Q.(i) Complete the following reaction and suggest a suitable mechanism for the reaction :
$CH_3CH_2OH \xrightarrow{H^+,\ 443\ K}$

The key to this question lies in understanding why a mixture deviates from Raoult’s law. Raoult’s law assumes that the intermolecular forces between unlike molecules (A–B) are the same as those between like molecules (A–A and B–B). When that’s true, the vapour pressure is simply the weighted average of the pure components — an ideal solution.

But real solutions often break this assumption. If A–B interactions are weaker than A–A and B–B, molecules escape more easily, vapour pressure is higher than expected — that’s positive deviation. If A–B interactions are stronger, molecules are held together more tightly, vapour pressure drops — that’s negative deviation.

Now, phenol ($\ce{C6H5OH}$) has a hydroxyl group (–OH) that can act as a hydrogen bond donor and acceptor. Aniline ($\ce{C6H5NH2}$) has an amino group (–NH2) that can also donate and accept hydrogen bonds. When you mix them, the –OH of phenol can form a hydrogen bond with the –NH2 of aniline. This new O–H···N interaction is stronger than the original O–H···O hydrogen bonds in pure phenol or the N–H···N bonds in pure aniline. The result: the mixture is more “sticky,” vapour pressure is lower, and you get negative deviation.

Let’s walk through the options one by one.

1. Option (A): Polar covalent bond — A polar covalent bond is a permanent bond within a molecule (like O–H or N–H). Mixing two liquids doesn’t create new covalent bonds between them; it creates intermolecular forces. So this is wrong.

2. Option (B): Non-polar covalent bond — Same issue. Covalent bonds are intramolecular. Even if a bond were non-polar, it wouldn’t form between separate molecules upon mixing. This is also wrong.

3. Option (C): Intermolecular hydrogen bond — This is exactly what happens. The –OH of phenol and the –NH2 of aniline form a hydrogen bond between molecules. This stronger attraction lowers the vapour pressure, causing negative deviation. This is correct. …

Step 1 — Cleavage of anisole by HI → A

Phenyl methyl ether (anisole) is $\mathrm{C_6H_5\!-\!O\!-\!CH_3}$. With HI, the ether oxygen is first protonated, and then $\mathrm{I^-}$ attacks — but which carbon?

The iodide attacks the methyl carbon, not the aryl carbon, for two reasons:

1. The C(aryl)–O bond has partial double-bond character (the oxygen lone pair is delocalised into the ring), making it short and strong.
2. Nucleophilic substitution at an $sp^2$ aryl carbon is not feasible under these conditions, whereas the methyl carbon is an ideal, unhindered $\mathrm{S_N2}$ centre.

The aromatic product asked for is phenol ($\mathrm{CH_3I}$ is not aromatic).

Step 2 — Nitration of phenol → B

The $-\mathrm{OH}$ group is a powerful activating, ortho/para-directing substituent: its lone pair is donated into the ring by resonance, greatly enriching the o- and p- positions in electron density. With concentrated $\mathrm{HNO_3}$ the ring is therefore nitrated repeatedly, at both ortho positions and the para position:

$\mathrm{C_6H_5OH} \xrightarrow{\text{conc. } \mathrm{HNO_3}} \mathrm{C_6H_2(NO_2)_3OH}$ …

Step 1 — Why phenols are acidic at all.

Phenol loses $\mathrm{H^+}$ to give the phenoxide ion, whose negative charge is delocalised into the benzene ring by resonance. Anything that further stabilises this negative charge makes the phenol more acidic; anything that intensifies the charge makes it less acidic.

Step 2 — (i) p-Methoxyphenol.

The $-\mathrm{OCH_3}$ group is a strong electron-donating group by resonance (+R): the oxygen lone pair is pushed into the ring. This increases the electron density on the ring and hence on the phenoxide oxygen, destabilising the anion. It is therefore the least acidic — even less acidic than phenol itself.

Step 3 — (ii) p-Methylphenol (p-cresol).

$-\mathrm{CH_3}$ is only a weak electron-releasing group (+I effect and hyperconjugation). It also destabilises the phenoxide, but far less than $-\mathrm{OCH_3}$ does, because it has no lone pair to donate by resonance. So it is more acidic than (i) but still less acidic than phenol.

Step 4 — (iii) p-Nitrophenol. …