Given `C_(6)H_(13)CI overset("alc.KOH")underset((-HCI))(to) (B) +(C)`
The molecular formula of (B) and (C) is `C_(6)H_(12)`. Thus on ozonolysis of each alkene, the two products must have six carbon atoms.
One alkene gives `CH_(3)CHO " and "(CH_(3))_(2)CHCHO.` Thus, the hydrocarbon should have the structure,
`CH_(3)CH=CHCH(CH_(3))_(2) " ".........(B)`
Other alkene gives `C_(2)H_(5) CHO " and " CH_(3)COCH_(3).` Thus, the hydrocarbon should have the structure,
`C_(2)H_(5)CH=C(CH_(3))_(2) " "..........(C)`
Since (B) and (C) are formed from (A) by dehydro- halogenation , the sturcture of (A) is : ,
`CH_(3)CH_(2)-underset(CI)underset(|)CHCH(CH_(3))_(2)overset("alc.KOH")(to)`
`underset((B))(CH_(3) -CH =CHCH(CH_(3))_(2)) + underset((C))(CH_(3)CH_(2)CH=C(CH_(3))_(2))`
As it is `2^(@)` halide it slowly reacts with the `AgNO_(3)` to form white ppt.