Answer :

C = 69.77% ie ==5.88;

H=11.63% ie==11.63;


O=(100-(69.77+11.63))=18.6% ie==1.16


Molecular Ratio will be 5.88:11.63:1.16=5:10:1.


Empirical formula is C5H10O, molecular weight will be =86.


Hence molecular formula will be same.


It does not reduce tollen’s reagent, it is not an aldehyde. The compound is addition product of sodium hydrogen sulphite it must be a ketone, It is giving positive iodoform test it is ethyl ketone. On vigorous oxidation, it forms ethanoic acid and propanoic acid the given compound is pentane-2-one.



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