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Course: Organic chemistry > Unit 11
Lesson 2: Formation of carboxylic acid derivativesAmide formation from acyl chloride
Amide formation from acyl chloride. Created by Sal Khan.
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- at5:00min, why can't the Cl give an electron to the amines hydrogen to form HCl (which will than react with the other amine)?
(7 votes)- Remember that halogens like chlorine have 7 valence electrons. The Cl leaving group in this reaction took an electron with it when it got kicked off by the secondary amine attacking the acyl chloride. So Cl now has 8 valence shell electrons. The extra electron not only gives Cl a negative charge but it ALSO completes Cl's valence shell (octet rule). Cl is now suddenly much more stable than it was before and thus relatively unreactive, versus the secondary amide which has a nucleophilic lone pair of electrons on the nitrogen that readily react with any proton in the mix.
If you threw Cl- in with just H+ (protons), of course it would form HCl salt like you indicated. However, in this reaction, the proton isn't free floating. It is still bonded to the molecule. Something stronger than a stable chloride ion with a complete valence shell is required to tear the hydrogen away from the molecule. The secondary amine is the much better nucleophile.(10 votes)
- Why does the second molecule of dimethyl amine function as a base, and not a nucleophile attacking the electrophilic carbon on the acyl chloride?(2 votes)
- The H on the acylammonium ion is much more positive than the carbonyl carbon, so the dimethylamine preferentially attacks the H.(3 votes)
- I read that when using NH3 for similar reactions, you need to use 2 equivalents of it. Why is that?(2 votes)
- Yes, you can just add NH3. If you do instead of the NH(CH3)2 then it is the same reaction as shown, just replace the methyl groups with hydrogen. The result is an acetamide. (The orange looking one at the end on the top left.)(2 votes)
- At4:15, doesn't the amine need to get deprotonated first and then the Cl get kicked off? Wouldn't the protonated amine be a better leaving group than the Cl?(1 vote)
- How would you know? You have to look at a pKa table. Then you can compare the strengths of the conjugate bases. The pKa of NH4+ (ammonium) is 10. the pka of HCl is -6 or -7. Therefore, Cl- is 1x10^17 times better as a leaving group than a protonated amine. Basically, there's no comparison.(3 votes)
- I am synthesizing 2- aminobenzothiazole with 4 chloroaniline . the reaction is between NH2 and Cl. but this reaction is not happening. why ?(2 votes)
- chemical test of carboxylic acid(2 votes)
- who studied the process of ammonolysis ?(2 votes)
- 6:20why wouldn't the Cl anion take the hydrogen. why was is necessary that the dimethyl amine take it?(1 vote)
- Chloride is not a very good base, which makes sense as it is the conjugate base of a strong acid (HCl)(2 votes)
- According to my book - aminolysis takes place in a solvent of H2O (and NaOH (if not two amines/ammoniums)).
Problem is that hydrolysis of Acyl Halides creates carboxylic acid. (NaOH to remove HCl + Water)
If you want to make an amide through Carboxyl Acid it requires you to modify your OH group with DCC (According to my book) following with amine (which we'll probably have).
So why according to your video and my book not mentioned that you'll probably have a mixed product of amide and carboxyl acid?(1 vote) - If i had an amide and wanted to convert it to an amine containing same number of carbons as before, would I be wrong to use reagent Zn-Hg/HCl ? If yes, why?(1 vote)
Video transcript
Let's think about what might
happen if we had a molecule of butanoyl chloride. So one, two, three,
four carbons. That's where the but-
comes in, and then it's a butanoyl chloride. So it's an acyl halide or
it's an acyl chloride. Let me write this down
just to give us some practice with namings. This is butanoyl chloride. We saw a couple of videos ago
that acyl halides, you just count the main number
of carbons. One, two, three, four carbons. That's where butan-
comes from. And then this is an acyl
chloride, so the acyl chloride part, this part, gives
us the -oyl. Or I guess you could look
at even the carbonyl. We could even say this whole
part over here, we know it's a carboxylic acid derivative. It's in an acyl chloride. So that's why we have the -oyl
here and the chloride. So let's think about what would
happen if we have a molecule of butanoyl chloride or
if we had a solution, even better, of butanoyl chloride,
and for every molecule of butanoyl chloride, we had two
molecules of dimethylamide, or sorry, dimethylamine. Let me draw that. So dimethyl, so two
methyl groups. One, two, and then we have one
hydrogen, and then we're going to have two molecules of this
for every one molecule of our butanoyl chloride. So let me copy and paste this. So edit, copy. And edit, paste. So let's think about what would
happen in a solution with these reactants in
this ratio right here. And just as a bit of review,
this right here is dimethyl. We kind of have a tie for our
longest chain attached to the nitrogen: dimethylamine. You could pick one of these as
the longest chain and then it would actually be
N-methylmethylamine. Either one, this is the
one that you're more likely to see. But this video isn't about
naming, this video is to think about what's likely to happen. So what are we dealing with? We have amine here and then
we have an acyl chloride. And if we look at our hierarchy
of stability that we looked at in the last video,
