Question #186e8

1 Answer
Sep 2, 2016

Answer:

Good question. The boiling points of the alkanes should increase in the order, #"2,2,4-trimethylheptane"#, #"2-methylheptane"#, #"octane"#.

Explanation:

The longer, and straighter the chain (i.e. the less unbranched) the greater should be the interaction between chains, and thus the greater the boiling point.

Now of course I did not know these boiling points off the top of my head, but I can account for their order:

#"2,2,4-trimethylheptane,"# #99# #""^@C#

#"2-methylheptane,"# #116# #""^@C#

#"octane", 125# #""^@C#

Longer, less branched chains have greater opportunity for intermolecular dispersion forces, and this is certainly reflected in the boiling point. That the longest chain alkane, #"octane"#, is the most involatile supports our analysis. And even though #"2,2,4-trimethylheptane"# is the most massive molecule, branching prevents as effective intermolecular interaction as exists in the other pair.

The boiling point of #n-"heptane"# is #98.5# #""^@C# (this I did know off the top of my head, because I found it to be a highly useful laboratory solvent), which is comparable to the boiling point of
#"2,2,4-trimethylheptane"#. Once again, branching serves to decrease the boiling point.