let's talk about the formation of enamines and related reactions here so secondary amines will
form those enamines with aldehydes and ketones we're going to require activation with just a
little bit of acid trace or catalytic amount we're gonna require anhydrous
conditions because the reaction is highly reversible we need the water remove water
and keep this thing water free so please don't invoke H3O+ as your acid a simple h+ is fine
to indicate or just the name of the acid HCl something like that to indicate that we are
not adding additional water just a catalytic acid these compounds are typically not isolated
but they are a useful synthetic intermediates so what happens when we put a secondary amine
into a flask so there's our amine secondary amine into a flask with trace acid and a ketone or an
aldehyde well there's a fast almost it's like a titration between a strong acid and a weak base
goes to completion and what you end up doing is you end up forming a trace amount of protonated
amine a trace amount of chloride just due to the stoichiometry of HCl you've still got plenty of
amine left over and you still have plenty of ketone left over okay so you've still got the
two things you need to react and you've got this trace protonated secondary amine there that's
gonna be important when we consider the mechanism in a moment alright the first what can we do with
this reaction we've got an aldehyde reacting with our secondary amine here we've got pyrrolidine is
commonly used and it will settle on like we said the enamine form and that looks like this with
the aldehyde if it was a ketone we just have that extra side and it and interestingly enough if it
is a unsymmetrical ketone an unsymmetrical ketone like say two butanone we would prefer reactions
of this to occur from the less substituted side we're gonna see that with alkylations
but for now let's just have a look here at what the enamine looks like for a ketone just two
carbons on either side so the Alpha carbon either side could take that alkene alright well what do
we do with these things once we've got them if we're not gonna isolate them well in the presence
of trace acid we do have a reducing agent sodium cyano borohydride so NaCNBH3 with a little workup
of H2O will give us a secondary or a tertiary amine from reduction reductive amination as it's
called of this enamine okay so ultimately we get that tertiary amine same thing with the ketone
we can do reductive amination again reduce the double bond we end up with a just a more highly
substituted tertiary amine at that point okay we can do other things with enamines but let's look
at the mechanism of the formation of this enamine first okay so if we take our ketone for example
many texts will say and that's no problem here that the amine is a strong enough nucleophile to
add directly and what that will do is that will create this doubly ionic intermediate here with
a positive and a negative charge at the same time that a base from solution can come in or
excuse me that a the trace acid then comes in which we just said was the protonated secondary
amine and donates that proton fairly efficiently it's microscopically reversible but it's going to
favor the products here quite dramatically to get to this next intermediate I still highlight that
bond okay and so that's fine others argue that that is a as a very challenging step both due to
the limited nucleophilicity of an amine versus a more powerful nucleophile like a hydroxide or an
alkoxide and then you've got our grignard reagent even and then you've got the argument that the
intermediates a little bit high in energy because of the double charge so some people will say that
this first step and again perfectly fine to use either one of these from my end that this first
step is better explained by hydrogen bonding between the protonated secondary amines so
there would be a formal charge of plus one well if it's sharing it's a delta plus and then that
puts a delta plus on the carbonyl so a hydrogen bond forms between the protonated amine and the
nucleophile and what that does it activates the electrophile the ketone or the aldehyde
and make sure we've got our secondary amine here since we're doing enamine catch me on
that so two R group's there instead of just the one for a primary amine okay and so that assists
the addition here and what ends up happening is we form this bond in one step okay so this
hydrogen bond interaction allows us to do you directly to this intermediate in one step okay
so either way you do it I'm perfectly fine with either one I think probably the second way is
more realistic but the rate limiting step again is probably going to be the elimination of our on our
leaving group so it's tough to say tough to say on that one there's a competition there and from
my reading it hasn't been resolved 100% just yet so from this intermediate we've got a good a poor
leaving group in hydroxides so we want to make sure that we fix that so we're gonna use a base
from solution and we are going to deactivate the amine and again that should be a second
R group because it is a secondary amine so sorry about that folks please make sure that's amended
there we want to deactivate that leaving group and then in a subsequent step we've just reformed that
trace acid again we want to or actually a full equivalent of acid therefore every one of those
intermediates we get and then activate hydroxide to water we're still reversible at this point
now we've got an excellent leaving group there's our two R groups on our secondary amine we kick
off the leaving group and that leaves us with not the most stable structure in the world because
we're going to have a plus one formal charge on nitrogen well that's why the enamine a little
bit more stable and so our base from solution now we'll come in and essentially tautomerize
this to the enamine okay and so that's why we get our enamine that's how we get our enamine
the whole process is equilibrium ultimately we drive this thing by that elimination of water
byproduct here we remove water as we form the amine using experimental set up techniques like
a dean-stark trap or a drying agent all right so there's making our enamine well how do we use it
well we can use an enamine for alkylations of the alpha carbon so once we've made it we can
reform that in carbon double bond alpha carbon reaches out to a primary halide this is just an
sn2 reaction and so we want primaries best but secondaries sort of okay because we've got a
weaker base here so it's not the strongest base in the world but E2 starts to creep in at that
point we'd rather stick with a primary amine with enolates we have to stick with a primary
amine what do we get out of this step we get that charged form again but one two three additional
carbons okay so what do we do with that thing well that's wait a second we'll address it in a moment
all right well we can also do a beta addition so these are again good nucleophiles enamines are but
weak bases so they're gonna be good substitution nucleophiles are also going to be soft bases soft
nucleophiles meaning they prefer beta addition to an alpha beta unsaturated ketone or as you might
call it or an aldehyde as you might call it a 1,4 addition the conjugate addition another name so we
attack that beta position kick over kick up and we get an enolate that enolate and we can go through
an intermediate here so we just form this 1 2 3 we just added 1 2 3 1 2 3 carbons and then an oxygen
so we got an enolate and so it will tautomerize back to an aldehyde here but we're still left with
this charged intermediate 1 2 3 plus our aldehyde that we need to deal with ok so what we do is we
hydrolyze that hydrolysis of an enamine and so we could consider this as being from the enamine
form by tautomerization so if you did have start with the enamine you would just talk tautomerize
by using a base from solution here it is water so we know exactly what it is our hydronium
even to get to this form so once we're in that form we then have attack by our water molecule
how many carbons one two three four five six one two three four
five six now we've got a tetrahedral intermediate
waters a good leaving group so we want to reduce the leaving group ability of water using another
water molecule or a base from solution here remove that proton but now we've got two bad
leaving groups so we want to activate the nitrogen using the acid we just formed and
we've already got acid in solution anyway as a aqueous acid workup don't forget our carbons
now the nitrogen is a better leaving group we can form the carbonyl here that lone pair let
nitrogen lead is a plus 1 formal charge there now we can reduce that to zero and at this point
what encourages this along is all the acid in solution we just eliminated this amine it's a
good it's a decent nucleophile so we're a little worried about it here so we're
going to use the Le Chatelier's principle by having excess water we're also going to
use the fact that this nitrogen is going to get protonated in order it will neutralize
itself essentially killing off its nucleophile ability leaving us with our alkylation product
okay so ends up being a ketone there from that enamine that we hydrolyzed