Fundamentholfundamenthol

Properties of Alkyl Halides (Halo Alkanes) & Aryl Halides (Halo Arenes)

ChemistryHaloalkanes And HaloarenesFor JEE aspirants

The properties of alkyl halides come from the polar C-X bond: the carbon carries a partial positive charge, so nucleophiles attack it and the halide leaves. Alkyl halides therefore undergo nucleophilic substitution ( or ), elimination to alkenes, and reactions with metals such as Na and Mg. Aryl halides are far less reactive because their C-X bond has partial double-bond character. The physical and chemical properties of alkyl halides and haloarenes are among the most tested areas of this chapter in JEE Main and NEET.

Key Formulas - Quick Reference
  1. Reactivity for the same R:
  2. : rate ; reactivity
  3. : rate ; reactivity
  4. Boiling point: ; falls with branching
  5. Dipole moment:
  6. , ; ,
  7. Aqueous KOH gives an alcohol (substitution); alcoholic KOH gives an alkene (elimination)
  8. Wurtz:
  9. Grignard:
  10. Chlorobenzene to phenol needs 623 K and 300 atm; - or - groups make it much easier

1. Nature of the C-X Bond

A halogen is more electronegative than carbon, so the C-X bond is polar: the carbon carries a partial positive charge () and the halogen a partial negative charge (). As the halogen gets larger from F to I, the bond gets longer and weaker.

BondBond length (pm)Bond enthalpy (kJ/mol)Dipole moment of (D)
C-F1394521.847
C-Cl1783511.860
C-Br1932931.830
C-I2142341.636
Exam Trick

has the highest dipole moment, not . Dipole moment is charge multiplied by distance. Fluorine has the larger charge, but the C-F bond is so short that the product ends up slightly smaller than for C-Cl.

2. Physical Properties

2.1 Haloalkanes

  • Boiling point and halogen: for the same alkyl group, RI > RBr > RCl > RF. A larger, heavier halogen gives stronger van der Waals forces.
  • Boiling point and size: for the same halogen, the boiling point rises with molecular mass. boils lower than .
  • Boiling point and number of halogens: (313 K) < (334 K) < (350 K).
  • Boiling point and branching: among isomers, more branching gives a more compact, nearly spherical molecule with less surface contact, so the boiling point falls.
  • Solubility: only very slightly soluble in water, because the energy released by new attractions to water is less than the energy needed to break water's hydrogen bonds. They dissolve readily in organic solvents.
  • Density: bromo, iodo and polychloro compounds are denser than water. Density rises with the number and atomic mass of the halogen atoms.
Isomer of StructureBoiling point (approx.)
1-Chlorobutane351 K
2-Chlorobutane341 K
1-Chloro-2-methylpropane342 K
2-Chloro-2-methylpropane324 K

The straight-chain isomer boils highest and the most branched (tertiary) isomer lowest.

2.2 Haloarenes

  • Boiling point: iodobenzene > bromobenzene > chlorobenzene.
  • Solubility: insoluble in water, acids and bases, because they cannot form hydrogen bonds with water. They are soluble in organic solvents.
  • Density: all are heavier than water; iodo > bromo > chloro.
  • Melting points of dichlorobenzenes: the three isomers boil at nearly the same temperature, but the para isomer melts 70-100 K higher than the ortho and meta isomers.
Property1,2-Dichlorobenzene1,3-Dichlorobenzene1,4-Dichlorobenzene
Boiling point453 K446 K448 K
Melting point256 K249 K325 K
Why the para isomer melts higher: it is symmetrical, so its molecules pack more closely in the crystal lattice. More energy is needed to break this lattice.

