Question

(B) Write the structures of A, B, and C in the following sequence of reactions:
\includegraphics[width=1\linewidth]{chi25.png}

Hint:

In this reaction sequence, DIBAL-H is used to reduce nitriles to aldehydes, and aldol condensation with NaOH under heat leads to the formation of β-hydroxy aldehydes or ketones, which can undergo further transformations to form the final product.

Answer 1

(a) Reaction Sequence: 1. CH\(_3\)COOH \( \xrightarrow{\text{SOCl}_2} \) CH\(_3\)COCl: Acetic acid (CH\(_3\)COOH) reacts with thionyl chloride (SOCl\(_2\)) to form acetyl chloride (CH\(_3\)COCl), which is an acyl chloride. 2. CH\(_3\)COCl \( \xrightarrow{H_2, \text{Pd-BaSO}_4} \) CH\(_3\)CH\(_2\)NH\(_2\): The acetyl chloride undergoes a reduction reaction using hydrogen (H\(_2\)) in the presence of a palladium catalyst on barium sulfate (Pd-BaSO\(_4\)) to form ethylamine (CH\(_3\)CH\(_2\)NH\(_2\)). 3. CH\(_3\)CH\(_2\)NH\(_2\) \( \xrightarrow{H_2\text{N-NH}_2} \) C: The ethylamine (CH\(_3\)CH\(_2\)NH\(_2\)) reacts with hydrazine (H\(_2\)N-NH\(_2\)) to form a product C, which is likely a hydrazone or another nitrogen-containing compound. 

(b) Reaction Sequence: 1. CH\(_3\)CN \( \xrightarrow{1. \text{DIBAL-H}, 2. H}_2\text{O} \) A: Methyl cyanide (CH\(_3\)CN) reacts with DIBAL-H (diisobutylaluminum hydride) to reduce the nitrile group (CN) to an aldehyde (CH\(_3\)CHO), which is compound A. 2. A \( \xrightarrow{\text{NaOH, heat}} \) B: The aldehyde A undergoes an aldol condensation in the presence of sodium hydroxide (NaOH) and heat, forming a product B, which is likely a β-hydroxy aldehyde or ketone. 3. B \( \xrightarrow{C} \) C: The compound B undergoes further reactions, likely a dehydration or other transformation to form the final product C.