we saw that when an amine reacts or a carboxylic
derivative from an amine, which is an amide, is
much more stable than an acyl chloride. This is the least stable. So if we're starting with an
acyl chloride and we have the ingredients for an amide, this
reaction will probably end up with amide. So let's see how we
can get there. What we know from previous
videos when we studied amines, that they are pretty good
nucleophiles, so they could have a nucleophile attack
on this carbonyl carbon right over here. So you can imagine this amine
right here, this nitrogen, gives an electron to this
carbonyl carbon. This carbonyl carbon can then
let go of this electron to this top oxygen over here. It was already hogging
it to begin with. And so the next step of our
reaction would look like this. What was the butanoyl chloride
will now look like one, two, three, four carbons. Now you only have
a single bond. This oxygen up here now
has a negative charge. Since it took an electron from
this central carbon right over here, you are still bonded
to this chlorine. And actually let me do that in
a different color so we can keep track of it. You are still bonded to this
chlorine atom and then the nitrogen has been bonded. So this bond I'll drawn green
because we're giving this green electron to form the bond,
and then the rest of the dimethylamine is still
over there. Let me do it in that
same color. The rest of the dimethylamine,
so you have CH3, CH3 and then you just have your hydrogen. And then this nitrogen gave away
an electron, so it has a positive charge. Now, the next step, and this
is chlorine, it's fairly electronegative. It's a not-so-bad
leaving group. It will want to hog this
electron that is on the central carbon. So you can imagine in the next
step that this oxygen gives back an electron to reform the
carbonyl group at the exact same time that this chlorine
takes an electron and forms a chloride anion. So the next step in this will
look like-- so the molecules will now look like--
you have there one, two, three, four carbons. That's one of the bonds to that
oxygen, but now you've now formed another bond. That electron was given back to
that central atom and now you have the bond to the
nitrogen, this bond right here, and then the nitrogen is
bound to one methyl group, two methyl groups, and a hydrogen. And it had given away an
electron so it has a positive charge, and the chloride anion
has been bumped off. The chloride anion, it took an
electron from that central carbon, so now it has
a negative charge. Now, we still have that other
dimethylamine molecule, or for every one of these, we should
have two of these. We've only use one of them, so
let me bring the other one into the mix, and this one could
just be in the mix to make this other molecule neutral
now, and this could go in either direction. This part right here could
go in either direction. This guy's not so weak or it's
not a bad base, so this guy could give an electron to this
hydrogen proton, and then the hydrogen proton's electron, or
that hydrogen atom's electron, can be taken back by this
nitrogen to make it neutral. And I'll make this go
in both directions. So this could go in
both directions. So then this molecule over here
will look like one, two, three, four carbonyl group. And let me color code it the
same way just so we know what parts are which parts. We have this green bond to this
nitrogen, which has now lost a proton. It was able to take
back an electron, so it is now neutral. It is bonded to one,
two methyl groups. This dimethylamine
took a hydrogen. So let me draw it over here. So it's nitrogen one, two, and
then it already had one hydrogen and now it's gaining
another hydrogen. It's gaining this hydrogen
right over here. So now it's bound to
that hydrogen. It gave an electron to get that
hydrogen nucleus so it now has a positive charge. And, of course, we can't forget
about that chloride anion that's floating around. And these might be attracted
to each other. One's positive, one's negative,
form a salt. So what are we left with? And then once again, this
is a little bit of practice of naming. We started with an acyl chloride
and we ended with an amide because we now have this
nitrogen group attached. So what is this? This has one, two, three, four
carbons so it's butan-, but it's an amide. So this part tells us that we're
dealing with an amide. It's butanamide, but to specify
what type of amide, we see that we have two methyl
groups attached to the nitrogen there, so we would call
this-- let me put another color here. We have one, two methyl groups
attached to the nitrogen, so we would call this N comma
N-dimethylbutanamide. This N comma N let's us know
that the methyl groups are attached here as opposed to on
the butyl group, I guess we could view it this way. So we end up with
N,N-dimethylbutanamide, which is an amide, one of the most
stable carboxylic acid derivatives. And then this right here,
this salt, what is this? This right here is dimethyl,
and this is now no longer dimethylamine. We now are a positively charged
cation, so this is dimethylammonium. We get that from the fact
that we have four bonds. We have a positive charge. This dimethylammonium, and then
we have this negative anion, dimethylammonium
chloride. I'll just go to the next line. Dimethylammonium chloride. But I really just wanted to show
you a mechanism here of how you could go from a less
stable or one of the least stable carboxylic acid
derivatives to one of the most stable, going from an acyl
chloride, which butanoyl chloride was, to an amide.