+5HI{{O}_{3}}+C{{O}_{2}}+{{H}_{2}}O\] \[\therefore \] molar ratio is 5 : 3


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