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Organic chemistry
Course: Organic chemistry > Unit 7
Lesson 3: Reactions of alcohols- Oxidation of alcohols I: Mechanism and oxidation states
- Oxidation of alcohols II: Examples
- Biological redox reactions
- Protection of alcohols
- Preparation of mesylates and tosylates
- SN1 and SN2 reactions of alcohols
- Formation of nitrate esters
- Preparation of alkyl halides from alcohols
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Preparation of alkyl halides from alcohols
How to convert an alcohol into an alkyl halide. Created by Jay.
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- Why are we using pyridine with SOCl2 and not with PBr3 ?(7 votes)
- Pyridine helps in the cleavage of O-Cl bond in SOCl2, which is used in formation of haloalkane....but in PBr3 the bond isnt very strong thus cleavage doesn't need catalysis(3 votes)
- Atwhy would it be an SN2 if that is a tertiary carbon and SN2 rxn only happens in primary and secondary carbons? 4:08(8 votes)
- its not a tertiary carbon, its given at the start that its a secondary or primary carbon(6 votes)
- What would happen if it was a tertiary alcohol that was combined with PBr3? Would it simply be no reaction or would there be some type of elimination?(4 votes)
- SN1 is a possible reaction which will form a tertiary halide. Elimination is possible but generally if elimination is required, heat is added. If heat is not shown, the major product can be assumed to be substitution product (SN1)(2 votes)
- is there any way to make an alkyl halide from a tertiary alcohol (by an Sn1 reaction)?(2 votes)
- OH- is a bad leaving group, because it's a very strong base. The reaction hapens but only if you neutralize it ( as water is stable and neutral, it's a good leaving group). Sorry if I wrote something wrongly, I'm not fluent in english.(4 votes)
- What about preparing an alkyl halide with a fluorine atom? What do you add to an alcohol to form that?(3 votes)
- you can simply do halide exchange after the final step. (Finkelstein reaction)(1 vote)
- Atit's kicking off the chlorine. Why does this happen if the O has a +1 formal charge and is more electronegative than chlorine? 2:37(2 votes)
- The other oxygen has a -1 formal charge, and as it forms a double bond with the sulfur, it kicks off the chlorine.(2 votes)
- then it would be the same mechanism when pbr3 is replaced with pcl5?(2 votes)
- NO ! PCl5 has no vacant orbital for the bond formation between PCl5 and R-OH. So NO REACTION! :)(2 votes)
- In a previous video about Ts and Ms we saw that if an alkohol attacks the TsCl-Molecule the elektrons kick off to the chloratom an not to one of the oxygeneatoms attached so the sulfur. Why is it different in this mechanism. How could I predict which way it´s gonna go? Thanks(2 votes)
- At, why didn't the bond electrons move to the chlorine atom instead of oxygen atom? Isn't chlorine more electronegative? 1:30(1 vote)
- Oxygen is more electronegative than chlorine(3 votes)
- why does the oxygen has a positive charge with it is bonded to SOCl2 or PBr3?(2 votes)
- In the first case, we have an O bonded to a C, an H, and an S. In the second case. we have an O bonded to a C, an H, and a P. In each case, there are three bonding pairs and one lone pair of electrons in the valence shell of the O atom. According to the roles for calculating formal charge, the O atom "owns" all of the lone pair electrons and half of the bonding electrons, for a total of five electrons. Since an isolated O atom has six valence electrons, the O atom has lost an electron. It has a formal charge of +1.(1 vote)
Video transcript
In this video we're
going to see how to prepare alkyl
halides from alcohols. And so if we start with this
alcohol over here on the left, and we add SOCl2, which is
called thionyl chloride, and pyridine to it. We're going to substitute a
chlorine atom for the OH group. And this mechanism occurs
via an SN2 type mechanism, which means that it's
only going to work with primary or
secondary alcohols. And you will get
inversion of configuration if you have a chirality center
present in your final product. So let's take a look
at the mechanism. And we'll start
with our alcohol. And so the oxygen is going
to have to leave somehow. But by itself OH is not
the best leaving group. And so we're going to react this
alcohol with thionyl chloride to convert it into a
better leaving group. And so if we draw the dot
structure for thionyl chloride, we would have to sulfur double
bonded to an oxygen here. And then the sulfur is
also bonded to chlorine. So I'll go ahead and put in
those lone pairs of electrons on the chlorines, like this. And if you count up
your valence electrons, it turns out you need two more. And those go on the sulfur. It's OK for sulfur to
violate the octet rule, since it is in the
third period now. So a lone pair of
electrons on oxygen is going to form a bond
with our sulfur atom, which would therefore kick
these electrons in here off onto the top oxygen. So if we go ahead
and draw what we get from that first
step of our mechanism, now our oxygen is
bonded to our sulfur. The oxygen is also a
bonded to a hydrogen. One lone pair of electrons
formed that new bond, so one pair of electrons
is left behind. Which would give this oxygen
a plus 1 formal charge. Connected to the
sulfur, this top oxygen here had two lone
pairs of electrons. Picked up one more
lone pair, which gives it a negative
1 formal charge. And this sulfur is still
bonded to chlorine. So we can go ahead
and draw those in. And we can go ahead and draw
that lone pair of electrons on that sulfur like that. And so in the next
step of the mechanism, we're going to reform the double
bond between oxygen and sulfur. So these electrons are
going to kick in here. And these electrons would
kick off on to that chlorine. So when we draw the
next intermediate here, we would now have our oxygen,
still bonded to a hydrogen, still with a plus 1
