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Phthalic anhydride reacts with resorcinol in the presence of concentrated ${{H}_{2}}S{{O}_{4}}$ to give:
(A) Phenolphthalein
(B) alizarin
(C) coumarin
(D) fluorescein

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Answer
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Hint: Phthalic acid is a dicarboxylic acid. It reacts with a dehydrating agent i.e. sulphuric acid. Phthalic acid loses water on heating and acts as a dehydrating agent. The reaction of phthalic anhydride with resorcinol results in the formation of organic dye. This dye has a wide application as a fluorescent material in the laser application.

Complete Solution :
We have been provided that phthalic acid with resorcinol in the presence of concentrated sulphuric acid,
Phthalic acid is an aromatic dicarboxylic acid, with formula ${{C}_{6}}{{H}_{4}}{{(C{{O}_{2}}H)}_{2}}$ .It is an isomer of isophthalic acid and terephthalic acid. Resorcinol is an organic compound with the formula ${{C}_{6}}{{H}_{4}}{{(OH)}_{2}}$. It is one of three isomeric benzenediols.
- When these two compounds react together in presence of concentrated sulphuric acid,
Sulphuric acid acts as a dehydrating agent because a substance that absorbs moisture from its surroundings is called a dehydrating agent. Sulphuric acid readily protonated ${{H}_{2}}O$ leading to the formation of hydronium ions.
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- Phthalic acid reacts with resorcinol in the presence of concentrated ${{H}_{2}}S{{O}_{4}}$ to give fluorescein. Phthalic acid gets dehydrated on treatment with concentrated ${{H}_{2}}S{{O}_{4}}$ to form phthalic anhydride. Phthalic anhydride further reacts with resorcinol to form fluorescein.
- So, we can say that Phthalic anhydride reacts with resorcinol in the presence of concentrated ${{H}_{2}}S{{O}_{4}}$ to give fluorescein.
So, the correct answer is “Option D”.

Note: Note that Adolf Von Baeyer first synthesized fluorescein in the laboratory. He prepared it from the phallic anhydride and resorcinol in presence of zinc chloride through the Friedel-Crafts reaction. There is one more method for synthesis using methane sulphonic acid . This gives a high yield.
Sulphuric acid is not used as a drying agent ${{H}_{2}}S$ because it reacts with it to form sulphur.
$\text{ }{{H}_{2}}S{{O}_{4}}+{{H}_{2}}S\to 2{{H}_{2}}O+S{{O}_{2}}+S\text{ }$