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Organic chemistry
Course: Organic chemistry > Unit 6
Lesson 4: Alkene reactionsAlkene halogenation
Halogenation is a reaction that occurs when one or more halogens are added to a substance. Halogens comprise the seventh column in the periodic table and include fluorine, chlorine, bromine, iodine, and astatine. The resulting product of a halogenation reaction is known as a halogenated compound. Created by Jay.
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NO WHERE on the internet can seem to answer this question. Actually I've found two scholarly sources that contradict each other. At3:31, you see the cyclic halonium ion is formed. However lets address the stereochemistry of the next halogen attack. Lets say that the second carbon has a methyl group attached to it instead of a hydrogen. This makes carbon 2 a secondary carbon. Now Markovnikovs rule states that "the halogen of a hydrogen halide(H-X) will attach itself to the carbon bearing the greater number of carbons. However this is an X-X molecule not an H-X. My question is, will the halogen follow Markovnikov's rule and attack carbon 2? Or does halogenation not follow Markovnikov's rule?(12 votes)
The most general statement of Markovnikov's rule is, "when an unsymmetrical reagent adds to an unsymmetrical alkene, the more positive part of the attacking reagent agent goes to the carbon that has the greater number of hydrogen atoms". Halogen is a symmetrical reagent, so Markovnikov's rule does not apply.(30 votes)
- At1:40, why must those electrons also attack? Why can't we just be left with a halogen ion and 1-halogen-ethane?(10 votes)
- We could. If the molecule is ethane, the attack of the π electrons would give a halide ion and a 1° 2-haloethyl carbocation. But the 1° cation is less stable than the cyclic halonium ion, so the reaction does not use that pathway.(8 votes)
I checked out other halogens in my chemistry book and perfluorocarbons came up. Oxygen easily dissolves in them apparently and they use it to help premature babies breathe--in essence, it's a liquid people can breathe in, so why aren't they using this in swimming pools yet?(3 votes)- Before birth, a fetus' lungs are filled with fluid. While inside the mother, a fetus does not use the lungs to breathe — all oxygen comes from the blood vessels of the placenta. Perfluorocarbons can dissolve large amounts of gases like oxygen and carbon dioxide compared to water. They can dissolve enough oxygen and carbon dioxide to sustain oxygenation and ventilation for a human. If a premature baby has trouble breathing, it is used to having its lungs filled with fluid, so perfluorocarbons perfused with oxygen are a convenient way to get oxygen into the baby's system.
If one were to fill a pool with a suitable perfluorocarbon it would, in theory, be possible to swim and breathe in the liquid, but there would be a few problems that would arise.
1. You would have to have a means of oxygenating the liquid. This could be as easy as a waterfall, fountain, or just bubbling air through the liquid. You would also be removing carbon dioxide by doing this, so you would kill two birds with one stone.
2. Perfluorocarbons are quite dense, usually between 1.5 - 2 times denser than water. Therefore, you would need a lot of additional weight to sink into the liquid.
3. Breathing liquid requires more energy than breathing air, because liquids are more viscous than air, so it would be exhausting to do this for very long.
4. Lastly, you would need a lot of money. At approximately $1000/L, a
40 000 L pool would cost $40 million to fill. And it would evaporate, so you would need to refill it often.
Can you afford such a pool?(7 votes)
- Are enantiomers racemic mixtures as well?(3 votes)
- No... A racemic mixture is one in which you end up with a 50/50 mix of two right/left handed enantiomers. Enantiomers are mirror images of one another, but are not always formed as racemic mixtures. This is why it's important to understand ways in which to form products in desired quantities, as you may want to isolate a specific enantiomer. Using a solvent that forms a racemic mixture creates quite a challenge to your yields.(3 votes)
- At3:44, why are the halogens added to the alkene by anti addition? Jay explains it (using the reaction mechanism) as the second halogen simply attacking from the bottom, BUT there's already a carbon/group/hydrogen bonded there - so does the carbon/group/hydrogen have to shift to make way for the halogen? Or is this simply a matter of the halogens wanting to be as far away from each other as possible due to electron repulsion (avoiding gauche interaction)? In that case, what if the group already bonded there is larger and provides more repulsion than the halogen?(2 votes)
- The way my chemistry book explains this is by pointing out that the trans product is the one that's actually observed when this reaction occurs. Therefore, we have to have a model that produces the actual result that we see in the lab. Syn addition does not explain why we end up with trans products in the lab.
If you're asking about the mechanism itself rather than it's origin, you have to consider that in the bromonium (if you're using Br, it's bromonium. He calls it a cyclic halonium ion, which applies to all of the halogens in the first example.) creation, you see that partial carbocation behavior and pi bond attack ties your"top" halogen to both carbons, and won't allow the "bottom" halogen access from the top side. It has to attack from underneath.(3 votes)
- If there is two double bonds in a molecule going through halogenation or even hydrohalogenation will the two double bonds both be reduced to single bonds? Also if an alkyne is going through hydrohalogenaton or halogenation does it only get reduced to a double bond unless stated that there is excess reagant?(2 votes)
- You always get a mixture of products. It's a matter of probabilities because it depends on which molecules happen to collide.
If you are adding 1 mol of HX to a diene, you will some molecules that have added 1 HX, some that have added 2 HX, and some that have added no HX.
But the majority of the product molecules will have added 1 HX.
If you are adding 2 mol HX, by far the major product will be the one that has added 2HX.
The same thing happens with alkynes.
Using 1 mol of HX gives a mixture of products, with the major product having added 1 HX.
Using 2 mol of HX, the major product will have added 2HX, with much smaller amounts of unreacted alkyne and monoadduct.(3 votes)
What solvent is used for Halogenation reactions?(2 votes)
why do we need CCl₄ as a solvent ...what's the job of CCl₄ in this mechanism?(1 vote)- CCl₄ is simply an inert solvent. Three reasons for using it are:
1. Ease of handling. Bromine, in particular, is much easier to handle in solution.
2. Controlling the rate of reaction. Decreasing the concentrations decreases the rate. The pure reactants may react so fast that the reaction gets out of control.
3. Ease of removal. Its low boiling point makes it easy to remove from the higher-boiling alkyl halides.(4 votes)
At7:56, why aren't they Diastereomers?(1 vote)- They aren't diastereomers because they are enantiomers.
Diastereomers are optical isomers that aren't enantiomers.(2 votes)
would the same reaction take place if there was Cl2 instead of Br2?(1 vote)
Yes. The speed of the reaction is likely to change but all halogens follow the same mechanism paths.(2 votes)
3:31
Video transcript
Let's take a look at the
halogenation of alkenes. So on the left,
I have my alkene. And I'm going to
add a halogen to it, so something like
bromine or chlorine. And I can see that
those 2 halogen atoms are going to add
anti to each other-- so they'll add on opposite sides
of where the double bond used to be. Let's take a look
at the mechanism so we can figure out why
we get an anti addition. So I start with my
alkene down here. And I'm going to show
the halogen approaching that alkene. So it's going to
approach this way. And I put in my lone
pairs of electrons. Like that. If I think about that
halogen molecule, I know that it's nonpolar. Because if I think
about the electrons in the bond between
my two halogens here, both halogen atoms,
of course, have the exact same
electronegativity. So neither one is
pulling more strongly. And so overall, the
molecule is nonpolar. However, if the pi
electrons in my alkene-- so I'm going to say that these
electrons right here are my pi electrons-- if those
electrons get too close to the electrons in
blue, they would, of course, repel them-- since
electrons repel each other, since they're like charge. And if the electrons in magenta
repel the electrons in blue, the electrons in
blue would be forced closer to the top
halogen, like that, giving the top halogen a
partial negative charge and leaving the bottom halogen
with a partial positive charge because it's losing a little
bit of electron density. Now, I could think about
that bottom halogen as acting like an electrophile
because it wants electrons. And so in this mechanism,
the pi electrons are going to function
as a nucleophile. And the pi electrons are going
to attack my electrophile, like that. At the same time,
these electrons over here-- this electron
pair on the left side of the halogen-- is going
to attack this carbon. And the electrons
in blue are also going to kick off onto the
top halogen, like that. So let's go ahead
and draw the result of all those electrons
moving around. So now, I have
carbon singly-bonded to another carbon, like that. The electrons in magenta formed
a new bond between the carbon on the right and my
halogen, like that. And these electrons over here,
I'm going to mark in red. So this lone pair of
electrons on my halogen are going to form a bond
with the carbon on the left, like that. And that halogen still has two
lone pairs of electrons on it. So I'm going to put in those
two lone pairs of electrons. And that halogen has a
plus 1 formal charge. This is called a
cyclic halonium ion, and it's been proven to
occur in this mechanism. If I think about that
positively charged halogen, halogens are very
electronegative, and they want electrons. So the electrons in
magenta, let's say, are going to be pulled a little
bit closer to that halogen, which would leave this
carbon down here losing a little bit of
electron density, giving it a partial
positive charge. And so that's going to
function as our electrophile in the next step. And our nucleophile is going
to be the halide anion created in the previous step. Right? So we had a halogen that had
3 lone pairs of electrons around it. It picked up the
electrons in blue. Right? So now, it has 4 lone
pairs of electrons-- 8 total electrons-- giving it
a negative 1 formal charge, meaning it can now
function as a nucleophile. So if I think about this
cyclic halonium ion here, the halogen on top is going
to prevent the nucleophile from attacking from the top. It's going to have to
attack from below here. So this negatively
charged halide anion is going to nucleophilic
attack this electrophile here-- this carbon. And that's going to kick
these electrons in magenta off onto this halogen here. So let's go ahead
and draw the results of that nucleophilic attack. All right. So now, I'm going to have
my 2 carbons still bonded to each other like that. And the top halogen has
swung over here to the carbon on the left. It used to have 2 lone
pairs of electrons. It picked up the
electrons in magenta. So that's what the carbon
on the left will look like. The carbon on the right is
still bonded to 2 other things. And the halide anion
had to add from below. So now we're going to have
this halogen down here. Like that. And so now we
understand why it's an anti addition of
my 2 halogen atoms. Let's go ahead
and do a reaction. So we're going to start with
cyclohexane as our reactant here. And we're going to
react cyclohexane with bromine-- so Br2. Now, if I think about the
first step of the mechanism, I know I'm going to form
a cyclic halonium ion. So I'm going to draw that ring. And I'm going to show the
formation of my cyclic halonium ion. It's called a bromonium ion. All right. So I'm going to form
a ring like this. And the bromine's going to
have 2 lone pairs of electrons. It's going to have a plus
1 formal charge, like that. And also, I know
for my mechanism I'm going to form a bromide
anion at the same time. So I'm going to have
a negatively charged bromide anion, like that. When I think about where
that bromide anion is going to attack, I know
that it's going to attack one of
these 2 carbons here. So it could attack
the one on the left. It could attack the
one on the right. Let's go ahead and start
with the carbon on the right and draw the product. So if a lone pair of
electrons-- the bromide anion-- attack this carbon
right here, that would kick these electrons
off onto the bromine. And we could go ahead and
draw the results of that. So I would have my ring. And the bromine on top
is going to swing over to the carbon on the left here. So now this bromine is
going to go look like that. And the bromide anion has
added from below the plane of the ring, like that. So that's one possible product. The bromide anion could also
attack the bromonium ion from the left side. So this lone pair of electrons
could attack this carbon, which would kick these
electrons off onto the bromine. And so we could
go ahead and draw the result of that
nucleophilic attack. So in this case, the top
bromine would swing over to the carbon on the right. And it would pick up an
extra lone pair of electrons. And the bromide
anion, again, added from below the plane
of the ring, like that. Now, these 2 molecules are
actually different molecules. Let's go ahead and
redraw them so it's a little bit easier to see. And we're going to stare down
this way at the molecule. And this is the
top of your head. So let's go ahead and
redraw that molecule. So this is the same molecule
as if we're looking down on it. This carbon right here
would be this carbon. And I could see there's
a bromine coming out at me in space. So I'm going to put
a bromine coming out at me in space at that carbon. I move to this carbon over here. That's this carbon. So there must be a bromine
going away from me in space at that carbon. So that's one possible product. On the right, we
do the same thing. All right. So we're going to put
our eye right here. We're going to look down. This is the top. So I can look at
this carbon first. So I can see there's a
bromine down at that carbon. So I go ahead and draw
my cyclohexane ring. And at this carbon,
there's now a bromine down. And of course, at
this carbon over here, there's a bromine
coming out at me. So I represent
that with a wedge. And these are my products. All right. And if you've already
had stereochemistry, you know that these 2 products
are enantiomers to each other. They're actually
different molecules that are non-superimposable
mirror images. So we can see that the
absolute configurations have been reversed. So if I think about
this carbon right here, bromine coming out at
me, bromine going away from me. And this one down here,
bromine going away from me. And for this one, a
bromine coming out at me. So that's the
halogenation reaction.