Preparation of Haloalkanes and Haloarenes | JEE Notes

Four routes cover nearly every question: from alcohols, from hydrocarbons, by halogen exchange and from diazonium salts.

From alcohols

Reagent	Product	Why use it
HCl with anhydrous $Z n C l_{2}$ (Lucas reagent)	RCl	Works fast for tertiary alcohols. Reactivity: $3^{\circ} > 2^{\circ} > 1^{\circ}$ .
Concentrated HBr or HI	RBr, RI	The acid protonates OH so that water can leave.
$P C l_{5}$ , $P C l_{3}$	RCl	Phosphorus halides replace OH.
Red P with $B r_{2}$ or $I_{2}$	RBr, RI	Makes $P B r_{3}$ or $P I_{3}$ on the spot.
$S O C l_{2}$ in pyridine	RCl	$S O_{2}$ and HCl escape as gases, leaving a pure product.

R O H + S O C l_{2} \to R C l + S O_{2} ↑ + H C l ↑

Phenols do not work. The C-O bond of phenol has partial double-bond character, so OH cannot be replaced this way.

From hydrocarbons

Alkanes react by a free-radical chain: initiation splits $X_{2}$ with light or heat, propagation passes the radical along, and termination joins two radicals. The result is a mixture, so this is a poor route for one pure halide.

The radical chain in chlorination of methane

Initiation. A single-barbed arrow moves one electron. Homolysis gives two chlorine atoms, each with an unpaired electron.

Propagation, first step. The radical count stays the same: one goes in and one comes out.

Propagation, second step. The chlorine atom made here goes back to the first step, so one initiation can run many cycles.

Termination. Two radicals join and the chain stops. Traces of ethane are the fingerprint of this mechanism.

The product share depends on how many hydrogens of each type there are and on how reactive each type is. Morrison and Boyd give relative rates per hydrogen of about 5 : 3.8 : 1 for chlorine at room temperature ( $3^{\circ} : 2^{\circ} : 1^{\circ}$ ) and about 1600 : 82 : 1 for bromine at 127 °C. Chlorine depends on counting. Bromine depends on radical stability.

Try the simulator: Radical halogenation: who wins?

Worked example: counting products

2-Methylbutane is monochlorinated at 25 °C. Use $3^{\circ} : 2^{\circ} : 1^{\circ} = 5 : 3.8 : 1$ per hydrogen. What share is 2-chloro-3-methylbutane?

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There are four kinds of hydrogen. Multiply the count of each by its reactivity.

Hydrogens	Count × rate	Product	Share
Two $C H_{3}$ on C-2	$6 \times 1 = 6$	1-Chloro-2-methylbutane	27.8%
The CH on C-2	$1 \times 5 = 5$	2-Chloro-2-methylbutane	23.1%
The $C H_{2}$	$2 \times 3.8 = 7.6$	2-Chloro-3-methylbutane	35.2%
The end $C H_{3}$	$3 \times 1 = 3$	1-Chloro-3-methylbutane	13.9%

The total is 21.6, so the answer is $7.6 / 21.6 = 35.2 %$ . Note that no tertiary product wins here, even though tertiary hydrogens are the most reactive.

Alkenes add HX by Markovnikov's rule. HBr with peroxide adds the other way, and this works only for HBr. $B r_{2}$ in $C C l_{4}$ adds to give a vicinal dibromide, which is the usual test for unsaturation.

Allylic and benzylic positions are special, because the radical there is resonance-stabilised. N-Bromosuccinimide (NBS) brominates exactly these carbons, and heat with $C l_{2}$ chlorinates a side chain. The JEE Advanced syllabus names allylic and benzylic halogenation.

Halogen exchange

R C l + N a I \frac{d r y a c e t o n e}{} \to R I + N a C l ↓ (Finkelstein)

The equilibrium would normally sit part way. NaI dissolves in dry acetone, but NaCl and NaBr do not, so they precipitate and pull the reaction forward.

Watch the animation: Finkelstein: precipitation drives it

R B r + A g F \to R F + A g B r ↓ (Swarts)

Swarts uses metallic fluorides such as AgF, $H g_{2} F_{2}$ , $C o F_{2}$ or $S b F_{3}$ .

Haloarenes

Electrophilic substitution. Toluene with $C l_{2}$ and $F e C l_{3}$ in the dark gives ortho and para chlorotoluene. Heat or light with no Lewis acid gives side-chain chlorination instead. This choice is a favourite question.

Worked example: choosing conditions

You have toluene. Name the conditions for (a) benzyl chloride and (b) 4-chlorotoluene. (c) Which of these two gives phenylacetic acid through a Grignard reagent and $C O_{2}$ ?

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(a) $C l_{2}$ with heat or light and no Lewis acid: a radical attacks the side chain. (b) $C l_{2}$ with $F e C l_{3}$ in the dark: an electrophile attacks the ring, and para is separated from ortho. (c) Benzyl chloride, because $C_{6} H_{5} C H_{2} M g C l$ adds one carbon to give $C_{6} H_{5} C H_{2} C O O H$ . The other gives 4-methylbenzoic acid. Always count carbons.

Iodination is reversible, so it needs an oxidising agent such as $H N O_{3}$ or $H I O_{4}$ to remove HI. Fluorine is too reactive for direct use.

From diazonium salts.

Reagent on $A r N_{2}^{+}$	Product	Name
$C u_{2} C l_{2}$ /HCl, $C u_{2} B r_{2}$ /HBr	ArCl, ArBr	Sandmeyer
Cu/HCl, Cu/HBr	ArCl, ArBr	Gattermann
CuCN/KCN	ArCN	Sandmeyer only
KI	ArI	No copper needed
$H B F_{4}$ , then heat	ArF	Balz-Schiemann

Making haloalkanes and haloarenes

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