CBSE Class 12 Chemistry Haloalkanes and Haloarenes Important Questions Set 07

Very Short Answer Questions (1 Mark)

Question 1. Why is iodination of benzene difficult ? (HP Board 2011)
Answer:
The reasons are : (i) I₂ is the least reactive of all the halogens because C–I bond formed is much weaker than C–Cl and C–Br bonds.
(ii) The reaction is reversible because HI produced during the reaction, being a reducing agent reduces iodobenzene thus formed back to benzene.
C₆H₆ + I₂ ⇌ C₆H₅I + HI
Therefore, to push the reaction in the forward direction, iodination is usually carried out in presence of an oxidising agent such as nitric acid or iodic acid (HIO₃) which oxidises HI to I₂. Iodination can also be carried out in presence of mercuric oxide which eliminates HI as insoluble mercuric iodide
2 HI + 2 HNO₃ → 2 NO₂ + 2 H₂O + I₂ ;
5 HI + HIO₃ → 3 I₂ + 3 H₂O
HgO + 2HI → HgI₂ + H₂O

Teacher's Note:
1. The oxidising agent is needed to remove HI; it does not make iodine itself more reactive towards benzene.

Question 2. Write a chemical reaction in which the iodide ion replaces the diazonium group in a diazonium salt. (CBSE 2008)
Answer:
C₆H₅N₂⁺Cl⁻ (Benzenediazonium chloride) + KI (aq) \( \xrightarrow{\text{Warm}} \) C₆H₅I (Iodobenzene) + N₂ + KCl

Teacher's Note:
1. No copper catalyst is needed for iodide; Sandmeyer conditions (Cu₂X₂) are used for Cl and Br.

Question 3. What effect should the following resonance of vinyl chloride have on its dipole moment ?
CH₂=CH–Cl: ↔ ⁻:CH₂–CH=Cl⁺:
Answer:
This resonance effect decreases the dipole moment of vinyl chloride relative to ethyl chloride.

Teacher's Note:
1. In the second structure the electron shift is towards carbon, opposite to the −I pull of Cl, so the two effects partly cancel.

Question 4. Why is vinyl chloride less reactive than ethyl chloride ?
Answer:
Vinyl chloride can be regarded as a resonance hybrid of the following two structures.
CH₂=CH–Cl: ↔ ⁻:CH₂–CH=Cl⁺:   (Vinyl chloride)     CH₃CH₂–Cl (Ethyl chloride)
As a result, C–Cl bond in vinyl chloride has some double bond character and hence more difficult to break than the C–Cl bond in ethyl chloride which has pure single bond character.

Teacher's Note:
1. The sp² carbon in vinyl chloride also holds the C–Cl bond more tightly than the sp³ carbon of ethyl chloride.

Question 5. Out of chlorobenzene and chloromethane which is more reactive towards nucleophilic substitution reactions.
Answer:
Chlorobenzene can be regarded as a resonance hybrid of five structures (three of them having C=Cl⁺ double bond). As a result, C–Cl bond in chlorobenzene has some double bond character and hence more difficult to break than C–Cl bond in chloromethane (CH₃–Cl) which has pure single bond character.
Therefore, chloromethane is more reactive than chlorobenzene towards nucleophilic substitution reactions.

Teacher's Note:
1. Aryl halides resist nucleophilic substitution for three reasons: resonance, sp² carbon and repulsion of the nucleophile by the electron-rich ring.

Question 6. The C–Cl bond length in chlorobenzene is shorter than C–Cl bond length in CH₃–Cl. (CBSE 2016)
Answer:
Chlorobenzene can be regarded as a resonance hybrid of five structures. As a result, C–Cl bond in chlorobenzene has some double bond character and hence shorter (169 pm) than C–Cl bond in CH₃–Cl (177 pm) which has pure single bond character.
Alternatively, in CH₃–Cl, Cl is attached to a sp³⁻hybridized carbon while in chlorobenzene it is attached to a sp²⁻hybridized carbon. Since a sp²⁻hybridized orbital is smaller in size as compared to a sp³⁻hybridized orbital of carbon, therefore, C–Cl bond in chlorobenzene is shorter (169 pm) than C–Cl bond length in CH₃–Cl (177 pm).

Teacher's Note:
1. More s-character (sp² = 33%, sp³ = 25%) means a smaller orbital and a shorter, stronger bond.

Question 7. The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride. (CBSE 2016)
Answer:
In chlorobenzene, Cl is attached to a sp²⁻hybrid carbon while in cyclohexyl chloride, it is attached to a sp³⁻hybrid carbon.
The sp²⁻hybrid carbon due to greater s-character is more electronegative than a sp³⁻hybrid carbon. Therefore, sp²⁻hybrid carbon of C–Cl bond in chlorobenzene has less tendency to release electrons to the Cl atom than a sp³⁻hybrid carbon in cyclohexyl chloride. As a result, C–Cl bond in chlorobenzene is less polar than in cyclohexyl chloride. In other words, chlorobenzene has lower dipole moment (1.69 D) as compared to that of cyclohexyl chloride (1.86 D).

Teacher's Note:
1. Resonance in chlorobenzene also pushes electron density from Cl into the ring, which further lowers the C–Cl bond polarity.

Question 8. Grignard reagent of C₆H₅Cl can be prepared in THF but not in ether. Explain why ?
Answer:
Due to +R-effect of Cl, the C–Cl bond has some double bond character and hence little more difficult to break. In other words, energy of activation for cleavage of C–Cl bond is little higher which cannot be supplied by the low boiling point (b.p. 308 K) of ether but can be easily supplied by the little higher b.p. (340 K) of THF.

Teacher's Note:
1. THF is also a better Lewis base than diethyl ether, so it stabilises the arylmagnesium halide formed.

Question 9. Arrange the following in order of increasing ease towards nucleophilic substitution.
4-chloronitrobenzene, chlorobenzene, 2, 4, 6- trinitrochlorobenzene, 2, 4-dinitrochlorobenzene.
Answer:
As the number of electron-withdrawing NO₂ groups increases at o- and p-positions, the stability of the intermediate carbanion increases and hence reactivity towards nucleophilic substitution increases in the same order, i.e.,
Chlorobenzene < 4-chloronitrobenzene < 2, 4-dinitrochlorobenzene < 2, 4, 6 trinitrochlorobenzene.

Teacher's Note:
1. This matches the falling temperatures needed for hydrolysis: 623 K for chlorobenzene, 443 K, 368 K and finally warm water for the trinitro compound.

Question 10. Give one example (with equation) of Wurtz-Fittig reaction.
Or Write the structure of the product when chlorobenzene is treated with methyl chloride in presence of sodium metal in dry ether. (CBSE 2018)
Answer:
Wurtz-Fittig reaction involves the reaction between an aryl halide and an alkyl halide in presence of metallic sodium in dry ether to yield alkyl substituted benzene. For example,
C₆H₅–Cl (Chlorobenzene) + 2Na + Cl–CH₃ (Methyl chloride) \( \xrightarrow{\text{Dry ether},\ \Delta\ (\text{Wurtz-Fittig reaction})} \) C₆H₅–CH₃ (Toluene) + 2 NaCl.

Teacher's Note:
1. Wurtz-Fittig: aryl + alkyl halide; Fittig: two aryl halides; Wurtz: two alkyl halides. Sodium and dry ether are common to all three.

Question 11. How will you convert : Chlorobenzene to biphenyl ? (CBSE 2016)
Answer:
Chlorobenzene on heating with Na in presence of dry ether undergoes Fittig reaction to give biphenyl.
2 C₆H₅–Cl (Chlorobenzene) + 2 Na \( \xrightarrow{\text{Dry ether},\ \Delta\ (\text{Fittig reaction})} \) C₆H₅–C₆H₅ (Biphenyl) + 2NaCl

Teacher's Note:
1. Balance it with two molecules of chlorobenzene and two sodium atoms, as two Na are needed to remove two Cl atoms.

Question 12. Write the product formed when chlorobenzene undergoes nitration. (CBSE 2018 (S))
Answer:
Chlorine is o, p-directing.
C₆H₅Cl (Chlorobenzene) \( \xrightarrow{\text{Conc. HNO}_3\text{/H}_2\text{SO}_4\ (\text{Nitration})} \) p-Nitrochlorobenzene (major) + o-Nitrochlorobenzene (minor)

Teacher's Note:
1. The para isomer is major because the bulky Cl hinders attack at the neighbouring ortho position.

Question 13. Arrange the following in increasing order of reactivity towards sulphonation with fuming sulphuric acid : benzene, toluene, methoxybenzene, chlorobenzene.
Answer:
Reactivity in electrophilic substitution reactions increases as the electron density in the benzene ring increases. Now since methoxy (–OCH₃) group is a stronger electron-donating group than methyl (–CH₃) group, therefore, methoxybenzene is more reactive than toluene and both are more reactive than benzene. Further, since chlorine has –I-effect, therefore, chlorobenzene is less reactive than even benzene. Thus, the overall reactivity towards sulphonation increases in the order :
chlorobenzene < benzene < toluene < methoxybenzene.

Teacher's Note:
1. Halogens are the odd case: o, p-directing (by +R) but ring-deactivating (by stronger −I).

Question 14. How is DDT prepared ?
Answer:
DDT is prepared by condensation of chlorobenzene with chloral, i.e., trichloroacetaldehyde, in presence of conc. H₂SO₄.
Cl₃C–CHO (Chloral) + 2 C₆H₅Cl (Chlorobenzene, 2 molecules, reacting at the p-position) \( \xrightarrow{\text{Conc. H}_2\text{SO}_4} \) Cl₃C–CH(C₆H₄Cl-p)₂ (DDT) + H₂O

Teacher's Note:
1. DDT is p,p′-dichlorodiphenyltrichloroethane: two para-chlorophenyl groups and a CCl₃ group on the same carbon.

Question 15. How will you test pure chloroform? (UP Board 2012)
Answer:
Chloroform is oxidised by air in presence of light to form poisonous gas phosgene and HCl as shown in Ans. to Q. 62.
When aq. AgNO₃ solution is added to the above solution, a white ppt. of AgCl is formed.
COCl₂ + H₂O → CO₂ + 2 HCl
HCl + AgNO₃ → AgCl↓ (White ppt.) + HNO₃
The formation of white ppt. confirms that chloroform is impure. If, however, no white ppt. is formed, it confirms that chloroform is pure.

Teacher's Note:
1. Pure CHCl₃ gives no precipitate with AgNO₃ because its C–Cl bonds are covalent and give no free Cl⁻ ions.

Question 16. A small amount of ethanol is usually added to chloroform bottles ? Why.
Answer:
Ethanol combines with the poisonous phosgene gas which may have formed during storage and converts it into harmless (non-toxic) diethyl carbonate
2C₂H₅OH (Ethanol) + COCl₂ (Phosgene) → O=C(OC₂H₅)₂ (Diethyl carbonate) + 2 HCl

Teacher's Note:
1. About 1% ethanol is added; it does not stop the oxidation, it only removes the phosgene that forms.

Question 17. Iodoform has antiseptic properties. Give one reason to support it.
Answer:
The antiseptic properties of iodoform are due to I₂ which is liberated when iodoform comes in contact with skin.

Teacher's Note:
1. Iodoform itself is not the antiseptic; the free iodine released from it is. Its unpleasant smell led to its replacement by other iodine formulations.

Short Answer Questions (2 Marks)

Question 18. Explain the mechanism of chlorination of benzene. (Karnataka Board 2012)
Answer:
Chlorination of benzene in presence of anhydrous FeCl₃ is an electrophilic substitution reaction. It occurs in the following steps :
Step 1. Formation of electrophile : Cl₂ + FeCl₃ → Cl⁺ + [FeCl₄]⁻
Step 2. Attack of electrophile : Cl⁺ attacks the benzene ring to form a resonance stabilized carbocation (arenium ion or σ-complex) : C₆H₆ + Cl⁺ \( \xrightarrow{\text{Slow}} \) [C₆H₆Cl]⁺
Step 3. Loss of proton : [C₆H₆Cl]⁺ + [FeCl₄]⁻ \( \xrightarrow{\text{Fast}} \) C₆H₅Cl + HCl + FeCl₃
The catalyst FeCl₃ is regenerated and the aromatic character is restored in the last step.

Teacher's Note:
1. The Lewis acid's job is only to produce the electrophile Cl⁺; it is given back in the final step.

Question 19. Complete the following reaction equations :
(i) C₆H₅N₂Cl + KI → ......... (CBSE 2008)
(ii) C₆H₅–N⁺≡NCl⁻ \( \xrightarrow{\text{CuBr/HBr}} \) (Hr. Board 2011)
(iii) C₆H₅N₂⁺Cl⁻ \( \xrightarrow{\text{Cu}_2\text{Cl}_2\text{/HCl}} \) (HP Board 2013)
Answer:
(i) C₆H₅N₂Cl + KI → C₆H₅I (Iodobenzene) + N₂ + KCl
(ii) C₆H₅–N⁺≡NCl⁻ \( \xrightarrow{\text{CuBr/HBr}} \) C₆H₅Br (Bromobenzene) + N₂
(iii) C₆H₅N₂⁺Cl⁻ \( \xrightarrow{\text{Cu}_2\text{Cl}_2\text{/HCl}} \) C₆H₅Cl (Chlorobenzene) + N₂

Teacher's Note:
1. In every diazonium replacement N₂ gas is a product; do not forget it when completing the equation.

Question 20. Why does p-dichlorobenzene have a higher m.p. than its o- and m-isomers. (CBSE 2013, 2019)
Answer:
p-Dichlorobenzene is more symmetrical than o- and m-isomers. Therefore, it fits more closely in the crystal lattice than the o- and m-isomers. As a result, the intermolecular forces of attraction are stronger in p-isomer and hence more energy is required to break its crystal lattice. Therefore, p-dichlorobenzene has a higher m.p. (323 K) than o-(256 K) and m-(249 K) isomers.

Teacher's Note:
1. Symmetry controls melting point (packing in the solid); boiling points of the three isomers are nearly the same.

Question 21. Chlorobenzene is extremely less reactive towards nucleophilic substitution reactions. Give two reasons for the same. (CBSE 2013)
or Why is chlorobenzene resistant to nucleophilic substitution reaction ? (CBSE 2019)
Answer:
(i) Resonance effect : Chlorobenzene is a resonance hybrid of the following five structures; in three of them (II, III, IV) the C–Cl bond is a double bond (C=Cl⁺) with a negative charge at the o- or p-position of the ring : (I) and (V) Kekulé structures C₆H₅–Cl ; (II), (III), (IV) ring carbanion with C=Cl⁺. As a result, C–Cl bond acquires partial double bond character and hence cannot be easily broken.
(ii) Hybridisation : In chlorobenzene, Cl is attached to a sp²⁻hybridized carbon while in alkyl halides it is attached to a sp³⁻carbon. Since sp²⁻carbon is more electronegative, it holds the electron pair of the C–Cl bond more tightly; the C–Cl bond in chlorobenzene (169 pm) is shorter and stronger than in CH₃Cl (177 pm) and hence is difficult to break.

Teacher's Note:
1. Bond length data (169 pm vs 177 pm) make a strong supporting point in this answer.

Question 22. Account for the fact that halogen in chlorobenzene is less reactive than in methyl chloride.
Answer:
(i) Due to resonance in chlorobenzene, the C–Cl bond acquires partial double bond character, whereas in methyl chloride it is a pure single bond. Hence C–Cl bond in chlorobenzene is stronger.
(ii) In chlorobenzene, the carbon of the C–Cl bond is sp²⁻hybridized while in methyl chloride it is sp³⁻hybridized. The sp²⁻carbon, having more s-character, forms a shorter and stronger bond with Cl (169 pm against 177 pm in CH₃Cl).
(iii) Phenyl cation formed by ionization of chlorobenzene is not stabilized by resonance, and the electron-rich benzene ring repels the attacking nucleophile.
Therefore, the halogen in chlorobenzene is much less reactive than in methyl chloride.

Teacher's Note:
1. Methyl chloride is the most reactive halide in S_N2 (no steric hindrance), so the contrast here is very large.

Question 23. Give reasons :
(i) C–Cl bond length in chlorobenzene is shorter than C–Cl bond length in CH₃–Cl. (CBSE 2016)
(ii) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride. (CBSE 2016)
Answer:
(i) Chlorobenzene is a resonance hybrid; due to resonance C–Cl bond acquires some double bond character and hence is shorter (169 pm) than C–Cl bond in CH₃–Cl (177 pm) which is a pure single bond. Further, in chlorobenzene Cl is attached to a sp²⁻carbon (smaller orbital) while in CH₃Cl it is attached to a sp³⁻carbon.
(ii) In chlorobenzene, Cl is attached to a sp²⁻hybrid carbon which, due to greater s-character, is more electronegative than the sp³⁻hybrid carbon of cyclohexyl chloride. Therefore, it has less tendency to release electrons to Cl and the C–Cl bond is less polar. Moreover, resonance shifts electron density from Cl towards the ring. Hence, chlorobenzene has lower dipole moment (1.69 D) than cyclohexyl chloride (1.86 D).

Teacher's Note:
1. Both parts rest on the same two ideas: partial double bond character from resonance and the sp² carbon.

Question 24. Account for the fact that C–X bond length in halobenzene is smaller than C–X bond length in CH₃–X. (CBSE 2013)
Answer:
(i) In halobenzene, the halogen is attached to a sp²⁻hybridized carbon while in CH₃–X it is attached to a sp³⁻hybridized carbon. Since a sp^2 orbital has more s-character (33%) and is smaller than a sp^3 orbital (25% s-character), the C–X bond in halobenzene is shorter.
(ii) Due to resonance, the lone pair of the halogen is delocalised into the ring and the C–X bond acquires partial double bond character, which further shortens it. For example, C–Cl bond length is 169 pm in chlorobenzene and 177 pm in CH₃Cl.

Teacher's Note:
1. A double bond is always shorter than a single bond between the same atoms, so partial double bond character means a shorter bond.

Question 25. Explain why alkyl halides undergo hydrolysis more easily as compared to aryl halides ?
Answer:
In alkyl halides, C–X bond is a pure single bond formed by a sp³⁻carbon and can be easily broken; nucleophiles such as OH⁻ easily attack the electron-deficient carbon.
In aryl halides, (i) due to resonance, C–X bond has partial double bond character, (ii) the carbon is sp²⁻hybridized so C–X bond is shorter and stronger, (iii) phenyl cation is unstable and (iv) the electron-rich ring repels the incoming OH⁻ ion.
Hence alkyl halides are hydrolysed easily by aqueous KOH, while chlorobenzene is hydrolysed only on heating with aqueous NaOH at 623 K and 300 atm (Dow's process).

Teacher's Note:
1. The harsh conditions of Dow's process (623 K, 300 atm) are good evidence to quote for the low reactivity of aryl halides.

Question 26. Arrange the following compounds in order of increasing reactivity towards nucleophilic substitution reactions :
(i) 2, 4-Dinitrochlorobenzene
(ii) Chlorobenzene
(iii) 2, 4, 6- Trinitrochlorobenzene
(iv) 4-Nitrochlorobenzene
Give reasons.
Answer:
Chlorobenzene < 4-Nitrochlorobenzene < 2, 4-Dinitrochlorobenzene < 2, 4, 6-Trinitrochlorobenzene
Reason : The –NO₂ group is a strong electron-withdrawing group. When present at o- and/or p-positions, it withdraws electrons from the ring, which helps the attack of the nucleophile, and it stabilises the intermediate carbanion by delocalising the negative charge. Greater the number of –NO₂ groups at o- and p-positions, more stable is the carbanion and hence greater is the reactivity.

Teacher's Note:
1. Hydrolysis temperatures make a quick check: 623 K, 443 K, 368 K and warm water for 0, 1, 2 and 3 nitro groups.