3. Chemical Properties: Overview

Haloalkanes are a highly reactive class of aliphatic compounds because of the polar C-X bond. Their reactions involve breaking the C-X bond, so the weaker the bond, the faster the reaction. For a given alkyl group:

The chemical reactions of haloalkanes are of four types:

  1. Nucleophilic substitution reactions
  2. Elimination (dehydrohalogenation) reactions
  3. Reactions with metals
  4. Reduction reactions

4. Nucleophilic Substitution Reactions

A nucleophile (an electron-rich species) attacks the electron-deficient carbon and the halide ion leaves:

The reaction can follow either of two mechanisms, or .

4.1 The mechanism (unimolecular)

The rate depends only on the concentration of the alkyl halide: rate . The reaction takes place in two steps:

  1. Slow step: the alkyl halide ionises into a carbocation and a halide ion. This step decides the rate.
  2. Fast step: the carbocation immediately combines with the nucleophile.
SN1 mechanism of tert-butyl bromide Step 1: tert-butyl bromide slowly ionises into a planar tertiary carbocation and a bromide ion; this is the rate-determining step, so the rate depends only on the alkyl halide. Step 2: a nucleophile quickly attacks the flat carbocation from either face. A side panel ranks carbocation stability: tertiary, then secondary, then primary, then methyl. Step 1: ionisation (slow) H3C C Br CH3 CH3 2-bromo-2-methylpropane slow RDS C H3C CH3 CH3 + planar sp2 carbocation + Br− Step 2: attack by Nu− (fast) Nu− from either face H3C C Nu CH3 CH3 RATE LAW rate = k[R-X] first order (unimolecular) CATION STABILITY 3° (CH3)3C+ 2° (CH3)2CH+ 1° CH3CH2+ methyl CH3+ more alkyl groups: +I and hyperconjugation
Figure 1: mechanism. The slow step forms a carbocation, so the more stable the carbocation, the faster the reaction ().

Since the slow step forms a carbocation, reactivity follows carbocation stability: . Allylic and benzylic halides are also very reactive by , because their carbocations are stabilised by resonance.

4.2 The mechanism (bimolecular)

The rate depends on the concentrations of both the alkyl halide and the nucleophile: rate . The nucleophile attacks from the side opposite the halogen (backside attack). The C-Nu bond forms while the C-X bond breaks, in a single step, through a transition state in which carbon is partly bonded to both.

SN2 mechanism with backside attack and Walden inversion Hydroxide ion attacks (R)-2-bromobutane from the side opposite the bromine atom. In one concerted step a trigonal bipyramidal transition state forms with partial bonds to both OH and Br, then bromide leaves and the carbon inverts like an umbrella to give (S)-butan-2-ol. Br H H3C C2H5 (R)-2-bromobutane δ+ δ− HO − one step concerted HO δ− Br δ− H H3C C2H5 ‡ 5 groups on C, 3 in one plane Transition state Backside attack Br− leaves HO H CH3 C2H5 (S)-butan-2-ol WALDEN INVERSION wind umbrella turns inside out
Figure 2: mechanism. The nucleophile enters at to the leaving group, so the configuration is inverted (Walden inversion).

The three other groups on carbon flip over like an umbrella in the wind, so the configuration is inverted (Walden inversion). Bulky groups block the backside attack, so the order of reactivity is .

4.3 Factors that decide or

Factor
Rate law, first order, second order
StepsTwo, via a carbocationOne, via a transition state
Nature of alkyl halide (plus allylic, benzylic)
NucleophileWeak nucleophile, low concentrationStrong nucleophile, high concentration
SolventPolar protic (water, alcohols)Polar aprotic (acetone, DMF, DMSO)
StereochemistryMostly racemisationComplete inversion
RearrangementPossible (carbocation)Not possible

4.4 Stereochemistry of substitution

  • gives inversion. An optically active alkyl halide gives a product with the opposite configuration, as in Figure 2, where (R)-2-bromobutane gives (S)-butan-2-ol.
  • gives racemisation. The carbocation is planar, so the nucleophile can attack from either face with nearly equal chance. An optically active substrate gives a nearly 50:50 mixture of both configurations, which is optically inactive.
JEE Advanced

Allylic and benzylic halides react fast by both mechanisms: their carbocations are resonance-stabilised (), and the adjacent system also stabilises the transition state. Vinylic and aryl halides react by neither under normal conditions. Neopentyl halides are primary but very slow in , because the bulky tert-butyl group blocks the backside.

