Match the entries of column I with appropriate entries of column II and choose the correct option out of the four options given.

(a)-(i), (b)-(iv), (c)-(ii), (d)-(iii)

The correct answer is option 4. (a)-(i), (b)-(iv), (c)-(ii), (d)-(iii).

(a) C₆H₅—OH + Br₂/H₂O →(i) 2,4,6-tribromophenol:

The reaction of phenol with bromine is known as bromination of phenol. Solvent has great influence on the reaction. In different solvents, different products are obtained. Phenol reacts with bromine water to give 2,4,6-tribromophenol. In water, ionization is facilitated. Phenol gets ionized to form phenoxide ion, which is even better ortho, para-directing. Bromine also gets ionized to a larger extent to form a large number of bromonium ions. And Bromine ions are highly stabilized in ionic solvents. So the formation of strong o, p-directing group and stabilization of \(Br^+\) enhances the formation of tribromophenol.

(b) C₆H₅—OH + Br₂/CS₂ →(iv) p-bromophenol and o-bromophenol:

The reaction of phenol with bromine is known as bromination of phenol. Solvent has great influence on the reaction. In different solvents, different products are obtained. Phenol reacts with bromine in a presence of Carbon disulphide to form a mixture of o-bromophenol and p-bromophenol.

Phenol reacts with bromine in a presence of Carbon disulphide to form a mixture of o-bromophenol and p-bromophenol. Among which p-bromophenol is major. In \(CS_2\) ionisation is not facilitated that much. As it is a non-polar solvent it doesn't the \(Br^+\) on. Also \(−OH\) group is moderately o, p-directing. So as usual para-product is more formed.

(c) C₆H₅—OH + dil. HNO₃(15^oC) →(ii) p-nitrophenol and o-nitrophenol:

When phenol (C₆H₅OH) reacts with dilute nitric acid (HNO₃) at low temperature \((15^oC)\), it undergoes nitration, resulting in a mixture of ortho-nitrophenol (2-nitrophenol) and para-nitrophenol (4-nitrophenol).

This reaction typically results in the formation of a mixture of ortho and para isomers, with the para isomer generally being the major product due to steric reasons. However, the exact ratio of ortho to para products can depend on various factors such as the concentration of nitric acid, temperature, and reaction time.

(d) C₆H₅—OH + conc. H₂SO₄/conc. HNO₃ →(iii) 2,4,6-trinitrophenol:

Phenol first reacts with concentrated sulphuric acid which converts it to phenol-2,4-disulphonic acid and then with concentrated nitric acid to get 2,4,6-Trinitrophenol. It is because \(-OH\) is a strong activating group towards electrophilic substitution. So it forms 2,4,6-Trinitrophenol instead of ortho or para nitro phenols.