Haloalkanes and Haloarenes

What to remember

• Haloalkanes (R–X) react mainly by nucleophilic substitution. SN2 is a one-step reaction with inversion of configuration; SN1 is a two-step reaction through a carbocation and gives racemisation. SN2 order: CH₃X > 1° > 2° > 3°. SN1 order: 3° > 2° > 1° > CH₃X.
• Haloarenes are very unreactive towards nucleophilic substitution because of resonance (partial double bond in C–X), the sp² carbon, and the unstable phenyl cation. A –NO₂ group at the ortho or para position makes the reaction easy; at the meta position it has no such effect.
• Halogens on a benzene ring are deactivating but ortho-para directing (–I effect wins on reactivity, +R effect decides the position).
• Know the named reactions with their reagents: Finkelstein (NaI, dry acetone), Swarts (AgF, Hg₂F₂, CoF₂ or SbF₃), Sandmeyer (Cu₂Cl₂/HCl or Cu₂Br₂/HBr), Wurtz (Na, dry ether), Fittig, Wurtz-Fittig, Saytzeff elimination, Grignard reagent formation (Mg, dry ether).

Classification and nomenclature

A haloalkane has a halogen atom on an sp³ carbon. A haloarene has a halogen atom directly on an sp² carbon of a benzene ring.

Classification by the number of halogen atoms: mono-, di-, tri- and polyhalogen compounds. In a dihalide, the two halogens on the same carbon are called geminal (gem) and those on adjacent carbons are called vicinal (vic).

Classification by the carbon bearing the halogen:

• Primary (1°), secondary (2°) and tertiary (3°) alkyl halides, where the carbon bonded to X is attached to one, two or three other carbons.
• Allylic halide: X is on an sp³ carbon next to a C=C double bond, for example CH₂=CH–CH₂Cl.
• Benzylic halide: X is on an sp³ carbon next to a benzene ring, for example C₆H₅CH₂Cl.
• Vinylic halide: X is on an sp² carbon of a C=C bond, for example CH₂=CHCl.
• Aryl halide: X is on a benzene ring, for example C₆H₅Cl.

For haloarenes the common names use ortho (1,2), meta (1,3) and para (1,4).

Nature of the C–X bond

Halogen is more electronegative than carbon. The C–X bond is polar, with a partial positive charge on carbon (δ⁺) and a partial negative charge on halogen (δ⁻). Going from F to I the atom becomes larger, so the bond becomes longer and weaker.

The C–I bond is the longest and weakest, so iodides are the most reactive and fluorides the least reactive in substitution reactions.

Physical properties

• Haloalkanes have higher boiling points than alkanes of similar mass because of stronger dipole-dipole and van der Waals forces.
• For the same alkyl group the boiling point order is R–I > R–Br > R–Cl > R–F (larger halogen, larger surface and greater polarisability).
• Among isomeric haloalkanes the boiling point falls with branching: 1° > 2° > 3°. For example 1-chlorobutane boils higher than 2-chloro-2-methylpropane.
• Density increases with the number of carbon atoms, the number of halogens and the atomic mass of the halogen. Bromo, iodo and polychloro compounds are denser than water.
• Haloalkanes are only slightly soluble in water. The attraction formed between haloalkane and water is weaker than the hydrogen bonds that must be broken in water. They dissolve well in organic solvents.
• Among dihalobenzene isomers, the para isomer has the highest melting point because its symmetrical shape fits well in the crystal lattice.

Preparation of haloalkanes

Direct iodination of an alkane is reversible, because the HI formed reduces the alkyl iodide back to the alkane. It works only in the presence of an oxidising agent such as HNO₃ or HIO₃, which destroys HI. Direct fluorination is too violent.

Preparation of haloarenes

• 1. Electrophilic substitution: benzene + Cl₂ or Br₂ with Fe or FeCl₃ (Lewis acid) → chlorobenzene or bromobenzene. Toluene gives a mixture of ortho and para products. Iodination needs an oxidising agent such as HNO₃; fluoro compounds are not made this way.
• 2. Sandmeyer reaction: a benzenediazonium salt reacts with Cu₂Cl₂/HCl or Cu₂Br₂/HBr and gives chlorobenzene or bromobenzene with release of N₂. With Cu powder and HCl or HBr (Gattermann reaction) the same products form, but the yield is lower.
• 3. Iodobenzene is made from the diazonium salt by simply warming with KI; no copper is needed.

