Question

What is Z in the following reaction:
 $ C{{H}_{3}}COOH\xrightarrow[\Delta ]{N{{H}_{3}}}X\xrightarrow{NaOBr}Y\xrightarrow{NaN{{O}_{2}}+HCl}Z. $
(A) $ C{{H}_{3}}-C\left( =O \right)-H $
(B) $ C{{H}_{3}}-C{{H}_{2}}-OH $
(C) $ C{{H}_{3}}-O-C{{H}_{3}} $
(D) $ C{{H}_{3}}-C{{H}_{2}}-Cl $

Answer 1

Hint: We know that the diazonium salts are the organic compounds consisting of triple bonds in between Nitrogen atoms as well as have either an alkyl or aryl benzene ring on the other side. The diazonium salts are the intermediate phase between the azo dye’s compounds and are known to be popular coloring agents.

Complete answer:
To understand the process of diazonium ions formation, you must first understand the process involved in diazotization or the operation of diazonium salt synthesis. Once sodium nitrite $ \left( NaN{{O}_{2}} \right) $ gets treated with $ HCl $ it forms because of the positive charge displacement on the nitrogen atoms, the electron rich nucleophiles can only couple with that of the terminal nitrogen, instead of the inner nitrogen, yields azo dyes. The complete reaction is given by;
 $ C{{H}_{3}}COOH\xrightarrow[\Delta ]{N{{H}_{3}}}\underset{\left[ X \right]}{\mathop{C{{H}_{3}}-C\left( =O \right)-N{{H}_{2}}}}\, $
These are termed salts because of the double nitrogen (diazo), which is generally found in ionic salts where chloride molecules replace the nitrogen atom. The overall process involved in the creation of diazonium salts is quite easy.
 $ \underset{\left[ X \right]}{\mathop{C{{H}_{3}}-C\left( =O \right)-N{{H}_{2}}}}\,\xrightarrow[\begin{smallmatrix}
 Hoffmann \\
 Degradation
\end{smallmatrix}]{NaOBr}\underset{\left[ Y \right]}{\mathop{C{{H}_{3}}-N{{H}_{2}}}}\, $
When the nitro group present in an amine is reduced, the amine activates the aromatic ring to lead electrophilic substitution to ortho and para locations. This is where the aryl amine gets converted into diazonium ions into different groups and can be used for the next-in-line reactions.
 $ \underset{\left[ Y \right]}{\mathop{C{{H}_{3}}-N{{H}_{2}}}}\,\xrightarrow{NaN{{O}_{2}}+HCl}\underset{\left[ Z \right]}{\mathop{C{{H}_{3}}-O-C{{H}_{3}}}}\,. $
Therefore, the correct answer is option C.

Note:
Remember that the diazo coupling reactions are used across many fields, such as the formation of synthetic dying agents for colours like yellow, red, orange, etc. Here they are called azo dyes because of their colour-pertaining abilities and may exist in cis and trans formations.