Haloarenes

When a mixture of an aryl halide and an alkyl halide is heated with sodium in dry ether, an alkylbenzene is formed (Wurtz–Fittig reaction). For example, bromobenzene and bromomethane give toluene:

$$\mathrm{C_6H_5Br + 2\,Na + CH_3Br \xrightarrow{\text{dry ether},\ \Delta} C_6H_5CH_3 + 2\,NaBr}$$

When benzene diazonium chloride is warmed with cuprous chloride dissolved in HCl, the diazo group is replaced by Cl and chlorobenzene is formed (Sandmeyer's reaction):

$$\mathrm{C_6H_5N_2^+Cl^- \xrightarrow{Cu_2Cl_2/HCl,\ \Delta} C_6H_5Cl + N_2\uparrow}$$

When benzene is treated with an alkyl halide in the presence of anhydrous $AlCl_3$, an alkyl group is substituted into the ring. With chloromethane, toluene is formed:

$$\mathrm{C_6H_6 + CH_3Cl \xrightarrow{\text{anhyd. }AlCl_3} C_6H_5CH_3 + HCl}$$

1. Preparation of chlorobenzene (any two)

(a) Direct chlorination of benzene. Benzene reacts with chlorine in the dark at room temperature, with a Lewis acid catalyst (Fe or $FeCl_3$):

$$\mathrm{C_6H_6 + Cl_2 \xrightarrow{FeCl_3,\ \text{dark, cold}} C_6H_5Cl + HCl}$$

(b) Sandmeyer's reaction. Benzene diazonium chloride is warmed with $Cu_2Cl_2$/HCl:

$$\mathrm{C_6H_5N_2Cl \xrightarrow{Cu_2Cl_2/HCl} C_6H_5Cl + N_2}$$

(Gattermann's reaction with Cu powder/HCl also gives chlorobenzene.)

2. Why chlorobenzene gives ortho and para products

The Cl atom has two opposing effects on the ring:

• –I effect: Cl is electronegative and pulls electrons out of the ring through the σ-bond. This lowers the electron density of the whole ring, so chlorobenzene reacts more slowly than benzene (Cl is deactivating).
• +R (+M) effect: a lone pair on Cl is delocalised into the ring. The resonance structures put a negative charge at the ortho and para positions only:

The real molecule is a hybrid of I–V, so the C–Cl bond has partial double-bond character and the ortho and para carbons carry extra electron density.

So the ortho and para positions have more electron density than the meta positions. The electrophile ($E^+$) attacks there, and the carbocation formed by o/p attack is extra-stabilised because Cl can share its lone pair with the positive centre. Since –I outweighs +R, Cl is deactivating but ortho-para directing. For example, nitration gives o- and p-chloronitrobenzene (para is major because of less steric hindrance):

$$\mathrm{C_6H_5Cl + HNO_3 \xrightarrow{\text{conc. }H_2SO_4} \textit{o-}ClC_6H_4NO_2 + \textit{p-}ClC_6H_4NO_2\ (\text{major}) + H_2O}$$

3. Action on chloral

Chlorobenzene is heated with chloral in the presence of conc. $H_2SO_4$. The insecticide DDT is formed:

$$\mathrm{2\,C_6H_5Cl + CCl_3CHO \xrightarrow{\text{conc. }H_2SO_4} (p\text{-}ClC_6H_4)_2CH{-}CCl_3 + H_2O}$$

Identification

Reactions

(i) A to B (Sandmeyer's reaction):

$$\mathrm{\underset{(A)}{C_6H_5N_2Cl} \xrightarrow{Cu_2Cl_2/HCl} \underset{(B)}{C_6H_5Cl} + N_2}$$

(ii) B to C. Chlorobenzene is reduced with Ni–Al alloy in alkali (NaOH):

$$\mathrm{\underset{(B)}{C_6H_5Cl} + 2[H] \xrightarrow{Ni\text{-}Al/NaOH} \underset{(C)}{C_6H_6} + HCl}$$

(iii) C to D (Friedel–Crafts alkylation):

$$\mathrm{\underset{(C)}{C_6H_6} + CH_3Cl \xrightarrow{\text{anhyd. }AlCl_3} \underset{(D)}{C_6H_5CH_3} + HCl}$$

(iv) B with chloral gives the insecticide DDT:

$$\mathrm{2\,C_6H_5Cl + CCl_3CHO \xrightarrow{\text{conc. }H_2SO_4} \underset{\text{DDT}}{(p\text{-}ClC_6H_4)_2CH{-}CCl_3} + H_2O}$$

Step 1: Phenol is heated with zinc dust, which reduces it to benzene:

$$\mathrm{C_6H_5OH + Zn \xrightarrow{\Delta} C_6H_6 + ZnO}$$

Step 2: Benzene is chlorinated with $Cl_2$ and anhydrous $FeCl_3$ in the dark:

$$\mathrm{C_6H_6 + Cl_2 \xrightarrow{FeCl_3} C_6H_5Cl + HCl}$$

Step 3: Chlorobenzene is heated with chloral and conc. $H_2SO_4$ to give DDT:

$$\mathrm{2\,C_6H_5Cl + CCl_3CHO \xrightarrow{\text{conc. }H_2SO_4} \underset{\text{DDT}}{(p\text{-}ClC_6H_4)_2CH{-}CCl_3} + H_2O}$$

Step 1: Diazotisation. Aniline is treated with sodium nitrite and HCl at 0–5 °C (273–278 K):

$$\mathrm{NaNO_2 + HCl \longrightarrow HNO_2 + NaCl}$$ $$\mathrm{C_6H_5NH_2 + HNO_2 + HCl \xrightarrow{0\text{–}5\,^\circ C} \underset{\text{benzene diazonium chloride}}{C_6H_5N_2Cl} + 2\,H_2O}$$

Step 2: Sandmeyer's reaction. The diazonium salt is warmed with cuprous chloride in HCl:

$$\mathrm{C_6H_5N_2Cl \xrightarrow{Cu_2Cl_2/HCl} C_6H_5Cl + N_2\uparrow}$$

Identification

Reactions and conditions

(i) Preparation of A (Sandmeyer's reaction). Aniline is diazotised with $NaNO_2$/HCl at 0–5 °C. The diazonium salt is then warmed with $Cu_2Cl_2$/HCl:

$$\mathrm{C_6H_5NH_2 \xrightarrow[0\text{–}5\,^\circ C]{NaNO_2/HCl} C_6H_5N_2Cl \xrightarrow{Cu_2Cl_2/HCl} \underset{(A)}{C_6H_5Cl} + N_2}$$

(ii) A to B. Reduction with Ni–Al alloy in NaOH:

$$\mathrm{\underset{(A)}{C_6H_5Cl} + 2[H] \xrightarrow{Ni\text{-}Al/NaOH} \underset{(B)}{C_6H_6} + HCl}$$

(iii) B to C. Friedel–Crafts alkylation:

$$\mathrm{\underset{(B)}{C_6H_6} + CH_3Cl \xrightarrow{\text{anhyd. }AlCl_3} \underset{(C)}{C_6H_5CH_3} + HCl}$$

(iv) C to D. Toluene is oxidised with acidified ceric oxide ($CeO_2/H_2SO_4$). This mild oxidising agent stops at the aldehyde stage:

$$\mathrm{\underset{(C)}{C_6H_5CH_3} + 2[O] \xrightarrow{CeO_2/H_2SO_4} \underset{(D)}{C_6H_5CHO} + H_2O}$$

A) i) Identification of (A) and (B)

Chlorobenzene is formed from (A) with $Cu_2Cl_2/H^+$ (Sandmeyer's reaction), so (A) = benzene diazonium chloride. Reduction of chlorobenzene with Ni–Al alloy and alkali gives (B) = benzene.

$$\mathrm{\underset{(A)}{C_6H_5N_2Cl} \xrightarrow{Cu_2Cl_2/HCl} C_6H_5Cl + N_2}$$ $$\mathrm{C_6H_5Cl + 2[H] \xrightarrow[\Delta]{Ni\text{-}Al/NaOH} \underset{(B)}{C_6H_6} + HCl}$$

ii) a) (A) boiled with water

Phenol is formed and nitrogen gas is given off:

$$\mathrm{C_6H_5N_2Cl + H_2O \xrightarrow{\Delta} C_6H_5OH + N_2 + HCl}$$

b) (A) treated with aniline

A coupling reaction takes place at the para position of aniline (weakly acidic medium, 0–5 °C). The product is the yellow dye p-aminoazobenzene:

$$\mathrm{C_6H_5N_2Cl + C_6H_5NH_2 \longrightarrow \underset{\text{p-aminoazobenzene}}{C_6H_5{-}N{=}N{-}C_6H_4{-}NH_2(p)} + HCl}$$

B) $C_6H_5Cl$ heated with chloral

With conc. $H_2SO_4$, two molecules of chlorobenzene condense with chloral to give the insecticide DDT (p,p'-dichlorodiphenyltrichloroethane):

$$\mathrm{2\,C_6H_5Cl + CCl_3CHO \xrightarrow{\text{conc. }H_2SO_4} (p\text{-}ClC_6H_4)_2CH{-}CCl_3 + H_2O}$$

Compound [A] is chlorobenzene, $C_6H_5Cl$.

a) Identification of B and C

[B] = DDT (p,p'-dichlorodiphenyltrichloroethane). Two molecules of chlorobenzene condense with chloral (2,2,2-trichloroethanal) at their para positions:

$$\mathrm{2\,C_6H_5Cl + CCl_3CHO \xrightarrow{\text{conc. }H_2SO_4} \underset{\text{DDT (B)}}{(p\text{-}ClC_6H_4)_2CH{-}CCl_3} + H_2O}$$

[C] = 4-chloroacetophenone (p-chloroacetophenone), with some ortho isomer. It is formed by Friedel–Crafts acylation with acetyl chloride:

$$\mathrm{C_6H_5Cl + CH_3COCl \xrightarrow{\text{anhyd. }AlCl_3} \underset{\text{p-chloroacetophenone (C, major)}}{p\text{-}ClC_6H_4COCH_3} + HCl}$$

b) Why chlorobenzene reacts with aq. NaOH less readily than chloroethane

Chloroethane is hydrolysed easily by aqueous NaOH. Chlorobenzene needs about 623 K and 300 atm (Dow's process), because:

1. Resonance: a lone pair on Cl is delocalised into the ring, so the C–Cl bond gets partial double-bond character. It becomes stronger and harder to break.

The real molecule is a hybrid of I–V, so the C–Cl bond has partial double-bond character and the ortho and para carbons carry extra electron density.

2. Hybridisation: in chlorobenzene the C–Cl carbon is $sp^2$ hybridised (more s-character, more electronegative). The bond is shorter (169 pm, against 177 pm in $CH_3Cl$ and about 178 pm in $C_2H_5Cl$) and stronger.
3. Unstable phenyl cation: the $C_6H_5^+$ ion is not stabilised by resonance, so the $S_N1$ route does not work.
4. Repulsion and blocking: the electron-rich π-cloud of the ring repels the incoming $OH^-$, and the ring blocks backside attack, so the $S_N2$ route does not work either.

$$\mathrm{C_6H_5Cl + NaOH \xrightarrow{623\,K,\ 300\,atm} C_6H_5OH + NaCl}$$

c) A from benzene diazonium chloride (Sandmeyer's reaction)

$$\mathrm{C_6H_5N_2Cl \xrightarrow{Cu_2Cl_2/HCl} C_6H_5Cl + N_2}$$

When benzene diazonium chloride is warmed with copper powder and HCl, chlorobenzene is formed:

$$C_6H_5N_2^+Cl^- \xrightarrow{Cu\ powder,\ HCl} C_6H_5Cl + N_2$$