Rate towards SN Ar reaction

In my organic chem notes,the following order for rate/ease of SN Ar reaction is given


[![flourobenzene> chlorobenzene>bromobenzene>iodobenzene in terms of rate of reaction (SN Ar)](https://i.stack.imgur.com/hx1U3.jpg)](https://i.stack.imgur.com/hx1U3.jpg)


The reason given is The -I effect which stabilizes the carbanion so produced in the reaction.
Turns out,my organic chem sir gave us 2 mechanisms for SN Ar
they are attached below
[![it shows mechanism one](https://i.stack.imgur.com/0CVJq.jpg)](https://i.stack.imgur.com/0CVJq.jpg)
[![it shows mechanism 2(involves formation of benzyne)](https://i.stack.imgur.com/0hIpa.jpg)](https://i.stack.imgur.com/0hIpa.jpg)


Since he introduced mechanism 1 first it seemed pretty intuitive
But if I consider mechanism 2,then the rates of reaction must be inverse(order of leaving ability)
My doubt primarily lies in the fact that 1)If the rate order is true,then why do we have two different mechanisms for two similar(more or less) reactions?
2)If one of the mechanism is incorrect,then which one is correct?
Sorry for my poor vocabulary and please explain the above(what seems to me to be a) fallacy.




Both the mechanisms you've mentioned exist, but they both have some kinetic difficulties. Your first mechanism (called $\ce{ArSN2}$) involves breaking the aromatic character of the ring, while your second mechanism (a form of $\ce{E1cB}$ leading to a benzyne intermediate) involves removal of a minimally acidic hydrogen.



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> [![enter image description here](https://i.stack.imgur.com/uz2TD.png)](https://i.stack.imgur.com/uz2TD.png)
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So a good, strong base favours the second mechanism, while a deactivating group ortho or para to the leaving group (such as a nitro substituent) that would stabilise the carbanion intermediate, favours the first mechanism.


References: [*Nucleophilic aromatic substitution, wikipedia*](https://en.wikipedia.org/wiki/Nucleophilic_aromatic_substitution)


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