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My question is due to a discrepancy I found in reactivity and organic conversions. In reactivity of organic compounds it has been established that an alkyl group activates the benzene ring, and, hence for example, methylbenzene is more reactive than benzene.

Then consider the following situation. Suppose that there are two benzene molecules in a sample, and we convert one of them to toluene (methylbenzene) using Friedel–Crafts alkylation. Then we would see the following change in the mixture:

Benzene alkylation

Now, consider we carry out alkylation once more. As toluene is more reactive than benzene, it will react first forming dimethylbenzene. By continuing this logic we would get hexamethylbenzene as the major product regardless of the amounts we add.

So, how is that we carry out conversions with Friedel–Crafts alkylation regarding toluene as the major product?

Feel free to correct me if my logic is incorrect and if so please specify what should be corrected.

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    $\begingroup$ Indeed the reaction would be hard to control. A "dirty" way for obtaining a somewhat good yield of toluene would be that of using a large excess of benzene (both as a reactant and as a solvent), and a smaller amount of alkyl halide: this way, chances are that the reactants "meet" a benzene molecule, instead of an "active" toluene molecule $\endgroup$
    – user32223
    Commented Sep 14, 2018 at 3:51
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    $\begingroup$ I agree. @The-Vinz you should make this an answer $\endgroup$
    – Waylander
    Commented Sep 14, 2018 at 9:05
  • $\begingroup$ @The_Vinz the solution is indeed using an excess of benzene. It is not a dirty way but the usual way of dealing with this kind of problem. $\endgroup$ Commented Sep 14, 2018 at 14:13
  • $\begingroup$ Can deactivating benzene by sulphonation work here? $\endgroup$
    – Carrick
    Commented Dec 14, 2018 at 4:27
  • $\begingroup$ @The_Vinz Can you please explain how one will get hexamethylbenzene. Isn't methyl group ortho para directing? Surely friedel craft wont add methyl group to meta positions? Isn't it? $\endgroup$ Commented Jan 8, 2020 at 4:08

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The solution to this problem of having chance of poly-alkylation of Benzene rings by Friedel-Crafts alkylation has a general way, and the way is to first perform Friedel-Crafts Acylation and then reduction of the introduced Carbonyl carbon to saturated carbon atom.

Now, proceeding via this way helps to avoid all the problems that arise in a Friedel-Crafts alkylation. Firstly, after the first step, in the main product, we are having one Benzene ring attached to a carbonyl carbon , which deactivates the ring for participating in further electrophillic aromatic substitution and reduces any chances of poly-acylation which upon reduction also doesn't give any poly-alkylated product. Secondly, Friedel-Crafts acylation helps to reduce the possibility of rearrangement which occurs very often in the alkylation step. That's because the positive charge is not on any primary,secondary or tertiary carbon atom but on the acyl carbon which will not undergo any changes in its carbon skeleton structure. So, we can add the unrearranged carbon chain to the Benzene ring and reduce to get less-substituted carbon chains attached to the ring.

To solve your problem, I would rather prefer that you first react Benzene with a mixture of $\ce{CO}$ and $\ce{HCl}$ in presence of anhydrous $\ce{AlCl3}$ (Gatterman-Koch Aldehyde Synthesis) to get Benzaldehyde. Then you can perform a Clemmensen reduction ($\ce{Zn -Hg / conc. HCl}$) or a Wolff-Kishner reduction ($\ce{H2N-NH2/KOH}$) to get Toluene as the exclusive product.

Please note that, you can always control the reaction conditions to get Toluene by Friedel-Crafts alkylation only, still there remains a good amount of chance of getting di-substituted product as those kinds of controls need to be optimised and difficult to carry out and you have to perform purification techniques anyways. However, if you proceed via acylation and reduction, your side products will be in lesser amount and product can be easily obtained.

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    $\begingroup$ It is much better to use an excess of benzene as @The_Vinz has said and to perform only one separation through distillation. Note also that benzaldehyde is made from toluene and not the other way round $\endgroup$ Commented Sep 14, 2018 at 14:19
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    $\begingroup$ We can definitely do that. But this is a kind of standard and nicer way to do it so I preferred to say this. $\endgroup$
    – Soumik Das
    Commented Sep 14, 2018 at 14:24
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    $\begingroup$ I would not call $\ce{CO}$, $\ce{Hg}$ and $\ce{NH2-NH2}$ nice; quite the opposite $\endgroup$ Commented Sep 14, 2018 at 14:27
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    $\begingroup$ Meh, it's not like anyone's gonna make toluene from benzene, practical details aren't that important for getting good grade on test, and that's very test-like not practical example. $\endgroup$
    – Mithoron
    Commented Sep 14, 2018 at 14:45
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    $\begingroup$ true - no-one does this in the real world $\endgroup$
    – Waylander
    Commented Sep 14, 2018 at 19:20

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