Reactions of haloarenes

Why the halogen is hard to replace

You have seen that a haloalkane loses its halogen to a nucleophile with ease. Boil chlorobenzene with aqueous sodium hydroxide and nothing happens. Four reasons explain this.

1. Resonance. A lone pair on chlorine overlaps with the ring. Structures III, IV and V of the picture show a partial double bond between carbon and chlorine. The C-Cl bond is shorter, about $169\ \text{pm}$ against $178\ \text{pm}$ in chloromethane, and harder to break.
2. Hybridisation. The carbon is sp2. An sp2 carbon has more s character than an sp3 carbon, so it holds the shared electrons more tightly. This also shortens and strengthens the bond.
3. A phenyl cation is unstable. $S_{\text{N}}1$ would need a phenyl cation. Its empty orbital lies in the plane of the ring, and it cannot be stabilised by the ring's pi electrons.
4. Repulsion. The ring is rich in electrons, and it repels an incoming nucleophile. Backside attack, as in $S_{\text{N}}2$, is also blocked by the ring.

Vinylic halides, such as chloroethene, are unreactive for the same reasons.

When substitution does happen

Drastic conditions. The Dow process turns chlorobenzene into phenol. Chlorobenzene is heated with aqueous sodium hydroxide at $623\ \text{K}$ under $300\ \text{atm}$ to give sodium phenoxide. Acidification gives phenol.

$\text{C}_{6}\text{H}_{5}\text{Cl}\ +\ \text{NaOH}\ \xrightarrow{623\ \text{K},\ 300\ \text{atm}}\ \text{C}_{6}\text{H}_{5}\text{ONa}$, then $\text{C}_{6}\text{H}_{5}\text{ONa}\ +\ \text{HCl}\ \rightarrow \ \text{C}_{6}\text{H}_{5}\text{OH}\ +\ \text{NaCl}$.

Nitro groups in the ortho and para positions. A nitro group at the ortho or para position makes the replacement easier. 4-Nitrochlorobenzene reacts with aqueous sodium hydroxide at $443\ \text{K}$. 2,4-Dinitrochlorobenzene reacts at $368\ \text{K}$. 2,4,6-Trinitrochlorobenzene reacts with warm water.

The reaction has two steps, addition and then elimination. The nucleophile adds to the carbon that holds chlorine, and a negative intermediate forms. This is the slow step. The ortho or para nitro group takes the charge by resonance, as the second form in the picture shows, so the intermediate is stabilised. Then chloride leaves. A nitro group in the meta position cannot take the charge in this way, so it does not help.

Electrophilic substitution

Chlorine is a strange substituent. It withdraws electrons through the sigma bond, and this slows the reaction, so chlorobenzene reacts more slowly than benzene. It also has lone pairs that donate into the ring by resonance. This donation puts extra charge on the ortho and para positions. So the incoming group goes to the ortho and para positions. Halogens are deactivating but ortho-para directing.

The reactions of chlorobenzene follow the usual list:

• Halogenation: chlorine with $\text{FeCl}_{3}$ gives 1,2- and 1,4-dichlorobenzene, mostly the para one.
• Nitration: concentrated nitric and sulphuric acids give 2- and 4-nitrochlorobenzene.
• Sulphonation: concentrated sulphuric acid gives 4-chlorobenzenesulphonic acid, mainly.
• Friedel-Crafts alkylation and acylation: an alkyl or acyl chloride with $\text{AlCl}_{3}$ gives the ortho and para products, with para the main one.

Move the slider. Methyl is an activating ortho-para director. Chlorine is deactivating and ortho-para. The nitro group is deactivating and meta. Look at the readout for chlorine: one para position is about five times as reactive as one ortho position. The ortho product is held back by the size of the chlorine atom, so the para isomer is the major product. The percentages are typical values from the textbook of Morrison and Boyd.

Reactions with metals

The Wurtz reaction. Two molecules of a haloalkane, sodium and dry ether give an alkane with the carbon chains joined: $2\text{CH}_{3}\text{CH}_{2}\text{Br}\ +\ 2\text{Na}\ \xrightarrow{\text{dry}\ \text{ether}}\ \text{CH}_{3}\text{CH}_{2}\text{CH}_{2}\text{CH}_{3}\ +\ 2\text{NaBr}$.

The Wurtz-Fittig reaction. A haloarene and a haloalkane are joined. $\text{C}_{6}\text{H}_{5}\text{Br}\ +\ \text{CH}_{3}\text{Br}\ +\ 2\text{Na}\ \xrightarrow{\text{dry}\ \text{ether}}\ \text{C}_{6}\text{H}_{5}\text{CH}_{3}\ +\ 2\text{NaBr}$. If two different halides are mixed, three products are possible: two symmetrical and one mixed.

The Fittig reaction. Two molecules of a haloarene give a biaryl: $2\text{C}_{6}\text{H}_{5}\text{Br}\ +\ 2\text{Na}\ \xrightarrow{\text{dry}\ \text{ether}}\ \text{C}_{6}\text{H}_{5}\text{-}\text{C}_{6}\text{H}_{5}\ +\ 2\text{NaBr}$. This is biphenyl.

Grignard reagents. Magnesium turnings react with a haloalkane or haloarene in dry ether to give an organomagnesium halide, $\text{RMgX}$. It is the Grignard reagent.

$\text{R}\text{-}\text{X}\ +\ \text{Mg}\ \xrightarrow{\text{dry}\ \text{ether}}\ \text{RMgX}$

The reactivity of the halide runs from iodide, to bromide, to chloride. Fluorides do not form the reagent under normal conditions. The carbon of a Grignard reagent is partly negative, like a carbanion, so it is a strong base and a strong nucleophile.

• It must be kept dry. Water takes a proton from it at once and gives an alkane: $\text{CH}_{3}\text{MgBr}\ +\ \text{H}_{2}\text{O}\ \rightarrow \ \text{CH}_{4}\ +\ \text{Mg}(\text{OH})\text{Br}$. Any compound with an O-H or N-H bond does the same. This is why the ether must be free of water and alcohol.
• With carbon dioxide it gives a carboxylic acid, with one more carbon, after hydrolysis.
• With aldehydes and ketones it gives alcohols. You will meet this in the next chapters.

Reduction. A haloalkane is reduced to an alkane by zinc and HCl, by lithium aluminium hydride, or by hydrogen over nickel: $\text{R}\text{-}\text{X}\ +\ 2[\text{H}]\ \rightarrow \ \text{R}\text{-}\text{H}\ +\ \text{HX}$.

Try it in Python

The program takes the percentages of the three nitration products of three substituted benzenes. It divides each by the number of positions of its kind to compare the reactivity of one position.

Common mistakes

• Saying chlorine activates the ring because it is ortho-para directing. It deactivates. Reactivity and orientation are different questions.
• Expecting chlorobenzene to react like a haloalkane with sodium hydroxide. It needs $623\ \text{K}$ and $300\ \text{atm}$.
• Putting the nitro group in the meta position when you want to activate. It must be ortho or para to the halogen.
• Using a wet solvent for a Grignard reagent. The reagent is destroyed at once.

Remember these

• Aryl halides resist nucleophilic substitution: resonance, sp2 carbon, unstable phenyl cation and repulsion.
• The Dow process: chlorobenzene and NaOH at $623\ \text{K}$, $300\ \text{atm}$, then acid, gives phenol.
• Nitro groups at ortho and para positions help ($443\ \text{K}$ with one, $368\ \text{K}$ with two, warm water with three).
• Halogens deactivate the ring and direct to ortho and para. The para product usually wins.
• Wurtz: R-R. Wurtz-Fittig: Ar-R. Fittig: Ar-Ar. Grignard: $\text{RMgX}$ in dry ether, destroyed by water.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Give one reason why chlorobenzene does not react with aqueous NaOH at 373 K.

2. What is produced when chlorobenzene is heated with sodium hydroxide at 623 K and 300 atm, and the product is acidified?

3. What is the main product of nitrating chlorobenzene?

4. Why must a Grignard reagent be made in dry ether?

References

• NCERT, Chemistry Part II for Class XII, the chapter Haloalkanes and Haloarenes, on the reactions of haloarenes.
• Morrison and Boyd, Organic Chemistry, the chapters on electrophilic aromatic substitution (the orientation data) and on aryl halides.
• OpenStax, Organic Chemistry, Chapters 10 (Organohalides, Grignard reagents) and 16 (Chemistry of Benzene: Electrophilic Aromatic Substitution).