CBSE Class 12 Chemistry Haloalkanes and Haloarenes Important Questions Set 01

Very Short Answer Questions (1 Mark)

Question 1. Write the IUPAC name of the following compound: [CBSE (AI) 2013]

Answer:
2,5-Dichlorotoluene / 1,4-Dichloro-2-methylbenzene.

Teacher's Note:
1. Number the ring so that the substituents get the lowest set of locants: {1,2,4} beats {1,3,4}.

Question 2. Draw the structure of the compound, 4-tert-butyl-3-iodoheptane.
Answer:
CH₃–CH₂–CH(I)–CH(C(CH₃)₃)–CH₂–CH₂–CH₃
i.e., a seven-carbon chain with I on C-3 and a tert-butyl group, –C(CH₃)₃, on C-4.

Teacher's Note:
1. After drawing, count the main chain again – “heptane” must have exactly seven carbons in the chain.

Question 3. Out of the two halides (a) and (b) shown below, which is an example of allylic halide? [CBSE (AI) 2017]

Answer:
Halide (a), in which X is attached to the carbon next to the C=C double bond (3-halocyclohexene), is an allylic halide.

Teacher's Note:
1. In (b) X is on the doubly bonded carbon itself, which makes it a vinylic halide.

Question 4. Out of C₆H₅–CHCl₂ and C₆H₅–CH₂CH₂Cl, which is an example of a benzylic halide? [CBSE (AI) 2017]
Answer:
C₆H₅–CHCl₂

Teacher's Note:
1. Benzylic means the halogen is on the sp³ carbon directly attached to the ring; in C₆H₅CH₂CH₂Cl it is one carbon too far.

Question 5. Wurtz reaction fails in case of tert-alkyl halides. Why?
Answer:
This is because tert-alkyl halides prefer to undergo dehydrohalogenation in the presence of sodium metal instead of undergoing Wurtz reaction.
(CH₃)₃C–Br (tert-Butyl bromide) + 2Na →^−NaBr (CH₃)₃C⁻Na⁺ (tert-Butyl sodium)
(CH₃)₃C⁻Na⁺ + H–CH₂–C(CH₃)₂–Br → (CH₃)₃CH (Isobutane) + CH₂=C(CH₃)–CH₃ (Isobutylene) + NaBr

Teacher's Note:
1. The tert-butyl carbanion acts as a strong base and pulls off a β-hydrogen, so elimination wins over coupling.

Question 6. What happens when bromine attacks CH₂=CH–CH₂–C≡CH? [CBSE (AI) 2012]
Answer:
The reddish brown colour of bromine is discharged.

Teacher's Note:
1. Decolourisation of bromine water is the standard test for unsaturation (C=C or C≡C).

Question 7. Identify the chiral molecule in the following pair: CH₃CH(Cl)CH₂CH₃ and ClCH₂CH₂CH₂CH₃ [CBSE (AI) 2014]
Answer:
CH₃CH(Cl)CH₂CH₃ (2-chlorobutane) is the chiral molecule.

Teacher's Note:
1. C-2 of 2-chlorobutane carries four different groups (H, Cl, CH₃, C₂H₅); no carbon of 1-chlorobutane does.

Question 8. Out of chlorobenzene and cyclohexyl chloride, which one is more reactive towards nucleophilic substitution reaction and why? [CBSE 2019 (56/2/1)]
Answer:
Cyclohexyl chloride is more reactive than chlorobenzene towards nucleophilic substitution reactions. Due to resonance the C–Cl bond in chlorobenzene acquires partial double bond character which makes the substitution of chlorine more difficult in chlorobenzene than in cyclohexyl chloride where the C–Cl bond has pure single bond character.

Teacher's Note:
1. Partial double bond character makes the C–Cl bond stronger and shorter – mention this keyword in board answers.

Question 9. Why is t-butyl bromide more reactive towards S_N1 reaction as compared to n-butyl bromide? [CBSE 2019 (56/4/1)]
Answer:
Due to higher stability of tertiary carbocation than primary carbocation.

Teacher's Note:
1. The 3° carbocation is stabilised by +I effect and hyperconjugation of three alkyl groups.

Question 10. Give one chemical test to distinguish between C₂H₅Br and C₆H₅Br.
Answer:
Hydrolysis of C₂H₅Br with aqueous KOH followed by acidification with dil. HNO₃ and subsequent treatment with AgNO₃ gives light yellow ppt. of AgBr whereas C₆H₅Br does not give this test.

Teacher's Note:
1. Acidify with HNO₃ before adding AgNO₃, otherwise excess KOH gives a brown Ag₂O precipitate.

Question 11. Give reason: Chloroform is stored in closed dark coloured bottles completely filled so that air is kept out. [CBSE Delhi 2010]
Answer:
This is because chloroform is slowly oxidised by air in the presence of light to an extremely poisonous gas phosgene.
2CHCl₃ (Chloroform) + O₂ →^Light 2COCl₂ (Phosgene) + 2HCl

Teacher's Note:
1. Both light (dark bottle) and air (filled to the top) are kept away; a little ethanol is also added to destroy any phosgene formed.

Teacher's Note:
1. Remember: gem = same carbon (geminal), vic = neighbouring carbons (vicinal).

Question 13. Write the structure of 1-Bromo-4-chlorobut-2-ene. [CBSE Delhi Set-1 2017]
Answer:
BrCH₂CH = CHCH₂Cl

Teacher's Note:
1. Number the chain so the double bond gets the locant 2; the halogens then sit on the two end carbons, C₁ and C₄.

Question 14. Write the structure of 2,4-dinitrochlorobenzene. [CBSE Delhi Set-2 2017]
Answer:
Cl is at C₁ of the benzene ring, with NO₂ groups at C₂ (ortho) and C₄ (para) to it:

Teacher's Note:
1. In names like this, the chloro carbon is C₁ and the nitro locants are counted from it.

Question 15. Write the structure of 3-Bromo-2-methylprop-1-ene. [CBSE Delhi Set-3 2017]
Answer:
BrCH₂(CH₃)C = CH₂

Teacher's Note:
1. Draw the parent prop-1-ene first (CH₂=CH–CH₃), then put CH₃ on C₂ and Br on C₃.

Question 16. Write the structure of an isomer of compound C₄H₉Br which is most reactive towards S_N1 reaction. [CBSE OD 2016]
Answer:
(CH₃)₃C–Br, i.e., 2-Bromo-2-methylpropane

Teacher's Note:
1. S_N1 reactivity follows carbocation stability, so pick the tertiary isomer.

Short Answer Questions (2 Marks)

Question 17. Draw the structure of major monohalo product in each of the following reactions:
(i) C₆H₁₁–OH (cyclohexanol) →^SOCl₂
(ii) C₆H₅–CH₂–CH=CH₂ + HBr →^Peroxide [CBSE Delhi 2014]
Answer:
(i) C₆H₁₁–OH + SOCl₂ → C₆H₁₁–Cl (chlorocyclohexane) + SO₂ + HCl
(ii) C₆H₅–CH₂–CH=CH₂ + H–Br →^Peroxide C₆H₅–CH₂–CH₂–CH₂–Br

Teacher's Note:
1. Peroxide reverses the orientation for HBr only (anti-Markovnikov, free radical); without peroxide Br would go to the middle carbon.

Question 18. Which one of the following compounds is more easily hydrolysed by KOH and why?
CH₃CHClCH₂CH₃   or   CH₃CH₂CH₂Cl [CBSE (AI) 2012]
Answer:
Due to +I effect of alkyl groups, the 2° carbonium ion CH₃–C⁺H–CH₂–CH₃ derived from sec. butyl chloride is more stable than the 1° carbonium ion CH₃–CH₂–C⁺H₂ derived from n-propyl chloride. Therefore sec. butyl chloride gets hydrolysed more easily than n-propyl chloride under S_N1 conditions.

Teacher's Note:
1. State the mechanism (S_N1) along with the carbocation stability, otherwise the comparison is incomplete.

Question 19. How the following conversion can be carried out? Ethanol to But-2-yne
Answer:
CH₃CH₂OH (Ethanol) →^+SOCl₂ CH₃CH₂Cl (Chloroethane) →^{KOH (alc.), heat} CH₂=CH₂ (Ethene) (Dehydrohalogenation)
CH₂=CH₂ →^Br₂/CCl₄ BrCH₂–CH₂Br (Electrophilic addition)
BrCH₂–CH₂Br →^{NaNH₂ in liq. NH₃, 196 K} HC≡CH →^{NaNH₂ in liq. NH₃, 196 K} Na⁺C⁻≡C⁻Na⁺ (Disodium acetylide)
Na⁺C⁻≡C⁻Na⁺ →^{CH₃Cl (excess)} CH₃–C≡C–CH₃ (But-2-yne)

Teacher's Note:
1. Count carbons first: ethanol has 2, but-2-yne has 4, so two CH₃ groups must be added – via the acetylide ion and CH₃Cl.

