jeez that is a huge atar aim - how are you tracking for thatWould they have 2 H-bonds per molecule cause of the two hydrogens on nitrogen or is it only 1 H bond cause nitrogen only has 1 lone pair?
says the one with a 98 aim and getting 69 in math papersjeez that is a huge atar aim - how are you tracking for that
At the moment its possible just gotta grind rlly hard, loljeez that is a huge atar aim - how are you tracking for that
I always thought it was 2 (1 per hydrogen):
H bonds form between an H (attached to N, O, F --electroneg difference needs to be great enough) and so the H's of an amine can form a hydrogen bond each...?
update just found this diagram if it helps???
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but... whilst it may be an invalid concept it still makes it much easier to deduce what will be more attracted to what - so what's the issue?This is not a valid question. Perhaps you have forgotten the fundamental nature of hydrogen bonding. Hydrogen bonds only last for between one to ten picoseconds. They are highly transitory and subject to statistical distribution and therefore cannot be counted as such. All we can say is there is a certain probability of an average number of hydrogen bonds per molecule, and this gives rise to increased attractive forces between molecules on average. We know how many charge centres are located on the molecule, but we can't say for sure how many hydrogen bonds a molecule will have at any given instant.
Experiments and calculations done on liquid water showed that although each water molecule is capable of forming four hydrogen bonds, in fact this is highly dependent on time and temperature, and at 25oC water forms about 3.5 hydrogen bonds per molecule.
We can say for sure that primary amines are more strongly attracted than secondary amines, for the same molecular weight, but you can't count the bonds that will occur on any given molecule; this is simply an invalid concept.
As the diagram in SadCeliac's post (reproduced below) shows, for an H-bond you need both the hydrogen bond donor (the H atom bound to the very electronegative N, O, or F atom) and the hydrogen bond acceptor (the lone pair on the second very electronegative atom). So, a primary amine like methanamine can, with two donors but only one receptor, can only average one H-bond per molecule. There will be some molecules with both H's involved in an H-bond, and some with none, but any average of one per molecule is reasonable.Would they have 2 H-bonds per molecule cause of the two hydrogens on nitrogen or is it only 1 H bond cause nitrogen only has 1 lone pair?
In the common year 11 question showing the BP trends of the group 4, 5, 6, and 7 hydrides against molar mass (where HF, H2O, and NH3 all have much higher BPs than any of the other species due to H-bonding), the BP(H2O) = 100 oC is greater than both BP(HF) = 20 oC and BP(NH3) = -33 oC because only water has 2 donors and 2 acceptors and is thus capable of double the H-bonding of HF (1 donor, 3 acceptors) or NH3 (3 donors, 1 acceptor). The graph is available on the Doc Brown Chemistry notes website at https: //docbrown.info/page07/equilibria8f.htm .
I've been taught that amides have higher BP than carboxylic acids of similar molar mass.All hydrogen bonds are not the same. We can assign a bond enthalpy to the hydrogen bonds, and typical enthalpies can range from -1 or -2 kJ/mol up to as much as -25 kJ/mol. You can talk about the potential for more hydrogen bonding between molecules. The difference between amide and carboxylic acid hydrogen bonding revolves around the strength of the hydrogen bonding; differing electronegativities of the oxygen and the nitrogen atoms give rise to different bond enthalpy.
Carboxylic acids have much higher boiling points than amides of equivalent molecular weight due to the higher hydrogen bond enthalpy.