5
$\begingroup$

List the following alkanes in order of increasing boiling point.

A. $\ce{CH3(CH2)4CH3}$
B. $\ce{(CH3)2CHCH(CH3)2}$
C. $\ce{CH3CH2CH(CH3)CH2CH3}$

Answer: $C<B<A$, low to high.

Why is it that B has not a lower boiling point than C? Is it not more branched? I believe branching decreases surface area, leading to less intermolecular interactions, and to a higher boiling point.

$\endgroup$
4
$\begingroup$

B has more branching and this means that molecules of B pack less efficiently and so there are fewer Van der Waals forces between them and so B has the lowest boiling point. On the other hand A is a linear molecule and so packs efficiently, has stronger VdWs forces and so has the highest boiling point. As @DavePhD's data shows $\ce{B < C < A}$

$\endgroup$
4
$\begingroup$

$A$ n-hexane, boiling point 69 degrees C.

$B$ 2,3 dimethyl butane, boiling point 58 degrees C.

$C$ 3-methyl pentane, boiling point 63 degrees C.

$B<C<A$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.