Nucleophilic substitution in haloalkanes

A nucleophile (Nu⁻) replaces the halide ion: R–X + Nu⁻ → R–Nu + X⁻. The halide is called the leaving group.

The cyanide and nitrite ions are ambident nucleophiles; each can attack through two different atoms. KCN and KNO₂ are largely ionic, so the more electronegative atom of the ion (C of CN⁻ is the usual site in KCN; O of NO₂⁻ in KNO₂) attacks, giving R–CN and R–ONO. AgCN and AgNO₂ are covalent, so the lone pair on N is used and the products are R–NC and R–NO₂.

SN2 (substitution nucleophilic bimolecular)

• One step. The nucleophile attacks from the side opposite to the leaving group (back-side attack), through a transition state in which carbon is bonded partly to both Nu and X.
• Rate = k[R–X][Nu⁻]; second order overall.
• The configuration at the carbon is inverted (Walden inversion), like an umbrella turned inside out.
• Bulky groups block the back-side attack, so the order is CH₃X > 1° > 2° > 3°.

SN1 (substitution nucleophilic unimolecular)

• Two steps. Step 1 (slow, rate-determining): R–X → R⁺ + X⁻. Step 2 (fast): R⁺ + Nu⁻ → R–Nu.
• Rate = k[R–X]; first order.
• The carbocation is planar, so the nucleophile can attack from either face. An optically active halide gives a racemic mixture.
• The order is 3° > 2° > 1° > CH₃X, following the stability of the carbocation (hyperconjugation and +I effect). Polar protic solvents favour SN1.
• Allylic and benzylic halides react fast by both mechanisms because their carbocations are resonance-stabilised.

For the same alkyl group, the rate of substitution follows the leaving-group order R–I > R–Br > R–Cl > R–F.

Stereochemistry basics

• A carbon bonded to four different groups is a chiral (asymmetric) carbon. 2-bromobutane, CH₃CHBrCH₂CH₃, has one chiral carbon.
• Enantiomers are non-superimposable mirror images. They rotate plane-polarised light by equal angles in opposite directions, dextro (+) or laevo (–).
• A racemic mixture contains equal amounts of the two enantiomers. It is optically inactive because the rotations cancel (external compensation).
• Retention of configuration means the spatial arrangement around the stereocentre is unchanged during the reaction (no bond to that carbon is broken). Inversion means the arrangement is turned to the opposite one. These terms describe geometry only; they do not tell whether the sign of rotation changes from (+) to (–).

Elimination and reactions with metals

Elimination (dehydrohalogenation). A haloalkane with a β-hydrogen, heated with alcoholic KOH, loses H and X from adjacent carbons and forms an alkene. Aqueous KOH favours substitution; alcoholic KOH favours elimination.

Saytzeff (Zaitsev) rule. When more than one alkene can form, the more highly substituted alkene is the major product. 2-bromobutane + alcoholic KOH → but-2-ene (major) + but-1-ene (minor).

Grignard reagent. R–X + Mg (dry ether) → R–MgX. Moisture must be absent, because RMgX + H₂O → R–H + Mg(OH)X. The carbon bonded to Mg behaves as a carbanion.

Wurtz reaction. 2 R–X + 2 Na (dry ether) → R–R + 2 NaX. Example: 2 CH₃Br + 2 Na → CH₃CH₃ + 2 NaBr.

Reactions of haloarenes

Low reactivity in nucleophilic substitution. Three reasons:

• 1. Resonance: the lone pair on halogen delocalises into the ring, giving the C–X bond partial double bond character. The C–Cl bond in chlorobenzene is about 169 pm, shorter than about 177 pm in chloromethane, so it is harder to break.
• 2. The carbon is sp² hybridised. It is more electronegative than sp³ carbon, so it holds the bond pair more tightly.
• 3. The phenyl cation is not stabilised by resonance, so SN1 is not possible; the electron-rich ring also repels an incoming nucleophile in SN2.

Chlorobenzene gives phenol only under drastic conditions: aqueous NaOH, 623 K and 300 atm, then acidification.