4.5 Important nucleophilic substitution reactions

No.ReagentProductProduct type
1Aqueous KOH or moist Alcohol
2Sodium alkoxide, Ether (Williamson synthesis)
3Alcoholic KCNNitrile (alkyl cyanide), major
4Alcoholic AgCNIsocyanide (carbylamine), major
5Alcoholic , sealed tubePrimary amine (then 2°, 3° amines)
6 in ethanolNitroalkane
7Alkyl nitrite
8Sodium mercaptide, Thioether
9NaSH or KSHThiol (thioalcohol)
10Sodium alkynide, Higher alkyne
11Silver carboxylate, Ester
12 in dry ether (hydride ion)Alkane

When the alkyl halide is in excess, the amine formed reacts further, replacing the remaining H atoms on nitrogen to give secondary and tertiary amines (and finally a quaternary ammonium salt).

Ambident nucleophiles: cyanide and nitrite ions can attack through two different atoms. KCN and are ionic, so the free ion attacks through C (giving a nitrile) or O (giving an alkyl nitrite). AgCN and are mainly covalent, so the carbon or oxygen is tied up with silver and the lone pair on nitrogen attacks instead, giving an isocyanide or a nitroalkane.
Exam Trick

"Silver switches the atom." KCN gives , AgCN gives . gives , gives .

5. Elimination Reactions (Dehydrohalogenation)

When haloalkanes are heated with alcoholic KOH, a hydrogen atom from the -carbon and the halogen from the -carbon are removed, and an alkene forms. This is called -elimination.

Dehydrohalogenation: E2 mechanism and Saytzeff rule Top: hydroxide removes a hydrogen from the beta carbon of bromoethane while the carbon-hydrogen electrons form the new double bond and bromide leaves, all in one step, giving ethene. Bottom: 2-bromobutane with alcoholic KOH gives but-2-ene as the major product (80 percent) and but-1-ene as the minor product (20 percent). E2: one step, base removes β-H C C H H H H H Br β α HO − alc. KOH heat H2C CH2 + H2O + Br− Saytzeff rule: more substituted alkene H3C CH2 CH CH3 Br alc. KOH H3C CH CH CH3 but-2-ene, 80% H3C CH2 CH CH2 but-1-ene, 20%
Figure 3: -elimination with alcoholic KOH. The H is lost from a -carbon, and the more substituted alkene is the major product (Saytzeff rule).
  • Order for alkyl groups: tertiary > secondary > primary. Tertiary halides give the most substituted alkenes, which are the most stable and form fastest.
  • Order for halogens: RI > RBr > RCl.
  • Saytzeff (Zaitsev) rule: when elimination can happen in two directions, the major product is the more substituted alkene (the one with fewer H atoms on the double-bond carbons). 2-Bromobutane gives but-2-ene (80%) and but-1-ene (20%).
Exam Trick

Aqueous for alcohol, alcoholic for alkene. Aqueous KOH mainly substitutes (OH replaces X). Alcoholic KOH mainly eliminates (gives an alkene). With a tertiary halide, a strong base such as an alkoxide also gives mainly the alkene.

6. Reactions with Metals

6.1 Wurtz reaction (sodium)

Haloalkanes react with sodium in dry ether to give alkanes with double the number of carbon atoms:

6.2 Grignard reagents (magnesium)

Haloalkanes react with magnesium in dry ether to form alkylmagnesium halides, called Grignard reagents:

Grignard reagents are organometallic compounds (they contain a carbon-metal bond). The C-Mg bond is highly polar, with carbon carrying a partial negative charge, so they are very reactive. Any proton donor (water, alcohols, amines) converts them to hydrocarbons:

This is why Grignard reagents must be prepared and used under strictly anhydrous conditions.

