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Preparation of Alkyl Halides from Alcohol Video Lecture - Class 12
FAQs on Preparation of Alkyl Halides from Alcohol Video Lecture - Class 12
| 1. What is the purpose of converting alcohols into alkyl halides? |  |
Ans. The conversion of alcohols into alkyl halides is done to introduce a halogen atom (such as chlorine, bromine, or iodine) into the alcohol molecule. This transformation is important in organic synthesis as alkyl halides serve as versatile intermediates for the preparation of a wide range of organic compounds.
| 2. How can alkyl halides be prepared from alcohols? | |
Ans. Alkyl halides can be prepared from alcohols through a substitution reaction known as nucleophilic substitution. This can be achieved by treating the alcohol with a hydrogen halide (HX) or a reactive halogenating agent such as thionyl chloride (SOCl2) or phosphorus tribromide (PBr3).
| 3. What factors influence the reactivity of alcohols in the preparation of alkyl halides? | |
Ans. The reactivity of alcohols in the preparation of alkyl halides is influenced by several factors, including the nature of the alcohol (primary, secondary, or tertiary), the strength of the hydrogen halide or halogenating agent, and the reaction conditions (temperature, solvent, etc.). Generally, primary alcohols react more readily than secondary alcohols, while tertiary alcohols may require harsher reaction conditions.
| 4. Can all alcohols be converted into alkyl halides? | |
Ans. Not all alcohols can be easily converted into alkyl halides. For example, alcohols with acidic hydrogen atoms (such as phenols) or alcohols that are hindered by bulky substituents may not undergo the substitution reaction efficiently. Additionally, some alcohols may require the use of specialized reagents or reaction conditions to achieve the desired conversion.
| 5. Are there any alternative methods to prepare alkyl halides from alcohols? | |
Ans. Yes, there are alternative methods to prepare alkyl halides from alcohols. One such method is the reaction of alcohols with phosphorus halides (such as phosphorus trichloride or phosphorus tribromide) to form alkyl halides. Another method involves the use of sulfuryl chloride (SO2Cl2) or oxalyl chloride (COCl)2 as reagents for the conversion of alcohols into alkyl chlorides. These alternative methods offer different reaction conditions and selectivities for specific applications.
Text Transcript from Video
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.
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.
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