In #"isobutane"#, #"methylpropane"#, why does the methyl group HAVE to substitute at #C2# with respect to the propyl chain...?

1 Answer
anor277
Apr 18, 2017

Well, because if it were at #"C3"# it would be #"n-butane"#.

Explanation:

All the butanes have the formula #C_4H_10#, and #"n-butane"# is #"H"_3"CCH"_2"CH"_2"CH"_3#. And #"isobutane"#, is #"H"_3"CCH(CH"_3")CH"_3#. The methyl group can only be substituted at the one position if there is a propyl chain.

#"Isobutane"# has a boiling point of #-11.7# #""^@C#; #"n-butane"# has a boiling point of #-1.0# #""^@C#. Can you account for this difference on the basis of structure?

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