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Organic chemistry
Course: Organic chemistry > Unit 9
Lesson 5: Directing effectsOrtho-para directors III
Moderate and weak activation, weak deactivation. Created by Jay.
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- At, Jay says that the halogen can give its pair of electrons to form a pi bond but if that halogen is so electronegative (as Jay says previously) then why does it give them anyway? Is it because electrons are highly active and doesn't stay where they are, so they are forming the pi bond even for a slightest moment? 4:33(5 votes)
- Cl isn't giving away its electrons. It is partially sharing its electrons through the π system. This delocalization of electrons lowers the total energy of the system despite the unhappiness of the Cl at losing some of its control over these electrons.(11 votes)
- Fromto 7:30, Jay says that Chlorine and Nitrogen have approximately the same electronegativity. Is there an authoritative (or even reliable) electronegativity chart (or period table showing electronegativity) that is worth memorizing? When I googled for it, different charts have different electronegativity values for the same element. 8:00
Also, how is the electronegativity value calculated in the first place?(1 vote)- I don’t think any electronegativity chart is worth memorizing, but you should memorize trends in electronegativity values.
There are different charts because different people created their own scales.
The one most commonly used is the Pauling electronegativity scale.
He developed it using differences in bond energies and then scaling the numbers so H had a value of 2.2 and F a value of 4.0.
There have been some minor adjustments since then.(5 votes)
- At, Jay says induction must be more important than resonance. Is induction in general a more important factor than resonance when determining the affect of a substituent on electron density in a benzene ring? Or does he mean that in that induction is stronger than resonance in this specific example? 5:35(1 vote)
- It's difficult to always predict whether induction or resonance will be the more important factor. In this specific case induction is the more important effect, but in other cases resonance may be more important. Often, we look at experimental results and induce which factor must have been more important, induction or resonance, based on the results (so it's an after-the-fact explanation, rather than a prediction).(4 votes)
- but resonance is more important factor than inductive effect. So here, how is it that inductive effect is given more priority?(2 votes)
- This is my question, too. I'm sending this video to my lecturer and asking them to explain, because I was told that resonance has a greater effect than induction as well.(1 vote)
- , R-groups( in this case -CH3) are electron-donating, isn't the electron density being donated to the ring through the bond between the ring and the methyl group? 2:30(2 votes)
- .
When the benzene ring attacked some of electrophiles [ ether ortho or para position ] the carbon where methyl group is attached will be in it a positive charge [ due to resonance ].
When we arrived to this situation , both methyl group and benzene will share with each other one electron [ from any hydrogen that attached to methyl group ], or to be more specific, they will share the positive charge.
This process make alkyl groups as activator groups.
Good Luck.(1 vote)
- may we not consider the halogen as a weak activator because it donates an electron to our benzene ring then why we are calling it de-activator??(1 vote)
- Being particularly electronegative, halogens pull electron density away from the carbon they are attached to, which overall lowers the reactivity of the benzene ring. However, due to the minor contributing (as the halogens prefer not to give up electrons and the benzene ring prefers not to break its aromaticity) resonance structure drawn in the video, there is a semblance of additional electron density available at adjacent carbons due to the presence of the halogen, leading to ortho-para direction. Remember that resonance structures are like snapshots of potential states for a molecule, and not separate states that molecules exist in: they actually exist as a tapestry of all of the resonance structures, with some states more preferable (stable) being represented more of the time.(2 votes)
- there was a question that said:
consider the three isotopes of dimethylbenzene
if it is chlorinated with one chlorine atom
it now forms only one isotope
what isotope is it
(is the answer 3-chloromethylbenzene because that is the only isotope for activating groups, but methyl is a deactivating group, and if you consider that you could have para or ortho chloromethylbenzene)
im so confused(1 vote)- Isotopes refer to atoms of the same element that have different masses due to differing numbers of neutrons, so I'm assuming you actually meant isomers.
I think your question is asking something like this:
1) You started with one of the three possible isomers of dimethylbenzene.
2) You chlorinated this compound in such as way as to get only a single chlorine added.
3) You then determined that this new compound had only a single isomer.
4) What is that isomer.
To answer this question start by drawing out the three possible isomers of dimethylbenzene.
Then draw the result of adding a single chlorine to each possible place on each of those isomers. For one of them you will then see that all products are actually the same structure. This is the one you want!