Effect of –NO₂ group. An –NO₂ group at the ortho or para position withdraws electrons and stabilises the negatively charged intermediate by resonance, so substitution becomes easier.

A –NO₂ group at the meta position does not give this resonance stabilisation, so it does not help the reaction in the same way.

Electrophilic substitution. Halogen withdraws electrons by the inductive effect, so the ring is less reactive than benzene (deactivated). Its lone pair is donated by resonance to the ortho and para positions, so the incoming group goes mainly to ortho and para. Examples: chlorination, nitration, sulphonation and Friedel-Crafts reactions of chlorobenzene give mainly para and some ortho product; the para isomer is the major one.

Fittig reaction. 2 Ar–X + 2 Na (dry ether) → Ar–Ar + 2 NaX. Two bromobenzene molecules give biphenyl.

Wurtz-Fittig reaction. Ar–X + R–X + 2 Na (dry ether) → Ar–R + 2 NaX. Bromobenzene + CH₃Br + 2 Na → toluene + 2 NaBr.

Grignard reagent. Bromobenzene + Mg (dry ether) → C₆H₅MgBr; with water it gives benzene.

Polyhalogen compounds

NEET traps

• 1. Chlorobenzene is less reactive than chloromethane in nucleophilic substitution, not more.
• 2. SN2 inversion does not mean the sign of rotation always changes from (+) to (–). Inversion describes the geometry.
• 3. SN1 gives racemisation only when the halide is optically active and the carbon is chiral; a 3° halide that is not chiral gives no stereo-change to notice.
• 4. KCN gives R–CN but AgCN gives R–NC; KNO₂ gives R–ONO but AgNO₂ gives R–NO₂.
• 5. Aqueous KOH gives an alcohol (substitution); alcoholic KOH gives an alkene (elimination).
• 6. Halogens are ring-deactivating yet ortho-para directing. Do not call them meta directing.
• 7. A meta –NO₂ group does not activate chlorobenzene towards nucleophilic attack; ortho and para do.
• 8. In the Finkelstein reaction the solvent must be dry acetone; the precipitated NaCl or NaBr drives the reaction.
• 9. Free-radical chlorination of an alkane is not a clean preparation because many products form.
• 10. Anti-Markovnikov addition of HBr needs peroxide; HCl and HI do not show the peroxide effect in the same way.

One-liners

• 1. In the C–X bond carbon is δ⁺ and halogen is δ⁻.
• 2. Bond length order: C–F < C–Cl < C–Br < C–I.
• 3. Bond enthalpy order: C–F > C–Cl > C–Br > C–I.
• 4. Boiling point order for the same alkyl group: RI > RBr > RCl > RF.
• 5. Boiling point falls as branching increases in isomeric haloalkanes.
• 6. Thionyl chloride is preferred for converting an alcohol to a chloroalkane because by-products are gases.
• 7. The Swarts reaction forms alkyl fluorides.
• 8. The Sandmeyer reaction uses cuprous halide with a diazonium salt.
• 9. SN2 shows second-order kinetics and Walden inversion.
• 10. SN1 shows first-order kinetics and racemisation.
• 11. The Saytzeff rule gives the more substituted alkene as the major product.
• 12. Grignard reagent must be made in dry ether.
• 13. Fittig joins two aryl halides; Wurtz-Fittig joins an aryl halide with an alkyl halide.
• 14. Chloroform is stored in dark brown bottles to prevent formation of phosgene.
• 15. DDT and freons are harmful to the environment; DDT accumulates in fats and freons deplete ozone.

Practice questions

1. Which reagent converts ethanol into chloroethane with only gaseous by-products?

   1. HCl with anhydrous ZnCl₂
   2. SOCl₂
   3. PCl₃
   4. PCl₅
   Answer

   B. SOCl₂

   SOCl₂ gives SO₂ and HCl, both gases, so the chloroalkane is obtained pure.

2. In the Finkelstein reaction of CH₃Br with NaI in dry acetone, the solid that separates out is

   1. NaI
   2. AgBr
   3. CH₃I
   4. NaBr
   Answer

   D. NaBr

   NaBr is insoluble in dry acetone and precipitates, driving the reaction forward.

