What reason would you say for N,N-dimethylaniline to be optically inactive - due to nitrogen inversion or delocalisation of lone pair due to resonance with the phenyl ring?

As per the logic mentioned in this answer verified by ron!, it seems reasonable to say that nitrogen inversion has some impact on the optical inactivity.

@GuruVishnu forget about nitrogen inversion

What's the reason to forget that?

With such a bulky group we can't think about inversion

Compared to ordinary amines the aniline inversion has a lower activation energy as described in the answer linked above. It's because when the N moves from sp3 to sp2, it's energetically favourable for the bulky groups to move apart. So the peak of the free energy diagram is reduced. I might be wrong. But this is what the answer linked above states. After all, ron too said that this is fine.

@Yuvraj That's what I too thought in the beginning. The energy required to move the bulky group is higher, but the answer in the above question based on MSC seems to state otherwise.

If I hadn't seen ron's comment I'd simply conclude that inversion doesn't have any role.

To be frank it look funny to me that one answer this question using analogy of inversion of nr3

@GuruVishnu

Let us start The nitrogen in aniline is somewhere between sp3 and sp2 hybridized, probably closer to the sp2 side. We are correctly taught that the nitrogen in simple aliphatic amines is pyramidal (sp3 hybridized)

We can assess the nitrogen hybridization by measuring its barrier for pyramidal inversion. If a trigonal nitrogen is sp2 hybridized, the barrier will be zero. On the other hand, in aliphatic amines where the nitrogen is sp3 hybridized the inversion barrier is typically around 4−5 kcal/mol

In aniline this barrier is very low, somewhere around 1−2 kcal/mol. This indicates that the nitrogen in aniline is not quite planar,

Which I feel is not sufficient to invert it?

@GuruVishnu

@Yuvraj You have interpreted things in a slightly top-down fashion I feel

First of all, if the barrier of inversion is going down, it actually means that inversion is going to be easier than the case of aliphatic amines, and so the inversion should be kinetically faster

@YusufHasan ah!

In aniline itself, if I assume your values of inversion barrier are correct, then that inversion will be more prominent than the aliphatic case, since the system already has a certain degree of planarity, so it will actually find it easier to approach the planar transition state

Have you heard about the Hammond Postulate?

No

Hammond's postulate is actually a postulate in organic chemistry which states that the geometry of the transition state resembles either the reactants or the products, and of the two, it is closer to the one which has lesser difference in energy from the transition state

Let me show a few examples

Suppose you have some endothermic reaction

You already know that in an endothermic reaction, products have have higher energy than reactants, right?

Yes

So, out of the two diagrams I have shown above, what would you expect the energy of the transition state to be closer to: products or the reactants? In other words, can you rationalize why the first diagram is marked as incorrect, and the 2nd as correct?

Energy stability of intermediate

Peak in curve of intermediate Is quite flate

Which make it more favorable for the rectant

Yeah I mean, the intermediate of an endothermic reaction is a high energy species, and so it will have lesser energy difference with the products than the reactants, as the products have higher energy than the reactants..So, the hump should be closer to the products than reactants in this case

My answer is correct?

@YusufHasan

Yeah kindof...Let’s just look at a concrete example to solidify things: the conversion of an alkene to a carbocation by HCl. What Hammond’s postulate says is that the transition state will more closely resemble the product higher in energy. In this case, that’s the carbocation. So the structure of the transition state more closely resembles the carbocation than the alkene.

I guess things should be clearer now

Yes

Yes, so now think about what you said earlier: in aniline, you see a greater degree of planarity due to resonance than an aliphatic amine

Yes

And for inversion to happen in both cases, the transition state is trigonal planar

Yes

So, if you apply Hammond's postulate, in which of the two cases (1) aliphatic amine (2) aniline, do you think the transition state will resemble the reactant more?

feel free to look at the diagrams above and think

Aniline

Yes, so there you have it...This will tell you that inversion barrier should actually go down, and so it should become easier to invert in aniline.

One doubt?

Yeah?

If the things are planner would it be easier to turn them?

if things are planar then it will be easier to reach a planar transition state, and after that it will simply go to the other side

Think of it this way: The top of a reaction coordinate diagram, that is, the peak of the hill(Transition state) is an unstable equilibrium, right?

