21
$\begingroup$

I was attempting to compare the boiling points of iso-hexane and 3-methylpentane.

Boiling points of organic compounds depend upon three factors according to Master Organic Chemistry – 3 Trends That Affect Boiling Points. They are:

  1. The relative strength of the four intermolecular forces is: Ionic > Hydrogen bonding > dipole dipole > van der Waals dispersion forces. The influence of each of these attractive forces will depend on the functional groups present.

  2. Boiling points increase as the number of carbons is increased.

  3. Branching decreases boiling point.

The number of carbons are same in both 2-methylpentane and 3-methylpentane and so that trend doesn't make a difference. Since there are no functional groups present, the only force acting between two molecules would be van der Waals dispersion forces and this depends upon the surface area of the molecule.

3-Methylpentane is more symmetric than 2-methylpentane and so would form a more spherical structure than iso-hexane. For a fixed volume, the surface area decreases the more spherical the shape becomes. Due to this, the more symmetric molecule (3-Methylpentane) should show a lower boiling point.

However, the boiling point of 3-methylpentane is $\pu{63 ^\circ C}$ whereas for isohexane it is $\pu{60 ^\circ C}$ which contradicts the above statement, according to which iso-hexane should have had the greater boiling point.

Thinking there was a flaw in my assumptions, I compared the boiling points of 3-methylheptane and 4-methylheptane.

This showed the same trend as my statement provided. 3-Methylheptane has a boiling point of $\pu{120 ^\circ C}$ and 4-methylheptane has a boiling point of $\pu{117 ^\circ C}$.

Going further, comparing 4-methylnonane($\pu{165.7 ^\circ C}$) and 5-methylnonane($\pu{165.1 ^\circ C}$), we see that the difference between the boiling points decreased to $\pu{0.6 ^\circ C}$ from $\pu{3 ^\circ C}$ but still can be predicted using the above statement.


Update

With reference to J. Am. Chem. Soc. 1929, 51 (5), 1540–1550, the values of boiling point and density measured for the isomers (only taking straight chain and single branched isomers) were as follows for:

  1. Hexanes

\begin{array}{|c|c|} \hline \text{IUPAC Name} &\text{Boiling point }(\pu{^\circ C})& \text{Density}(\pu{g ml-1)} \\ \hline n\text{-Hexane} & \pu{68.95} & 0.6595\\ \text{2-Methylpentane} & \pu{60.2}&0.6542 \\ \text{3-Methylpentane} & \pu{63.2} &0.6647 \\ \hline \end{array}

  1. Heptanes

\begin{array}{|c|c|} \hline \text{IUPAC Name} &\text{Boiling point }(\pu{^\circ C})& \text{Density}(\pu{g ml-1)} \\ \hline n\text{-Heptane} & \pu{98.4} & 0.6836 \\ \text{2-Methylhexane} & \pu{90.0} & 0.6789 \\ \text{3-Methylhexane} & \pu{91.8} & 0.6870 \\ \hline \end{array}

  1. Octanes

\begin{array}{|c|c|} \hline \text{IUPAC Name} &\text{Boiling point }( \pu{^\circ C}) & \text{Density}(\pu{g ml-1)}\\ \hline n\text{-Octane} & \pu{124.6} & 0.702 \\ \text{2-Methylheptane} & \pu{116.0} & 0.6985 \\ \text{3-Methylheptane} & \pu{122.2} & 0.707 \\ \text{4-Methylheptane} & \pu{118.0} & 0.722 \\ \hline \end{array}

  1. Decanes

\begin{array}{|c|c|} \hline \text{IUPAC Name} &\text{Boiling point }( \pu{^\circ C}) & \text{Density}(\pu{g ml-1)}\\ \hline n\text{-Decane} & \pu{174.0} & 0.730 \\ \text{2-Methylnonane} & \pu{160.0} & 0.724 \\ \text{3-Methylnonane} & \pu{166.9} & 0.735\\ \text{5-Methylnonane} & \pu{166.2} &0.732 \\ \hline \end{array}


In all these cases, it seems as though the 2-methyl isomer (the "iso-" isomer) has the lowest boiling point.

Therefore, it seems as though the answer for the question about comparing the boiling point of 2-Methylpentane and 3-Methylpentane lies in the reason for why an "iso-" alkane has the lowest boiling point for an n-carbon alkane.

Is there a reason for the "iso-" isomer having the lowest boiling point in the single branched methyl isomers?

Why is the comparison of the boiling points of 3-Methylpentane and 2-Methylpentane anomalous? (This would be answered since 2-Methylpentane is the "iso-" isomer of hexane)

$\endgroup$
  • 1
    $\begingroup$ Page 937 of this ancient paper and references therein might contain an answer. $\endgroup$ – Aniruddha Deb Aug 21 at 17:19
  • 3
    $\begingroup$ Actually 2-methylpentane can be considered as symmetric as 3-methylpentane - both have plane of symmetry, it's just not perpendicular, but overlaps with longest chain. $\endgroup$ – Mithoron Aug 21 at 19:54
  • 3
    $\begingroup$ @Aditya How do you decide the "apparently less van der waal" forces? that is the whole question. $\endgroup$ – Safdar Aug 22 at 15:42
  • 1
    $\begingroup$ @RahulVerma A mix of the two or are you saying that there are more factors in play? $\endgroup$ – Safdar Aug 23 at 15:02
  • 1
    $\begingroup$ There is a flaw anyway. More spherical implies also more tight packing. It seem there is a contradictory passage in your question. What exactly does your statement in bold suggest? $\endgroup$ – Alchimista Aug 24 at 11:31
1
$\begingroup$

I suggest that an entropic term explains the little difference observed. While this aspect is normally important to justify what isomer melts at lower temperature, in principle it can be invoked to justify why an isomer having a longer - not too much - branch boils at a bit higher temperature.

In this case, which might be rare as for a balancing between enthalpy factors it is needed to make this entropic term of decisive importance, the $3$ carbons chain can access a considerable space - both in physical as well conformational terms - precluded to the $2$ carbons ones in liquid phase.

Considering $\Delta G$ of the liquid to gas transition would imply a slightly higher $T$ for 2-Methylpentane.

| improve this answer | |
$\endgroup$
  • $\begingroup$ why just the entropic terms, can we not use the fact stated by the OP that the packing of 3me- pentane would be better than 2 me-pentane having a slightly higher effect on the vander waals attraction counter to the claim that it decreases the surface area? $\endgroup$ – user98209 Aug 25 at 20:56
  • 1
    $\begingroup$ I have also proposed that fact in a comment. But it would be better called "effective surface". The point is that should not be anomalous anymore but seen in most analogue terms. I am convinced that the more compact shape of the 2 isomer as compared to the more flexible propyl branch is responsible for a higher entropy content in liquid 2-methylpentane... @I am Alita $\endgroup$ – Alchimista Aug 25 at 21:32

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.