An alkene with molecular formula C6H12[A] on ozonolysis produces only one product [B] which given 2,4−DNP test positive but lodoform test negative. What will be the number of α−Hatoms present in that alkene [A]?
Open in App
Solution
The alkene is CH3−CH2−CH=CH−CH2−CH3
as if gives only one product on ozonolysis.
i.e.CH3−CH2−CHO
It is an aldehyde and hence undergoes 2,4−DNP test positive. But it doesn't undergo idoform reaction.
The number of ∝−H atom present in the alkene are 4.