formal charge like that. And now our sulfur is
double bonded to our oxygen again with two lone pairs
of electrons on the oxygen. The sulfur is also bonded
to one chlorine now, so one of the chlorines left. And we can go ahead and
draw in that chlorine. So one of the
chlorines left here. It's a negatively
charged chloride anion. And then still
there's a lone pair of electrons on our
sulfur like that. So at this part
of the mechanism, the pyridine comes along. So if we go ahead and draw the
dot structure for pyridine. It's a base, and so it
looks like a benzene ring, except we have a
nitrogen here instead. And there'd be a lone pair of
electrons on this nitrogen. And so that lone
pair of electrons going to function as
a Bronsted-Lowry base and take this proton
here on the oxygen. And that would kick
these electrons back off onto this oxygen. So when we go ahead and
draw that-- let's go ahead and get some more room
here-- so what would we get? We would now have our
carbon bonded to our oxygen. Our oxygen now has two lone
pairs of electrons around it. And we have our sulfur,
and our chlorine, and our lone pair of
electrons on the sulfur. And now we've made a
better leaving group. So this is a better
leaving group than the OH was
in the beginning. And if we think about an
SN2 type mechanism now, we know that the bond
between carbon and oxygen is polarized, right? Oxygen being more
electronegative, it will be partially negative. And this carbon here
be partially positive. And so now we can think
about our SN2 type mechanism. Our nucleophile will be
this chloride anion up here that we formed in the mechanism. So that's going to
be the nucleophile, and it's going to attack our
partially positive carbon. An SN2 type mechanism. So as the chloride attacks,
this stuff on the right is going to leave. So the electrons in magenta are
actually going to move in here, and then these
electrons are going to kick off onto that chlorine. So when we draw the
product, we can go ahead and show the chlorine
has now added on to our carbon on the left. And on the right, if
you follow the movement of those electrons, they're
going to form sulfur dioxides. So SO2. And also the chloride anions,
so the Cl minus, like that. And so we've done it. We've substituted our
chlorine atom for the OH and formed an alkyl halide. So this is just a better way
of forming an alkyl chloride from an alcohol. So if we look at
an example, we'll just take something
like ethanol here. And if we react ethanol with
thionyl chloride, SOCl2, and we had some pyridine as our base. We're going to replace the OH
with our chlorine like that. And so once again, if
we look at our alcohol, this is a primary alcohol. And so primary
alcohols work the best because there's decreased
steric hindrance. And we don't have to worry
about stereochemistry, since we don't have
any chirality centers in our product. Let's look at a way
to form alkyl bromide. So we just formed
an alkyl chloride. Let's look at the
general reaction for forming an
alkyl bromide here. So I go ahead and
have my alcohol. And I react that with
phosphorus tribromide, PBr3. The OH group is going
to leave and I'm going to put a
bromine in its place. And once again, this mechanism
is an SN2 type mechanism. So primary or secondary
alcohols only. And possible inversion
of configuration for your products, depending on
whether chirality centers are present or not. So another SN2 mechanism. And again, we need to
use phosphorus tribromide because the OH group is
not the best leaving group. When we look at this
mechanism here, let's go ahead and show that lone pair of
electrons better, like that. We have phosphorus tribromide. So I'm going to go to
draw the dot structure. So we would have
these bromines here with lone pairs of electrons. And there's three of them. So I'll go ahead and
put in those bromines. And we still have two
more valence electrons to account for, and those would
go on our phosphorus like that. So the first step,
it's analogous to our previous mechanism. Lone pair of electrons
on oxygen are going to for a bonds with phosphorus. And that would kick
these electrons off onto one of the bromines,
so I just chose that one. It doesn't matter
which one you choose. And so when we show
the result of that, we would now have our oxygen
bonded to a phosphorus. The oxygen is still
bonded to a hydrogen. There's still a lone pair
of electrons left behind on this oxygen, which gives this
oxygen a plus 1 formal charge. And the phosphorus is now
bonded to only two bromines. So we can show the phosphorus
bonded to only two bromines here. So I can go and put those in. And I can put in the lone pair
of electrons on phosphorus as well. And so we lost one
of our bromines, and that formed a bromide anion. So we go ahead and draw
in our bromide anion here. And once again, we've made
a better leaving group. So all the stuff
here on the right is a better leaving
group than the OH. So we could think about our
SN2 type mechanism, where once again, this
carbon right here is going to be electrophilic. So it once it wants
negatively charged electrons, which it could
get from the bromide anion. So nucleophilic
attack, and then that would kick these electrons in
here off on to your oxygen. And we can go ahead
and show our products. So when we draw the product,
all we need to do now is show we have substituted. The bromine is now attached
to the carbon like that. And our other product would be a
hydrogen attached to an oxygen, and the oxygen would be
attached to a phosphorus, and the phosphorus is attached
to two bromines like that. So we've formed
our alkyl bromide. So if we just show
a quick example. Once again, we'll
start with ethanol. If we wanted to convert
ethanol into ethyl bromide, a all we would have to do
is add phosphorus tribromide like that. And of course we're going to
replace the OH with a bromine. And so that's one way to prepare
alkyl halides from alcohols.