Question

Write the steps involved in the process of conversion from ethanoic acid to benzene.

Answer 1

Reaction:

• The conversion of ethanoic acid to benzene involves nine steps.
• These reactions involve Wurtz reaction, bromination, and elimination reactions.
• Ethanoic acid is nothing but acetic acid.
• Ethanoic acid is a saturated straight chain compound.
• Benzene is an unsaturated cyclic compound.

Steps:

• The ethanoic acid is reacted with a base to form sodium acetate.

$$\begin{gathered} {\mathrm{CH}}_3\mathrm{COOH}\left(\mathrm{aq}\right)+\mathrm{NaOH}\left(\mathrm{aq}\right)\to {\mathrm{CH}}_3\mathrm{COONa}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) \\ (\mathrm{Ethanoic}\mathrm{acid})(\mathrm{Sodium}\mathrm{hydroxide})(\mathrm{Sodium}\mathrm{acetate})\left(\mathrm{water}\right) \end{gathered}$$

• Then the sodium acetate is heated with sodium hydroxide and calcium oxide to give methane as the main product and sodium carbonate.

$$\begin{gathered} {\mathrm{CH}}_3\mathrm{COONa}\left(\mathrm{aq}\right)\underset{∆}{\overset{\mathrm{NaOH}\&\mathrm{CaO}}{\to }}{\mathrm{CH}}_4\left(\mathrm{aq}\right)+{\mathrm{Na}}_2{\mathrm{CO}}_3\left(\mathrm{s}\right) \\ (\mathrm{Sodium}(\mathrm{Methane})(\mathrm{Sodium} \\ \mathrm{acetate})\mathrm{carbonate}) \end{gathered}$$

• The methane formed is heated with chlorine to give methyl chloride.

$$\begin{gathered} {\mathrm{CH}}_4\left(\mathrm{g}\right)+{\mathrm{Cl}}_2\left(\mathrm{l}\right)\stackrel{∆}{\to }{\mathrm{CH}}_3\mathrm{Cl}\left(\mathrm{aq}\right)+\mathrm{HCl}\left(\mathrm{g}\right) \\ \left(\mathrm{Methane}\right)\left(\mathrm{Chlorine}\right)(\mathrm{Methyl}(\mathrm{Hydrogen} \\ \mathrm{chloride})\mathrm{chloride}) \end{gathered}$$

• Now the methyl chloride formed reacts with another molecule of methyl chloride and sodium metal in dry ether to give ethane, this reaction is known as the Wurtz reaction.

$$\begin{gathered} {\mathrm{CH}}_3\mathrm{Cl}\left(\mathrm{aq}\right)+{\mathrm{CH}}_3\mathrm{Cl}\left(\mathrm{aq}\right)+2\mathrm{Na}\left(\mathrm{s}\right)\stackrel{\mathrm{Dry}\mathrm{ether}}{\to }{\mathrm{CH}}_3-{\mathrm{CH}}_3\left(\mathrm{g}\right)+2\mathrm{NaCl}\left(\mathrm{aq}\right) \\ (\mathrm{Methyl}\mathrm{chloride})(\mathrm{sodium}(\mathrm{Ethane})(\mathrm{sodium}\mathrm{chloride}) \\ \mathrm{metal}) \end{gathered}$$

• The ethane reacts with chlorine giving ethyl chloride.

$$\begin{gathered} {\mathrm{CH}}_3-{\mathrm{CH}}_3\left(\mathrm{g}\right)+{\mathrm{Cl}}_2\left(\mathrm{l}\right)\to {\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Cl}\left(\mathrm{aq}\right)+\mathrm{HCl}\left(\mathrm{g}\right) \\ \left(\mathrm{Ethane}\right)\left(\mathrm{Chlorine}\right)(\mathrm{Ethyl}\mathrm{chloride})(\mathrm{Hydrogen}\mathrm{chloride}) \end{gathered}$$

• The ethyl chloride on reaction with alcoholic potassium hydroxide gives ethene, water, and potassium chloride as products.

$$\begin{gathered} {\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Cl}\left(\mathrm{aq}\right)+\mathrm{alc}.\mathrm{KOH}\to {\mathrm{CH}}_2={\mathrm{CH}}_2\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}+\mathrm{KCl} \\ (\mathrm{Ethyl}(\mathrm{Potassium}\left(\mathrm{Ethene}\right)\left(\mathrm{water}\right)(\mathrm{Potassium} \\ \mathrm{chloride})\mathrm{hydroxide})\mathrm{chloride}) \end{gathered}$$

• Ethene on reaction with bromine in carbon tetrachloride solvent gives dibromo ethane.

$$\begin{gathered} {\mathrm{CH}}_2={\mathrm{CH}}_2\left(\mathrm{aq}\right)+{\mathrm{Br}}_2/{\mathrm{CCl}}_4\left(\mathrm{aq}\right)\to \mathrm{CHBr}-\mathrm{CHBr}\left(\mathrm{l}\right) \\ \left(\mathrm{Ethene}\right)(\mathrm{Bromine}/(\mathrm{dibromo}\mathrm{ethane}) \\ \mathrm{carbon}\mathrm{tetrachloride}) \end{gathered}$$

• The dibromo ethane on reaction with alcoholic potassium hydroxide gives ethyne as a product.

$$\begin{gathered} \mathrm{CHBr}-\mathrm{CHBr}\left(\mathrm{aq}\right)+\mathrm{KOH}\left(\mathrm{alc}\right)\to \mathrm{HC}\equiv \mathrm{CH}\left(\mathrm{g}\right)+\mathrm{KBr}\left(\mathrm{aq}\right)+\mathrm{HOBr}\left(\mathrm{l}\right) \\ (\mathrm{dibromo}\mathrm{ethane})(\mathrm{Potassium}(\mathrm{Ethyne})(\mathrm{Potassium}(\mathrm{Hypobromous} \\ \mathrm{hydroxide})\mathrm{bromide})\mathrm{acid}) \end{gathered}$$

• The ethyne when passed over red hot iron at $873\mathrm{K}$ gives benzene.

$$\begin{gathered} 3\mathrm{CH}\equiv \mathrm{CH}\left(\mathrm{aq}\right)\underset{873\mathrm{K}}{\overset{\mathrm{Red}\mathrm{hot}\mathrm{iron}}{\to }}{\mathrm{C}}_6{\mathrm{H}}_6\left(\mathrm{l}\right) \\ \left(\mathrm{Ethyne}\right)\left(\mathrm{Benzene}\right) \end{gathered}$$