Complete: $C_6H_5N_2^+Cl^-$ reacts with:
Complete: $C_6H_5N_2^+Cl^-$ reacts with:
(a) $CuCN$ (b) $HBF_4$ (c) $H_3PO_2 + H_2O$
1 Answer
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· 2mo ago
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(a) Sandmeyer reaction with CuCN:
$$C_6H_5N_2^+Cl^- + CuCN \rightarrow C_6H_5CN (benzonitrile) + N_2$$
(b) Balz–Schiemann reaction:
$$C_6H_5N_2^+Cl^- + HBF_4 \rightarrow C_6H_5N_2^+BF_4^- \xrightarrow{\Delta} C_6H_5F (fluorobenzene) + N_2 + BF_3$$
(c) Deamination (replacement by H):
$$C_6H_5N_2^+Cl^- + H_3PO_2 + H_2O \rightarrow C_6H_6 (benzene) + N_2 + HCl + H_3PO_3$$
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Discussion (4)
NA
Underrated solution. The way you set it up makes it almost obvious.
SJ
Thanks a ton, I was stuck on this exact problem for an hour.
AS
Why do we take the positive value only in the last step?
E
Brilliant explanation, the substitution step is what I kept missing.