6.3 Alkyllithium (lithium)

Alkyllithium compounds behave as strong bases and are used like Grignard reagents.

6.4 Tetraethyl lead (sodium-lead alloy)

Tetraethyl lead (TEL) was once used as an anti-knock additive in petrol.

7. Reduction of Haloalkanes

Haloalkanes can be reduced to alkanes in three common ways.

(a) Hydrogen with nickel:

(b) Zinc-copper couple in alcohol, which produces nascent hydrogen:

(c) Lithium aluminium hydride in dry ether, where a hydride ion replaces X (reaction 12 in the table of Section 4.5).

8. Why Aryl Halides Are Less Reactive Than Alkyl Halides

Aryl halides are much less reactive than haloalkanes towards nucleophilic substitution. The main reasons are:

  1. Resonance: a lone pair on the halogen is in conjugation with the electrons of the ring. In three of the resonance structures (II, III, IV below) the C-X bond is a double bond, so the real C-X bond has partial double-bond character. It is shorter and stronger than in alkyl halides and is hard to break.
  2. Hybridisation: the carbon holding X is in haloarenes but in haloalkanes. An orbital has more s-character, holds electrons closer to the nucleus, and gives a shorter, stronger bond. The C-Cl bond length is 169 pm in chlorobenzene against 177 pm in chloromethane.
  3. Lower polarity: the carbon is more electronegative than an carbon, so the C-X bond in haloarenes is less polar, and the carbon is a weaker target for nucleophiles.
  4. Unstable phenyl cation: an route would need a phenyl cation, which resonance cannot stabilise.
  5. Repulsion: the electron-rich ring repels an approaching electron-rich nucleophile, and the ring blocks backside attack.
Resonance in chlorobenzene Five resonance contributors of chlorobenzene in two rows. Structures I and V are the two Kekule forms. Curly arrows show a chlorine lone pair moving into the ring in I, then the negative charge moving from the ortho carbon in II to the para carbon in III and the other ortho carbon in IV, and back onto chlorine to give V. In II, III and IV the C-Cl bond is a double bond and chlorine carries a positive charge. Cl Cl Cl Cl Cl (I) (II) (III) (IV) (V) II, III, IV: C=Cl double bond, negative charge at ortho and para carbons + − + − + −
Figure 4: Resonance in chlorobenzene gives the C-Cl bond partial double-bond character, so it is shorter, stronger and harder to break than in alkyl halides.

9. Reactions of Aryl Halides

9.1 Nucleophilic substitution (drastic conditions)

Replacement by OH: chlorobenzene is heated with aqueous NaOH at 623 K and 300 atm (Dow process). Sodium phenoxide forms, and acidification gives phenol.

Replacement by CN: bromobenzene heated with anhydrous CuCN in pyridine or DMF at 470 K gives cyanobenzene (phenyl cyanide).

Replacement by : chlorobenzene reacts with aqueous ammonia in the presence of at high temperature and pressure to give aniline.

9.2 Effect of nitro groups (activated substitution)

An electron-withdrawing group such as at the ortho or para position makes substitution much easier. The nucleophile adds first, and the negative charge of the intermediate (Meisenheimer complex) spreads onto the oxygen atoms of the nitro group. A nitro group at the meta position cannot do this.

