[anythingin]

I did an experiment in class and found that different isomers gave me different enthalpies of combustion. Why is this? I used Butan-1-ol, Butan-2-ol, 2-methylpropan-1-ol and 2-Methylpropan-2-ol.

Thanks.

[AWK]

Only enantiomers show the same enthalpy of combustion.

[anythingin]

Thanks for your reply. Why do non-enantiomers isomers have a different enthalpy of combustion?

[AWK]

http://web.chem.ucla.edu/~harding/IGOC/E/enthalpy_of_combustion.html
https://wiki2.org/en/Heat_of_formation_group_additivity+Newton

[Enthalpy]

Since the combustion products are the same, the difference in released heat results from the different heat of formation of the isomers, which is quite normal.

A first trivial case is when the bonds differ. The same C, H, O, N can be grouped as a nitro or a nitramine compound, with a high heat of formation, or as ketones, alcohols, amides... with a lower heat of formation, because O makes stronger bonds with C and H than with N. Or think of benzene versus di(cyclopropenyl), they're isomers too.

The four compounds in the example are less trivial because each has one alcohol function. Only its position differs, together with optional branching. The differences are then smaller but measurable, which tells that bond energies depend on neighbour atoms - for instance, the polarization of C-O depends on the other atoms bonded with C and with O.

An other important reason, often forgotten, is that the rotations and vibrations of a molecule depend on the isomer. With RT=2.5kJ/mol at room temperature, this is an important contribution, as big as the observed differences. In liquids and solids, the differences in the heat of vaporization and melting among the isomers is also as important.

This is a hard limitation to the estimation of a heat of formation through the bond energy: taking all effects into account is painstaking, and the result is still inaccurate.