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Question: How is ethyl amine prepared from methyl iodide?...

How is ethyl amine prepared from methyl iodide?

Explanation

Solution

The conversion of methyl iodide into ethyl amine begins with a double displacement reaction followed by the reduction of the formed product Reaction of Methyl iodide with potassium cyanide (KCN) take place to give ethane nitrile or methyl cyanide and small amount of potassium iodide. Then the reductions of Ethane nitrile or methyl cyanide with sodium and alcohol will take place to form ethyl amine.

Complete answer:
First of all let us understand about both the given compounds. Let us start with ethylamine. Ethyl amine is an organic compound with chemical formula C2H5−NH2{C_2}{H_5} - N{H_2}. It is a colourless gas having strong order such as ammonia. It is a base. Ethyl amine is produced majorly by the reaction of ethyl alcohol with ammonia i.e.

CH3−CH2−OH+NH3→CH3−CH2−NH2+H2OC{H_3} - C{H_2} - OH + N{H_3} \to C{H_3} - C{H_{ 2}} - N{H_2} + {H_2}O
 (Ethanol) (Ethyl amine)\text { (Ethanol) (Ethyl amine)}

Ethyl amine is used in herbicides and is very commonly used in rubber products. It is also used in medicines such as tranquilizer.
Now let us discuss whether methyl iodide or iodomethane is an organic compound with the chemical formula of CH3IC{H_3}I . It is dense and volatile liquid. Iodomethane is used in herbicides, insecticide, nematicide etc. Iodomethane has high acute toxicity for inflation and ingestion.
Now we have to count methyl iodide into ethyl amine. Ethyl amine contains a carbon atom more than the methyl iodide. So we have to increase the carbon chain in methyl iodide This can be done by treating methyl iodide either potassium nitrite which leads to the formation of ethane-nitrile.

CH3−I+KCN→CH3−C≡N+KIC{H_3} - I + KCN \to C{H_3} - C \equiv N + KI
 (Methyl iodide) (Ethane-nitrile)\text { (Methyl iodide) (Ethane-nitrile)}
Now the formula ethane-nitrile undergoes mediums reduction with NaNa metal in alcoholic medium to give the required ethyl aminei.e.
CH3−C≡N+4H→Mendues Reduction Na/C2H5O4CH3−CH2−NH2C{H_3} - C \equiv N + 4H\xrightarrow[{Mendues{\text{ Reduction }}}]{{Na/{C_2}{H_5}{O_4}}}C{H_3} - C{H_2} - N{H_2}
(Ethyl amine)
Hence, in this we can convert methyl iodide into ethyl amine.

So, the correct answer is Option C.

Note: Conversion of methyl iodide can also take place by treating methyl iodide to sodium metal to give wurtz reaction. The product then formed is treated with chlorine in the presence of sunlight which leads to the formation of ethyl chloride. Ethyl chloride then reacts with ammonia to give ethyl amine. The reaction is:

2CH3−2Na→wartsdry ether CH3−CH3→−HCLCL2/hvC2H5CL→−HCL NH3C2H5NH22C{H_3} - 2Na\xrightarrow[{warts}]{{dry{\text{ ether }}}}C{H_3} - C{H_3}\xrightarrow[{ - HCL}]{{C{L_2}/hv}}{C_2}{H_5}CL\xrightarrow[{ - HCL}]{{{\text{ N}}{{\text{H}}_3}}}{C_2}{H_5}N{H_2}