Sciencemadness Discussion Board

Basic Organic Chm Doubts

Claisen - 9-3-2011 at 12:21

I have two doubts:

Why does methylation of toluene in the presence of anh. AlCl3 give m-xyelene as opposed to p- or o- product?

When 2 Bromo 1 Methyl cyclohexane is treated with CH3ONa in methanol solvent, does the reaction proceed with SN1 or SN2 and why?

Claisen - 9-3-2011 at 13:37

Got one more-

When Ph-CH=CH-CO-Ph is treated with 1) PhMgX/H3O+ 2) Ph-Li/H3O+
Why are the products obtained different (when carbanions produced from reagents are same)?

[Edited on 9-3-2011 by Claisen]

Bot0nist - 9-3-2011 at 13:41

Try the short answer thread my friend. Best way to get a quick answer for something like this, IMO.

Chordate - 9-3-2011 at 14:04

Unless I am taking crazy pills or you are leaving something out, I think that friedel crafts alkylation of toluene will give a mixture of all three products, with para predominating, then ortho, and a very small amount of the meta product.

As far as your second question, the answer is yes, but I imagine that isn't satisfying.

You have a a secondary halide, which can allow both mechanisms, but it is more hindered to backside attack than a straight secondary alkane, and just how much it is hindered also depends on the cis/trans conformation of the ring. You have a strong base, which can favor E2/SN2. You have a polar protic solvent, which favors the SN1/E1 Mechanisms. And to top it off you have a neighboring tertiary center so carbocation rearrangement is possible if your substitution reaction is slow.

And that says nothing of possible hydride shifts. If this is a homework problem, then whoever came up with it is a jerk.

The order of reactivity here should be something like E1>SN1>E2>SN2.

Claisen - 9-3-2011 at 16:33

@Botonist: Can you provide me the link for that thread?

@Chordate: Its given that the meta product is formed in majority (first question).

I think it will favour SN1 because as you said backside attack for SN2 is hindered and there is a polar solvent too for the stabilisation of the intermediates.

[Edited on 10-3-2011 by Claisen]

DJF90 - 9-3-2011 at 19:38

The freidel crafts alkylation of toluene should give predominantly the o,p-xylenes, with small formation of m-xylene. However its likely you'll end up with a mess in practice, as the xylenes will be more reactive than the toluene, and further alkylation will occur. I believe the thermodynamic product in with sufficient alkylating agent will be 1,3,5-trimethylbenzene.

As for the second question, the reaction product will almost definately depend upon the relationship between the stereogenic centers. A cis relationship will likely favour an E2 mechanism, giving 1-methylcyclohex-1-ene as the product. (I hope you're up to scratch with your stereoelectronics!). The trans isomer is likely subject to E1 or Sn1 pathways, although the intermediate cation can undergo a 1,2-hydride shift to give a tertiary carbocation, which may further undergo proton removal to give the same product as before, or trapping of a solvent molecule to give 1-methoxy-1-methylcyclohexane. And I'm pretty sure I haven't covered all the options.

As for your final question, since you do not give the respective products, I don't really know what to explain, but I suspect it is an issue of 1,2- vs 1-4 addition to the enone. Typically lithium dialkylcuprates are used for conjugate/micheal/1,4- addition as they are "soft nucleophiles", and prefer to attack the extended pi-framework, although a more detailed explanation is that the reaction operates along a Cu(I)-Cu(III) mechanism. In this case however its a tough rationalisation to say if the grignard reagent would go 1,2- or 1,4-, but seeing as the two reagents give different products I guess its decided that the grignard takes the latter route. With organometallics, despite the fact you're able to discuss them as synthons for "R-", the free carbanion isnt actually found in solution, and the metal plays a vital role. The more electropositive the metal is, the harder the "R-" nucleophile is.

I hope this helps.

Claisen - 10-3-2011 at 04:59

Hi DJF90!
For the first question, the reason provided for the major meta product is that the reaction involves lesser activation energy (btw methylation is caused using CH3Cl)

"Typically lithium dialkylcuprates are used for ...."

