Write the reactions for the final alkylation product of aniline with an excess of methyl iodide in the presence of sodium carbonate solution.
(N/A) Aniline reacts with methyl iodide $(CH_3I)$ to form $N,N$-dimethylaniline.
When $N,N$-dimethylaniline reacts with an excess of methyl iodide,it forms $N,N,N$-trimethylanilinium iodide $([C_6H_5N(CH_3)_3]^+I^-)$.
In the presence of sodium carbonate $(Na_2CO_3)$ solution,the iodide ion is replaced by the carbonate ion to form $N,N,N$-trimethylanilinium carbonate,represented as $[C_6H_5N(CH_3)_3]_2CO_3$.
When $N,N$-dimethylaniline reacts with an excess of methyl iodide,it forms $N,N,N$-trimethylanilinium iodide $([C_6H_5N(CH_3)_3]^+I^-)$.
In the presence of sodium carbonate $(Na_2CO_3)$ solution,the iodide ion is replaced by the carbonate ion to form $N,N,N$-trimethylanilinium carbonate,represented as $[C_6H_5N(CH_3)_3]_2CO_3$.
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