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Phenol, when it first reacts with concentrated sulphuric acid and then with concentrated nitric acid, gives
Phenol contains the strongly activating, ortho-para directing group $$-OH$$. If it is nitrated directly with $$\text{conc}\;HNO_3/H_2SO_4$$, uncontrolled substitution occurs and the very strongly activated ring is converted to 2,4,6-trinitrophenol (picric acid). To obtain a mono-nitration product we first block the para position by sulphonation.
Step 1 : Sulphonation
Phenol is treated with hot concentrated sulphuric acid (473 K).
$$\text{Phenol} \xrightarrow[\;473\text{ K}\;]{\text{conc }H_2SO_4}$$ gives mainly $$p$$-phenol-sulphonic acid, with a small amount of the ortho isomer:
$$\ce{ \underset{phenol}{C6H5OH} + H2SO4 \; \longrightarrow \; \underset{p\text{-phenol sulphonic acid}}{HO-C6H4-SO3H} }$$
The bulky, electron-withdrawing $$-SO_3H$$ group now occupies the para position. It is strongly deactivating and meta directing, so during the next electrophilic substitution the incoming electrophile avoids the para position (already occupied) and the meta position with respect to $$-SO_3H$$ becomes the ortho position with respect to $$-OH$$.
Step 2 : Nitration
The sulphonic acid is now nitrated with concentrated nitric acid (usually a nitrating mixture, $$HNO_3 + H_2SO_4$$).
$$\text{Phenol-SO}_3H \xrightarrow{\text{conc }HNO_3}$$
Because $$-SO_3H$$ is meta directing, $$NO_2$$ enters the position meta to $$-SO_3H$$, which is orthogonal (ortho) to the original $$-OH$$ group. Simultaneously, under the hot acidic conditions, the $$-SO_3H$$ group is removed (desulphonation). Thus the net result of “sulphonation → nitration → desulphonation” is introduction of a single $$NO_2$$ group at the ortho position to $$-OH$$.
$$\ce{HO-C6H4-SO3H \; \xrightarrow[H2SO4]{HNO3} \; HO-C6H4-NO2 + SO3 + H2O}$$
Product obtained
The compound formed is o-nitrophenol (2-nitrophenol).
Hence, phenol subjected first to concentrated $$H_2SO_4$$ and then to concentrated $$HNO_3$$ yields o-nitrophenol.
Option B which is: o-nitrophenol
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