




To identify compounds A and B in the reaction sequence, we need to analyze each step of the given reactions.
The initial compound is bromobenzene. The reaction sequence involves two main steps:
In this mechanism, due to steric hindrance, the nitro group preferably occupies a para position. Therefore, compound A is para-bromonitrobenzene.
Thereby, compound B is phenol.
From the options, the correct answer is shown in the image below:
Thus, the compounds A and B in the reaction sequence are para-bromonitrobenzene and phenol, respectively.
The reaction mechanism involves nitration followed by hydrolysis:
\[ \text{Bromobenzene} \xrightarrow{\text{Conc. HNO}_3} \text{Bromonitrobenzene (A)} \]
Then:
\[ \text{Bromonitrobenzene (A)} \xrightarrow{\text{NaOH, HCl}} \text{p-Bromophenol (B)} \]

Amines are usually formed from amides, imides, halides, nitro compounds, etc. They exhibit hydrogen bonding which influences their physical properties. In alkyl amines, a combination of electron releasing, steric and H-bonding factors influence the stability of the substituted ammonium cations in protic polar solvents and thus affect the basic nature of amines. Alkyl amines are found to be stronger bases than ammonia. Amines being basic in nature, react with acids to form salts. Aryldiazonium salts, undergo replacement of the diazonium group with a variety of nucleophiles to produce aryl halides, cyanides, phenols and arenes.
Which of the following amine(s) show(s) positive carbamylamine test? 
Consider the following sequence of reactions : 
Molar mass of the product formed (A) is ______ g mol\(^{-1}\).

In the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by: