question_answer

  The product \[(X)\] on oxidation with \[HI{{O}_{4}}\] gives :

(i) HCHO

(ii) \[C{{H}_{3}}-CHO\]

(iii) \[C{{O}_{2}}\]

(iv) \[~HCOOH\]

A)  I & ii                           

B)  i & iv

C)  i, iii & iv                      

D)  only iv

Correct Answer: B

Solution :

\[\xrightarrow{Hl{{O}_{4}}}\,HCOOH\,+2\,C{{H}_{2}}=O\]