Question

What is the action of following on benzoic acid?
( i ) \[N{H_2}\]/ $\Delta $
( ii ) Nitrating mixture
(iii ) ${P_2}{O_5}$/ $\Delta $

Answer 1

Hint:As we know that the electron releasing groups such as ,etc. tend to decrease the acid strength of benzoic acid while the electron withdrawing groups such as $ - Cl, - N{O_2},$etc., tend to increase the strength of benzoic acid. ${P_2}{O_5}$ is a dehydrating agent.

Complete step-by-step answer: Benzoic acids ( ${C_6}{H_5}COOH$ ) is an organic compound in which carboxyl group is attached to the benzene ring . This compound is either white or colourless solid formula. It’s preparation is easy to form in the laboratory. Benzoic acid is practically insoluble in water .
 ( i ) As we all know that when benzoic acid reacts with $ - N{H_3}$in presence of heat it results in the formation of ammonium benzoate. On heating ammonium benzoate a water molecule releases which results in the formation of benzamide.
${C_6}{H_5}COOH\,\,\,\xrightarrow[\Delta ]{{N{H_3}}}\,\,\,\,{C_6}{H_5}CON{H_{2\,}}\, + {H_2}O$
( ii ) As we all know that when benzoic acid reacts with a Nitrating mixture it results in a compound named as $3 - $Nitrobenzoic acid due to the electrophilic substitution reaction. As we know $ - COOH$ is a meta directing group that is why after niration $ - N{O_2}$ group attaches to the meta position of benzoic acid.
${C_6}{H_5}COOH\,\,\xrightarrow[{MIXTURE}]{{NITRATING}}\,\,\,{C_7}{H_5}N{O_4}$
( iii ) As we all know that ${P_2}{O_5}$ is a strong dehydrating agent . And because of its presence , when it reacts with benzoic acid it releases water molecules and results in the formation of BENZOIC ANHYDRIDE.
${C_6}{H_5}COOH\,\,\,\xrightarrow[{{P_2}{O_5}}]{\Delta }\,\,\,{({C_6}{H_5}CO)_2}O + {H_2}O$

Note:Benzoic acid is used in manufacture of perfumes, dyes , topical medications and insect repellents too. It is used only in well ventilated areas. It naturally occur in many plants and in animals. Ortho isomers of benzoic acid are the strongest of all the isomers.