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Organic chemistry
Course: Organic chemistry > Unit 6
Lesson 5: Naming and preparing alkynesAlkyne acidity and alkylation
How terminal alkynes can act as weak acids and react with alkyl halides, leading to their alkylation. Created by Jay.
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- How would you name the chemical at the end of the video ()? 7:40(7 votes)
- The benzene ring takes priority over the triple bond here, which makes this either an alkylbenze or a phenylalkyne structure. Since the chain has fewer carbon atoms than the ring (5:6), the ring is the parent compound and the chain is the substituent, so the IUPAC name would be pent-2-ynylbenzene. (The -yn- is the part that tells us it's a triple bond, and the -yl just signifies alkYL group, so if it were a double instead of triple bond in the chain, it would be pent-2-enylbenzene; and if there were only single bonds it would just be pentylbenzene. Be careful there, because seeing the -yl can be a little tricky sometimes and make you think triple bond when it's not!)
If the chain portion contained either more carbon atoms or higher priority groups (like a carboxylic acid or hydroxide group at the end, for example), then the benzene ring would be treated as a substituent and we would use phenyl-[whatever] instead of [whatever]-benzene. {*That's just a loose, general rule; benzene nomenclature is very weird so there are many exceptions, most of which are kind of random and used only for historical reasons. Be ready for lots of memorizing in that chapter! Flash cards, ftw!}
He probably didn't cover it here because it seems they assume most people watch the videos in order, and you won't see nomenclature quite that complicated until later.
But in case I haven't yet confused anyone enough, I'm wondering now about the common names you could use here. It seems like benzyl butyne might work (?), but I'm not sure about this part; just making an educated guess now. I'm pretty sure everything else in this post is accurate though. Hope it helps!(9 votes)
- at, the alkylated species sholud have an inverted configuration because of Sn2 mechanism, rite? 4:20(10 votes)
- I was going to ask the same the thing. Yes, I believe that's an error. Backside attack always results in inversion. The lone hydrogen drawn with slashes/dashes should swing around & end up where the leaving group (X) started out.(4 votes)
- At, why is that proton acidic? 6:38(4 votes)
- The carbon atom is sp hybridized (50% s character). Since s electrons are held more tightly to the carbon nucleus, they are further from the hydrogen nucleus. The H atom is not tightly held, so it is more easily removed. The proton is acidic.(4 votes)
- has sodium amide something to do w/ amides? how is it an amide as we saw int the fn group video?(2 votes)
- Confusingly, the name amide is given to two different, unrelated structure types. An amide can refer to a carbonyl-based functional group, O=C-NR2, where a carbon has both a double bond to an oxygen and a single bond to a nitrogen (and a single bond to some alkyl group). An amide can also refer to a species with a negatively charged nitrogen, such as R2N^(-1), where R is an alkyl group.(3 votes)
- Atwhy doesn't the Na +cation instantly stick to the alkylnide -anion? Thank you:) 5:00(2 votes)
- I believe this reaction is taking place in water. The ions are solvated (hydrated) - each ion is interacting with multiple water molecules (ion-dipole interactions), which shields the ions from each other and reduces their interaction.
I think this video covers this:
https://www.khanacademy.org/science/chemistry/states-of-matter-and-intermolecular-forces/introduction-to-intermolecular-forces/v/solubility(2 votes)
- what happens to sodium cation from the sodium amide molecule? at aboutjay says that the Na+ and the carbanion will interact what does that mean? 5:00(2 votes)
- atyou said there was a primary carbon. Would'nt that be a methyl carbon because there is only one carbon? 2:40(1 vote)
- primary carbon is carbon attached to only one carbon(2 votes)
- what happened to the chloride on the last reaction?(1 vote)
- Lets supouse we are in a Lab. and we want to do the first reaction of the video, if we had a solution of NaNH2 and we want to form a strong base to react with the terminal alkyne, do we just have to add water to the NaNH2?(1 vote)
- At, the professor mentioned that one molar NaNH2 is used to generate carbanion with -1 formal charge. However, what if you use 2 molar equivalent of NaNH2, will it be possible to generate two carbanions with -2 formal charge? Thank you. 4:48(1 vote)
Video transcript
Terminal alkynes can
function as weak acids if you react them with
a very strong base. So something like sodium amide,
so this in NH2 minus over here came from Na plus
NH2 minus, so sodium amide, which is a
very strong base. And so if the amide
anion functions as a base, a lone pair of
electrons and this nitrogen, it's going to take
this proton right here. This is the acidic proton
on terminal alkynes, and that leaves these
electrons in here to kick off onto your carbon. So if I take NH2 minus, and
that picks up an H plus, well, that would form NH3. So now I have nitrogen with
three hydrogens attached to it and one lone pair of electrons. So when sodium amide
functions as a base, it forms ammonia as its product. What is our other product? So we had our
carbon triple bonded to another carbon
with an R group here and then a lone pair of
electrons on this carbon. So these electrons
right here were the electrons in this bond. So those electrons, and
they're moved onto that carbon. That gives this carbon a
negative one formal charge, like that. So we form a
carbanion here, also called an alkynide
anion for this portion. So this is an alkynide
anion, a carbanion in here. It's a relatively
stable conjugate base, because the electrons,
these two electrons here, are housed in an SP
hybridized orbital, which has a lot of S character to