Nucleophilic aromatic substitution by addition-elimination Hydroxide ion adds to the carbon bearing chlorine in 1-chloro-4-nitrobenzene, forming a resonance-stabilised Meisenheimer complex whose negative charge spreads onto the oxygen atoms of the para nitro group. Loss of chloride restores aromaticity and gives 4-nitrophenol. Cl NO2 HO − Step 1: addition (slow) slow RDS HO Cl N O O − + − sp3 carbon (ring aromaticity lost) Meisenheimer complex resonance HO Cl N O O + − − charge reaches O only from o- or p-NO2 fast Cl− leaves Step 2: elimination HO NO2 4-nitrophenol (after H3O+)
Figure 5: (addition-elimination). An group at ortho or para stabilises the carbanion, so the reaction needs much milder conditions than for chlorobenzene (443 K instead of 623 K).
SubstrateConditions with NaOH, then Product
Chlorobenzene623 K, 300 atmPhenol
1-Chloro-4-nitrobenzene443 K4-Nitrophenol
1-Chloro-2,4-dinitrobenzene368 K2,4-Dinitrophenol
1-Chloro-2,4,6-trinitrobenzeneWarm water, 328 K2,4,6-Trinitrophenol (picric acid)

The more nitro groups at ortho and para positions, the milder the conditions needed.

9.3 Reactions with metals

Magnesium: aryl bromides and iodides form Grignard reagents in dry ether.

Fittig reaction: aryl halides react with sodium in dry ether, and two aryl groups join.

Wurtz-Fittig reaction: an aryl halide and an alkyl halide react together with sodium in dry ether to give an alkylbenzene.

Lithium: aryl halides form aryllithium compounds, which behave like Grignard reagents.

Copper powder (Ullmann reaction): iodobenzene heated with copper powder in a sealed tube gives biphenyl.

9.4 Reduction

An aryl halide is reduced to the parent hydrocarbon by nickel-aluminium alloy in the presence of alkali.

9.5 Electrophilic substitution in the ring

Aryl halides undergo the usual electrophilic substitution reactions of benzene, but two effects of the halogen act together:

  • Deactivating: the strong (electron-withdrawing inductive) effect of the halogen lowers the electron density of the whole ring, so haloarenes react more slowly than benzene.
  • Ortho-para directing: the (resonance) effect raises electron density at the ortho and para positions (see structures II, III and IV in Figure 4), so the new group enters there. The para product is usually major because the ortho position is crowded.
ReactionReagentMinor product (ortho)Major product (para)
Halogenation, Fe or anhyd. 1,2-Dichlorobenzene1,4-Dichlorobenzene
Nitrationconc. + conc. 1-Chloro-2-nitrobenzene1-Chloro-4-nitrobenzene
Sulphonationconc. , heat2-Chlorobenzenesulphonic acid4-Chlorobenzenesulphonic acid
Friedel-Crafts alkylation, anhyd. 1-Chloro-2-methylbenzene1-Chloro-4-methylbenzene
Friedel-Crafts acylation, anhyd. 2-Chloroacetophenone4-Chloroacetophenone

10. Solved Examples

Solved Example 1
Chloroform, , is polar, yet it is not soluble in water. Explain.
Solution:

Chloroform molecules cannot form hydrogen bonds with water. To dissolve, chloroform would have to break the strong hydrogen bonds between water molecules, and the weak attractions it forms with water do not release enough energy to pay for this. So it stays insoluble in spite of its polarity.

Solved Example 2
Arrange in increasing order of density: , , , .
Solution:

Density rises as more heavy chlorine atoms replace hydrogen: < < < .

Solved Example 3
Why are the melting and boiling points of alkyl halides higher than those of the corresponding alkanes?
Solution:

Alkyl halides have a higher molecular mass and a polar C-X bond. Their molecules are held by stronger van der Waals forces and dipole-dipole attractions, so more energy is needed to separate them. For the same alkyl group, the order is RI > RBr > RCl > RF.

Solved Example 4
Why are alkyl halides very reactive?
Solution:

The C-X bond is polar, so the carbon carries a partial positive charge and is easily attacked by nucleophiles. Halide ions are weak bases, which makes them good leaving groups, and the C-X bond (especially C-Br and C-I) is weaker than C-H or C-C bonds. Together these make the C-X bond easy to break.

Solved Example 5
Which alkyl halide is the most reactive?
(A)
(B)
(C)
(D)
Solution:

Answer: (D). The C-I bond is the weakest (234 kJ/mol) and iodide is the best leaving group, so ethyl iodide reacts fastest.

Solved Example 6
Which alkyl halide is hydrolysed by the mechanism?
(A)
(B)
(C)
(D)
Solution:

Answer: (B). Methyl bromide has no bulky groups to block backside attack and cannot form a stable carbocation, so it reacts only by . Benzyl and allyl bromides can also react by , and tert-butyl bromide reacts by .

Solved Example 7
Which compound reacts faster with HCl? (i) or (ii) or
Solution:

The faster reaction is the one that forms the more stable carbocation.

(i) 2-Methylpropene, . Protonation gives a 3° carbocation, while propene gives only a 2° carbocation.

(ii) 2-Methylbuta-1,3-diene, . Protonation gives a carbocation that is both 3° and allylic, so it is further stabilised by resonance. 2-Methylbut-1-ene gives a 3° carbocation without that extra resonance.

Solved Example 8
The order of reactivity of alkyl halides towards elimination is:
(A) 3° > 2° > 1°
(B) 2° > 1° > 3°
(C) 3° > 1° > 2°
(D) 1° > 2° > 3°
Solution:

Answer: (A). Tertiary halides give the most substituted, most stable alkenes, which form fastest.

Solved Example 9
In the sequence ethylamine X Y Z, the end product Z is:
(A) methylamine
(B) acetamide
(C) ethylamine
(D) propylamine
Solution:

X is ethanol and Y is chloroethane. Ammonia replaces Cl by , so Z is ethylamine (C).

Solved Example 10
Why should Grignard reagents be prepared under anhydrous conditions?
Solution:

Grignard reagents react with water, which is a proton donor, and are decomposed to hydrocarbons:

Even traces of moisture destroy the reagent, so dry ether and dry apparatus are used.

Solved Example 11
Predict the major product when is treated with .
Solution:

is a very strong base. It removes HBr from the vinylic bromide (dehydrohalogenation), creating a triple bond:

The product is diphenylacetylene (1,2-diphenylethyne).

Solved Example 12
Alkyl iodides darken on standing. Why?
Solution:

The C-I bond is the weakest carbon-halogen bond. In light, alkyl iodides slowly decompose and release free iodine. The iodine dissolves in the remaining alkyl iodide and gives it a brown colour. Alkyl iodides are therefore stored in dark bottles.

Solved Example 13
Benzyl chloride is more reactive than chlorobenzene towards nucleophilic substitution. Explain.
Solution:

In benzyl chloride, , chlorine is on an carbon, so its lone pairs cannot conjugate with the ring and the C-Cl bond is an ordinary single bond. Also, loss of gives a benzyl carbocation that is stabilised by resonance. In chlorobenzene the lone pairs on chlorine are in conjugation with the ring, so the C-Cl bond has partial double-bond character and is hard to break. So benzyl chloride is far more reactive.

Solved Example 14
It is difficult to remove the halogen from an aromatic ring by attack of KOH. Why?
Solution:

Because of resonance (Figure 4), the C-X bond in haloarenes has partial double-bond character and is strong. The Cl of chlorobenzene is replaced by OH only under very drastic conditions, about 623 K and 300 atm.

Solved Example 15
1-Chloro-2,4-dinitrobenzene reacts with dilute NaOH to give sodium 2,4-dinitrophenoxide. This transformation proceeds through:
(A) electrophilic addition
(B) a benzyne intermediate
(C) activated nucleophilic substitution
(D) an oxirane
Solution:

Answer: (C). The two nitro groups at the ortho and para positions withdraw electrons and stabilise the negatively charged intermediate, so the hydroxide ion can replace Cl under mild conditions.

Solved Example 16
X. The compound X is:
(A) phenol
(B) benzene
(C) o- and p-chlorophenol
(D) benzol
Solution:

Answer: (B). Nickel-aluminium alloy in alkali reduces the aryl halide to the parent hydrocarbon, benzene.

