




The reaction given involves the treatment of the compound with bromine (Br2) in the presence of sodium hydroxide (NaOH) under heat (Δ). This is a classic example of the Hofmann Bromamide Degradation Reaction.
Amide Group Conversion: The amide group (-CONH2) undergoes degradation in the presence of Br2/NaOH, forming a primary amine (-NH2).
Retention of Other Groups: The ester group (-COOCH3) remains intact during this reaction as it does not participate in the Hofmann degradation.
Formation of the Product: The resulting product is a compound where the -CONH2 group is replaced by a -NH2 group, while the ester group remains unchanged.
The major product formed in this reaction is:
Cyclohexanone derivative with an amine group (-NH2) and an ester group (-COOCH3) intact.
Calculate the potential for half-cell containing 0.01 M K\(_2\)Cr\(_2\)O\(_7\)(aq), 0.01 M Cr\(^{3+}\)(aq), and 1.0 x 10\(^{-4}\) M H\(^+\)(aq).



For the circuit shown above, the equivalent gate is:
Let \( f : \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that \[ (\sin x \cos y)(f(2x + 2y) - f(2x - 2y)) = (\cos x \sin y)(f(2x + 2y) + f(2x - 2y)), \] for all \( x, y \in \mathbb{R}. \)
If \( f'(0) = \frac{1}{2} \), then the value of \( 24f''\left( \frac{5\pi}{3} \right) \) is: