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Organic chemistry
Course: Organic chemistry > Unit 9
Lesson 4: Electrophilic aromatic substitutionElectrophilic aromatic substitution mechanism
The general reaction and mechanism of electrophilic aromatic substitution. Created by Jay.
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how do electrons move from one position to another during resonance, what is the driving force during resonance(6 votes)
Actually, in the molecule, electrons do not move during resonance. What we see is a probability cloud of electron density that is always higher at some atoms than at other atoms. We try to explain this by drawing resonance structures that put the extra electron density on the appropriate atoms.
We draw the structures and move the electrons. The electron movements usually start with a lone pair on an atom moving between that atom and the atom to which it is σ bonded, or with a pair of π electrons moving between the atoms that are σ bonded to the atoms on either side of the π bond.
The driving force is that this spreading out or delocalization leads to a lower energy.(24 votes)
why we call complex as sigma complex(3 votes)
Called sigma complex as the two pi bonds are spread out across the aromatic ring making the ring seem to be 1 and 1/2 bond closely resembling one bond, also known as a sigma bond hence sigma complex(3 votes)
- Why benzene undergo electrophilic substitution type of reaction but alkene undergo electrophilic addition reaction although both are unsaturated compound?(2 votes)
- because benzene is a stable aromatc compund and the intermediate formed in case of alkene is unstable for benzene.(4 votes)
- Can you please explain how we know which compound is more electrophilic than the other? For example, ranking propanal, 1-propanol and 1-chloropropane in increasing electrophilicity?(2 votes)
Electrophilicity is a result of deficiency of electrons at a particular atom in the compound. For example, BF3 is an electrophile, that is, it's octet is not complete, and hence is in search of electron donors to complete the octet. F- acts as such a donor and forms BF4-. AlCl3, Br+, NO+ are a few examples, where, owing to either their structures or to the very high electronegativity (desire for electrons) of the central atom, there is a deficiency of electrons because of which these act as electrophiles. Having said that, you can assess the electrophilicity of the compounds in your question, considering the carbon attached to the functional group as the electrophilic site. Since a double bonded carbon is very electronegative, the carbon to which it is bonded is severely deficient in electrons, as compared to -OH group which is more electronegative than -CL group. Hence, Propanal>propanol>cvhloropropane. However, if you consider the carbon adjacent to the carbon doubly bonded to oxygen in the aldeheyde to be the electrophilic site, the order might change.(4 votes)
- Does this reaction categories as a Sn1 or Sn2 reaction?(1 vote)
- It is neither. It is in its own category: SEAr for Substitution, Electrophilic, Aromatic.(4 votes)
- After drawing all of the resonance intermediates, how do you know which one to choose to proceed with?(1 vote)
- hi
tyresome has given a great answer.
Just wanted to add an extra point to go with it.
There is no one fixed resonance structure that could depict the entire molecule correctly on its own. The actual molecule exists as a mixture of its resonance structures.
Hope it helps.(2 votes)
- why in sigma complex carbon contains +charge
+charge appear on it at that time when it loses its one electron.Does it lose its electron?
I am confused over here can you please explain it for me ?(1 vote)- The two electrons in the π bond are used to attack the electrophile.
The carbon that is bonded to the electrophile doesn't lose any electrons, but the carbon ortho to it does lose an electron, and that gives it a positive charge.(2 votes)
why secondary carbocations are more stable than primary carbocations?(1 vote)
Carbocations have a stability pattern of: 3° >2° >1°> methyl. Where more alkyl groups bonded to a carbocation make it more stable. This is due to two effects, through the inductive effect and hyperconjugation of the unhybridized p orbital. The positively charged carbon at the center of a carbocation is electron-deficient without a completed octet and is stabilized by additional electron density.
The inductive effect is electron density donation through sigma bonds from the alkyl groups to the positively charged carbon.
Hyperconjugation is electron density donation from the alkyl group’s neighboring sp3 hybrid orbitals into the unhybridized p orbital of the positively charged carbon.
Hope that helps.(2 votes)
What is Ipso substitution? We have only been taught the desulphonation reaction, where the -SO3H group gets substituted by -H. My question is- Does this reaction happen on its own (ipso facto)? I searched online and haven't gotten a clear answer... Usually, they show the reaction in presence of water, which makes sense: a proton donor should be there... So, I'm confused. Not about the reaction but about the conditions under which it occurs.(1 vote)- Hello Mihir. The ipso position refers to the position (carbon) on the aromatic ring which bears the substituent taking part in the reaction of interest. For benzene rings, the position names are ipso, ortho, meta and para.(2 votes)
I don't understand why do we study mechanisms in organic chemistry. Like we don't have mechanisms in inorganic chemistry (as far as i have studied).(1 vote)
There are mechanisms in inorganic chemistry, but they often aren't as involved as these.
