Mechanism for reaction of C6H5NH3 with NaOH?

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SUMMARY

The reaction of C6H5NH3 (anilinium ion) with NaOH results in the formation of C6H5NH2 (aniline) through a deprotonation mechanism. The positively charged nitrogen in C6H5NH3 loses a proton, facilitated by NaOH, which acts as a base. This process generates water (H2O) as a leaving group. It is essential to include the counterion for the amine salt in the mechanism for clarity.

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nimbuscloud
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Hello again!
I have a question regarding the reaction of C6H5NH3 with NaOH to make C6H5NH2. How would a mechanism be written to describe this process? I know that C6H5NH3 goes from having a positive charge on the nitrogen and after it reacts with NaOH, C6H5NH2 has a neutral charge but not sure how to write a mechanism to describe it. Would the NaOH just deprotonate C6H5NH3 and form H2O as a leaving group? Any help would be appreciated!
Thanks!
 
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Yeah, but you should include the counterion for the amine salt.
 
Thanks!
 

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