Question

Benzene can be obtained by heating either benzoic acid with X or phenol with Y. X and Y are ________ respectively.
A. zinc dust and soda lime
B. soda lime and zinc dust
C. zinc dust and sodium hydroxide
D. soda lime and copper

Answer 1

Phenol to benzene is a radical reaction. While benzoic acid to benzene is a decarboxylation reaction. This occurs at low temperature and high temperature.

Complete step by step answer:
$\xrightarrow{{\text{Y}}}$
This reaction can only be done by heating phenol with zinc dust. The product formed will be benzene and zinc oxide. When phenol reacts with zinc dust, it releases proton which forms phenoxide ion. The proton then accepts an electron from zinc and produces a hydrogen radical and zinc cation. When phenoxide ion is heated, it forms phenyl radical. This phenyl radical reacts with the hydrogen radical and forms benzene. The complete reaction is given below:
${{\text{C}}_6}{{\text{H}}_5}{\text{OH}} \to {{\text{C}}_6}{{\text{H}}_5}{{\text{O}}^ - } + {{\text{H}}^ + }$
${{\text{H}}^ + } + {\text{Zn}} \to {{\text{H}}^ \bullet } + {\text{Z}}{{\text{n}}^ + }$
${{\text{C}}_6}{{\text{H}}_5}{{\text{O}}^ - }\xrightarrow{\Delta }{{\text{C}}_6}{{\text{H}}_5}^ \bullet + {{\text{O}}^{ \bullet - }}$
${{\text{C}}_6}{{\text{H}}_5}^ \bullet + {{\text{H}}^ \bullet } \to {{\text{C}}_6}{{\text{H}}_6}$
${\text{Z}}{{\text{n}}^ + } + {{\text{O}}^ - } \to {\text{Z}}{{\text{n}}^{2 + }} + {{\text{O}}^{2 - }} \to {\text{ZnO}}$
Zinc dust oxidizes itself to ${\text{ZnO}}$ and reduces phenol to benzene.
Therefore Y is zinc dust.
Soda lime is the mixture of ${\text{NaOH}}$ and ${\text{CaO}}$. Reaction of benzoic acid with soda lime is called Oakwood reaction. Sodium hydroxide removes ${\text{C}}{{\text{O}}_2}$ in benzoic acid molecules which produce alkane, i.e benzene. And the ${\text{C}}{{\text{O}}_2}$ reacts with sodium hydroxide to give sodium carbonate. The function of ${\text{CaO}}$ is that it makes ${\text{NaOH}}$ less reactive. The reaction is given below:
$$ $+ {\text{NaOH}} + {\text{CaO}} \to$ $+ {\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3}$
Hence, X is soda lime and Y is zinc dust.
So, the correct answer is option B.

Additional information Since ${\text{C}}{{\text{O}}_2}$ is removed, it is called a decarboxylation reaction. Soda lime is also used for decarboxylation of sodium salts of aromatic compounds. This reaction is called the Duma reaction.

Note:
At lower temperatures, when soda lime is used, benzoic acid forms sodium benzoate and then forms benzene and sodium carbonate. At higher temperatures, it directly forms benzene. This reaction is similar to pyrolysis.