Benzene and phenol
Read our general notes on Risk Assessment
How does the position of a functional group affect a molecule’s boiling point, e.g. between 2-nitrophenol and 4-nitrophenol? Could it be to do with the packing of molecules or the electron distribution affecting hydrogen bonds?
Igloo writes … This is a good question.
However, since both nitrophenols which you mention are solids, I would prefer to compare their melting points, rather than their boiling points. Yes, the melting point of a solid does depend on the manner in which the molecules pack, i.e. their crystal structure, and this in turn depends on the magnitude of the attractive forces operating between the molecules (the intermolecular attractions).
If you draw out the displayed formulae for 2-nitrophenol and 4-nitrophenol, you will see that for 2-nitrophenol, there is the possibility of an internal hydrogen bond forming between the hydrogen atom of the hydroxyl (OH) group and one of the oxygen atoms of the nitro group. This internal hydrogen bond is known as an intramolecular force, as opposed to an intermolecular force.
However, with 4-nitrophenol the functional groups are on opposite sides of the benzene ring, and so the hydroxyl hydrogen atom is too far away from the nitro group to make this intramolecular attraction possible, so, instead, hydrogen bonds are formed between molecules (intermolecular forces). In other words, a hydrogen bond forms between the hydroxyl group hydrogen atom of one molecule and the oxygen atom of the nitro group of a neighbouring molecule, and between the oxygen atom of the nitro group and the hydroxyl group hydrogen atom of another molecule on the other side.
This means that intermolecular bonding in 4-nitrophenol is far stronger than it is in 2-nitrophenol, which displays intramolecular bonding as well.
In general terms, solid isomers (with the same number of electrons) will have similar melting points, but, as you can see, in the case of 2-nitrophenol and 4-nitrophenol, there are other factors at work.
In other situations, where intramolecular forces are irrelevant, the melting points might differ because the shapes of the molecules are different. Molecular shapes which allow greater alignment of molecules - and therefore a greater surface area in contact - are going to allow greater Van der Waals forces to operate, and therefore the melting point of such structures is likely to be greater. Overall then, each case needs to be considered on its merits.
Risk assessment
Before attempting any practical work based on the advice and suggestions on this website, you must do the following. Identify any hazards, assess the risks from these hazards, and then decide appropriate control measures to reduce the risks. You must have these approved by those in authority in your school or college laboratory. Do not rely on what is said on this website.
For further guidance see our tutorial on Risk Assessment.
Rate this page or react
Share your views on this page, 2 ratings so far
, rated at 
updated: 21 August 2003
