Qu: enthalpy change of combustion and isomers of alcohols
 

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The short answer is that propan-2-ol is a more stable molecule...

The longer answer ought to tell why!

The electronegative oxygen of the -OH group creates more of a dipole when it is at the end of the molecule, relatively destabilising it with respect to the 2 isomer.

In the case of 2- propanol the central OH creates a dipole in which the partially positive carbon atom joined to the -OH group is stabilised by the +I (inductive) effect of two neighbouring methyl groups.

Whenever charge density is reduced in a molecule the molecule becomes relatively more stable. As the products are identical in both combustions then the fact that 1-propanol releases more energy tells us that it starts off from higher energy than 2-propanol.

:) Thanks so much for replying. Just to clarify a few points...

Why would propna-1-ol start off from higher energy than propan-2-ol? Surely all fuels would start off the same...when no fuel has been burnt, no energy would have been lost?

If propan-2-ol is more stable, it would take more energy to break its bonds, however weak (instantaneous dipoles?) in comparison to propan-1-ol. So when the bonds reform and energy is given out, propan-1-ol is more exothermic. Is this right?

Again, thankyou.

Fuels don't 'contain' energy and they don't all start out from the same energy level.

The energy is released by the formation of the bonds in the products - in this case the CO2 and the H2O. It is the strength of these bonds being created that releases the energy (bond formation is exothermic and releases energy)

Each combusting fuel has to break the atoms apart before forming the products and breaking these bonds in the propanol requires energy. The more stable a molecule is the more energy required to break it apart.

So propan-2-ol starts off from a position of greater stability (lower energy) and so the energy required to break its bonds is greater than for the propan-1-ol.

If more energy is required to break it apart but the same energy is released on forming the products bonds then the actual observed energy released will be less.

This can be expressed simply by a Hess's law diagram showing the propan-2-ol at a lower level than the propan-1-ol but the products at the same level (which in turn are at an even lower level in the diagram as the overall reaction is exothermic). The drop from propan-2-ol to the products is less than the drop from propan-1-ol to the products.

Summary
The OVERALL energy experienced by the experimenter is what is left after the endothermic breaking apart of the reactant molecules and the exothermic release of energy by the formation of the product molecules.

Another way to express this idea is by looking at the actual bond energies.

Even though the two molecules apparently have the same numbers and types of bonds, as explained in the previous post, the positive inductive effect (+I) of the two methyl groups makes the electron density more even in the case of the 2-isomer, hence the bonds are stronger and require more energy to break.

E(overall) = (sum of bond energies of the reactants) - (sum of bond energies of the products)

If the concept of electron induction by methyl groups is unfamiliar to you just accept the fact that alkyl groups have a tendency to 'push' electrons towards electronegative elements or positive charges in dipoles. This is called the +I (inductive) effect.

The OH creates a dipole between the oxygen and the neighbouring carbon in which the carbon carries a partial positive charge. It is this charge that destabilises the molecule in the 1- isomer with respect to the 2- isomer. n the 1- isomer there is only one alkyl group 'pushing ' electrons towards it whereas in the 2- isomer there are two alkyl groups performing the same function.

Thanks, thats been a great help.

Just wondering what you do in life? (job wise?)

see the signature :wink: