OH^- being a strong base is a bad leaving group. When H₂SO₄ is added, protonation of alcohol occurs i.e -OH is converted into \(-\stackrel{+}{\mathrm{O}} \mathrm{H}_{2}\)

In aq. solution, KOH is almost completely ionised to give OH^- ions which being a strong nucleophile brings about a substitution reaction to form alcohols. Further in aq. solution, OH^- ions are highly solvated (hydrated).
This solution reduces the basic character of OH^- ions which fail to abstract a hydrogen from the -carbon of the alkyl halide to form an alkene.
However an alcoholic solution of KOH contains alkoxide (RO^-) ions which being a much stronger base than OH^- ions preferentially abstracts a hydrogen from the carbon of the alkyl halide to form alkene.

(a) This is because pcc is a mild oxidising agent and can oxide methanol to methanal only while KMnO₄ being strong oxidising agent oxidises it to methanoic acid.
(b) Because esterification rxn is reversible and presence of base (pyridine) neutralises HCl produced during reaction thus promoting forward reaction.

(i) Phenol to benzoquinone.

(ii) Propanone to 2-Methylpropan-2-ol.
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(iii) Propene to propan-2-ol.

(c), esters having the same number of carbon atoms in the alkyl and acyl group, i.e., CH₃CH₂ and COCH₃ group in CH₃CH₂OCOCH_3.

(i) Teacher preferred denatured alcohol over absolute alcohol as it would be unfit for drinking. As commercial alcohol is mixed with copper sulphate to give it a colour and pyridine to give a foul smell, which makes absolute alcohol unfit for drinking and called as denatured alcohol. 
(ii) As the teacher was ordering ethanol for school laboratory, maintaining discipline, repsonsible behaviour and social obligation were values kept in her mind, while taking decision.

(i) Since compound (A) on treatment with CHCI3 and KOH (i.e., Reimer-Tiemann reaction), gives two products Band C, therefore, A must be phenol and Band C must be o-hydroxybenzaldehyde and P: hydroxybenzaldehyde respectively or vice-versa.
(ii) Since both Band C on distillation with Zn dust give the same compound (D), therefore, D must be benzaldehyde.
(iii) Since oxidation of D gives E with M.F. C7R602, therefore, E must be benzoic acid.
(iv) Since sodium salt of E, i.e.benzoic acid upon heating with soda-lime gives compound (F), therefore.
(F) must be benzene. Zn dust distillation of A would also give benzene (F).

In the acidic medium, the oxygen atom of the oxirane ring gets protonated. The presence of positive charge on the oxygen atom weakens the C-O bond more on the side of C² since the partial positive charge created on C₂ will be stabilized by the + I- effect of the two CH₃ groups in the transition state (T.S.). Consequently, the attack of the nucleophile, i.e., H₂O preferentially ocurson C₂ yielding 2-methyl propane- 1, 2-diol.