Question 27. What is a nucleophilic substitution reaction ? Write such a reaction, one each of a monohaloalkane and a haloarene. (Manipur Board 2011)
Answer:
A reaction in which an atom or group (usually a halogen) present in a molecule is replaced by a nucleophile (an electron-rich species) is called a nucleophilic substitution reaction.
Monohaloalkane : CH₃CH₂Br + KOH(aq) → CH₃CH₂OH + KBr
Haloarene : C₆H₅Cl + NaOH \( \xrightarrow{623\ \text{K},\ 300\ \text{atm}} \) C₆H₅ONa \( \xrightarrow{\text{H}^+} \) C₆H₅OH (Dow's process)

Teacher's Note:
1. Show the harsh conditions over the arrow for the haloarene; that contrast is what the examiner looks for.

Question 28. What do you understand by :
(i) nucleophilic substitution and
(ii) electrophilic substitution reactions ?
Illustrate by taking one example in each case.
Answer:
(i) Nucleophilic substitution : A reaction in which a nucleophile (electron-rich species) replaces another atom or group (leaving group) from a molecule. For example,
CH₃–Br + OH⁻ → CH₃–OH + Br⁻
(ii) Electrophilic substitution : A reaction in which an electrophile (electron-deficient species) replaces an atom (usually H) from a molecule, especially an aromatic ring. For example, nitration of chlorobenzene :
C₆H₅Cl + HNO₃ \( \xrightarrow{\text{Conc. H}_2\text{SO}_4} \) p-ClC₆H₄NO₂ (major) + o-ClC₆H₄NO₂ (minor) + H₂O

Teacher's Note:
1. Haloalkanes typically show nucleophilic substitution; haloarenes typically show electrophilic substitution on the ring.

Question 29. What happens when : (a) Methyl chloride is treated with alcoholic KOH.
(b) Ethyl chloride is treated with alcoholic KOH.
(c) Chloroform is heated with Ag powder? (Jharkhand Board 2012)
Answer:
(a) Methyl chloride has no β-hydrogen atom and hence cannot undergo dehydrohalogenation to form an alkene. Only nucleophilic substitution occurs to give methanol : CH₃Cl + KOH → CH₃OH + KCl
(b) Ethyl chloride undergoes dehydrohalogenation to form ethene : CH₃CH₂Cl + KOH (alc.) \( \xrightarrow{\Delta} \) CH₂=CH₂ + KCl + H₂O
(c) Chloroform on heating with silver powder gives acetylene : 2CHCl₃ + 6Ag \( \xrightarrow{\Delta} \) HC≡CH + 6AgCl

Teacher's Note:
1. Elimination needs a hydrogen on the carbon next to the C–X carbon; check for a β-H before writing an alkene.

Question 30. What happens when :
(i) Chlorobenzene is subjected to hydrolysis. (Uttarakhand Board 2012)
(ii) Ethyl chloride is treated with aqueous KOH? (Uttarakhand Board 2012 ; CBSE 2013)
(iii) Chloroethane is treated with silver nitrite.
Answer:
(i) Chlorobenzene is hydrolysed to phenol only under drastic conditions, i.e., on heating with aqueous NaOH at 623 K and 300 atm (Dow's process) :
C₆H₅Cl + NaOH \( \xrightarrow{623\ \text{K},\ 300\ \text{atm}} \) C₆H₅ONa \( \xrightarrow{\text{H}^+} \) C₆H₅OH
(ii) Ethanol is formed : CH₃CH₂Cl + KOH (aq) → CH₃CH₂OH + KCl
(iii) Nitroethane is formed (AgNO₂ is covalent and attack occurs through N-atom) : CH₃CH₂Cl + AgNO₂ → CH₃CH₂NO₂ + AgCl

Teacher's Note:
1. AgNO₂ gives the nitro compound (C–N bond); KNO₂ gives the alkyl nitrite (C–O bond).

Question 31. Write one use of each of the following :
(i) Chloroform (Uttarakhand Board 2012)
(ii) Iodoform (CBSE 2011 ; Uttarakhand Board 2012)
(iii) Freon (Uttarakhand Board 2012 ; Hr. Board 2012 ; Pb. Board 2012)
(iv) DDT (CBSE 2011 ; Uttarakhand Board 2012 ; Hr. Board 2012 ; UP Board 2012)
Answer:
(i) Chloroform : It is used as a solvent for fats, alkaloids, iodine, etc. and in the production of the freon refrigerant R-22.
(ii) Iodoform : It is used as an antiseptic.
(iii) Freon : Freons (e.g., CCl₂F₂) are used as refrigerants in refrigerators and air conditioners and as propellants in aerosols.
(iv) DDT : It is used as an insecticide, especially against mosquitoes (which spread malaria).

Teacher's Note:
1. Pair each use with its drawback for long answers: freons deplete ozone, DDT is non-biodegradable and accumulates in food chains.

Short Answer Questions (3 Marks)

Question 32. (a) Write the IUPAC names of all the possible structural isomers of C₄H₉Br. Point out optically active isomer, if any.
(b) Write the structural formulae and IUPAC names of two optically active halides containing five carbon atoms in their molecules.
Answer:
(a) (i) CH₃CH₂CH₂CH₂Br : 1-Bromobutane (ii) CH₃CH₂C*H(Br)CH₃ : 2-Bromobutane (iii) (CH₃)₂CHCH₂Br : 1-Bromo-2-methylpropane (iv) (CH₃)₃CBr : 2-Bromo-2-methylpropane.
Of these, only 2-bromobutane contains a chiral carbon (attached to H, Br, CH₃ and C₂H₅) and hence it is optically active.
(b) CH₃CH₂CH₂C*H(Cl)CH₃ : 2-Chloropentane
(CH₃)₂CH–C*H(Cl)–CH₃ : 2-Chloro-3-methylbutane

Teacher's Note:
1. Mark the chiral carbon with an asterisk and list its four different groups; examiners give credit for this check.

Question 33. Define : (a) optical activity (b) asymmetric carbon (c) enantiomers (d) racemates (Maharashtra Board 2012)
Answer:
(a) Optical activity : The property of certain substances to rotate the plane of plane polarised light when it is passed through their solutions is called optical activity. Substances which rotate it to the right (clockwise) are dextrorotatory (+) and those which rotate it to the left (anticlockwise) are laevorotatory (–).
(b) Asymmetric carbon : A carbon atom which is attached to four different atoms or groups is called asymmetric or chiral carbon, e.g., C-2 in butan-2-ol, CH₃C*H(OH)CH₂CH₃.
(c) Enantiomers : Stereoisomers which are non-superimposable mirror images of each other are called enantiomers, e.g., (+)- and (–)-butan-2-ol. They have identical physical properties except the direction of rotation of plane polarised light.
(d) Racemates : An equimolar (50 : 50) mixture of the two enantiomers of a compound is called a racemate or racemic mixture. It is optically inactive due to external compensation, e.g., (±)-butan-2-ol.

Teacher's Note:
1. Enantiomers rotate light by equal amounts in opposite directions; that is exactly why their 1 : 1 mixture shows zero rotation.

Question 34. (a) Give two reasons for low reactivity of aryl halides towards nucleophilic substitution reactions. (Kerala Board 2012)
(b) Vinyl chloride is unreactive in nucleophilic substitution reactions. Explain.
Answer:
(a) (i) Resonance effect : In aryl halides, the lone pair of electrons on the halogen is delocalised into the benzene ring. As a result, C–X bond acquires partial double bond character and hence is difficult to break.
(ii) Difference in hybridisation : In aryl halides, the halogen is attached to a sp²⁻hybridized carbon which is more electronegative than a sp³⁻carbon. Therefore, C–X bond is shorter and stronger than in alkyl halides and is difficult to break.
(Further, phenyl cation is highly unstable, so S_N1 is not possible, and the electron-rich ring repels the attacking nucleophile.)
(b) Vinyl chloride is a resonance hybrid of two structures : CH₂=CH–Cl: ↔ ⁻:CH₂–CH=Cl⁺:. As a result, C–Cl bond acquires some double bond character. Moreover, Cl is attached to a sp²⁻hybridized carbon. Therefore, C–Cl bond in vinyl chloride is stronger and difficult to break; hence it is unreactive in nucleophilic substitution reactions.

Teacher's Note:
1. The same two reasons (resonance and sp² carbon) explain the low reactivity of both aryl and vinyl halides.

Question 35. Why are haloarenes more stable than haloalkanes and undergo electrophilic substitution at ortho- and para-positions ?
Answer:
Stability : In haloarenes, due to resonance the C–X bond acquires partial double bond character and the halogen is attached to a sp²⁻carbon. Thus, the C–X bond in haloarenes is stronger than in haloalkanes and hence haloarenes are more stable.
o, p-orientation : Chlorobenzene is a resonance hybrid of the following five structures; in three of them (II, III, IV) the C–Cl bond is a double bond (C=Cl⁺) with a negative charge at the o- or p-position of the ring : (I) and (V) Kekulé structures C₆H₅–Cl ; (II), (III), (IV) ring carbanion with C=Cl⁺. Thus, electron density is relatively more at o- and p-positions than at m-position. Hence, the electrophile attacks preferably at o- and p-positions.
However, due to the strong –I effect of the halogen, the ring as a whole is deactivated and electrophilic substitution in haloarenes occurs slowly as compared to benzene.

Teacher's Note:
1. Remember the pair: −I effect decides the rate (slower than benzene); +R effect decides the position (ortho/para).

Question 36. Why are haloarenes less reactive than haloalkanes towards nucleophilic substitution reactions ? (HP Board 2011, 2013 ; MP Board 2013 ; Hr. Board 2013 ; CBSE 2008, 2010)
Answer:
(i) Resonance effect : In haloarenes, the lone pair of the halogen is delocalised into the ring; the C–X bond acquires partial double bond character and is difficult to break.
(ii) Difference in hybridisation of carbon : In haloalkanes, carbon of C–X bond is sp^3 while in haloarenes it is sp²⁻hybridized. The sp²⁻carbon is more electronegative; the C–X bond is shorter and stronger.
(iii) Instability of phenyl cation : The phenyl cation is not stabilized by resonance, so S_N1 mechanism is ruled out.
(iv) Repulsion between nucleophile and electron-rich arenes : The electron-rich benzene ring repels the electron-rich nucleophile; also back side attack is blocked by the ring, so S_N2 is not possible.

Teacher's Note:
1. Two reasons are usually enough for 2 marks: resonance and sp² hybridisation.

Question 37. How are the following conversions carried out ?
(i) 2-Propanol into 1-bromopropane
(ii) n-Propyl chloride to n-propylamine
(iii) 2-Chlorobutane to 2-butanol
(iv) Methyl bromide to ethyl bromide
(v) Carbon tetrachloride to freon
(vi) Aniline into fluorobenzene
(vii) Aniline into bromobenzene
(viii) Aniline into chlorobenzene
(ix) Chloroform into diethyl carbonate
(x) Methyl bromide to acetic acid
(xi) Chlorobenzene into aniline
(xii) Chlorobenzene into toluene
(xiii) Propene to 1-bromopropane.
(xiv) Ethene to bromoethane (Uttarakhand Board 2012)
(xv) Benzene to biphenyl (Uttarakhand Board 2012, Hr. Board 2013)
(xvi) Ethyl bromide to ethylbenzene (Karnataka Board 2012)
(xvii) Chlorobenzene to phenol (Manipur Board 2012)
Answer:
(i) CH₃CH(OH)CH₃ \( \xrightarrow{\text{Conc. H}_2\text{SO}_4,\ 443\ \text{K}} \) CH₃CH=CH₂ \( \xrightarrow{\text{HBr, peroxide}} \) CH₃CH₂CH₂Br
(ii) CH₃CH₂CH₂Cl + NH₃ (alc., excess) \( \xrightarrow{\Delta} \) CH₃CH₂CH₂NH₂ + HCl
(iii) CH₃CH(Cl)CH₂CH₃ + KOH (aq) \( \xrightarrow{\Delta} \) CH₃CH(OH)CH₂CH₃ + KCl
(iv) CH₃Br \( \xrightarrow{\text{Mg, dry ether}} \) CH₃MgBr \( \xrightarrow{\text{(i) HCHO, (ii) H}_2\text{O/H}^+} \) CH₃CH₂OH \( \xrightarrow{\text{PBr}_3} \) CH₃CH₂Br
(v) 3CCl₄ + 2SbF₃ \( \xrightarrow{\text{SbCl}_5} \) 3CCl₂F₂ (Freon-12) + 2SbCl₃ (Swarts reaction)
(vi) C₆H₅NH₂ \( \xrightarrow{\text{NaNO}_2/\text{HCl},\ 273\text{--}278\ \text{K}} \) C₆H₅N₂⁺Cl⁻ \( \xrightarrow{\text{HBF}_4} \) C₆H₅N₂⁺BF₄⁻ \( \xrightarrow{\Delta} \) C₆H₅F + BF₃ + N₂ (Balz-Schiemann reaction)
(vii) C₆H₅NH₂ \( \xrightarrow{\text{NaNO}_2/\text{HCl},\ 273\text{--}278\ \text{K}} \) C₆H₅N₂⁺Cl⁻ \( \xrightarrow{\text{CuBr/HBr}} \) C₆H₅Br + N₂
(viii) C₆H₅NH₂ \( \xrightarrow{\text{NaNO}_2/\text{HCl},\ 273\text{--}278\ \text{K}} \) C₆H₅N₂⁺Cl⁻ \( \xrightarrow{\text{Cu}_2\text{Cl}_2\text{/HCl}} \) C₆H₅Cl + N₂
(ix) 2CHCl₃ + O₂ \( \xrightarrow{h\nu} \) 2COCl₂ + 2HCl ;   COCl₂ + 2C₂H₅OH → O=C(OC₂H₅)₂ (Diethyl carbonate) + 2HCl
(x) CH₃Br + KCN → CH₃CN + KBr ;   CH₃CN + 2H₂O \( \xrightarrow{\text{H}^+} \) CH₃COOH + NH₃
(xi) C₆H₅Cl + 2NH₃ \( \xrightarrow{\text{Cu}_2\text{O},\ 475\ \text{K},\ 60\ \text{atm}} \) C₆H₅NH₂ + NH₄Cl
(xii) C₆H₅Cl + 2Na + ClCH₃ \( \xrightarrow{\text{Dry ether}} \) C₆H₅CH₃ + 2NaCl (Wurtz-Fittig reaction)
(xiii) CH₃CH=CH₂ + HBr \( \xrightarrow{\text{Peroxide}} \) CH₃CH₂CH₂Br
(xiv) CH₂=CH₂ + HBr → CH₃CH₂Br
(xv) C₆H₆ + Br₂ \( \xrightarrow{\text{Anhyd. FeBr}_3} \) C₆H₅Br + HBr ;   2C₆H₅Br + 2Na \( \xrightarrow{\text{Dry ether}} \) C₆H₅–C₆H₅ + 2NaBr (Fittig reaction)
(xvi) C₆H₆ + CH₃CH₂Br \( \xrightarrow{\text{Anhyd. AlCl}_3} \) C₆H₅CH₂CH₃ + HBr (Friedel-Crafts alkylation)
(xvii) C₆H₅Cl + NaOH \( \xrightarrow{623\ \text{K},\ 300\ \text{atm}} \) C₆H₅ONa \( \xrightarrow{\text{H}^+} \) C₆H₅OH (Dow's process)

Teacher's Note:
1. For each conversion first compare the number of carbons in the starting material and product; a change tells you whether a C–C bond-forming step is needed.
2. Reagents and conditions written over the arrows carry marks; never leave the arrows bare.

Question 38. Give at least one characteristic test which can distinguish between
(i) Chlorobenzene (C₆H₅Cl) and cyclohexyl chloride (C₆H₁₁Cl)
(ii) Chlorobenzene and benzyl chloride
(iii) n-Propyl bromide and isopropyl bromide
(iv) Chlorobenzene and n-hexyl chloride.
(v) CH₂ = CH–CH₂Br and CH₃CH₂CH₂Br
(vi) Ethyl bromide and bromobenzene
(vii) o-Chlorotoluene and benzyl chloride
(viii) o-Chlorobenzyl bromide and o-bromobenzyl chloride
(ix) Benzyl chloride and benzyl bromide.
(x) Bromobenzene and benzyl bromide
(xi) C₂H₅Br and C₂H₅Cl
Answer:
(i) Boil each compound with aq. NaOH, acidify with dil. HNO₃ and add AgNO₃ solution. Cyclohexyl chloride gives a white ppt. of AgCl while chlorobenzene does not (its C–Cl bond is not hydrolysed under these conditions).
(ii) Same test : benzyl chloride gets hydrolysed easily and gives a white ppt. of AgCl, while chlorobenzene gives no ppt.
(iii) Hydrolyse each with aq. KOH and carry out the iodoform test (I₂ + NaOH) on the alcohol formed. Isopropyl bromide gives propan-2-ol which gives a yellow ppt. of iodoform; n-propyl bromide gives propan-1-ol which does not.
(iv) Boil with aq. NaOH, acidify with dil. HNO₃ and add AgNO₃. n-Hexyl chloride gives a white ppt. of AgCl; chlorobenzene does not.
(v) Add bromine water (or Br₂ in CCl₄). CH₂ = CH–CH₂Br (allyl bromide), being unsaturated, decolourises it; CH₃CH₂CH₂Br does not.
(vi) Boil with aq. KOH, acidify with dil. HNO₃ and add AgNO₃. Ethyl bromide gives a pale yellow ppt. of AgBr; bromobenzene does not.
(vii) Same test : benzyl chloride gives a white ppt. of AgCl; o-chlorotoluene (Cl on the ring) does not.
(viii) Boil with aq. NaOH, acidify with dil. HNO₃ and add AgNO₃. Only the side-chain halogen is hydrolysed. o-Chlorobenzyl bromide gives a pale yellow ppt. of AgBr (partially soluble in NH₄OH) while o-bromobenzyl chloride gives a white ppt. of AgCl (soluble in NH₄OH).
(ix) Same test : benzyl chloride gives a white ppt. of AgCl soluble in ammonia; benzyl bromide gives a pale yellow ppt. of AgBr partially soluble in ammonia.
(x) Same test : benzyl bromide gives a pale yellow ppt. of AgBr; bromobenzene does not.
(xi) Make the Lassaigne's extract of C₂H₅Br and C₂H₅Cl separately. Acidify them with dil. HNO₃ and add AgNO₃ solution. C₂H₅Br gives pale yellow ppt. of AgBr partially soluble in ammonia while C₂H₅Cl gives white ppt. of AgCl soluble in ammonia

Teacher's Note:
1. Haloalkanes give no ppt. with AgNO₃ directly because the halogen is covalently bonded; it must first be released as X⁻ by boiling with alkali.
2. Always acidify with HNO₃ before adding AgNO₃, otherwise excess OH⁻ gives a brown ppt. of Ag₂O.

Question 39. Complete the following reaction equations :
(i) CH₃CH₂Br + H₂ \( \xrightarrow{\text{Ni}} \) (Hr. Board 2011)
(ii) CH₃CH₂Br \( \xrightarrow{\text{Alc. KCN}} \) (West Bengal 2012)
(iii) CH₃CH₂Br \( \xrightarrow{\text{Alc. AgCN}} \) (West Bengal 2012)
(iv) CH₃CH₂CH₂Br \( \xrightarrow{\text{Alc. KOH},\ \Delta} \) (UP Board 2012)
(v) CH₃CH₂Br + KOH (aq) → (Hr. Board 2012)
(vi) C₆H₅Cl + H₂SO₄ \( \xrightarrow{\Delta} \) ........ (Hr. Board 2011)
(vii) C₆H₅–Cl + Na → (Hr. Board 2011)
(viii) C₆H₅Cl + CH₃COCl \( \xrightarrow{\text{Anhyd. AlCl}_3} \) (Assam Board 2012)
Answer:
(i) CH₃CH₂Br + H₂ \( \xrightarrow{\text{Ni}} \) CH₃CH₃ + HBr
(ii) CH₃CH₂Br + KCN \( \xrightarrow{\text{alc.}} \) CH₃CH₂CN (Propanenitrile) + KBr
(iii) CH₃CH₂Br + AgCN \( \xrightarrow{\text{alc.}} \) CH₃CH₂NC (Ethyl isocyanide) + AgBr
(iv) CH₃CH₂CH₂Br + KOH (alc.) \( \xrightarrow{\Delta} \) CH₃CH=CH₂ + KBr + H₂O
(v) CH₃CH₂Br + KOH (aq) → CH₃CH₂OH + KBr
(vi) C₆H₅Cl + H₂SO₄ \( \xrightarrow{\Delta} \) p-ClC₆H₄SO₃H (4-Chlorobenzenesulphonic acid, major) + o-ClC₆H₄SO₃H (minor) + H₂O
(vii) 2C₆H₅–Cl + 2Na \( \xrightarrow{\text{Dry ether}} \) C₆H₅–C₆H₅ (Biphenyl) + 2NaCl (Fittig reaction)
(viii) C₆H₅Cl + CH₃COCl \( \xrightarrow{\text{Anhyd. AlCl}_3} \) p-ClC₆H₄COCH₃ (4-Chloroacetophenone, major) + o-ClC₆H₄COCH₃ (minor) + HCl

Teacher's Note:
1. Electrophilic substitution of chlorobenzene always gives a para (major) + ortho (minor) mixture; write both.

Question 40. Give a brief account of the following with one example of each.
(i) Markovnikov's rule (Manipur Board 2012)
(ii) Kharasch effect
(iii) Swarts reaction (Hr. Board 2012, 2013)
(iv) Finkelstein reaction (Hr. Board 2012, 2013)
(v) Sandmeyer reaction (Hr. Board 2012, 2013)
(vi) Balz-Schiemann reaction.
(vii) Wurtz reaction (CBSE 2007 ; AP Board 2012 ; Hr. Board 2012, 2013 ; Manipur Board 2012)
(viii) Dow's process (Hr. Board 2012)
(ix) Friedel-Crafts reaction (Hr. Board 2010 ; Manipur Board 2010 ; CBSE 2010)
(x) Friedel-Crafts acylation.
(xi) Friedel-Crafts alkylation. (AP Board 2012 ; MP Board 2012)
(xii) Wurtz-Fittig reaction (CBSE 2007 ; HP Board 2011, 2012 ; Hr. Board 2013)
Answer:
(i) Markovnikov's rule : When an unsymmetrical reagent (HX) adds to an unsymmetrical alkene, the negative part of the reagent goes to that carbon of the double bond which carries the lesser number of hydrogen atoms. CH₃CH=CH₂ + HBr → CH₃CH(Br)CH₃
(ii) Kharasch effect (peroxide effect) : In presence of peroxides, addition of HBr to an unsymmetrical alkene takes place contrary to Markovnikov's rule. CH₃CH=CH₂ + HBr \( \xrightarrow{\text{Peroxide}} \) CH₃CH₂CH₂Br
(iii) Swarts reaction : Alkyl fluorides are prepared by heating alkyl chlorides or bromides with metallic fluorides such as AgF, Hg₂F₂, CoF₂ or SbF₃. CH₃Br + AgF → CH₃F + AgBr
(iv) Finkelstein reaction : Alkyl iodides are prepared by heating alkyl chlorides or bromides with NaI in dry acetone. C₂H₅Br + NaI \( \xrightarrow{\text{Dry acetone}} \) C₂H₅I + NaBr
(v) Sandmeyer reaction : A diazonium salt on treatment with cuprous chloride or cuprous bromide dissolved in the corresponding halogen acid gives the haloarene. C₆H₅N₂⁺Cl⁻ \( \xrightarrow{\text{Cu}_2\text{Cl}_2\text{/HCl}} \) C₆H₅Cl + N₂
(vi) Balz-Schiemann reaction : Benzenediazonium chloride is treated with fluoroboric acid to give the diazonium fluoroborate which on heating gives fluorobenzene. C₆H₅N₂⁺Cl⁻ + HBF₄ → C₆H₅N₂⁺BF₄⁻ \( \xrightarrow{\Delta} \) C₆H₅F + BF₃ + N₂
(vii) Wurtz reaction : Alkyl halides react with sodium in dry ether to form alkanes containing double the number of carbon atoms. 2CH₃Br + 2Na \( \xrightarrow{\text{Dry ether}} \) CH₃–CH₃ + 2NaBr
(viii) Dow's process : Chlorobenzene is heated with aqueous NaOH at 623 K and 300 atm to form sodium phenoxide which on acidification gives phenol. C₆H₅Cl + NaOH \( \xrightarrow{623\ \text{K},\ 300\ \text{atm}} \) C₆H₅ONa \( \xrightarrow{\text{H}^+} \) C₆H₅OH
(ix) Friedel-Crafts reaction : Introduction of an alkyl or acyl group into the benzene ring by treating it with an alkyl halide or acyl halide in presence of anhydrous AlCl₃. It is of two types, alkylation and acylation (see (x) and (xi)).
(x) Friedel-Crafts acylation : C₆H₅Cl + CH₃COCl \( \xrightarrow{\text{Anhyd. AlCl}_3} \) p-ClC₆H₄COCH₃ (major) + o-ClC₆H₄COCH₃ (minor) + HCl
(xi) Friedel-Crafts alkylation : C₆H₅Cl + CH₃Cl \( \xrightarrow{\text{Anhyd. AlCl}_3} \) p-ClC₆H₄CH₃ (major) + o-ClC₆H₄CH₃ (minor) + HCl
(xii) Wurtz-Fittig reaction : A mixture of an alkyl halide and an aryl halide reacts with sodium in dry ether to give an alkylarene. C₆H₅Cl + 2Na + ClCH₃ \( \xrightarrow{\text{Dry ether}} \) C₆H₅CH₃ + 2NaCl

Teacher's Note:
1. Learn each named reaction as a pair: the reagent/condition and the type of product (e.g., Finkelstein = NaI/dry acetone = alkyl iodide).
2. Peroxide effect applies only to HBr; Swarts and Finkelstein are both halogen exchange reactions.

Question 41. Identify the compounds X, Y and Z in each of the following sequence of reactions :
(i) C₂H₄ \( \xrightarrow{\text{HBr}} \) X \( \xrightarrow{\text{aq. KOH}} \) Y \( \xrightarrow{\text{I}_2\text{.NaOH}} \) Z
(ii) C₂H₅Br \( \xrightarrow{\text{alc. KOH}} \) X \( \xrightarrow{\text{Br}_2} \) Y \( \xrightarrow{\text{alc. KOH}} \) Z
(iii) C₆H₆ \( \xrightarrow{\text{Cl}_2\text{/Fe}} \) X \( \xrightarrow{\text{CuCN, pyridine},\ \Delta} \) Y \( \xrightarrow{\text{H}^+,\ \text{H}_2\text{O}} \) Z
(iv) C₆H₅NH₂ \( \xrightarrow{\text{NaNO}_2,\ \text{HCl},\ 273\text{--}278\ \text{K}} \) X \( \xrightarrow{\text{CuBr/HBr}} \) Y \( \xrightarrow{(\text{CH}_3\text{CO})_2\text{O} + \text{Anhyd. AlCl}_3} \) Z (major product)
(v) C₆H₆ \( \xrightarrow{\text{CH}_3\text{Cl} + \text{Anhyd. AlCl}_3} \) X \( \xrightarrow{\text{Cl}_2\ (1\ \text{mole}),\ \text{sunlight}} \) Y \( \xrightarrow{\text{KOH}} \) Z
Answer:
(i) X = CH₃CH₂Br (Bromoethane), Y = CH₃CH₂OH (Ethanol), Z = CHI₃ (Iodoform)
(ii) X = CH₂=CH₂ (Ethene), Y = BrCH₂CH₂Br (1, 2-Dibromoethane), Z = CH₂=CHBr (Vinyl bromide)
(iii) X = C₆H₅Cl (Chlorobenzene), Y = C₆H₅CN (Benzonitrile), Z = C₆H₅COOH (Benzoic acid)
(iv) X = C₆H₅N₂⁺Cl⁻ (Benzenediazonium chloride), Y = C₆H₅Br (Bromobenzene), Z = p-BrC₆H₄COCH₃ (4-Bromoacetophenone)
(v) X = C₆H₅CH₃ (Toluene), Y = C₆H₅CH₂Cl (Benzyl chloride), Z = C₆H₅CH₂OH (Benzyl alcohol)

Teacher's Note:
1. In (ii) one molecule of alc. KOH removes only one HBr from the dibromide, giving vinyl bromide; acetylene needs a second, stronger base step (NaNH₂).
2. In (v) sunlight means side-chain substitution, so Y is benzyl chloride, not chlorotoluene.

Question 42. Identify A and B in each of the following sequences :
(i) CH₃–CH = CH₂ \( \xrightarrow{\text{HBr}} \) A \( \xrightarrow{\text{Aq. KOH}} \) B (Jharkhand Board 2012)
(ii) C₆H₅Cl \( \xrightarrow{\text{NaOH},\ 623\ \text{K, Pressure}} \) A \( \xrightarrow{\text{H}^+/\text{H}_2\text{O}} \) B (Jharkhand Board 2012)
(iii) C₆H₁₁–Br (Cyclohexyl bromide) + Mg \( \xrightarrow{\text{Dry ether}} \) A \( \xrightarrow{\text{H}_2\text{O}} \) B (Assam Board 2012)
(iv) C₆H₅NH₂ \( \xrightarrow{\text{NaNO}_2/\text{HCl},\ 273\text{--}278\ \text{K}} \) A \( \xrightarrow{\text{CuBr/HBr}} \) B (Manipur Board 2012)
Answer:
(i) A = CH₃CH(Br)CH₃ (2-Bromopropane), B = CH₃CH(OH)CH₃ (Propan-2-ol)
(ii) A = C₆H₅ONa (Sodium phenoxide), B = C₆H₅OH (Phenol)
(iii) A = C₆H₁₁MgBr (Cyclohexylmagnesium bromide), B = C₆H₁₂ (Cyclohexane)
(iv) A = C₆H₅N₂⁺Cl⁻ (Benzenediazonium chloride), B = C₆H₅Br (Bromobenzene)

Teacher's Note:
1. Grignard reagents react with water (any compound with active H) to give the alkane; this is why they are made in dry ether.

Long Answer Questions (5 Marks)

Question 43. (a) With the help of 'hybridisation of carbon atom of C–X bond' show that aryl halides are less reactive than alkyl halides.
(b) How does chlorobenzene react with :
(i) Lithium (ii) HNO₃ (Pb. Board 2012)
Answer:
(a) In alkyl halides, the carbon of the C–X bond is sp³⁻hybridized (25% s-character), while in aryl halides it is sp²⁻hybridized (33% s-character). Greater the s-character, smaller is the orbital and more electronegative is the carbon. Therefore, the sp²⁻carbon of aryl halides holds the electron pair of the C–X bond more tightly; the C–X bond is shorter and stronger (C–Cl : 169 pm in chlorobenzene and 177 pm in CH₃Cl) and less polar than in alkyl halides. Hence, it is difficult to break the C–X bond in aryl halides and they are less reactive than alkyl halides towards nucleophilic substitution.
(b) (i) Chlorobenzene reacts with lithium in dry ether to form phenyllithium :
C₆H₅Cl + 2Li \( \xrightarrow{\text{Dry ether}} \) C₆H₅Li + LiCl
(ii) Chlorobenzene on nitration with conc. HNO₃ in presence of conc. H₂SO₄ gives a mixture of p- and o-nitrochlorobenzene :
C₆H₅Cl + HNO₃ \( \xrightarrow{\text{Conc. H}_2\text{SO}_4} \) p-ClC₆H₄NO₂ (1-Chloro-4-nitrobenzene, major) + o-ClC₆H₄NO₂ (1-Chloro-2-nitrobenzene, minor) + H₂O

Teacher's Note:
1. Give the percentage s-character figures (25% vs 33%) and the bond lengths; numbers make the hybridisation argument convincing.

Question 44. Comment upon low reactivity of haloarenes. How can this low reactivity be increased ?
Answer:
Haloarenes are much less reactive than haloalkanes towards nucleophilic substitution reactions due to the following reasons :
(i) Resonance effect : The lone pairs of the halogen are delocalised into the benzene ring. As a result, the C–X bond acquires partial double bond character and is difficult to break.
(ii) Hybridisation : The carbon of the C–X bond is sp²⁻hybridized; the C–X bond is shorter and stronger than in haloalkanes.
(iii) Instability of phenyl cation : The phenyl cation is not stabilized by resonance, so S_N1 is not possible.
(iv) Repulsion : The electron-rich ring repels the attacking nucleophile, and back side attack (S_N2) is blocked by the ring.
Thus, chlorobenzene is converted into phenol only on heating with NaOH at 623 K and 300 atm.
Increasing the reactivity : The reactivity of haloarenes increases when electron-withdrawing groups such as –NO₂, –CN, etc. are present at ortho- and/or para-positions to the halogen. These groups withdraw electrons from the ring, help the attack of the nucleophile and stabilise the intermediate carbanion by resonance. Greater the number of such groups, greater is the reactivity :
C₆H₅Cl \( \xrightarrow{\text{NaOH},\ 623\ \text{K},\ 300\ \text{atm}} \) phenol ;   p-nitrochlorobenzene \( \xrightarrow{\text{aq. NaOH},\ 443\ \text{K}} \) p-nitrophenol ;
2, 4-dinitrochlorobenzene \( \xrightarrow{\text{aq. NaOH},\ 368\ \text{K}} \) 2, 4-dinitrophenol ;   2, 4, 6-trinitrochlorobenzene \( \xrightarrow{\text{warm H}_2\text{O}} \) 2, 4, 6-trinitrophenol (picric acid)

Teacher's Note:
1. A nitro group at the meta position has almost no activating effect because it cannot delocalise the negative charge of the carbanion.