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).
Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is: