Write the mechanism for the following 1 friedel craft alkylation 2 friedel craft acylation 3 nitration of benzene

Here are the mechanisms for the reactions you requested:

1. Friedel-Crafts Alkylation

This reaction involves the alkylation of an aromatic ring with an alkyl halide using a strong Lewis acid catalyst, typically $\text{AlCl}_3$.

• Step 1: Formation of the carbocation electrophile. The alkyl halide reacts with the Lewis acid to form a carbocation.

  $$\text{R-Cl} + \text{AlCl}_3 \longrightarrow \text{R}^+ + \text{AlCl}_4^-$$

  (Note: Carbocation rearrangements can occur if a more stable carbocation can be formed.)

• Step 2: Electrophilic attack on the benzene ring. The carbocation attacks the electron-rich benzene ring, forming a sigma complex (arenium ion), which temporarily disrupts the aromaticity.

  $$\text{C}_6\text{H}_6 + \text{R}^+ \longrightarrow \begin{bmatrix} \text{H} \\ \text{R} \end{bmatrix}\text{C}_6\text{H}_5^+$$

  (The positive charge is delocalized over the ring carbons ortho and para to the point of attack.)

• Step 3: Deprotonation and restoration of aromaticity. A base (e.g., $\text{AlCl}_4^-$) removes a proton from the sigma complex, restoring the aromaticity of the ring and regenerating the $\text{AlCl}_3$ catalyst.

  $$\begin{bmatrix} \text{H} \\ \text{R} \end{bmatrix}\text{C}_6\text{H}_5^+ + \text{AlCl}_4^- \longrightarrow \text{C}_6\text{H}_5\text{R} + \text{HCl} + \text{AlCl}_3$$

2. Friedel-Crafts Acylation

This reaction involves the acylation of an aromatic ring with an acyl halide (or anhydride) using a strong Lewis acid catalyst, typically $\text{AlCl}_3$.

• Step 1: Formation of the acylium ion electrophile. The acyl halide reacts with the Lewis acid to form a resonance-stabilized acylium ion.

  $$\text{R-CO-Cl} + \text{AlCl}_3 \longrightarrow \text{R-C}\equiv\text{O}^+ + \text{AlCl}_4^-$$

• Step 2: Electrophilic attack on the benzene ring. The acylium ion attacks the electron-rich benzene ring, forming a sigma complex (arenium ion).

  $$\text{C}_6\text{H}_6 + \text{R-C}\equiv\text{O}^+ \longrightarrow \begin{bmatrix} \text{H} \\ \text{CO-R} \end{bmatrix}\text{C}_6\text{H}_5^+$$

• Step 3: Deprotonation and restoration of aromaticity. A base (e.g., $\text{AlCl}_4^-$) removes a proton from the sigma complex, restoring the aromaticity of the ring.

  $$\begin{bmatrix} \text{H} \\ \text{CO-R} \end{bmatrix}\text{C}_6\text{H}_5^+ + \text{AlCl}_4^- \longrightarrow \text{C}_6\text{H}_5\text{CO-R} \cdot \text{AlCl}_3 + \text{HCl}$$

  (The ketone product forms a complex with $\text{AlCl}_3$.)

• Step 4: Hydrolysis. Aqueous workup is required to release the ketone product from its complex with $\text{AlCl}_3$.

  $$\text{C}_6\text{H}_5\text{CO-R} \cdot \text{AlCl}_3 + \text{H}_2\text{O} \longrightarrow \text{C}_6\text{H}_5\text{CO-R} + \text{Al(OH)}_3 + \text{HCl}$$

3. Nitration of Benzene

This reaction involves the introduction of a nitro group ($\text{-NO}_2$) onto an aromatic ring using a mixture of concentrated nitric acid and sulfuric acid.

• Step 1: Formation of the nitronium ion electrophile. Nitric acid is protonated by sulfuric acid, and then water is eliminated to form the nitronium ion ($\text{NO}_2^+$), which is the active electrophile.

  $$\text{HNO}_3 + \text{H}_2\text{SO}_4 \rightleftharpoons \text{H}_2\text{NO}_3^+ + \text{HSO}_4^-$$ $$\text{H}_2\text{NO}_3^+ \longrightarrow \text{NO}_2^+ + \text{H}_2\text{O}$$ $$\text{H}_2\text{O} + \text{H}_2\text{SO}_4 \rightleftharpoons \text{H}_3\text{O}^+ + \text{HSO}_4^-$$

  Overall:

  $$\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \longrightarrow \text{NO}_2^+ + \text{H}_3\text{O}^+ + 2\text{HSO}_4^-$$

• Step 2: Electrophilic attack on the benzene ring. The nitronium ion attacks the electron-rich benzene ring, forming a sigma complex (arenium ion).

  $$\text{C}_6\text{H}_6 + \text{NO}_2^+ \longrightarrow \begin{bmatrix} \text{H} \\ \text{NO}_2 \end{bmatrix}\text{C}_6\text{H}_5^+$$

• Step 3: Deprotonation and restoration of aromaticity. A base (e.g., $\text{HSO}_4^-$ or $\text{H}_2\text{O}$) removes a proton from the sigma complex, restoring the aromaticity of the ring.

  $$\begin{bmatrix} \text{H} \\ \text{NO}_2 \end{bmatrix}\text{C}_6\text{H}_5^+ + \text{HSO}_4^- \longrightarrow \text{C}_6\text{H}_5\text{NO}_2 + \text{H}_2\text{SO}_4$$