Which of the following is a by product of the reaction between fluoroborate diazonium salt and sodium nitrite in the presence of copper?

NaBF₄

The correct answer is option 1. \(NaBF_4\)

Let us delve into the reaction why \(NaBF_4\) is the correct by-product.

The reaction between a fluoroborate diazonium salt and sodium nitrite in the presence of copper generally involves the replacement of the diazonium group. Here's a more detailed look at the process:

Reaction Mechanism:

Starting Materials:

Fluoroborate diazonium salt: \((ArN_2^+ BF_4^-)\)

Sodium nitrite: \((NaNO_2)\)

Copper: \(Cu\)

Reaction Process: The diazonium salt can undergo a Sandmeyer-like reaction or similar copper-catalyzed decomposition. The diazonium group \((N_2^+)\) is replaced by another group through the action of the copper catalyst.

Products: During the reaction, the diazonium group is replaced, and \(BF_4^-\) (tetrafluoroborate) ion is released. This \(BF_4^-\) ion can combine with \(Na^+\) from sodium nitrite to form sodium tetrafluoroborate (NaBF4).

Balanced Reaction:

\(ArN_2^+ BF_4^- + NaNO_2 + Cu \rightarrow Ar + N_2 + NaBF_4 \)

The replacement of the diazonium group leads to the release of nitrogen gas \((N_2)\).

The \(BF_4^-\) ion from the fluoroborate diazonium salt pairs with \(Na^+\) ions from sodium nitrite, resulting in the formation of \(NaBF_4\).

Therefore, the by-product of the reaction is indeed: \(NaBF_4\)