1.

Inisomericaminecorrectorderhas lowestB. P

Answer»

`1gt 2^(@) GT 3^(@)`
`3^(@) gt 2^(@) gt 1^(@)`
`2^(@) gt ^(@) gt 1^(@)`
`1^(@) gt 3^(@) gt 2^(@)`

Solution :In isomericcompourds normalcompounds have higher B.Pthanbranched compound . BUTIN aminesactiveactivehydrogenatomsare different.
`1^(@)`amines= - `NH_(2)`group= 2 activeH - atoms.
`2^(@)`amines= ` gt`NHgroup =1 acitveH -atoms.
`3^(@)`amines N group= 0 activeH -atoms.
HenceB.Porderis `1^(@) gt2^(@)gt3^(@)`


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