
The reaction of organic compounds with hot potassium permanganate (KMnO4) is an example of oxidative cleavage. Aromatic hydrocarbons with side chains, when oxidized, can be converted to benzoic acid, provided that there is at least one benzylic hydrogen present. The process can be summarized as follows:
When a compound containing the structure of an alkyl benzene is treated with hot KMnO4, the side chain is oxidized all the way to a carboxyl group (-COOH), resulting in the formation of benzoic acid if any hydrocarbon chains containing benzylic hydrogens (i.e., C-H bonds directly attached to the carbon adjacent to the benzene ring) are present.
Now, consider the given options: we need to find out which compounds, when treated with hot KMnO4, will give benzoic acid:
1. Toluene or any compound with alkyl (methyl, ethyl, etc.) groups attached directly to the benzene ring.
2. Xylene (symmetrical with alkyl group on different positions), if at least one alkyl group has a benzylic hydrogen.
3. Anything that already has benzoic acid won't undergo further conversion to benzoic acid under these conditions.
Thus, out of the provided compounds in the figure, only those containing alkyl groups directly bonded to benzene with benzylic hydrogens will yield benzoic acid after reaction with hot KMnO4.
In conclusion, the total number of compounds that form benzoic acid when treated with hot KMnO4 is 5.
Chlorobenzene to biphenyl
Match the LIST-I with LIST-II

Choose the correct answer from the options given below:
Let \( y^2 = 12x \) be the parabola and \( S \) its focus. Let \( PQ \) be a focal chord of the parabola such that \( (SP)(SQ) = \frac{147}{4} \). Let \( C \) be the circle described by taking \( PQ \) as a diameter. If the equation of the circle \( C \) is: \[ 64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta, \] then \( \beta - \alpha \) is equal to:
Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
Let $ A \in \mathbb{R} $ be a matrix of order 3x3 such that $$ \det(A) = -4 \quad \text{and} \quad A + I = \left[ \begin{array}{ccc} 1 & 1 & 1 \\2 & 0 & 1 \\4 & 1 & 2 \end{array} \right] $$ where $ I $ is the identity matrix of order 3. If $ \det( (A + I) \cdot \text{adj}(A + I)) $ is $ 2^m $, then $ m $ is equal to: