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Example:
Why is the enthalpy change of combustion more negative for propan-1-ol
than propan-2-ol.
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.
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