Leave a comment if you want more help :-)(1 vote)
- By application of the concept presented in this video, I reasoned that inductive effect prevails in +:S(CH3CH2)2 when connected to benzene ring. While the lone pair on S can participate in resonance with the ring, S comes from a different period than C does and so, the overlap between 2pC and 3pS is poor (or at least weaker than the overlap between 2pC and 2pO, 2pC and 2pN, and 2pC and 2p F. Thus, resonance effect in +:S(Et)2 isn't that effective. A reason why the inductive effect might prevail is the difference in electronegativity between C and S, but this difference is negligible (2.55 vs. 2.58). So, which effect would prevail then? What do you think?(1 vote)
- in last line he said halo re op directors ?/////(1 vote)
- He mentioned that: -
(i) Due to the large inductive effect (due to the electronegativity of halogens), the electrons will be absorbed by the halogens (electron- withdrawing group)
(ii) However, since halogens have lone pair electrons which can be contributed to the ring via the overlapped- orbital, it can contribute more to the resonance effect.
(iii) Due to the strong inductive effect, when compared to the resonance effect, halogen benzene tends to be weak deactivator, they are, however, ortho para directing group due to the contributing resonance hybrids.
Consider the following points:-
(i) Going down the group (G 17), the electronegativity decreases.
(ii) But, due to the different size of the orbitals, (the orbitals of halogen get bigger going down the group)
(iii) Therefore, degree of overlapping will decrease as well.
(iv) This, in return, prevent the resonance effect > inductive effect
(v) So, halogens remain as weak deactivator group, although their electronegativity decreases going down the group.(1 vote)
- I can't Understand why NHCOR is an activating group although the oxygen withdraws the electrons leaving a positive charge on the N atom which should make it an electron withdrawer in return(1 vote)
Video transcript
In the previous video, we saw
how induction and resonance affect ortho-para directors
that are strong activators. This video is really
just a continuation of the previous video
so make sure you've seen that one first. In that video, we
saw that when you have an atom with a lone
pair of electrons that's right next to the
carbon on your benzene, ring that lone pair of electrons
can participate in resonance and it can be conjugated into
the pi system of your ring. That increases the
electron density of your ring, which
activates the ring towards electrophilic
aromatic substitution since the ring can be
a better nucleophile, and the positively-charged
sigma complex is better stabilized by the
increased electron density. And so that's where
the activated portion comes in for this molecule so
that lone pair of electrons can activate the ring towards
electrophilic aromatic substitution. However, it's only a moderate
activator and not a strong one like we saw on the
previous video. And the explanation for that
observation comes from the fact that this lone pair of
electrons can participate in resonance
outside of the ring. So if I go ahead and draw
a resonance structure here. I have my benzene ring. I now have my nitrogen
double bonded to this carbon. That carbon is
bonded to an R group. The nitrogen is
bonded to a hydrogen, and now my oxygen
has three lone pairs of electrons around it giving
it a negative 1 formal charge. The nitrogen now is a
plus 1 formal charge. So the electrons in
magenta moved in here to form a pi bond, and because
those electrons in magenta are tied up in resonance
outside of the ring, they can't donate as much
electron density to the ring, and so that's why it's only
a moderate activator instead of a strong one. Let's look at an example
of a weak activator so something like toluene here,
so a methyl group on a benzene ring. All of the carbons
on our benzene ring are SP2 hybridized so
all of those carbons have free P orbitals. Let me go ahead and
sketch a free P orbital on all of my carbons. So I have my six pi electron
system here in my benzene ring, and so those electrons are
delocalized in my pi system like that. I also have a
methyl group, right? So let's go ahead and sketch
in the methyl group here. So I have a carbon bonded
to three hydrogens, and it turns out one of
those carbon hydrogen sigma bonds can
actually interact with the pi system of our ring. So we could think
about this sigma bond right here interacting with
the pi system, and therefore, increasing or donating some
electron density to that ring. And since that sigma bond can
interact weakly and donate some electron
density to that ring, we can think about
the methyl group as being an activator
activating the ring towards electrophilic
aromatic substitution. So you could call this
interaction sigma conjugation, you could call it
hyperconjugation, whatever you want to call it, but there
is a little bit of electron donation to the pi
system, therefore, making it an activator. However, this sigma
conjugation is nowhere near as strong as the
full conjugation that we saw in
the previous video with the oxygen and nitrogen. And so since sigma conjugation
is not as strong of an effect, that's why this is
only a weak activator. Let's look at one more example
of an ortho-para director, and this one is the exception. This is an ortho-para director
that is a deactivator here. So a halogen on a
benzene ring will still direct substituents
ortho and para to it because of the resonance
structures that you can draw. However, it turns out to be
a deactivator most the time. The reactions are slower than
that of benzene on its own. Let's first examine