(You must have read about stable, unstable and neutral equilibrium)

Yes

Here in the second one, peak is high mean very unstable!

yes that is correct, that is why in endothermic reactions you supply energy as heat to make them happen

@YusufHasan thanks I understood now

so now, if u think about it, for a general reaction, the only tough task is to reach the top of the hill, because the top being an unstable equilibrium, it will immediately roll down to the product side right?

Yes

and in case of aniline, like u said...the task of reaching the top of the hill is easier energetically than aliphatic amines, so ultimately, product formation in the former will also be easier as i have explained above

hope it's clear

Hello folks.

hello\

@YusufHasan College opened? Semester closed?

@GuruVishnu Also I feel that the answer verified by ron is essentially correct, because if you look at Matthew Mahindaratne's answer below, it talks about calculating such barrier on a computational level. He mentions one of the important factors for such a calculations as the "Y−X−Y angle (α) of XY3" for some pyramidal molecule XY3. " Using any molecular modeling software, one can show that when bulkiness of Y increase the α increases". So the logic provided in the first answer matches with the

computational factor listed by Matthew, as steric bulk will definitely increased the Y-X-Y angle

@WilliamR.Ebenezer College not opened, semester is being wrapped up online, will be done by the 1st week of July

What about you?

@YusufHasan: Thanks! I'm reading your previous messages. The problem was, I thought N,N-dimethylaniline was optically inactive due to inversion whereas in MSC problem book, he stated that this is due to resonance and the N atom is sp2 hybridised.

@GuruVishnu Well again, if you say fundamentally, no qualitative logic is "correct" at it's core, but it simply has to explain fit the experimental data logically, which this answer does

@GuruVishnu MSC is using a semi-important factor, but I believe that inversion has the more major role to play

@YusufHasan Ok. Then I think it's fine. In that problem, we were asked to find the optically inactive out of a given set and not the reason for optically inactive.

@YusufHasan Thanks for the clarification! It's also in sync with my logic. I'm satisfied.

Yeah but still, reason is important to know...In my opinion, the major role is of inversion

Initially I thought of asking it on the main site but I felt that this cannot be framed in a way it doesn't fall into the HW category. Also it consumes a lot of time.

If needed, this was based on Qn. 13, page 75 (in 11th edition).

And its solution on the solution booklet.

This reminds me of Quindar tones:

It's interesting to note that 1,4-diiodobenze has a non-zero dipole moment as stated in this answer. Do anyone have any ideas on what might be the reason for this?

Also can the direction of dipole moment vector be ascertained if it's due to the polarizability of the large iodine atoms?

@GuruVishnu Here is my ad hoc explanation for the case. I think that due to steric hindrance with nearby hydrogen atoms the iodine might get out of the plane. Hence this may result a net dipole moment.

I cannot think of any other scenario.

But there is a flaw in this explanation because it suggests the existance of cis-trans stereoisomer in the molecule.

@YusufHasan Our semester was closed recently without endsems. Classes concluded long ago.

@WilliamR.Ebenezer Yeah, we also aren't having classes, but the profs are bent on giving assignments..You're lucky they wrapped up the sem, we are still in the endgame :(

@YusufHasan I know your agony. We were also going to have a chem class test and were given just 8 days to cram the entire Inorganic and Organic material. But the students mass-mailed our profs and they had to cancel it.

@WilliamR.Ebenezer Our profs are mad...mass mailing doesn't affect them much, but atleasst they are grading thru assignments, and not a full fledged endsem exam

@YusufHasan Oho... nice.

@GuruVishnu See first up, these small values of dipole moments themselves suggest the involvement of weak, pretty-close range molecular forces. So it is not exactly polarizability at play here, but in the case of diiodobenzene, the increased exposed surface area of the molecule due to two big I's and a benzene ring will lead to increase in Van der walls forces, or instantaneous dipoles b/w adjacent molecules. That's why again, due to the fluctuating nature of such forces, the value of dipole

moment is pretty close to zero, if we look at things in bulk and then average things out for a unit molecue\le

@WilliamR.Ebenezer So when is your next sem supposed to begin?

Any news on returning to campus?

@YusufHasan The last mail from the registrar says that they will start bring people back in Sept. But no (official) info about when the sem will start