Every question from past board papers on Diazonium salts in synthesis, newest paper first. Open any question for its full worked answer.
Given below are two statements :
Statement I : Benzenediazonium salt is prepared by the reaction of aniline with nitrous acid at 273 - 278 K. It decomposes easily in the dry state.
Statement II : Insertion of iodine into the benzene ring is difficult and hence iodobenzene is prepared through the reaction of benzenediazonium salt with KI.
In the light of the above statements, choose the most appropriate answer from the options given below :
Identify the product in the following reaction :
Benzenediazonium chloride (\(\mathrm{C_6H_5N_2^+Cl^-}\)) with (i) Cu\(_2\)Br\(_2\)/HBr (ii) Mg/dry ether (iii) H\(_2\)O \(\rightarrow\) Product
Which of the following sequence of reactions is suitable to synthesize chlorobenzene ?
The product formed from the following reaction sequence is
[benzonitrile] \(\xrightarrow[\text{(iii) } H_2O]{\text{(i) } LiAlH_4, H_2O;\ \text{(ii) } NaNO_2 + HCl}\)
The reagent 'R' in the given sequence of chemical reaction is :
2,4,6-tribromoaniline \(\xrightarrow{NaNO_2, HCl,\ 0-5^\circ C}\) 2,4,6-tribromobenzenediazonium chloride \(\xrightarrow{R}\) 1,3,5-tribromobenzene
A given nitrogen-containing aromatic compound \(A\) reacts with Sn/HCl, followed by \(HNO_2\) to give an unstable compound \(B\). \(B\), on treatment with phenol, forms a beautiful coloured compound \(C\) with the molecular formula \(C_{12}H_{10}N_2O\). The structure of compound \(A\) is
