Exam three number seven is the synthesis of cinnamic acid starting with benzene and it's
based on a homework problem and a couple of examples from our chapter and so if we look
at the opening box we have the option of only using reagents that contain two or fewer carbons
and we're limited to three up to three additional steps here for our second set of reactions,
but we got a couple of options to get to benzaldehyde so if we remembered from our reading a particular
reaction, the Gatterman-Koch reaction where we could take carbon monoxide, HCl under high
pressure and an aluminum chloride copper one chloride catalyst mixture, we could do this
all in one step and so each part of this the carbon monoxide, the HCl, the pressure and
the catalyst one point each for a maximum total here four points okay or we could have
used the alternative option also worked as an example in our reading where we have to
start with a Friedel-Crafts reaction to form a methyl benzene and then or toluene
excuse me and then we have bromine and light to brominate this position we could also have
used NBS and light or we could have used NBS a peroxide and heat as well for this step
two and then we could take that primary halide and replace it with an alcohol via a SN2 reaction
okay so that's a good candidate even though hydroxide is a strong base it's a primary
halide it'll prefer to go SN2 and then finally we could oxidize this up with sodium hypochlorite
in acetic acid at zero degrees okay I would allow in case we don't really understand or
have the knowledge of chromates and chromic acid in pyridines and chlorochromate taking
this species all the way up to benzoic acid that's something that we would not have normally
encountered so would allow PCC as well alternatively you could propose the Swern oxidation as well
each of these being worth an additional of one point each each of these four steps, but
this step four we could have alternatively used PCC or the Swern oxidation which would
involve dimethyl sulfide, oxalyl chloride at low temperature it's negative 60 degrees
followed by a workup with triethylamine so we could have used those options as well though
in practice PCC would not be a good option okay, but again it would be a good thinking
on your part to think of it as a gentle oxidizing agent, now alright so from here with benzaldehyde
we're allowed three additional steps to form cinnamic acid so based on the homework problems
and what we've seen in the chapter I can think of four ways that we can potentially do this
the two from the assigned homework problems are the malonic ester starting material and
the Perkin Synthesis which actually uses an anhydride across aldol with an ester hydrolysis
and across aldol with an oxidation of these I could see these three satisfying the requirements
of the problem and out of these three all three of these would be accepted the two in
the middle would actually work and the fourth would not work for a reason that we are not
aware of at this level of organic chemistry so let's talk about all four of these and
talk about what is going on so with a malonic ester we would be starting with a for synthesis
of a carboxylic acid that's something that the malonic ester is designed to do we would
have to have a starting material of a malonic ester and if we look at this it's got one,
two, three, four and five carbons at the minimum and if we try to make this from simpler materials
the very best we could do would be an enolate of an ester with a carbonate type species
here so that we could end up with our attack here tetrahedral intermediate kick off on
the methoxide we'd have our malonic ester the problem is each one of these has at least
three carbons each so we have to get rid of this one as one of our homework problems that
does make exactly cinnamic acid because of the carbon requirement and so that leaves
us with anhydrides and the Perkin synthesis and so let's have a look at that one okay
if we look at our starting material being acetic anhydride it's got four carbons, but
we could make acetic anhydride from a two carbon carboxylic acid okay materials containing
two carbons or less two of those would form an acetic anhydride with heating or with the
addition of a water removing agent like phosphorus pentoxide okay so we can satisfy the two carbon
requirement using this synthesis so let's look at that one first first thing up we would
need to make our anhydride using one of our steps we're limited to three, we could do
this just by heating or we could do it with the catalyst phosphorus pentoxide okay and
we would remove water in that method now from there as the homework problem would suggest
we use acetate to prevent the destruction so acetate is the carboxylate here sodium
acetate would work we can use acetate to prevent the destruction of our anhydride if acetate
attacks acetic anhydride just makes acetic anhydride and acetate right back so no problem
there we use that to form our enolate so ultimately in equilibrium we would form an enolate with
an alpha carbanion form that would be here and that could attack our benzaldehyde if
we add these three in solution okay so that enolate attacks benzaldehyde and it would
undergo spontaneous condensation to form this intermediate structure and so this should
look very familiar just like an aldol condensation just using acetic anhydride as a starting
material instead of an aldehyde or ketone so just like we saw in the book and then finally
also a homework problem portion is adding water to do the hydrolysis of this we could
also have used aqueous acid H3O plus to make sure we're in the protonated cinnamic acid
form in the final product so one step two step and three step gets us to cinnamic acid
okay so that's if we're using the Perkin synthesis like the homework problem, however, we could
also think of using a crossed aldol followed by ester hydrolysis so let's look and see
how that would look to us here if we are thinking backwards so this is not in any one of the
problems, but it's something you might have come up with if you're thinking backwards
allow me to use pH here as phenol to abbreviate this to make it a little faster we can imagine
the hydrolysis of an alpha beta unsaturated ester even though it would attack in the beta
position ultimately the final product of basic saponification or acidic hydrolysis would
be cinnamic acid in this through a series of mechanistic steps so we don't need for
this problem just hydrolyze the ester you could work that out you could have worked
that out, but hydrolysis of an ester is given in this form in our chapter so hydroxide or
with H3O plus okay to get to that ester we would have to have our say we're using a methyl
ester our methoxide and our aldehyde and then we would drip in this ester slowly and we
could form our beta-hydroxy ester that would be spontaneously
decompose or dehydrate into our alpha,beta-unsaturated ester we could also have used LDA addition
as well to get this to go, but since this doesn't have any alpha hydrogens we were able
to accomplish everything in just two synthetic steps since our process here would go all
the way to the alpha,beta-unsaturated ester so we're limited to three steps if you thought
of it as requiring three steps you could have used this as one of your steps maybe even
put heating there to indicate that dehydration and that's fine we could make this into exactly
three steps if we used LDA addition and so let me show you that as well well we're getting
a little long here, but we gotta make sure we understand this you can scrub ahead if
we're taking awhile and so let's have a look if we were to take our starting material or
excuse me we want our ester sorry our ester we could treat this with LDA this would make
our enolate of the ester and then we could add our compound slowly
to form our alpha,beta-unsaturated alkoxide here and then we could make our third step
adding a proton source to this okay and then we would end up with a spontaneous dehydration
and then we could treat it with hydrolysis conditions in order to proceed to the final
product so we could have done that as well for exactly three steps for our synthesis
now the last way we could perceive of doing this would be in a cross aldol we just did
a crossed aldol with an ester here adding to an aldehyde now let's do a crossed aldol
with the ketone adding and so we could say using acetic acid or excuse me acetone we
can say let's add LDA to make our enolate and then there's our first step let's add
benzaldehyde slowly to form here our alpha,beta-unsaturated system,
ah but then we would need two steps to protonate and then oxidize up the aldehyde so we can't
get there in four steps if we're using LDA, but what we can do instead is to a solution
already containing hydroxide we could add our ketones slowly and encourage the formation
directly of our unsaturated aldehyde this is just an aldol condensation with spontaneous
dehydration so that's one step and that's a proper a slow addition of the ketone as
well because we don't have any alpha hydrogens on our benzaldehyde okay so that's perfectly
legal and here's the part that we may or may not have thought of we could attempt to oxidize
our aldehyde with chromic acid and heat to the carboxylic acid okay and so if we do that
if seems like it should work, but in practice this goes all the way to benzoic acid and
so in reality it doesn't work however we don't know that that's beyond our study at this
point so for you to understand that the harsh oxidizing conditions to go to the carboxylic
acid here I'm okay if you use this route.
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