Muzz1969
Harmless
Posts: 37
Registered: 19-11-2014
Member Is Offline
Mood: No Mood
|
|
Help with Boiling Points
Hi everyone,
I'm looking for some help with intermolecular forces and there relation to boiling points. I have 3-methylbutanal and 2-methylpropanol. One has a
carboxylic and the other has and aldehyde function group.
My question is, is it the different function groups and only the different groups that change the boiling point of these two very similar molecules or
are there other factors involved.
|
|
|
smaerd
International Hazard
Posts: 1262
Registered: 23-1-2010
Member Is Offline
Mood: hmm...
|
|
Functional groups are a way of describing the structure of a molecule. In this case there are two factors to consider.
First lets look at the functional groups, we have an alcohol and an aldehyde(not a carboxylic acid, be careful!). The aldehyde can accept hydrogen
bonds at the carbonyl oxygen. It doesn't have a viable hydrogen bond donor however. Whereas the alcohol can accept and donate to form hydrogen bonds.
So from this perspective, we can say that the alcohol compound, 2-methylpropanol, should have the strongest attraction to other molecules of the same
variety compared to the aldehyde. I hope that makes sense. The other functional groups present on either molecule is the alkane functional group.
The alkane part of both molecules at first glance looks the same. Look closely though, the propanol is one carbon shorter in length than the butanal
compound. This means they aren't isomers and one has a greater molecular weight. Molecular weight also effects boiling point (takes more energy to
lift/boil something that is heavy). High molecular weight compounds also have more interactions. Consider a weak intermolecular force such as van der
waals(london dispersion) and look at methanes boiling point, verses octane. Large difference there.
So the intermolecular forces dictate a lot of boiling point phenomena, but so does the molecular weight, and orientation of bonds (different isomers
even stereoisomers can have stronger/weaker IM forces). In this case the orientation of the alkane chain bonds are essentially the same, so that can
sort of be ruled out for general considerations.
I hope this helps. Analyzing structure and gaining an intuition about chemicals is a skill that is very important!
[Edited on 23-7-2015 by smaerd]
|
|
|
Muzz1969
Harmless
Posts: 37
Registered: 19-11-2014
Member Is Offline
Mood: No Mood
|
|
Thanks for that , I'll begin to digest what you have said and shoot back some more questions. This is my first organic unit at uni.
|
|
|
Muzz1969
Harmless
Posts: 37
Registered: 19-11-2014
Member Is Offline
Mood: No Mood
|
|
Ok one question doesn't the O=C-OH on the 2-Methylpropanoic Acid make it a Carboxylic function Group and alcohol function group does not have the
double bond
[Edited on 23-7-2015 by Muzz1969]
|
|
|
Texium
Administrator
Posts: 4587
Registered: 11-1-2014
Location: Salt Lake City
Member Is Offline
Mood: PhD candidate!
|
|
Well,
in your first post you said 2-methylpropanol, but if that's actually what you meant, then yes, it would be a carboxylic acid.
|
|
|
Muzz1969
Harmless
Posts: 37
Registered: 19-11-2014
Member Is Offline
Mood: No Mood
|
|
Quote: Originally posted by zts16  | | Well,
in your first post you said 2-methylpropanol, but if that's actually what you meant, then yes, it would be a carboxylic acid. |
Sorry yes i made a typo on that my bad.
|
|
|
Muzz1969
Harmless
Posts: 37
Registered: 19-11-2014
Member Is Offline
Mood: No Mood
|
|
Sorry about my mistake, how does that now change things from the point of my original questiion.
|
|
|
![[*]](images/xpblue/default_icon.gif){kind=icon}