Organic chemistry leads on to synthesis (eg making molecules) at higher levels. It's helpful to us and others replicating works to know the outcomes of reactions and how the reaction proceeds.(2 votes)
Video transcript
Let's look at the
general reaction for electrophilic
aromatic substitution. So we start with
the benzene ring, and we react benzene
with a molecule that contains an
electrophile in there. And what happens in
electrophilic aromatic substitution. We're going to substitute
the electrophile for a proton on
our benzene ring. And so over here, we can
see the electrophile is now in place of that proton. So that's where
that that's where the electrophilic
part comes in this. And that's where the
substitution parts comes in. You're substituting an
electrophile for a proton. The aromatic comes
in because you are going to reform an aromatic
ring in your mechanism. Electrophilic
aromatic substitution requires a catalyst. And the point of a catalyst is
to generate your electrophile. So down here, you can
see that the catalyst is going to react to
produce the positively charged electrophile. So remember, electrophile
means loving electron. So if something is
positively charged, it's going to love electrons. We also formed this
catalyst complex over here, which is going to
factor into our mechanism. So now that we formed
our electrophile, let's look in more
detail as to what happens in electrophilic,
aromatic substitution. So we start with
our benzene ring. And I'm showing one of the
hydrogens on the benzene ring. It could be any of the six,
since they are all equivalent. And now we formed our
electrophile from our catalyst. So the pi electrons
in the benzene ring can be attracted to
the positively charged electrophile. Because negative charges are
attracted to positive charges. And so pi electrons
in your benzene ring are going to function
as a nucleophile, and those electrons are going
to attack the electrophile. So this is a nucleophile,
electrophile attack, where those pi
electrons are going to bond to that
electrophile there. So those pi electrons are
going to form a covalent bond with your electrophile. So let's go ahead and show that. So these pi electrons
didn't do anything. The hydrogen stays there. Now, I could show
the electrophile adding to either
of the two carbons on the side of the double bonds. So it could be that carbon. Or it could be this carbon. Since I've drawn this
hydrogen up here at the top, I'm going to go ahead and say
that the electrophile adds to the top carbon there. So there's my
electrophile there. Let me go ahead and
highlight the electrons that are forming
that covalent bond. So these pi electrons
here are the ones that are functioning
as a nucleophile. And those pi electrons are going
to form this bond right here. Now in forming that bond,
we're taking a bond away from this bottom carbon here. And so that bottom
carbon is going to be left with a positive
one formal charge. Therefore, we can draw
a resonance structure for this cation. So let's go ahead and show a
possible resonance structure here. So these pi electrons
could move over to here. And let's go ahead and draw what
would result if that happened. So now, we have these
pi electrons up here. We have our hydrogen. We have our electrophile. And the electrons moved
over to this position. Let me go ahead and
highlight those in magenta. So I'm saying that these pi
electrons right here moved over to here. And when those electrons
moved over to there, we're taking a bond away
from this carbon this time. So that is the
carbon that's going to get a plus 1 formal
charge like that. So we can draw another
resonance structure. So let's go ahead and do that. So we could take
these pi electrons and move them into here. So let's go ahead and show
what that would look like. So if those pi electrons moved
into there, we would now have, again, our hydrogen, our
electrophile, these pi electrons, and then these
pi electrons right here. So once again, let me go
ahead and highlight those. This time I'll use blue. These pi electrons are
going to move over to here. And once again, we're taking
a bond away from a carbon. This time, it's this
top carbon up here. So that's the
carbon that's going to get the plus one
formal charge like that. So these are all
resonance structures. And remember, the
actual cation would be a hybrid of these
resonance structures. And we call we call that
hybrid a sigma complex. So you have a positive
one formal charge de-localized over three
carbons in your sigma complex. So the next step in
the mechanism-- I'm just going to redraw
the first resonance structure that we did here. So I'm going to go and
redraw that down here. So let's go ahead and show
the first resonance structure. So in our first
resonance structure, we had our hydrogen
here, our electrophile already bonded to our ring. And we had a positive one
formal charge on this carbon right here. Well remember, the catalyst
had formed a complex. And I represent it like this. So something bonded to
your catalyst like that. So let's just go up here
and refresh our memory. So right up here, when we
generated our electrophile, we also generated this
catalyst complex up here. So y bonded to a catalyst, so
I have y bonded to a catalyst down here. And you could think about
this as functioning as a base. Or it's going to
accept a proton. So I could show these electrons
in here taking this proton. And if it takes
that proton, that leaves these electrons behind. And those electrons are
going to move in here to reform your benzene
ring and take away that positive one formal charge. So let's go ahead and show that. So we now have our
benzene ring back. And our electrophile is
now bonded to our ring. And the proton has left. So the electrophile
has completely substituted for that proton. Let's follow those
electrons again. So the electrons
in magenta in here, so those are the
ones that are going to move in here to reform
your aromatic ring. So deproteination
of the sigma complex restores the aromatic ring. And so we have a
stable product here. So the other product you
could think about this y here is now going to be
bonded to that proton. So you could have the y
here bonded to that proton. And you could highlight
those electrons. You could say that these
electrons right here are now these electrons. And, taking those electrons
away from the catalyst would of course
regenerate your catalyst. And so it's free to then
catalyze another reaction. And so this is the
general mechanism for electrophilic aromatic
substitution, which the reactions that
we're going to see are pretty much going to
follow this general mechanism.