How do you account for the following observations?
$(a)$ Though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants,yet in the manufacture of benzoic acid from toluene,we use alcoholic potassium permanganate as an oxidant. Why? Write a balanced redox equation for the reaction.
$(b)$ When concentrated sulphuric acid is added to an inorganic mixture containing chloride,we get a colourless pungent-smelling gas $HCl$,but if the mixture contains bromide,then we get red vapours of bromine. Why?

(N/A) In the manufacture of benzoic acid from toluene,alcoholic potassium permanganate is used as an oxidant for the following reasons:
$(i)$ In a neutral medium,$OH^-$ ions are produced in the reaction itself. As a result,the cost of adding an acid or a base is avoided.
$(ii)$ $KMnO_4$ and alcohol are homogeneous to each other since both are polar. Toluene and alcohol are also homogeneous to each other because both are organic compounds. Reactions proceed at a faster rate in a homogeneous medium than in a heterogeneous medium. Hence,in alcohol,$KMnO_4$ and toluene react at a faster rate.
The balanced redox equation for the reaction in a neutral medium is:
$C_6H_5CH_3 + 2KMnO_4 \to C_6H_5COOK + 2MnO_2 + KOH + H_2O$
$(b)$ When conc. $H_2SO_4$ is added to an inorganic mixture containing bromide,initially $HBr$ is produced. $HBr$,being a strong reducing agent,reduces $H_2SO_4$ to $SO_2$ with the evolution of red vapours of bromine.
$2NaBr + 2H_2SO_4 \to 2NaHSO_4 + 2HBr$
$2HBr + H_2SO_4 \to Br_2 + SO_2 + 2H_2O$
(Note: $Br_2$ is the red vapour.)
However,when conc. $H_2SO_4$ is added to an inorganic mixture containing chloride,a pungent-smelling gas $(HCl)$ is evolved. $HCl$,being a weak reducing agent,cannot reduce $H_2SO_4$ to $SO_2$.
$2NaCl + 2H_2SO_4 \to 2NaHSO_4 + 2HCl$