Question

Identify the following reaction, Acetaldoxime $\xrightarrow[{Alcohol}]{{Na}}A\xrightarrow[{}]{{NaN{O_2}}}B + {H_2}O + {N_2} \uparrow $
A) $C{H_3}C{H_2}C{H_2}OH$
B) ${C_2}{H_5}OH$
C) ${C_2}{H_5}Cl$
D) ${C_2}{H_5}N{H_2}$

Answer 1

Hint:‌ We will write what is given in the question and then we will know about the reactions happening and how the aldoximes and ketoximes reacts with the sodium, than we will know that those product will react with $NaN{O_2}$. Then we will choose the correct option.

Complete answer:
Step-1: We are given with acetaldoxime that will react with the sodium and alcohol. The product will react with $NaN{O_2}$ and HCl to form the compound we need.
Step-2: Aldoximes and the ketoximes react with the sodium and the alcohol to form the primary amine. they undergo the reduction. The acetaldoxime will react with Na and alcohol to produce the ethyl amine and the water.
The reaction is given below:
${C_2}{H_4}NOH + 4[H]\xrightarrow[{Alcohol}]{{Na}}C{H_3} - C{H_2} - N{H_2} + {H_2}O$
The Alcohol gives the hydride ion which reacts with the hydrogen ion and forms water; the extra hydrogen reduces the nitrogen oxygen bond and forms the simple ethyl amine.
Step-3: Now the ethyl amine needs to react with $NaN{O_2}$ and HCl . The HCl will create a diazonium chloride which is highly unstable. It will quickly react with water to make methyl alcohol as a major product. Water and the nitrogen gas will be the bi product of the reaction.
The reaction is given below:
\[C{H_3}C{H_2}N{H_2}\xrightarrow[{HCl}]{{NaN{O_2}}}{C_2}{H_5} - {N^ + } \equiv NC{l^ - }\xrightarrow{{{H_2}O}}{C_2}{H_5}OH + {N_2} + {H_2}O\]

Therefore, the correct option is option A.

Note: Ethanol is one of the most important compounds with the alcohol functional group. It is also used as a beverage. It is used in a lot of medicines and to make sanitizers. Its chemical composition is ${C_2}{H_5}OH$.