Its just an organo lithium compound - no copper.

Yes you are right about the Grignard reagent going 1-4 addition.
Can this be a possible reason that carbanion of lithium is a very strong/good nucleophile so it is less selective and more reactive going 1-2 addition?

DJF90 - 10-3-2011 at 05:36

I was commenting on the typical use of reagent for conjugate addition - its fairly uncommon for a grignard to go 1,4-.

The reason the RLi reagent goes 1,2- was mentioned... Li is a very electropositive element and so the R-Li bond, whilst still covalent, is very much polarised, such that the alkyl group is very delta negative, and thus is a harder nucleophile - as such it prefers to react at the harder site of the pi-system, that being the carbonyl group (as the carbon here is the most delta plus available).

Bot0nist - 10-3-2011 at 06:28

<a href="http://www.sciencemadness.org/talk/viewthread.php?tid=14239&page=9">Short answers thread.</a>

smuv - 10-3-2011 at 07:19

DJF90 that is a good answer to both questions, but I have just one thing to add. It is kind of a chemistry meme that the FC alkylation is unselective in cases like this because the product is slightly more active for electrophilic aromatic substitution. It is true that if you mix up chloromethane, aluminum chloride and toluene you will predominantly get trialkyl products; this product distribution, however, is mostly due to the fact that the reaction is heterogeneous when conducted in toluene (or any other common FC solvent). What happens is, there is a sort of reaction mass adsorbed into the aluminum chloride that causes conditions where transport between this mass and the bulk solution is very slow, causing overalkylation beause the product doesn't leave the catalyst. If the reaction is conducted under homogonous conditions, in nitrobenzene or nitromethane with an excess of toluene you can observe good yields of xylenes with little mesitylenes.

I don't want to take away from what you posted, but I just wanted to break this meme.

DJF90 - 10-3-2011 at 10:13

I wasnt aware that the selectivity issue was due to inhomogeniety of the reaction medium. Thanks for bringing it to my attention smuv.

Claisen - 10-3-2011 at 11:28

For my 1st question, I found a link which supports the formation of meta xylene as the major product.

http://www.chemguide.co.uk/organicprops/arenes/fc.html

"Friedel-Crafts alkylation of methylbenzene (toluene)

Again, the reaction is just the same with methylbenzene except that you have to worry about where the alkyl group attaches to the ring relative to the methyl group.

Unfortunately this time there is a problem! Where the incoming alkyl group ends up depends to a large extent on the temperature of the reaction.

At 0°C, substituting methyl groups into methylbenzene, you get a mixture of the 2-.3- and 4- isomers in the proportion 54% / 17% / 29%. That's a higher proportion of the 3- isomer than you might expect.

At 25°C, the proportions change to 3% / 69% / 28%. In other words the proportion of the 3- isomer has increased even more. Raise the temperature some more and the trend continues.

The reason for this is again beyond UK A level."

But no reason is provided here except that it is temperature dependent. Can somebody throw light on it?

Chordate - 10-3-2011 at 11:58

Hmm. I have heard of AlCL<sub>3</sub> and FeCL<sub>3</sub> being used used to isomerize para to meta substituents directly, because the meta product is more thermodynamically stable. I would imagine that this is what is going on, but had no idea the effect could be so strong at room temperature. Cool.

DJF90 - 10-3-2011 at 13:33

Written in green below your quoted comment:
Quote:

The problem in this case lies in the fact that the methyl groups attaching to the ring can fall off again and reattach somewhere else in the presence of the aluminium chloride. You can get equilibria set up between the various isomers.

The reason for the 2,4- directing effect of the methyl group in methylbenzene lies in the fact that the 2- and 4- isomers form faster than the 3- isomer. However, in this case, the 3- isomer is the most thermodynamically stable of the three. If you raise the temperature, or allow more time, the equilibria set up favour the most stable product.

Claisen - 10-3-2011 at 15:47

I seriously need to start wearing glasses :o
I got it now, thanks!