the inductive effect that the halogen has
on our benzene ring. So let me go ahead and
write induction here. If we think about the sigma
bond between the carbon on the ring and the halogen--
so this is our generic halogen here with an X-- halogens a
are relatively electronegative, more electronegative
than carbon so they're going to withdraw some electron
density from that ring. Whenever you withdraw electron
density from a benzene ring, you deactivate it towards
electrophilic aromatic substitution. So the inductive
effect would lead you to think that a
halogen is a deactivator. However, the halogen does have
lone pairs of electrons on it so you could also think
about the resonance effect. So let me just go ahead and
write the resonance effect here. And if I took one of these
lone pairs of electrons and moved them in
here to form a pi bond between that carbon
and that halogen, right? That would push these
electrons off onto this carbon so we could go ahead and draw
our resonance structure here. So we would have our halogen
now double bonded to our carbon. The halogen has two lone
pairs of electrons, a plus 1 formal charge, and we now
have a lone pair of electrons out on this carbon with a
negative 1 formal charge like that. So let me just go
ahead and highlight one of these lone
pairs of electrons can participate in
resonance, and that would increase the electron
density of your ring. And again, I'm not going to draw
the other resonance structures, but increasing the electron
density of your ring implies activation towards
electrophilic aromatic substitution. So we have these two
competing factors here. Induction says that
the halogen is electron withdrawing and
therefore a deactivator. Resonance suggests
that the halogen might be electron donating,
and therefore an activator. Experimental results show that
these halogens are actually deactivators here. So induction must be a
more important effect than resonance. Let's see if we can think about
why using different halogens. So let's start
with fluorine here. So if your halogen is fluorine. Fluorine is, of course,
extremely electronegative, and so you could
think about what the inductive effect if
the halogen is fluorine, it's withdrawing a lot
of electron density from that benzene
ring, deactivating the ring towards electrophilic
aromatic substitution and so the inductive
effect is very strong when fluorine is on
your benzene ring. When you think about
resonance, so the fluorine forming a pi bond with
a carbon on your ring. So let me show this
carbon right here forming a pi bond
with a halogen. I'm saying the halogen
is fluorine this time so I can go ahead and show
carbon bonded to fluorine. So we have P orbitals here,
which are overlapping, and so we can draw in
the P orbitals on carbon. Carbon and fluorine
are in the same period on the periodic table. Therefore, you can think
about their P orbitals being pretty much the
same size, and that allows for good overlap. And so the fluorine can
donate some electrons and increase the electron
density of your ring. And so because there is some
good overlap of your orbitals there, fluorobenzene
turns out to be the most reactive of your halogens
because of that overlap, but because fluorine
is so electronegative you could think about the
inductive effect winning most the time, and still
giving you a deactivator. However, there are some
exceptions for fluorobenzene, but, in general, we consider
the halogens to be deactivates. Let's look at
another halogen here. So for that example,
induction wins because of the large
electronegativity difference between fluorine and carbon. However, if the
electronegativity difference decreases, such as when
your halogen is something like chlorine, then you have
to come up with a little bit different explanation
because an atom like nitrogen has pretty much the same
electronegativity as chlorine, and so the inductive effect
would be the same for nitrogen as it is for chlorine,
approximately. And so there must be a
slightly different explanation as to why chlorine
is a deactivator. And if you think about
the resonance effect where your chlorine forming
a pi bond with your carbon. So let's go ahead and show
chlorine bonded to carbon here. Carbon being in the second
period of the periodic table has a P orbital of a certain
size, chlorine of course is in a different period,
it's in a third period, so it has a larger P orbital. And the size mismatch
of those P orbitals means that you don't get as
good of an overlap as you would with orbitals that
are the same size, and so therefore,
the chlorine can't donate as much electron
density via resonance. So resonance decreases
in importance because of the poor
overlap of those orbitals. And so that's a way of
thinking about the fact that the resonance
effect is decreased, and so therefore, you have
more of an inductive effect and the chlorine
is a deactivator. I was talking about comparing
chlorine to nitrogen, and the two, of course,
have very similar electronegativities, but
the nitrogen-- aniline in the previous video was
observed to react much, much faster. And again, we can think about
the size of those P orbitals. Nitrogen being in the
same period as carbon, and so you get better
overlap of those orbitals, the nitrogen is better able
to donate electron density to that ring, making the
nitrogen an activator here. So you could think about,
again, large differences in electronegativity
favoring induction. You could think about
size mismatch of orbitals, but overall, you need
to think about halogens as being weak
deactivators, but still ortho-para directors because
of the resonance structures that you can draw.