3. Which reagent is used in the Swarts reaction to convert an alkyl bromide into an alkyl fluoride?

   1. AgF
   2. KCN
   3. NaI in dry acetone
   4. Cu₂Cl₂
   Answer

   A. AgF

   Swarts reaction uses a metallic fluoride such as AgF, Hg₂F₂, CoF₂ or SbF₃.

4. Chlorobenzene is obtained from benzenediazonium chloride by heating with Cu₂Cl₂ and HCl. This reaction is called the

   1. Swarts reaction
   2. Finkelstein reaction
   3. Sandmeyer reaction
   4. Wurtz-Fittig reaction
   Answer

   C. Sandmeyer reaction

   Replacement of the diazonium group by Cl using cuprous chloride is the Sandmeyer reaction.

5. Among CH₃CH₂Cl, CH₃CH₂Br, CH₃CH₂I and CH₃CH₂F, the fastest nucleophilic substitution is shown by

   1. CH₃CH₂Cl
   2. CH₃CH₂I
   3. CH₃CH₂Br
   4. CH₃CH₂F
   Answer

   B. CH₃CH₂I

   I⁻ is the best leaving group because the C–I bond is the weakest.

6. The correct order of reactivity of the following chlorides towards SN2 reaction is

   1. CH₃Cl < CH₃CH₂Cl < (CH₃)₂CHCl < (CH₃)₃CCl
   2. (CH₃)₂CHCl < (CH₃)₃CCl < CH₃Cl < CH₃CH₂Cl
   3. CH₃CH₂Cl < CH₃Cl < (CH₃)₃CCl < (CH₃)₂CHCl
   4. (CH₃)₃CCl < (CH₃)₂CHCl < CH₃CH₂Cl < CH₃Cl
   Answer

   D. (CH₃)₃CCl < (CH₃)₂CHCl < CH₃CH₂Cl < CH₃Cl

   SN2 reactivity falls with steric hindrance: CH₃Cl > 1° > 2° > 3°.

7. Heating 2-bromobutane with alcoholic KOH gives mainly

   1. but-2-ene
   2. butan-2-ol
   3. butane
   4. but-1-ene
   Answer

   A. but-2-ene

   By the Saytzeff rule the more substituted alkene, but-2-ene, is the major product.

8. A Grignard reagent is prepared by treating an alkyl halide with magnesium in

   1. aqueous ethanol
   2. dilute HCl
   3. dry ether
   4. water
   Answer

   C. dry ether

   The reaction needs moisture-free ether because RMgX is destroyed by water.

9. Ethyl bromide reacts with AgNO₂ to give mainly

   1. CH₂=CH₂
   2. CH₃CH₂OH
   3. CH₃CH₂NO₂
   4. CH₃CH₂ONO
   Answer

   C. CH₃CH₂NO₂

   AgNO₂ is covalent; the lone pair on N attacks, giving the nitroalkane CH₃CH₂NO₂.

10. Ethyl bromide reacts with AgCN to give mainly

    1. CH₃CH₂NC
    2. CH₃CH₂OH
    3. CH₃CH₂CN
    4. CH₃CH₂NH₂
    Answer

    A. CH₃CH₂NC

    AgCN is covalent, so N attacks and the isocyanide CH₃CH₂NC forms (KCN would give CH₃CH₂CN).

11. Chloroform is stored in dark brown bottles filled to the brim to prevent its slow oxidation by air and light to

    1. dichloromethane
    2. phosgene
    3. carbon tetrachloride
    4. formaldehyde
    Answer

    B. phosgene

    Air and light oxidise CHCl₃ to the poisonous gas phosgene, COCl₂.

12. Triiodomethane is used as an

    1. anaesthetic
    2. refrigerant
    3. insecticide
    4. antiseptic
    Answer

    D. antiseptic

    CHI₃ liberates free iodine and acts as an antiseptic.

13. An equimolar mixture of two enantiomers is

    1. optically active and dextrorotatory
    2. optically active and laevorotatory
    3. optically inactive and called a racemic mixture
    4. optically inactive because it is a meso compound
    Answer

    C. optically inactive and called a racemic mixture

    Equal and opposite rotations cancel, so a racemic mixture is optically inactive.

14. Which of the following bonds has the greatest length?

    1. C–I
    2. C–Cl
    3. C–Br
    4. C–F
    Answer

    A. C–I

    Bond length rises down the group: C–I is the longest.

