I was looking through the CRC Handbook of Chemistry and Physics section on liquid viscosities, trying to explain trends in viscosities. In higher alkanes, the viscosity increases with the degree of branching, which is logical. However, I noticed for lower alkanes, the opposite is true. For example, 1-methylbutane has a viscosity of 0.286 mPa-s at 25 C, whereas its linear counterpart pentane has a viscosity of 0.30 mPa-s at 25 C. The same trend holds for alkanes with 6 carbons. These trends go against my chemical intuition and I cannot think of a sound explanation, which brings me to my question: why is the viscosity lower for branched vs. linear alkanes in the lower alkanes?
Viscosity is related to Van der Waals/London forces between molecules. This in turn is related to packing and surface area.
A branched 6-chain carbon, tumbling in solution cannot align as much of its surface area with another molecule as the linear hexane.
At some point the larger molecules are becoming able to be 'entangled' and form structures somewhat similar to knots. When the molecules get this large, viscosity increasing branching becomes more likely.
Finally, when you get to true polymers, the chains are so long that branching again begins exclude other molecules from structure and again linear molecules are more likely to have higher viscosities (although for polymers, these are solution viscosities).
$\begingroup$ I don't think I quite follow your line of reasoning. I understand that viscosity is related to vdW forces. When you say "A branched 6-chain carbon, tumbling in solution cannot align as much of its surface area with another molecule as the linear hexane," it seems like molecular alignment in linear hexane would decrease the viscosity, since molecules can more easily slide past one another. Also, linear polymers tend to have lower viscosities than branched ones, so I'm not sure I follow your last point, either. $\endgroup$– tcmoore3Mar 11, 2016 at 20:50
$\begingroup$ Think is if more like strips of velcro. The more they touch the stronger the bond is. $\endgroup$ Mar 12, 2016 at 16:04