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Introduction to Organic Chemistry

Treatment of ...

Question

Treatment of ammonia with excess of ethyl iodide will yield:

A

diethyl amine

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B

ethyl amine

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C

triethyl amine

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D

tetraethylammonium iodide

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Solution

The correct option is B tetraethylammonium iodide
The reaction takes place as follows:

$N H_{3} + C_{2} H_{5} I \to C_{8} H_{20} N^{+} I^{-}$

$C_{8} H_{20} N I$ is the chemical formula of tetraethylammonium iodide.

Hence, option D is correct.

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Similar questions

Q. Treatment of ammonia with excess ethyl chloride will yield ?

Q. Trearment of ammonia with excess ethyl choria will yield ?

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

Q. How is ethyl amine prepared from methyl iodide?

C_{5} H_{11} O_{2} N

, occurs in beet sugar molasses. It is water soluble solid that melts with decomposition at

300^{\circ} C

. It is unaffected by a base but reacts with hydrochloric acid to form a crystalline product

C_{5} H_{12} O_{2} N C l

. It can be made from glycine with methyl iodide or treatment of chloroacetic acid with trimethyl amine. Draw structure for betaine which will account for all properties given above.

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