Solved Example 17
Which is more reactive towards nucleophilic substitution: 1-chloro-2-nitrobenzene or 1-chloro-2,4,6-trinitrobenzene?
Solution:

1-Chloro-2,4,6-trinitrobenzene. It has three nitro groups at the ortho and para positions, all of which help spread the negative charge of the intermediate. It reacts even with warm water.

Solved Example 18
Give the principal organic products: (i) with / Fe (ii) 4-iodotoluene heated with copper
Solution:

(i) The group has a strong effect, so it is deactivating and meta-directing. The product is 1-chloro-3-(trichloromethyl)benzene.

(ii) Ullmann reaction: two aryl groups join.

The product is 4,4'-dimethylbiphenyl.

Solved Example 19
Arrange in decreasing order of reactivity: , , , .
Solution:

is fastest where the backside of the carbon is least crowded: > > > .

Practice Questions
  1. p-Dichlorobenzene is a solid at room temperature, while the o- and m-isomers are liquids. Why?Answer: the symmetrical para isomer packs more closely in the crystal, so its melting point (325 K) is much higher.
  2. Why are chloro compounds, rather than bromo compounds, used as industrial solvents?Answer: chloro compounds are cheaper, and their lower boiling points make them easier to evaporate and recover.
  3. Ethyl chloride is a gas, whereas ethyl iodide is a liquid at room temperature. Why?Answer: ethyl iodide has a much higher molecular mass and a larger, more polarisable iodine atom, so its van der Waals forces are stronger (boiling points about 285 K and 345 K).
  4. Arrange in increasing order of boiling point: bromobenzene, chlorobenzene, iodobenzene.Answer: chlorobenzene < bromobenzene < iodobenzene
  5. Why are haloarenes insoluble in water but soluble in benzene?Answer: they cannot form hydrogen bonds with water, but their intermolecular forces are similar to those of benzene ("like dissolves like").
  6. Vinyl chloride does not undergo nucleophilic substitution, but allyl chloride does. Explain.Answer: in vinyl chloride the C-Cl bond has partial double-bond character and the carbon is ; in allyl chloride the C-Cl bond is normal, and the allyl carbocation is resonance-stabilised.
  7. Give the products: (a) with (b) with .Answer: (a) mainly elimination, 2-methylpropene ; (b) substitution, tert-butyl methyl ether .
  8. with alcoholic KCN gives a mixture of isomeric products. Explain.Answer: loss of Cl gives an allylic carbocation whose positive charge is shared by C1 and C3, so cyanide attacks both, giving and .
  9. Arrange in increasing order of reactivity towards nucleophilic substitution: , , .Answer: < <
  10. Why does neopentyl bromide undergo nucleophilic substitution very slowly?Answer: the bulky tert-butyl group blocks backside attack (), and as a primary halide it cannot form a stable carbocation ().
  11. Benzene X Y Z. Identify X, Y and Z.Answer: X = chlorobenzene, Y = benzonitrile , Z = benzoic acid (acid hydrolysis of the nitrile).
  12. How would you distinguish chlorobenzene from hexyl chloride?Answer: boil with aqueous NaOH, acidify with , add . Hexyl chloride gives a white precipitate of AgCl; chlorobenzene does not.
  13. How would you distinguish from ?Answer: bromine in is decolourised by allyl bromide (it has a C=C) but not by 1-bromopropane.
  14. How would you distinguish p-bromobenzyl chloride from p-chlorobenzyl bromide?Answer: boil with aqueous NaOH, acidify with , add . Only the side-chain halogen is released: p-bromobenzyl chloride gives white AgCl, p-chlorobenzyl bromide gives pale yellow AgBr.
  15. X Y Z. Identify X, Y and Z.Answer: X = aniline, Y = benzenediazonium chloride, Z = iodobenzene.
  16. How will you obtain 2,6-dinitrophenol from chlorobenzene?Answer: sulphonate (conc. ) to block the para position, nitrate to put two groups ortho to Cl, replace Cl by OH with aqueous NaOH (activated substitution), then remove the group by heating with dilute acid.

