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This seemingly trivial question is as follows: Why is the boiling point of 1-butanol ($\ce {117.7 ^\circ C}$) higher than that of 2-butanol ($\ce {99 ^\circ C}$)?

The only reason I can think of is that the former molecule has a "straighter" chain than the other and a "straighter" chain has a greater surface area allowing for stronger van der Waals forces to exist between the molecules. As molecules in the liquid state do not pack themselves close to each other, ideas of "packing efficiency" would not be useful in explaining these boiling point differences. Perhaps, it is instructive for us to first observe if the trend is persistent for longer chains?

\begin{array}{|c|c|c|c|} \hline \text {No. of C} & \text{1-alcohol} & \text{2-alcohol} & \text{3-alcohol}\\ \hline \ce{5} & 138 & 119.3 & 116\\ \hline \ce{6} & 157 & 140 & 135\\ \hline \ce{7} & 175 & 159 & 156\\ \hline \ce{8} & 195 & 178.5 & 173\\ \hline \ce{9} & 214 & 194 & -\\ \hline \end{array}

As can be seen from the above table, where the boiling points are all sourced from Wikipedia, with units of $\ce { _^\circ C}$, the trend is indeed persistent. The alcohol isomer with the $\ce {-OH}$ group attached to the 2-position consistently exhibits lower boiling points of around $\ce {16}$ to $\ce {20}$ $\ce {^\circ C}$. Shifting the hydroxyl group further along the chain by one more position also lowers the boiling point of the molecule but to a lesser extent.

The explanation I can come up with for this is the same as aforementioned: By placing the hydroxyl group at the end, there is the maximum surface area for interaction of the alcohol molecules with each other. Furthermore, the hydroxyl groups, being placed at the ends, would cause the least amount of disruption to the hydrophobic interactions between the alkyl chains. Does anyone have better explanations for the above trends?

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    $\begingroup$ A detailed answer might even turn out somehow opposite as far molecular region are considered, but ending at the same and actual fact... You neglect that a source of interaction is H bond. Consider that and go through it again. Using the same concepts I am pretty sure you can find a rationale. Don't worry that liquid do not pack. They certainly do it with the nearest neighbours. $\endgroup$ – Alchimista Jan 2 at 10:54
  • $\begingroup$ @Alchimista I am well aware that alcohols can undergo hydrogen bonding. Are you suggesting that by having the hydroxyl group at the terminal carbon, hydrogen bonding between the molecules is facilitated? $\endgroup$ – Tan Yong Boon Jan 2 at 11:19
  • $\begingroup$ basically you have to look as you did in the question but bonding molecule first. This would make the molecules differently arranged. Then you look if the hydrocarbon chains can intercalate and how, etc. You shift the analysis done before for individual molecules to an H bound ensemble. $\endgroup$ – Alchimista Jan 2 at 11:45
  • $\begingroup$ If liquids didn't pack, there would be a tremendous jump in density at the mp. Some substances even expand on crystallisation! $\endgroup$ – Karl Jan 2 at 15:10
  • $\begingroup$ @Karl Hmmm... So what are you trying to imply by that comment? You mean to say that packing of the different alcohol isomers can possibly account for differences in the boiling points? $\endgroup$ – Tan Yong Boon Jan 2 at 15:15
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It does seem that placing the hydroxyl group at the terminal carbon maximises the hydrogen bonding interactions between alcohol molecules and also the van der Waal forces between them. This is best illustrated with a crude diagram I have made:

enter image description here

Finer differences due to the position of the hydroxyl group along the chain, such as the differences between the 2- isomer and the 3- isomer, are not explained here. My explanation does explain the markedly higher boiling point of the 1- isomer.

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  • $\begingroup$ Upvoted because it is most likely a very pictorial and rough description but it gives the sense of the situation and of the type of reasoning involved. It is what I have suggested to OP - who, by the way, was already on the right track. $\endgroup$ – Alchimista Jan 3 at 9:07

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