saorg-1

guys pls help it is urgent for me.

1. (d)

Notice first H₃O⁺ is used..first write product after attack of H₃O⁺.. then it will be obvious...

no need for mechanism as I think it is not directly there in syllabus...

But What you have to do is...just break bonds between carbons(when both of adjacent carbon has -OH..)and add one -OH each to both... then those carbons will have two -OH ... then u know what to do when a carbon has two -OH...(ketone or aldehyde)

when a -OH bearing carbon has -OH bearing carbons on both sides then there will be two -OH added to it...
so net three -OH so acid will be formed here..

I think it is used for conversion of organic halides to carbonyls.....though my chem is weak

that was the function of NaIO₄
i forgot to put up the image... here it is:

well NaIO4 is an oxidising agent ...

jus like KMnO4....

o o
\\ //
I
// \\
oNa o

this is the structure.....

when there are two adjacent OH bonds naa... they attack with the H's
of OH ...

and something something happens (i dun remember, bcoz its not necessary)

and they oxidise.......

thanx abi and didi have u tried the next qn.

for 2nd
ans shud b a

no deepan. 29th qn ans is a

msp i think u didnt read my post proprely enuf...