[jrpestana]

I'm trying to differentiate between inter-intramolecular bonding. I understand one is between two different molecules while the other in within a molecule...and I know intra creates a higher bp... but how do determine if a compound has one or the other. Some compound can participate in both so how do I tell which a specific compund is involved in?  

If two compounds have the same water solubility but one has a higher/lower bp would I assume intermolecular bonding to be the reason? (or intra, why one and not the other if they both create higher bp??)

If an ortho compound is less water soluble and has lower bp than a para compound would it be because of intra molecular bonding involving the para compound? (why couldn't it be inter??)

I can't figure out how to differentiate. Does one effect the boiling point only and the other effect both bp and solubility? I have a comp. final exam (Org. II) in two weeks and need to learn this. If anyone can shed some light... Thanks  

[Donaldson Tan]

Let's compare 2-hydroxyl-benzoic acid and 4-hydroxyl-benzoic acid.

Which has the higher  boiling point? Ans: 4-hydroxyl-benzoic acid

Why? Intra-molecular H-bonding betweent the OH group and the carboxylic acid group means 2-hydroxyl-benzoic acid would have less inter-molecular bonding sites for H-bonding and permnament-dipole-permnament-dipole interaction.

I hope this clears up your confusion.

[jrpestana]

Unfortunately, you made things worst. If someone can relate directly to my post I will be able to better understand. I thought intramol. meant a higher bp... so how does 2 hyd ben. have a higher bp... and HOW do I tell which has what?... can someone draw a simpler compound and mechanism that I can relate to. I would sincerely appreciate it.  

[Donaldson Tan]

When you boil a molecular substance, you are overcoming the intermolecular bonds. Hence, boiling point depends on the extent of intermolecular bonding.

the more sites for intermolecular bonding, the greater the extent.

We shall only consider H-bonding to determine which compound has the higher boiling point because H-bonding is the strongest among the inter- and intra- molecular bonds. H-bonding refers the attraction of the lone pair on F,O or N to a H attached to another F, O or N atom.

2-hydroxyl-benzoic acid exhibit intra-molecular bonding. The -OH group is situated adjacent to the -COOH group. The hydrogen of the -OH group is exposed to the oxygen atoms of the -COOH group, thus facilitating intra-molecular H-bonding.

In 4-hydroxyl-benzoic acid, the -OH and -COOH groups are projected away from each other. Hence, there can be no intra-molecular H-bonding between them.

consider 4-hydroxyl-benzoic acid:
number of lone pairs on -OH group for inter-molecular H-bonding = 2
number of lone pairs on -OH group for intra-molecular H-bonding = 0
number of lone pairs on -COOH group for inter-molecular H-bonding = 4
number of lone pairs on -COOH group for intra-molecular H-bonding = 0
number of H for inter-molecular H-bonding = 2
number of H for intra-molecular H-bonding = 0

consider 2-hydroxyl-benzoic acid:
number of lone pairs on -OH group for inter-molecular H-bonding = 2
number of lone pairs on -OH group for intra-molecular H-bonding = 0
number of lone pairs on -COOH group for inter-molecular H-bonding = 3
number of lone pairs on -COOH group for intra-molecular H-bonding = 1
number of H for inter-molecular H-bonding = 1
number of H for intra-molecular H-bonding = 1

From the above figures, you can see there are less intermolecular bonding sites for 2-hydroxyl-benzoic acid. Thus, 2-hydroxyl-benzoic acid exhibits less extensive inter-molecular bonding than 4-hydroxyl-benzoic acid. Thus 4-hydroxyl-benzoic acid must have the higher boiling point.

[arnyk]

Also with hydrocarbons, alkanes will have a higher BP than alkenes (with the same # of carbons).  Since the single bonds in alkanes allow the chains to "tangle up" with each other, this creates even stronger intermolecular forces.  The double bonds in alkenes actually weakens the forces as the section of the chain without the double bond will move towards another molecule (let's say to its right side) and thus it is moving even farther away from the one on its left weakening the entire system overall.  Think about it in terms of saturated and unsaturated  fats, the unsaturated (akenes, alkynes) are liquid at room temperate because the intermolecular forces are not strong enough to form the solid.  Saturated fats (alkanes) on the other hand are solid at room temperature because of the strength of their IMF's.  The forces between the hydrogens and carbons I think are also Van der Waals.

[jrpestana]

Ok... allow me to ask the following:

If I have compound A and compound B (w/ same molecular formula) both have the SAME exact solubility in water but A has a HIGHER bp than B, INTER must be involved and intra would be an unlikely reason for the increase in bp?

If I have compound A and compound B (w/ same molecular formula) and A is less soluble and has lower bp, would that be an indicative of INTRA or will it depend on the compound?

Thanks geodome and arnyk for your help.

[jrpestana]

geodome:

With regards to your last post. I can safely assume that if a have a 1-2 substituted and a 1-4 substituted compound, the 1-4 will have a higher bp because of they are greater accessible for INTERmolecular bonding. Using your post as an example, how will this effect solubility? Does the 1-4 have a higher/lower/same solubility compared to 1-2 as a result of this ability for greater intermolecular bonding?

[Mitch]

This all depends on the liquid packing of the substance, and I highly doubt there are any steadfast rules regarding this, just very genal genearalizations. If there were hard fast rules they would be in your book.

[Donaldson Tan]

With regards to your last post. I can safely assume that if a have a 1-2 substituted and a 1-4 substituted compound, the 1-4 will have a higher bp because of they are greater accessible for INTERmolecular bonding. Using your post as an example, how will this effect solubility? Does the 1-4 have a higher/lower/same solubility compared to 1-2 as a result of this ability for greater intermolecular bonding?

the 1,4-substituted compound would be more soluble in a polar solvent than the 1,2-substituted one.

mitch: i thought packing is normally considered to compare  the melting points of 2 solid substances. isn't liquid supposed to be rather unordered too (although much less than gas)?