hey guys it's professor Dave let's synthesize some alkynes
so when we're doing synthesis we may for whatever reason want to make an alkyne so let's
look at one way that we can make alkynes, and it's not really any new chemistry
we're just going to be doing E2 reactions. we already know how to do
dehydrohalogenation so we can use a double dehydrohalogenation we can basically do
this twice on a dihalide so we know if then if we have a halogen if we have a
bromine or something on an alkane we can do an E2 but if we have two we can do
this twice so let's look first at a situation where we have a vicinal
dihalide so a vicinal dihalide means that we have two halogens and they are
on adjacent carbons, so maybe an easy way to remember that is vicinal kind of
sounds like vicinity they are in the vicinity of one another they're adjacent
so this is a vicinal dihalide because we have halogens that are on adjacent
carbons and we're going to use a very strong base we're gonna use sodamide
and so this is an extremely strong base just as in the side right sodamide what
is the conjugate acid of sodamide that's that's ammonia and ammonia itself is a
base so the conjugate acid of this base is a base. so you'd better believe that
this is an extremely strong base and so we're gonna be able to do
hydrohalogenation here so let's say that we go ahead and take this guy here right
we've got implied hydrogens right there let's go ahead and take this one and
this is just an E2 right we already know about an E2 reaction so this is no oops
wait I did I did that the wrong way let's go back I want to kick off the
other one so let's write this hydrogen here so we're gonna grab this we're
gonna kick off that bromine and so that is what it's going to form this pi bond
here and we do want to point out that it is specifically going to be these
hydrogen's on the same carbon that bear the halogens because
they actually are more acidic they're more acidic than the right we would
think that maybe we could get one of these protons and make the pi bond here
but remember that these carbons have a slight partial positivity because
they're connected to halogens so if these carbons are slightly partially
positive whereas these are not these protons are necessarily going to be more
acidic because the sigma bond is is going towards a partially positive
carbon that is withdrawing that electron it's pulling that electron density
towards itself so that makes those protons more acidic so let's say we've
gotten this one we make that pi bond right here and we kick off that bromine
so that gets us here we've got the alkyne but we still have one more
bromine on there and we are gonna run this we're gonna use two equivalents so
if we're gonna do a double dehydrohalogenation, we're gonna
need two equivalents of base and so that's actually fine we can go ahead and
get this proton and now at this point it's quite obvious that this is gonna be
the one that we're gonna get right we certainly can't get one of these three
because number one we're not going to be able to make this cumulated I mean
that's not a very favorable situation to have two adjacent pi bonds like that but
second of all an sp2 proton is certainly more acidic than sp3 so it goes sp3 pKa
right around somewhere in the 50s sp2 drops down to somewhere in the 40s SP is
somewhere around 29 or something like that so it's really 10 orders of
magnitude each time so this is definitely the most acidic proton so we
are gonna do another D hydrohalogenation we're gonna grab that proton and we're
gonna make another PI bond and we're gonna kick that off and so that's how we
get to our alkyne remember when we draw alkynes we really do need to make this
in linear fashion right don't forget that this is a carbon this is a carbon
right the the carbons that are participating in the triple bond are SP
hybridized therefore they exhibit linear geometry so we cannot draw this any
longer in zigzag fashion if you draw another pi bond right there and leave
that zigzag you know connectivity wise it's correct but geometrically it's just
not accurate so we do want to draw our alkynes this way and so there's our
double dehydrohalogenation from a vicinal dihalide, we're just using
two equivalents of that strong base and we do end up with
this alkyne here. so now let's look at a an analogous situation this time let's
start with a geminal dihalide so what's a geminal dihalide
that is when the halogens there's two halogens where a dihalide two halogens
geminal on the same carbon so the way to remember that geminal think of Gemini
twins so they're kind of like twin halogens they're on the same mother
carbon so to speak so that's how we remember what a geminal dihalide is and
we need a specific situation where we have a geminal dihalide where the
halogens are alpha to the terminus meaning here's the terminal carbon and
they're one away so alpha is 1 over terminus alpha and that's where the
that's where the two halogens are so that's a geminal dihalide and this is
not really going to go that much different it's just that we're gonna
need excess we're gonna need excess base and so this time let's exclusively go
for these terminal protons here so let's take our NH2 minus and we're gonna do
this is just more sterically available so let's go ahead and get one of these
guys we're gonna make a pi bond we're gonna kick off one of those and so we're
gonna get two here so again mechanistically this is pretty much the
same we do have more we do have more base in there so let's go ahead and do
it again and kick off that other halogen and that's going to get us to our
terminal alkyne so we have a terminal alkyne but the thing is here's why we
need excess base there's not just one molecule in solution there's many and we
need to do 2d hydrohalogenation x' but anytime NH 2 - finds an alkyne in
solution right in there maybe an alkyne produced well while some other substrate
molecule has undergone only one or none neither of the dehydrogenation x' but if
base finds this remember we just said earlier that SP 3 protons SP 2 protons
SP protons we're going down in PKA SP protons are way
way more acidic than sp2 or sp3 so if strong base finds this it is going to
absolutely go ahead and get that proton and we're going to deprotonate that
terminal alkyne to get the acetylide ion that's why we need to perform this with
excess because if we use a if we use a stoichiometric amount right if we use
two equivalents we're gonna get a mixture right we're not gonna get
everything to double dehydrogenate because some of them are going to go
three right three reactions with base some of them are going to go - some of
them are going to go one so we need to use excess and push this all the way to
the acetylide ion so that we have all acetylide ions in solution so that's why
we need excess and then once we're at the acetylide ion we can just go ahead
and finish our alkyne synthesis we can just do some up some some aqueous acidic
workup right so just whatever whatever that is and we can go ahead and get just
our terminal alkyne right we can get a terminal alkyne that way we would just
protonate that this would just be our workup step at the end of the reaction
just throw some ammonium chloride in there whatever it is and we've got our
terminal alkyne there but I do want to show you one other interesting thing we
can do with an acetylide ion because this is certainly can certainly act as a
base but on a satellite ion because of the formal negative charge and because
of the shape it's almost bullet shaped because it's linear right so we have
this we have this carb anion and then we have a triple bond going back here so
there's it's not there's not a lot of there's no steric hindrance so it's
actually a very good nucleophile and we can actually do sn2 with an acetylide
ion so we could just as well as aqueous workup we could just do an sn2 with say
an alkyl halide that would be a common sn2 electrophile so why don't we just go
ahead and we'll go over here we'll attack there we'll kick that off and so
what we've done is we've attached a three carbon fragment right 1 2 3 right
that's these carbons 1 2 3 to the end of this acetylide ion so
would be the new bond this would be the new bond that was formed in doing so so
this red bond right there represents this arrow that's where the acetylide
ion attack this three carbon this three carbon molecule and so that's just
another thing that we can do and now we've got this symmetric alkyne we've
got this fully substituted alkyne so using a geminal dihalide that's alpha 2
the terminus we have two options we can go ahead and get our terminal alkyne
with just aqueous acidic workup or we can do an sn2 and we can get a fully
substituted alkyne and so these are some different ways that we can make alkynes
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