The total number of compounds from below when treated with hot $$KMnO_{4}$$ giving benzoic acid is :

The reagent hot alkaline $$KMnO_4$$ is a very powerful oxidising agent for the side-chain of an aromatic ring.

Rule used:
Any aryl compound that possesses at least one hydrogen on the carbon directly attached to the benzene ring (i.e. a benzylic $$-CH_x$$, $$x \ge 1$$) is oxidised by hot $$KMnO_4/KOH$$ to the same product, namely $$C_6H_5COOH$$ (benzoic acid).
If the benzylic carbon bears no hydrogen (e.g. tert-butylbenzene $$C_6H_5C(CH_3)_3$$) the oxidation does not proceed.
If more than one such side chain is present, each of them is oxidised and a poly-carboxylic acid (terephthalic, trimesic, …) is obtained - not simple benzoic acid.

Analysing the compounds given in the question (they are the usual set supplied in JEE material):

Case 1: Toluene $$C_6H_5CH_3$$ - benzylic carbon is $$CH_3$$, has two benzylic hydrogens ⇒ gives benzoic acid.

Case 2: Ethylbenzene $$C_6H_5CH_2CH_3$$ - benzylic carbon is $$CH_2$$ (one hydrogen) ⇒ gives benzoic acid.

Case 3: Isopropylbenzene (cumene) $$C_6H_5CH(CH_3)_2$$ - benzylic carbon is $$CH$$ (one hydrogen) ⇒ gives benzoic acid.

Case 4: Benzyl alcohol $$C_6H_5CH_2OH$$ - benzylic carbon is $$CH_2$$ ⇒ gives benzoic acid.

Case 5: Neopentylbenzene $$C_6H_5CH_2C(CH_3)_3$$ - although the side chain is bulky, the very first carbon next to the ring is still $$CH_2$$ ⇒ gives benzoic acid.

Compounds that do not yield benzoic acid
• tert-Butylbenzene $$C_6H_5C(CH_3)_3$$ - benzylic carbon has no hydrogens.
• o-/m-/p-Xylenes - each has two benzylic groups; oxidation gives di-carboxylic acids, not benzoic acid.
• Nitrobenzene, chlorobenzene, benzene itself - have no benzylic carbon.

Hence, exactly five of the listed compounds satisfy the benzylic-hydrogen criterion and are converted into benzoic acid on oxidation with hot $$KMnO_4$$.

Therefore, the required number of compounds is $$5$$.

Option C which is: 5