Common Mistakes to Avoid

Watch out
  • Swapping the stereochemistry: gives inversion, gives racemisation.
  • Assuming has the largest dipole moment. does.
  • Using aqueous KOH when an alkene is wanted. Elimination needs alcoholic KOH.
  • Writing Williamson synthesis with NaOH. It needs a sodium alkoxide, and a primary alkyl halide; a tertiary halide gives an alkene instead.
  • Mixing up KCN / AgCN (nitrile / isocyanide) and / (nitrite / nitro).
  • Treating neopentyl halides as fast in because they are primary. Steric hindrance makes them very slow.
  • Thinking halogens are activating because they are ortho-para directing. They deactivate the ring.
  • Expecting vinyl or aryl halides to give normal or reactions.
  • Forgetting that a meta nitro group does not activate an aryl halide towards substitution.

Frequently Asked Questions

What is the difference between SN1 and SN2 reactions?

happens in two steps through a carbocation; its rate depends only on the alkyl halide, it is fastest for tertiary halides and gives racemisation. happens in one step with backside attack; its rate depends on both the halide and the nucleophile, it is fastest for methyl and primary halides and gives inversion.

Why are aryl halides less reactive than alkyl halides towards nucleophilic substitution?

In aryl halides the halogen lone pair is in resonance with the ring, giving the C-X bond partial double-bond character. The carbon is also hybridised, making the bond shorter, stronger and less polar. In addition, the phenyl cation is unstable and the electron-rich ring repels nucleophiles.

Why does KCN give alkyl cyanides but AgCN gives isocyanides?

The cyanide ion can attack through carbon or nitrogen. KCN is ionic, so the free cyanide ion attacks through carbon and forms a stronger C-C bond, giving a nitrile. AgCN is mainly covalent, with carbon bonded to silver, so the nitrogen lone pair attacks and an isocyanide forms.

What is Saytzeff's rule?

Saytzeff's (Zaitsev's) rule says that in dehydrohalogenation, when two alkenes can form, the major product is the more substituted alkene, which has fewer hydrogen atoms on the double-bond carbons. For example, 2-bromobutane with alcoholic KOH gives about 80 percent but-2-ene and 20 percent but-1-ene.

Why are haloarenes ortho-para directing but deactivating?

The halogen's strong electron-withdrawing inductive effect lowers the electron density of the whole ring, so haloarenes react more slowly than benzene. Its resonance effect, however, increases electron density at the ortho and para positions, so electrophiles attack there. The para product is usually major because of steric hindrance at ortho.

What is the Wurtz-Fittig reaction?

In the Wurtz-Fittig reaction, an aryl halide and an alkyl halide are treated with sodium in dry ether, and the aryl and alkyl groups join to form an alkylbenzene. For example, chlorobenzene and chloromethane give toluene. When two aryl halides join with sodium, it is called the Fittig reaction.

Which properties of haloalkanes are most asked in NEET?

NEET questions often test the order of reactivity (RI > RBr > RCl), versus features, reagent-based products such as KCN versus AgCN, alcoholic versus aqueous KOH, and Grignard and Wurtz reactions. Physical property orders, such as boiling points and dipole moments, also appear as one-mark questions.

How do I decide between substitution and elimination in JEE Main questions?

Look at the substrate, the reagent and the conditions. Aqueous KOH and good nucleophiles with primary halides favour substitution. Alcoholic KOH, strong bulky bases, tertiary halides and heat favour elimination, with the Saytzeff alkene as the major product. JEE Main often combines this with carbocation stability and stereochemistry.

Previous year questions on Properties of Alkyl Halides (Halo Alkanes) & Aryl Halides (Halo Arenes)

22 questions from past papers, each with a step-by-step solution.

Show all 22 questions

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