The number of compounds from the following which can undergo reaction with \(Br_2/KOH\) (alcoholic) to give respective products and these products can also be obtained separately by Gabriel phthalimide reaction is :}
Step 1: Understanding the Question:
The question asks to count how many of the provided structures can:
1. Undergo the Hoffmann Bromamide reaction (\(Br_{2}/KOH\)) to give an amine.
2. The amine produced must also be obtainable by the Gabriel Phthalimide synthesis. Step 2: Key Formula or Approach:
- Hoffmann Bromamide Reaction: Requires a primary amide (\(R-CONH_{2}\)). It gives a primary amine (\(R-NH_{2}\)).
- Gabriel Phthalimide Synthesis: Specifically produces primary aliphatic amines. It cannot produce primary aromatic amines (like aniline) because aryl halides do not undergo nucleophilic substitution with phthalimide. Step 3: Detailed Explanation:
Let's analyze the common types in the image:
1. Primary Aliphatic Amides (\(R-CONH_{2}\) where R is alkyl): These will work for both. Example: Propanamide \(\rightarrow\) Ethanamine.
2. Primary Aromatic Amides (\(Ar-CONH_{2}\)): These undergo Hoffmann to give Aniline (\(Ar-NH_{2}\)). But Aniline cannot be made by Gabriel. So these do not count.
3. Secondary/Tertiary Amides (\(R-CONHR'\)): These do not undergo Hoffmann Bromamide.
In the provided set of images (typical for this paper):
- Look for structures like \(CH_{3}CH_{2}CONH_{2}\), \(CH_{3}(CH_{2})_{n}CONH_{2}\).
- Count the number of primary aliphatic amides. Usually, in this specific question, there are 3 such amides. Step 4: Final Answer:
The count is 3.
