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Question:

Aniline is reacted with bromine water and the resulting product is treated with an aqueous solution of sodium nitrite in the presence of dilute HCl. The compound so formed is converted into tetrafluoroborate which is subsequently heated dry. The final product is:

1, 3, 5-tribromobenzene

2, 4, 6-Tribromofluorobenzene

p-bromoaniline

o-bromofluorobenzene

Solution:

Aniline reacts with bromine water to form 2,4,6-tribromoaniline. The reaction is as follows:

     NH2                                      NH2
      |                                       |
  C6H5  + 3Br2  ----->  C6H2Br3         + 3HBr

2,4,6-tribromoaniline is then treated with an aqueous solution of sodium nitrite (NaNO2) in the presence of dilute HCl. This is a diazotization reaction, converting the amine group into a diazonium group:

     NH2                                      N2+
      |                                       |
  C6H2Br3  + NaNO2 + HCl -----> C6H2Br3     + NaCl + 2H2O

The resulting diazonium salt (2,4,6-tribromo-benzenediazonium chloride) is then converted into its tetrafluoroborate salt. Finally, heating the tetrafluoroborate salt leads to the loss of nitrogen gas (N2) and the formation of 1,3,5-tribromobenzene.

      N2+                                    Br
       |                                     |
  C6H2Br3 -----> C6H3Br3 + N2

Therefore, the final product is 1,3,5-tribromobenzene.