15. Bromobenzene, bromomethane and sodium in dry ether react to give mainly

    1. benzene
    2. toluene
    3. ethane
    4. biphenyl
    Answer

    B. toluene

    This is the Wurtz-Fittig reaction, which joins an aryl and an alkyl group: C₆H₅CH₃.

16. Which of the following isomeric chlorides has the highest boiling point?

    1. 2-chlorobutane
    2. 1-chloro-2-methylpropane
    3. 2-chloro-2-methylpropane
    4. 1-chlorobutane
    Answer

    D. 1-chlorobutane

    Straight-chain isomers have more surface contact; branching lowers boiling point, so 1-chlorobutane is highest.

17. Given below are two statements. Statement I: An optically active halide reacting by SN1 gives a racemic mixture. Statement II: SN2 reactions proceed with inversion of configuration.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    Planar carbocation is attacked from both faces in SN1; back-side attack in SN2 inverts the configuration. Both correct.

18. Given below are two statements. Statement I: Chlorobenzene is more reactive than chloromethane towards nucleophilic substitution. Statement II: The C–Cl bond in chlorobenzene has partial double bond character.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    Chlorobenzene is far less reactive because of resonance, so Statement I is wrong; Statement II is correct.

19. Given below are two statements. Statement I: C–X bond enthalpy decreases from C–F to C–I. Statement II: Among haloalkanes, the C–F bond is the most reactive in nucleophilic substitution.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    The first is correct; C–F is the strongest bond and least reactive, so II is wrong.

20. Given below are two statements. Statement I: Haloalkanes dissolve readily in water because they form hydrogen bonds with water. Statement II: Haloalkanes have higher boiling points than alkanes of comparable molecular mass.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    Haloalkanes cannot form hydrogen bonds with water and are only slightly soluble; II is correct because of dipole-dipole forces.

21. Which of the following statements are correct? 1. Alkyl iodides can be prepared from alkyl chlorides by the Finkelstein reaction. 2. Alkyl fluorides can be prepared by the Swarts reaction. 3. Direct iodination of alkanes with I₂ alone is an efficient method of preparing alkyl iodides.

    1. 1 and 2 only
    2. 1 and 3 only
    3. 2 and 3 only
    4. 1, 2 and 3
    Answer

    A. 1 and 2 only

    Statements 1 and 2 are correct. Iodination with I₂ alone is reversible because HI reduces the product, so 3 is wrong.

22. Given below are two statements. Statement I: A –NO₂ group at the meta position increases the reactivity of chlorobenzene towards nucleophilic substitution by aqueous NaOH. Statement II: A meta –NO₂ group stabilises the negative charge of the intermediate by resonance.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    B. Both Statement I and Statement II are incorrect

    Only ortho and para –NO₂ groups can delocalise the negative charge; meta cannot. Both statements are incorrect.

23. Which of the following statements are correct? 1. Freons deplete the ozone layer. 2. DDT is readily broken down in the body and does not accumulate. 3. Tetrachloromethane can damage the liver.

    1. 1 and 2 only
    2. 1 and 3 only
    3. 2 and 3 only
    4. 1, 2 and 3
    Answer

    B. 1 and 3 only

    DDT is not metabolised and accumulates in fatty tissue, so 2 is wrong; 1 and 3 are correct.

24. Given below are two statements. Statement I: A Grignard reagent reacts with water to form an alkane. Statement II: A Grignard reagent is best prepared in aqueous solution.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    RMgX + H₂O → RH + Mg(OH)X, so I is correct; it must be made in dry ether, so II is wrong.

25. Given below are two statements. Statement I: An SN2 reaction of (+)-2-bromobutane always gives a (–) product. Statement II: Inversion of configuration describes the spatial arrangement and does not predict the sign of optical rotation.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    The sign of rotation is not predicted by inversion or retention. Statement I is wrong, II is correct.

26. Assertion (A): tert-Butyl bromide undergoes SN1 reaction faster than methyl bromide. Reason (R): The tert-butyl carbocation is stabilised by hyperconjugation and the +I effect of three methyl groups.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    A. Both A and R are true and R is the correct explanation of A

    Greater carbocation stability explains faster SN1.

27. Assertion (A): The boiling points of alkyl halides follow the order RI > RBr > RCl. Reason (R): Alkyl halides are polar compounds.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    B. Both A and R are true but R is not the correct explanation of A

    Both true, but the order comes from increasing size and van der Waals forces, not from polarity alone.

28. Assertion (A): Ethyl chloride on heating with alcoholic KOH gives ethene. Reason (R): Alcoholic KOH favours substitution to give ethanol as the main product.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    C. A is true but R is false

    A is true; alcoholic KOH favours elimination, so R is false.

29. Assertion (A): Halogen atoms in chlorobenzene activate the ring towards electrophilic substitution. Reason (R): Halogens withdraw electrons by the inductive effect.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    D. A is false but R is true

    Halogens deactivate the ring, so A is false; R is true.

30. Assertion (A): SN1 reactions of haloalkanes are favoured by polar protic solvents. Reason (R): SN1 reactions occur in two steps.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    B. Both A and R are true but R is not the correct explanation of A

    Both true; solvent stabilisation of the ions is the real reason, so R does not explain A.

31. Match List I with List II. List I: a. Finkelstein reaction b. Swarts reaction c. Sandmeyer reaction d. Wurtz reaction List II: i. 2 R–X + 2 Na in dry ether → R–R ii. ArN₂⁺Cl⁻ + Cu₂Cl₂ → ArCl iii. R–Cl + NaI in dry acetone iv. R–Br + AgF

    1. a-i, b-ii, c-iii, d-iv
    2. a-ii, b-i, c-iv, d-iii
    3. a-iii, b-iv, c-ii, d-i
    4. a-iv, b-iii, c-i, d-ii
    Answer

    C. a-iii, b-iv, c-ii, d-i

    Finkelstein-NaI/acetone (iii); Swarts-metal fluoride (iv); Sandmeyer-Cu₂Cl₂ (ii); Wurtz-Na/ether (i).

32. Match the reagent (with CH₃CH₂Br) in List I with the main product in List II. List I: a. aqueous KOH b. alcoholic KOH c. KCN d. AgNO₂ List II: i. CH₃CH₂NO₂ ii. CH₃CH₂CN iii. CH₂=CH₂ iv. CH₃CH₂OH

    1. a-iv, b-iii, c-ii, d-i
    2. a-iii, b-iv, c-ii, d-i
    3. a-iv, b-iii, c-i, d-ii
    4. a-iv, b-ii, c-iii, d-i
    Answer

    A. a-iv, b-iii, c-ii, d-i

    Aqueous KOH gives the alcohol (iv); alcoholic KOH gives ethene (iii); KCN gives the nitrile (ii); AgNO₂ gives the nitroalkane (i).

33. Match the compound in List I with its use or effect in List II. List I: a. CHI₃ b. CCl₄ c. CCl₂F₂ d. DDT List II: i. insecticide ii. antiseptic iii. refrigerant that depletes ozone iv. fire extinguisher and cleaning solvent

    1. a-iv, b-ii, c-i, d-iii
    2. a-ii, b-iii, c-iv, d-i
    3. a-ii, b-iv, c-iii, d-i
    4. a-i, b-iv, c-iii, d-ii
    Answer

    C. a-ii, b-iv, c-iii, d-i

    CHI₃ antiseptic (ii); CCl₄ fire extinguisher (iv); CCl₂F₂ refrigerant (iii); DDT insecticide (i).

34. Match List I with List II. List I: a. SN1 b. SN2 c. Saytzeff elimination d. Wurtz-Fittig reaction List II: i. inversion of configuration ii. racemisation iii. formation of the more substituted alkene iv. formation of an alkylbenzene

    1. a-ii, b-i, c-iv, d-iii
    2. a-ii, b-i, c-iii, d-iv
    3. a-iii, b-i, c-ii, d-iv
    4. a-i, b-ii, c-iii, d-iv
    Answer

    B. a-ii, b-i, c-iii, d-iv

    SN1-racemisation (ii); SN2-inversion (i); Saytzeff-more substituted alkene (iii); Wurtz-Fittig-alkylbenzene (iv).