it, so it's relatively small. So those negatively charged
electrons are held a little bit more closely to the positively
charged nucleus of this carbon here. So that somewhat stabilizes
the conjugate base, which is the reason
why a terminal alkyne can function as an acid. So once we formed
our alkynide anion, we can use that alkynide anion
to do an alkylation reaction. So let's go ahead and redraw
that alkynide anion here. So I'm going to draw
this portion, this R group over here. And then we have our
carbon triple bonded to another carbon,
a negative charge on this carbon on
the right here. And this can now function
as a nucleophile. So a negatively charged anion
can function as a nucleophile. And if we react this alkyline
anion with an alkyl halide-- let's go ahead and draw
an alkyl halide here. So I'm going to
put hydrogen there, and I'll put my halogen
over here on the right, so putting my lone
pairs of electrons. Let's draw in one
R group right here and then a hydrogen over here. So here is my alkyl halides. And if you react a
strong nucleophile with an alkyl halide that
is not very sterically hindered-- this is a primary
alkyl halide right here-- you're going to get
an SN2 reaction. So think about this
being an SN2 reaction. We have an alkyl
halide, which has a polarized bond between
the carbon and the halogen. So if the halogen is
more electronegative, it's going to pull the
electrons and the bond between it and carbon
closer to itself. So this halogen ends up
being partially negative. This carbon, therefore, will be
partially positive, like that. So we have an electrophile. This carbon right here is
partial positive charge. It wants electrons. Of course, our nucleophile
has those electrons. So the lone pair of
electrons on our carbon can attack our electrophile,
so nucleophile attacks electrophile. And an SN2 mechanism, remember,
is a concerted mechanism, meaning the nucleophile
attacks the electrophile at the same time you're
leaving group is leaving here. So these electrons are going
to kick off onto your halogen. So let's go ahead
and draw the product. So now we would have an R
group, carbon triple bonded to another carbon. And now this is bonded
to yet another carbon. So we formed a carbon carbon
bond here in this reaction. And then this
hydrogen is up here. This R group is still
here, coming out at us, and then the hydrogen going
away from us like that in space. And then we have our
halogen over here with now four lone pairs
of electrons, a negative 1 formal charge. It is stable on its own
as an anion like that. So this is an alkylation
reaction, right? We put an alkyl group
onto our alkyne. So our alkyl group
consisted of this carbon, whatever this R group is here. And we formed a new
carbon carbon bond, so we? Alkylated our alkyne. Let's look at an example
of an acid base reaction followed by an alkylation. So let's start with
the acetylene, so the simplest alkyne. So we have carbon triple
bonded to another carbon and then two hydrogens
on either side and if we react that
with sodium amide here. So we know sodium amide
being a strong base, if we use one more
equivalent, it's going to take off one
of these acidic protons. So let's say it's the
proton on the right here, and so we're going
to lose that proton. So we're going to leave
those two electrons behind on this carbon, making this
carbon negatively charged. And then the positively
charged sodium ion is going to interact with that
negatively charged carbanion like that. So that's so that's the
first reaction, Formation of your alkynide anion. And then if you want
to do an alkylation, it's a separate reaction. You take this, and let's react
it with the ethyl bromide. So CH3CH2Br. If you think about
what's going to happen, the lone pair of
electrons on the carbon is going to attack this
carbon, the one that's bonded to your
halogen, like that. The halogen is going
to leave, and you're going to put this alkyl
group onto your alkyne. So you're going to end up with
an ethyl group on your alkyne. So let's go ahead and draw that. So we have hydrogen and
then carbon triple bonded to another carbon,
and then we have to put our alkyl group on there. So a CH2CH3, like that. So we've alkylated our alkyne. This is a very useful reaction
for organic synthesis. So let's take the
molecule we just made, and let's make
something else with it. So if I took this-- let me go
ahead and redraw it over here. So if I took this alkyne,
so we just formed this. And let's react it. Two steps. Let's first react it
with our base again. So let's use sodium
amide right here. And in our second
step, we'll react it with a primary alkyl halide. So let's go ahead and draw
a primary alkyl halide here, so that is our molecule. So we think to
ourselves, what happens? I have a strong base. I still have an acidic proton
left on my alkyne, right? So the proton over
here on the left. So that's what the
base is going to do. The base is going to take that
proton forming a negatively charged carbanion,
an alkynide anion. And then that anion is going to
be our nucleophile for an SN2 reaction. So when you're
thinking about it, these electrons in
here that are going to be on that carbon giving
a negative 1 formal charge are going to come
all the way over here and attack this carbon and
attach all of this alkyl group to our carbon. So let's go ahead and
draw the products of that. We're going to have
our benzene ring. So let's go ahead and draw
our benzene ring here, so let's put in our electrons
going around my benzene ring. And then on that
benzene ring is a CH2. So that CH2 is the red one
that we marked right here, and this is the alkyl group
that gets put onto your alkyne. So let's just go ahead and
finish drawing our alkyne here. So we have now our
triple bonds, right? Carbon triple bonded
to another carbon. And then our ethyl group. So CH2CH3. So you'll see in
later videos how we use the acidity
of terminal alkynes to alkylate when we do a few
different synthesis problems.