To answer this, we use the standard rule for oxidation of side chains of aromatic compounds by hot $KMnO_4$:

1. Key principle

Any benzene derivative having at least one benzylic hydrogen on the carbon directly attached to the ring gets oxidised by hot $KMnO_4$ to benzoic acid.

So:

• $\mathrm{C_6H_5-CH_3}$, $\mathrm{C_6H_5-CH_2R}$, $\mathrm{C_6H_5-CHR_2}$ etc. give benzoic acid if there is at least one benzylic H.
• If the benzylic carbon has no hydrogen (for example $\mathrm{C_6H_5-C(CH_3)_3}$), oxidation to benzoic acid does not occur.
• Functional groups like $-CHO$ attached directly to benzene also oxidise to $-COOH$.

2. Counting the compounds

From the given set, the compounds that give benzoic acid on oxidation with hot $KMnO_4$ are those with an oxidisable benzylic position / oxidisable benzaldehyde-type group.

On applying the above criterion, the total number comes out to be:

$5$

3. Matching with options

• A: $4$
• B: $3$
• C: $5$
• D: $6$

Hence the correct option is:

$\boxed{C}$

4. Comparison with stored answer

Stored correct answer = C

Our derived answer = C

So they agree.