35. Match the preparation in List I with the result in List II. List I: a. alcohol with SOCl₂ b. propene with HBr in presence of peroxide c. benzene with Cl₂/FeCl₃ d. ethane with Cl₂ under UV light List II: i. chlorobenzene ii. alkyl chloride with gaseous by-products iii. anti-Markovnikov product iv. mixture of products

    1. a-i, b-iii, c-ii, d-iv
    2. a-ii, b-i, c-iii, d-iv
    3. a-iii, b-ii, c-i, d-iv
    4. a-ii, b-iii, c-i, d-iv
    Answer

    D. a-ii, b-iii, c-i, d-iv

    SOCl₂-gaseous by-products (ii); peroxide effect-anti-Markovnikov (iii); Fe-catalysed chlorination-chlorobenzene (i); radical chlorination-mixture (iv).

36. Match the haloarene reaction in List I with the product in List II. List I: a. chlorobenzene + NaOH at 623 K, 300 atm, then H⁺ b. 1-chloro-2,4-dinitrobenzene + aqueous NaOH at 368 K c. 1-chloro-2,4,6-trinitrobenzene + warm water d. chlorobenzene + Na in dry ether List II: i. 2,4,6-trinitrophenol ii. biphenyl iii. phenol iv. 2,4-dinitrophenol

    1. a-iii, b-iv, c-i, d-ii
    2. a-iii, b-iv, c-ii, d-i
    3. a-iii, b-i, c-iv, d-ii
    4. a-iv, b-iii, c-i, d-ii
    Answer

    A. a-iii, b-iv, c-i, d-ii

    a gives phenol (iii); b gives 2,4-dinitrophenol (iv); c gives picric acid (i); d is Fittig reaction giving biphenyl (ii).

37. Optically pure (+)-2-bromobutane reacts completely by the SN1 pathway with water. The butan-2-ol formed is

    1. a pure enantiomer with retained configuration
    2. but-2-ene
    3. an optically inactive racemic mixture
    4. a pure enantiomer with inverted configuration
    Answer

    C. an optically inactive racemic mixture

    The planar carbocation is attacked equally from both faces, giving a racemic, optically inactive mixture.

38. A tertiary bromide of formula C₄H₉Br is heated with alcoholic KOH. The hydrocarbon formed is

    1. 2-methylpropan-2-ol
    2. but-2-ene
    3. but-1-ene
    4. 2-methylpropene
    Answer

    D. 2-methylpropene

    The tertiary bromide is 2-bromo-2-methylpropane; elimination gives only 2-methylpropene.

39. Which compound reacts fastest with aqueous NaOH at 368 K?

    1. chlorobenzene
    2. 1-chloro-2,4-dinitrobenzene
    3. 1-chloro-3-nitrobenzene
    4. 1-chloro-4-nitrobenzene
    Answer

    B. 1-chloro-2,4-dinitrobenzene

    Two nitro groups at ortho and para positions give the strongest stabilisation of the intermediate.

40. Propan-1-ol needs to be made from propene via 1-bromopropane. The reagent that adds HBr to propene to give 1-bromopropane is

    1. HBr in the presence of peroxide
    2. HBr in the absence of peroxide
    3. aqueous KBr
    4. Br₂ in CCl₄
    Answer

    A. HBr in the presence of peroxide

    HBr in the presence of peroxide adds in the anti-Markovnikov way, placing Br on the terminal carbon.

41. Nitration of chlorobenzene gives mainly

    1. 1-chloro-3-nitrobenzene only
    2. nitrobenzene
    3. 1-chloro-4-nitrobenzene with some 1-chloro-2-nitrobenzene
    4. 1,3-dinitrobenzene
    Answer

    C. 1-chloro-4-nitrobenzene with some 1-chloro-2-nitrobenzene

    Chlorine is ortho-para directing; the para product is the major one.

42. Bromobenzene is treated with magnesium in dry ether, and then with water. The final organic product is

    1. bromobenzene
    2. benzene
    3. phenol
    4. biphenyl
    Answer

    B. benzene

    C₆H₅MgBr + H₂O → C₆H₆ + Mg(OH)Br.

43. To obtain iodobenzene from benzenediazonium chloride, the reagent used is

    1. Cu₂Cl₂/HCl
    2. Cu₂Br₂/HBr
    3. AgF
    4. KI
    Answer

    D. KI

    Iodide replaces the diazonium group on simple warming with KI; no copper catalyst is needed.