Organic reaction and chirality

[bsrishu]

There is a compound 1-fluoro,1-chloro,1-bromo, propanone (a substituted ketone).
If we react it with an alkali, we know alkali will attack carbonyl carbon and chloro-fluoro-bromo- methyl carbanion will be the leaving group,
finally acetic acid and chloro-fluoro-bromo- methane will be formed. My question is whether the configuration of rectant's chiral carbon and product's chiral carbon will be same?
If yes, can the optical activity of the intial solution and final solution be different?
Please explain

 

[The-Exiled]

My question is whether the configuration of rectant's chiral carbon and product's chiral carbon will be same?

Do you know how to determine whether a compound is either R or S? Try it for both compounds and see what you get.

If yes, can the optical activity of the intial solution and final solution be different? Please explain

Not sure what you mean by "optical activity"? There is no way to tell whether either the R or S enantiomer will rotate plane polarised light to the right or left (D or L). This needs to be determined experimentally.

Try drawing out the reaction mechanism if your really stuck.

 

[bsrishu]

according to mechanism, carbanion will be the leaving group, as the carbanion will be tetrahedral, therefore i think the confg. Must not change.
But the exam in which i faced this ques, the answer was the other way.
According to them confg. Can change, i want to know how confg. Can change. Because a compound has a chance to change its confg. If it exist in some planar intermediate state.

 

[The-Exiled]

Yes it can change if the reaction is a SN2 type reaction.

I'm assuming the reaction was as follows:

CH₃-CH=O-C(BrClF) + Al₂(OH)₃ → C^-(BrClF) + HO-CH=O-CH₃

C^-(BrClF) + H₂O → CHBrClF + OH^-

(Maybe with a different base, I just Al₂(OH)₃ as an example)

The second reaction would indeed be a SN2 type reaction, which would have an intermediate planar compound (as you said) and results in an inversion from the R to S enantiomer (or vice versa)

This means that the enantiomer will be different to the original compound even though C^-(BrClF) would be have the same stereochemistry as the original compound (CH₃-CH=O-C(BrClF))

Let me know if that helped.

 

[bsrishu]

in SN2 how can the confg. Of attacking species (carbanion here) change.

 

[bsrishu]

about the optical activity, i want to confirm if the two solutions (with diff. Compounds) have same confg. (R or S) and same concentrations then will their ability (degree of rotation) to rotate the plane polarised will be same or not.

 

[sjb-2812]

What is the relative priority of the acyl group, compared to the halogens? What about a proton?

I don't see you performing an SN2 reaction per se when you protonate the carbanion in the second reaction?

As The-Exiled has mentioned, there is no real correlation between R/S and +/- (or D/L etc)

 

[bsrishu]

i am not saying that the reaction will be sn2, i was just replying to the-exiled's explanation that due to sn2 rxn confg. Can change (of the attacking species)

 

[sjb-2812]

Carbanions are configurationally stable (usually), cf the planar carbocations. However, if the stereocentre in the ketone is S, what would it be in the alkane?

 

[The-Exiled]

What is the relative priority of the acyl group, compared to the halogens? What about a proton?

Yes, I don't know what I was doing, haha. I have probably confused even you more. Consider what sjb-2812 has said here.

 

[epenguin]

I am unsure of this, for me it would be a tough question, but my instinct is to go for racemisation on the following basis:

A carbanion is isoelectronic to the corresponding amine. Amines invert very fast so a tertiary amine NR₁R₂R₃ is not normally resolvable into enantiomers, which however exist.
It is true that protonation of a carbanion is also very fast. But it is a bimolecular process, so I would think there is time for many inversions before the proton is added.

Except I wonder if the carbanion really has separate existence or is it almost a metaphor here, and the proton is added practically concertedly with the C-C bond breaking? In that case inversion could be expected. Or could it? - the proton could come from any direction?

Another question would be even if there is inversion (or retention) in this reaction, in the alkaline solution how much formation of carbanion BrClFC^- from the BrClFCH leading to racemisation of the product while the reaction is going on is there?

Unless the answers to these questions are known the student cannot feel very secure - well I don't. :shy:

 

[sjb-2812]

You raise an interesting point, but don't forget that phosphines (which are also isoelectronic) are usually configurationally stable. See e.g. the references at http://en.wikipedia.org/w/index.php?title=Special:Cite&page=Carbanion&id=472044878 which suggests that at low temperature they are stable, but that this is lost at elevated (relatively) temperature. I guess without a formal protocol for the reaction we may never really be sure which answer is intended; or indeed analysis of the mixture afterwards to see how much has gone "with retention" and how much not. This is not a black and white reaction - things do not just go one way and not the other.

I think "inversion" is perhaps the wrong word. Sure, the label of the centre changes, but this is more to do with the CIP priority of the groups, and it's not really an inversion like you would get with SN2 attack of iodide on 2-butylbromide, for instance.

 

[epenguin]

You raise an interesting point, but don't forget that phosphines (which are also isoelectronic) are usually configurationally stable. See e.g. the references at http://en.wikipedia.org/w/index.php?title=Special:Cite&page=Carbanion&id=472044878 which suggests that at low temperature they are stable, but that this is lost at elevated (relatively) temperature.

Yes, however the books in fact contrast amines and phosphines on this very point. I thought our case was closer to amines.

 

[MrSid]

The carbanion will behave differently depending on solvation of the anion (solvent effects); this is usually where the inversion can take over the course and effect racemization. a lot of handwaving can be invoked. Usually on tests it is best to be consistent with speculative handwaving, and use the experiment to explain the rationale with the simplest explanation that fits the facts.
Good luck!

 

[epenguin]

The carbanion will behave differently depending on solvation of the anion (solvent effects); this is usually where the inversion can take over the course and effect racemization. a lot of handwaving can be invoked. Usually on tests it is best to be consistent with speculative handwaving, and use the experiment to explain the rationale with the simplest explanation that fits the facts.
Good luck!

Thank you. That may be regarded as either confirming a horrible suspicion or else as a relief. In either case the student and I are licenced to consider ourselves not stupid, at worst ignorant of something rather specialised.

Here we unfortunately do not yet know the facts that could educate our handwaving. (Except the qualitative facts about inversion in amines and phoshines which I just accept, without delving into the QM calculations about this which if anyone has time for them will better rationalise.)

Some people will like, and some will not, that there is this educable handwavy chemical thinking that is not the same thing as computational physics.

I am sure you are right to stress solvent effects. Sovent effects are on the one hand a computational physics which has made much progress in recent years I hear, but on the other hand has a simple qualitative aspect, that given a mechanism you can qualitatively predict or rationalise effect of change of solvent and vice versa the effects give indications about mechanism.

If the student learns the answer or anyone else knows it will be interesting to hear.