Identify (a) in the following sequence of reactions. \(\mathrm{C}_{4} \mathrm{H}_{7} \mathrm{OCl}(\mathrm{a}) \stackrel{\mathrm{NH}_{3}}{\longrightarrow} \mathrm{C}_{4} \mathrm{H}_{9} \mathrm{ON} \stackrel{\mathrm{Br}_{2}+\mathrm{NaOH}}{\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}}\) (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COCl}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHClCHO}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{Cl}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}_{2} \mathrm{Cl}\)

Understanding functional groups is essential in organic chemistry as they dictate the chemical behavior of molecules. In our exercise, the focus is on identifying a specific functional group that changes during a reaction sequence.

Initially, we have a compound with the formula \(\mathrm{C}_{4}\mathrm{H}_{7} \mathrm{OCl}(\mathrm{a})\). The key to identifying the correct functional group involves looking at the reactants and the products. Given that the product includes an amide group \(\mathrm{-CONH}_{2}\), we can infer that the reactant must contain an acyl chloride group \(\mathrm{-COCl}\), which can react with ammonia to form the amide.

This process of elimination is powerful. It leads us to dismiss any option that doesn't have an acyl chloride functional group. The correct identification is fundamental to predict the behavior of the molecule in further chemical reactions, such as the ammonolysis and Hoffmann Bromamide Decomposition that we see later in the sequence.

Ammonolysis is a type of substitution reaction where ammonia \(\mathrm{NH}_{3}\) replaces a functional group in an organic compound. Typically, it involves the conversion of acyl chlorides into amides.

In our case, the compound with the acyl chloride functional group reacts with ammonia, leading to the formation of an amide. Here, ammonia acts as a nucleophile, attacking the carbon atom bonded to the chlorine in the acyl chloride group. The reaction follows:

• Initial Compound: \(\mathrm{R-COCl}\)
• Ammonia: \(\mathrm{NH}_{3}\)
• Product: \(\mathrm{R-CONH}_{2}\)

Through this reaction, the chlorine is substituted by an \(\mathrm{NH}_{2}\) group. The ability to identify an ammonolysis reaction is crucial as it is a common way to form amides from more reactive acyl chlorides in organic synthesis. Recognizing this reaction allows for the prediction of the type of compounds formed under specific conditions.

The Hoffmann Bromamide Decomposition is a reaction that breaks down an amide into an amine with one less carbon atom. It involves treating an amide with bromine and an alkali like sodium hydroxide.

In the step-by-step solution to the given exercise, the Hoffmann Decomposition converts the amide group into an amine. The reaction proceeds as follows:

• Initial Compound: \(\mathrm{R-CONH}_{2}\)
• Reagents: \(\mathrm{Br}_{2}\) and \(\mathrm{NaOH}\)
• Product: \(\mathrm{R-NH}_{2}\) (with one less carbon atom)

This reaction not only forms the amine but also releases nitrogen gas and a carboxylic acid salt as byproducts. In our exercise, if the original compound has the acyl chloride at the end of a carbon chain (as in option [a]), the resulting primary amine reflects the loss of the carbonyl carbon due to the Hoffmann Decomposition. Understanding this reaction is pivotal for students, as it exemplifies how rearrangements and decompositions can lead to simpler products with practical applications in synthetic organic chemistry.