Animated Solution for Chemistry - Organic Chemistry: Which of the following compound gives pink colour on reaction with phthalic anhydride in conc. ${\text{H}}_2{\text{SO}}_4$ followed by treatment with NaOH?

The Phthalein Dye Test is a classic and visually striking reaction in organic chemistry, often used to detect the presence of phenols. When a phenol is heated with phthalic anhydride in the presence of a dehydrating agent like concentrated sulfuric acid ($H_{2}S{O}_4$), it forms a colorless compound called a phthalein. Upon adding an alkali like NaOH, this compound undergoes a structural transformation into a highly conjugated quinonoid form, which exhibits an intense, brilliant color—most famously, the bright pink of phenolphthalein.

Electrophilic Aromatic Substitution

The reaction kicks off with the protonation of phthalic anhydride by the strong acid, $H_{2}S{O}_4$. This protonation generates a highly reactive, resonance-stabilized carbocation. This bulky electrophile is now on the hunt for an electron-rich aromatic ring.

Enter the phenol. The hydroxyl ($-OH$) group on the phenol ring is strongly activating and directs incoming electrophiles to the ortho and para positions. However, because the phthalic anhydride carbocation is exceptionally large and sterically demanding, attacking the ortho position (right next to the $-OH$ group) is highly unfavorable.

Consequently, the electrophilic attack occurs almost exclusively at the para position.

A Free Para Position

For the phthalein dye to successfully form, the phenol molecule must have its para position completely free and unsubstituted. If there is a substituent blocking this position, the bulky electrophile simply cannot attack, and the reaction halts.

Let's evaluate the given options based on this strict geometric requirement:

1. Option (a): This is p-cresol (4-methylphenol). The para position is occupied by a methyl group. It cannot form the dye. 2. Option (c): This is 2-isopropyl-4-methylphenol. Once again, the para position (position 4) is blocked by a methyl group. 3. Option (d): This is 4-methylbenzene-1,2-diol. The para position relative to the primary $-OH$ group is blocked.

Option (b) is 2-propylphenol. If we look at its structure, the propyl group is located at the ortho position, leaving the para position completely unobstructed.

When 2-propylphenol reacts with the phthalic anhydride carbocation, the electrophile smoothly attacks the free para position. Two molecules of 2-propylphenol condense with one molecule of phthalic anhydride, ultimately yielding a phthalein derivative that turns a beautiful pink upon treatment with NaOH.

Thus, the correct answer is 2-propylphenol!