Solveeit Logo

Question

Question: When excess of ethyl iodide is treated with ammonia, the product is....

When excess of ethyl iodide is treated with ammonia, the product is.

A

Ethylamine

B

Diethylamine

C

Triethylamine

D

Tetraethylammonium iodide.

Answer

Tetraethylammonium iodide.

Explanation

Solution

: NH3+C2H5I(Excess)[(C2H5)4N+]IQuaternarysaltNH_{3} + \underset{(Excess)}{C_{2}H_{5}I}\overset{\quad\quad}{\rightarrow}\underset{Quaternarysalt}{\lbrack(C_{2}H_{5})_{4}N^{+}\rbrack I^{-}}