Mechanism For Reaction Of C6H5NH3 With NaOH? - Physics Forums
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- Thread starter Thread starter nimbuscloud
- Start date Start date Mar 31, 2009
- Tags Tags Mechanism Reaction
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.
PREREQUISITES- Understanding of acid-base reactions and deprotonation.
- Familiarity with organic chemistry mechanisms.
- Knowledge of the properties of amines and amine salts.
- Basic grasp of reaction intermediates and leaving groups.
- Study the mechanism of deprotonation in organic reactions.
- Learn about the properties and reactions of amines, specifically C6H5NH2.
- Explore the role of counterions in acid-base reactions.
- Investigate the use of NaOH in various organic synthesis reactions.
Chemistry students, organic chemists, and anyone interested in reaction mechanisms involving amines and bases.
nimbuscloud Messages 4 Reaction score 0 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! Physics news on Phys.org- Swimming in a shared medium makes particles synchronize without touching
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