The number of compounds from the following which can undergo reaction with Br$$_2$$/KOH (alcoholic) to give respective products and these respective products can also be obtained separately by Gabriel phthalimide reaction is :

Hofmann bromamide degradation converts a primary amide into a primary amine containing one carbon atom less.

$$RCONH_2\xrightarrow{Br_2/KOH}RNH_2.$$

Therefore, only primary amides can undergo this reaction.

Gabriel phthalimide synthesis is used for the preparation of primary aliphatic amines. It is not suitable for the preparation of aromatic amines because aryl halides do not undergo nucleophilic substitution with potassium phthalimide under these conditions.

Hence, a compound must satisfy both of the following conditions:

• It should be a primary amide ((RCONH_2)).
• The amine obtained after Hofmann degradation should be a primary aliphatic amine.

Evaluating the given compounds:

• (C_6H_5CONH_2): Primary amide, but gives aniline (aromatic amine), so it does not qualify.
• (C_6H_5CH_2CONH_2): Primary amide, gives benzylamine, so it qualifies.
• (CH_3CONH_2): Primary amide, gives methylamine, so it qualifies.
• (C_6H_{11}CONHCH_2CH_3): Secondary amide, so it does not qualify.
• ((CH_3)_3CCONHCH_3): Secondary amide, so it does not qualify.
• (C_6H_{11}CONH_2): Primary amide, gives cyclohexylamine, so it qualifies.

Thus, 3 compounds satisfy both conditions.

Hence, the correct answer is 3.