synthesis of meta chloro propylbenzene

[budullewraagh]

this week's exam the last question asked for a synthesis of meta chloro propylbenzene.

 

in notes, halides were considered to be weak as electron donating and electron withdrawing groups, but basically mostly neutral between the two.

 

my proposed synthesis was the chlorination of benzene with AlCl3 as a catalyst, then meta addition of propane by using propyl chloride and AlCl3.

 

the answer key says to add propyl chloride with AlCl3 and then to add chlorine, reduce the C=O with NaBH4 (i would have just done a wolff-kishner) then eliminate with H2SO4 and hydrogenate with H2 on a Pd catalyst. other routes would involve using TsCl on the -OH and eliminating with a strong base, then hydrogenating with H2 on a Pd catalyst.

 

i recognize how the "correct" answer would yield more chlorine in the meta position, but even if each of those 5 steps have a 75% yield, the overall yield is only 23.7%. i believe that competition between ortho/para and meta addition of propyl chloride would result in at least 25% meta product, especially considering the fact that Cl is the second most electronegative halogen.

 

does anyone agree?

[akcapr]

i agree

[Yggdrasil]

Chlorine is an ortho/para dirrecting group, so if you chlorinate first, then use f-c alkylation, you would get primarily o-chloro-propylbenzene and p-chloro-propylbenzene. Although you would get some m-chloro-propylbenzene, you'd probably lose a significant amount separating it from the ortho and para forms.

 

So, while the correct answer takes more steps and may have a lower overall yield (I would question whether it does, though), it allows you to easily purify your product.

[Bluenoise]

I dissagree. Your method would produce no desired product at all.

First of all you'd get about 1% meta yeild, (Organic chemistry 3rd edition L.G. WADE). Secondly you can't use propyl chloride to add a proply group; the carbocation would rearrange to the most substituted position and you would get the addition of an isoproply group instead. You must instead us an propanoyl chloride group, then follow it with a clemmensen reduction (Zn(Hg) + Weak HCl).

 

I believe this is the proper procedure.

First use propanoyl chloride with AlCl3 and H20

Secondly do your chorination with Cl2 and AlCl3

Third clemmensen reduction with Zn(Hg) and aq HCl

[budullewraagh]

according to dr garry procter's lecture notes, halogens dont really direct ortho/para or meta.

 

keep in mind that this is a test. if a professor teaches his class something false, it's not the fault of the students for not knowing it is false.

 

the proposed answer doesnt even work. elimination wouldnt necessarily work; OH isnt a good leaving group and the aryl halide would eliminate to form a benzyne.

plus you don't H2/Pd an aryl halide. doesnt work. the "answer" certainly doesn't seem feasible from any practical standpoint.

[Bluenoise]

I agree the proposed answer is wrong. Hydrogenation just doesn't make any sense at all, I can't even begin to understand why they'd recomend hydrogenation, you'd just endup with substituted hexane.

 

I think maybe he ment to say that halogens don't really activate/deactivate very much? They're considered ortho/para directing and weak deactivators. My book here has an example of a nitration on chlorobenzene and there is only a 1% meta yeild.

 

I think that my previously posted solution would work well.

*edit*

Yes it will work (though the halogenation will proceed far slower then with benzene); it will still produce the product and almost entirely the meta as well.

[Yggdrasil]

according to dr garry procter's lecture notes' date=' halogens dont really direct ortho/para or meta.

 

keep in mind that this is a test. if a professor teaches his class something false, it's not the fault of the students for not knowing it is false.

 

the proposed answer doesnt even work. elimination wouldnt necessarily work; OH isnt a good leaving group and the aryl halide would eliminate to form a benzyne.

plus you don't H2/Pd an aryl halide. doesnt work. the "answer" certainly doesn't seem feasible from any practical standpoint.[/quote']

 

 

Maybe you misunderstood. When talking about electrophilic aromatic substitution, most people say that electron-donating, activating and ortho/para dirrecting are synonymous; similarly, electron-withdrawing, deactivating and meta dirrecting are synonymous. The halides present an exception to this rule. They are ortho/para dirrecting yet deactivating; hence, they exhibit properties of both electron-withdrawing groups (EWGs) and electron-donating groups (EDGs).

 

The reason has to do with the two types of bonds in an aromatic ring. The lone pairs on the halide are able to donate electrons into the pi electron system of the aromatic ring. Since these electrons can enter into the pi electron system, these electrons can help stabilize the positive charge generated when an electrophile attatches to an ortho/para position on the ring through resonance. However, the halide acts as an electron withdrawing group through the inductive effect in the sigma bonds. This reduces the overall nucleophilicity of the aromatic ring, making electrophilic aromatic substitution harder to achieve. Although other electronegative atoms (such as oxygen) exhibit this effect, the electron donation through the pi electron system far outweighs the electron withdrawal through the sigma bonds to make the substituents activating overall. For the halides, the electron donation into the pi electron system is too weak to outweigh the electron withdrawal through the sigma bonds. (The electron donation into the pi electron sytem still does occur, and it still helps stabilize the pi electron system through resonance as resonance is not affected by sigma bonds).

 

Also, Bluenoise, I think your solution is the best posted so far. The basic conditions of a Wolf-Kishner could promote nucleophilic aromatic substitution reactions and create side products, while the Clemmensen reduction would likely not create any side produces.

[budullewraagh]

"I agree the proposed answer is wrong. Hydrogenation just doesn't make any sense at all, I can't even begin to understand why they'd recomend hydrogenation, you'd just endup with substituted hexane."

 

i mentioned the problem of hydrogenation to my professor, but he said such a reaction would not occur. could you substantiate your claim?

[Bluenoise]

Hmmm maybe you're right the benzene might be too deactivated. I'll look into it later. I was more confused why he'd want to hydrogenate it in the first place.

 

*edit*

Right you could hydrogenate the benzene to cyclohexane but it would require high pressure and temperate which wouldn't be required for hydrogenation of the alkene.

 

Still hydrogenation isn't necessary in the first place. Do a clemmensen reduction instead.

[budullewraagh]

well, the benzene with chlorine substituent would be reduced to benzene via H2/Pd

[Bluenoise]

That is weird, is this question from a text? Or from your proff? Have you asked him about that?

[budullewraagh]

it's from my prof. btw, we just learned wolff-kishner, thioacetal reduction and clemmensen reduction today.

 

and do you think the aryl halide reduction via H2/Pd wouldn't work? one of my contacts cited a book that agreed with that statement