Question

Methylamine reacts with \[HN{{O}_{2}}\]to form:
A. \[C{{H}_{3}}-O-N=O\]
B. \[C{{H}_{3}}-O-C{{H}_{3}}\]
C. \[C{{H}_{3}}OH\]
D. \[C{{H}_{3}}CHO\]

Answer 1

Hint: In this question, the intermediate step is the formation of aliphatic diazonium salt plays an important role in the completion of the reaction.

Complete Step by Step Solution:
In step 1, methylamine reacts with nitrous acid to form aliphatic diazonium salts. Diazonium salts are one of the most versatile combinations of organic and inorganic components. Its general way of representation is \[R-N_{2}^{+}{{X}^{-}}\].The\[R\] is an organic group, generally an aryl group while \[X\]represents an ion.

Generally, diazonium salts have\[C{{l}^{-}}\],\[B{{r}^{-}}\], \[BF_{4}^{-}\]as\[X\]. The name of these salts is based on the presence of the \[N_{2}^{+}\] group or the diazonium group. The naming of these salts is done by adding the suffix diazonium to the parent hydrocarbon from which they are derived and then it is followed by the anion \[X\] such as bromide.


 \[C{{H}_{3}}-N{{H}_{2}}\xrightarrow{NaN{{O}_{2}}+HCl}C{{H}_{3}}-N_{2}^{+}C{{l}^{-}}\]
Since diazonium salts are unstable and are water-soluble. They further react with water.
     \[C{{H}_{3}}-N_{2}^{+}C{{l}^{-}}\xrightarrow{-{{H}_{2}}O}C{{H}_{3}}-OH+{{N}_{2}}\uparrow +HCl\]
When unstable diazonium salt reacts with water it forms a stable compound that is methanol.
Hence, the correct option is option C that is\[C{{H}_{3}}OH\].

Note: The formation of the intermediate reactant that is diazonium salt should be correct and the reaction formed from the intermediate step should be accurate to get the desired product from the given reactant.