(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) Methyl alcohol is easily oxidized to formaldehyde and then to formic acid which may cause blindness and death. 
(ii) Doctors give an intra venous infusion of diluted ethanol the enzyme responsible for oxidation of HCHO to acid is swamped allowing time for kidneys to excrete methanol. 
(iii) People should be educated not to drink cheap alcohol from unauthorized sources. It is not worth saving money when the life get endangered. If at all one has to drink, the stuff must be purchased from an authorized source.

(i) \({ CH }_{ 3 }-{ CH }_{ 2 }-CH={ CH }_{ 2 }+{ H }_{ 2 }O\overset { { H }^{ + } }{ \longrightarrow } { CH }_{ 3 }{ CH }_{ 2 }(OH){ CH }_{ 3 }\)
Since X forms bisulphate and 2, 4-dinitrophenylhydrazone, it means Xhas carbonyl group formed due to oxidation of alcohol. This oxidation is due to coating of metal (Cu).
 \({ CH }_{ 3 }{ CH }_{ 2 }(OH){ CH }_{ 3 }\longrightarrow { CH }_{ 3 }{ CH }_{ 2 }CO{ CH }_{ 3 }(ketone/X)+{ H }_{ 2 }I\)
(ii) Alcohol and X both give iodoform test. 
(iii) Alcohol and X can be separated by
(a) Distillation
(b) BisulpIilte formation only (X) forms Bisulphite complex (s) which can be decomposed.
(iv) Usha lacks critical awareness of chemistry.

(i) Since compound (A) with M.F. C₆H₆O gives characteristic colour with FeCI₃, therefore, (A) must be phenol.
(ii) Since compound (A), i.e., phenol reacts with CO₂ and NaOR under pressure (i.e., Kolbe's reaction) to give compound (B) which on acidification gives compound (C), therefore, (B) must be sodium salicylate and (C) must be salicylic acid.
(iii) Since (C) reacts with acetyl chloride to form a popular pain killer (D), therefore, (D) must be aspirin.

(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).

(i) Since compound A (C₄H₁₀O) does not react with Na metal or KMnO₄, it cannot be an alcohol.
(ii) Since compound A dissolves in cone, H₂SO₄, it may be an ether.
(iii) Since ether A on heating with excess of HI gives a single alkyl halide, therefore, ether (A) must be symmetrical. Now the only symmetrical ether having M.F. C₄HIOO is diethyl ether (CH₃CH₂OCH₂CH₃).

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.

The allyl carbocation is resonance stabilized and hence is as stable as a 30 carbocation
Therefore, on treatment with Lucas reagent, allyl alcohol gives white turbidity in less than 1 min.l-Propanol, on the other hand, does not react with Lucas reagent.

To begin with both ethanol and phenol react with PCI₅ to give the corresponding chlorophosphate esters.
Since in ethanol, C-O bond has only single bond character, therefore, nucleophilic attack by Cl^- ion occurs easily on chlorophosphate ester (I) to give chloroethane. However, in case of phenol, C-O bond has some double bond character due to resonance and hence difficult to break. Instead nucleophilic attack by C₆H₅OH occurs twice on the chlorophosphate ester (II) to form intermediate (III). Further, nucleophilic attack by C₆H₅OH on III does not occur due to steric hindrance. Instead (III) undergoes hydrolysis by water to form triphenylphosphate.

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