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Why is the salt (CH3)4N+ CL- less basic than the alkyl amine (CH3CH2)2NH?
Why is the salt (CH3)4N+ CL- less basic than the alkyl amine (CH3CH2)2NH?
NVRMND, I'm ******ed. But can anyone tell me why the trimethyl amine (CH3)3N is less basic than the dimethyl amine (CH3)2NH?
ok, so it turns out that this has to do with the solvation of the cation formed. so let's think about it:
a neutral amine: you've got a polar region around the nitrogen and a nonpolar region around the alkyl groups.
a protonated amine: you've got a charged region (which means it's effectively super polar) around the nitrogen and a nonpolar region around the alkyl groups.
in both cases, water wants to solvate the molecule at the polar/charged end. however, this "need" to solvate is much much more important when the amine is protonated because unsolvated cations are rather unstable.
ok, so now we have to look at what having more alkyl groups (trimethylamine) will do to this: if you have much more steric bulk around the molecule, it is more difficult for the water molecules to solvate the cation. thus, the trimethylamine cation is less stable so neutral trimethyl amine is more stable, which lowers the pKa (less basic). Dimethylamine has less steric bulk so the protonated form can be solvated more easily. However, it still is rather bulky, and if you'll notice, the pKa of methylamine is 10.66 while the pKa of dimethylamine is 10.73. They are effectively the same basicity, even though the dimethylamine has an extra donating methyl group, so this is another example of the solvation ability influencing basicity.
Excellent question, but for the record, this will never be on the MCAT because it's not a well known effect. Hope this helps... Lemme know if you have any questions
oh you just had to 1 up me haha
So can one not just use the knowledge of electron donating/electron withdrawing groups to figure out which is mroe acidic/less acidic? It works in the case of trimethyl vs dimethyl amines.
no, it doesn't. trimethyl amine, by that reasoning, should be more basic (more electron donating methyl groups) than dimethyl amine, but it's almost 1 pKa unit lower (less basic) than dimethylamine. steric effects on solvation ability are important for these molecules.
So can one not just use the knowledge of electron donating/electron withdrawing groups to figure out which is mroe acidic/less acidic? It works in the case of trimethyl vs dimethyl amines.
So can one not just use the knowledge of electron donating/electron withdrawing groups to figure out which is mroe acidic/less acidic? It works in the case of trimethyl vs dimethyl amines.
So an acyl chloride will be more acidic than a carboxilic acid because according to periodic trends the cloride ion is more electronegative than the oxygen conjugate of the OH on the carboxilic acid. I'm just trying to remember this stuff so idk if it's 100% correct, so feel free to tweek my recollection.... I won't take offense.
Yeah, I just think the idea is that I cannot just use this "trick" all the time, as tehre are exceptions to that rule. But MCAT does not test the exceptions, from what I hear.Oh,, one important thing... the answer i gave above is based on the fact that electron pair donar are considered as base and electron pair acceptors are acid. I believe you memorized that electron pair donating gps. decreases acidity,, well I would say it is better to understand why that happens.
Acid when gives proton then it remains in the anion form and electron donating grps. increases negative charge making it less stable and it reattaches itself to proton ( acidity decreases). opposite happens with electron pair withdrawing gps.
Sleepy425 is right. And the answer I gave on the basis of Electron pair donar tendency is also right.
No, that wasn't the point I meant. What I meant was that the mcat will test that, but it tends not to test you on exceptions to the rules for biochem/organic chem.I don't wana infrige on
MCAT regulations, but I actually had a question that involved basicity of two molecules. I can't be specific but the MCAT does test on these questions.