Question 20. Give reasons for the following:
(i) Chloroethane is insoluble in water.
(ii) Thionyl chloride method is preferred for preparing alkyl chlorides from alcohols. [CBSE 2019 (56/4/1)]
Answer:
(i) Chloroethane is unable to form hydrogen bonds with water. Hence, it is insoluble in water.
(ii) The byproducts of the reaction, i.e., SO₂ and HCl being gases escape into the atmosphere leaving behind alkyl chlorides in almost pure state.

Teacher's Note:
1. For (i), note that dissolving would need energy to break water’s H-bonds, which the weaker new attractions cannot repay.

Question 21. Differentiate between retention and inversion.
Answer:
If the relative configuration of the atoms/groups around a chiral centre in an optically active molecule remains the same before and after the reaction, the reaction is said to proceed with retention of configuration. On the other hand, if the relative configuration of the atoms/groups around a stereocentre in the product is opposite to that in the reactant, the reaction is said to proceed with inversion of configuration. For example,

Teacher's Note:
1. Retention and inversion describe relative configuration (positions of groups), not the sign of rotation (+/−), which may or may not change.

Question 22. Give reasons:
(i) C–Cl bond length in chlorobenzene is shorter than C–Cl bond length in CH₃–Cl.
(ii) S_N1 reactions are accompanied by racemisation in optically active alkyl halides. [CBSE Delhi 2016]
Answer:
(i) In chlorobenzene, C–Cl bond acquires partial double bond character while in methyl chloride, C–Cl bond has pure single bond character. As a result C–Cl bond in chlorobenzene is shorter than methyl chloride.

(ii) Carbocations are intermediate in S_N1 reactions. Carbocations being sp² hybridised are planar species, therefore, attack of nucleophile on it can occur from both front and rear with almost equal ease giving a racemic mixture.

Teacher's Note:
1. In (i) the sp² carbon of chlorobenzene (more s-character) also pulls the bond pair closer, adding to the shortening.

Question 23. (i) Which alkyl halide from the following pair is chiral and undergoes faster S_N2 reaction?
(a) CH₃CH₂CH₂CH₂Br   (b) CH₃CH₂CH(Br)CH₃
(ii) Out of S_N1 and S_N2, which reaction occurs with
(a) Inversion of configuration
(b) Racemisation [CBSE Delhi 2014]
Answer:
(i) CH₃CH₂CH(Br)CH₃, 2-bromobutane is a chiral molecule.
CH₃CH₂CH₂CH₂Br: primary halides undergo faster S_N2 reactions than secondary halides due to less steric hinderance.
(ii) (a) S_N2 reaction occurs with inversion of configuration.
(b) S_N1 reaction occurs with racemisation.

Teacher's Note:
1. The question asks two separate things: the chiral one is (b), but the faster S_N2 one is (a) – do not give a single answer.

Question 24. Predict the order of reactivity of the following compounds in S_N1 and S_N2 reactions:
C₆H₅CH₂Br, C₆H₅CH(C₆H₅)Br, C₆H₅CH(CH₃)Br, C₆H₅C(CH₃)(C₆H₅)Br
Answer:
C₆H₅C(CH₃)(C₆H₅)Br > C₆H₅CH(C₆H₅)Br > C₆H₅CH(CH₃)Br > C₆H₅CH₂Br (S_N1).
C₆H₅C(CH₃)(C₆H₅)Br < C₆H₅CH(C₆H₅)Br < C₆H₅CH(CH₃)Br < C₆H₅CH₂Br (S_N2).
Of the two secondary bromides, the carbocation intermediate obtained from C₆H₅CH(C₆H₅)Br is more stable than that obtained from C₆H₅CH(CH₃)Br because it is stabilised by two phenyl groups due to resonance. Hence, the former bromide is more reactive than the latter in S_N1 reaction. Phenyl group is bulkier than a methyl group. Thus, C₆H₅CH(C₆H₅)Br is less reactive than C₆H₅CH(CH₃)Br in S_N2 reactions.

Teacher's Note:
1. A phenyl group stabilises a carbocation (resonance) but hinders S_N2 (bulk) – the same group works in opposite directions in the two mechanisms.

Question 25. Give reasons for the following:
(i) p-nitrochlorobenzene undergoes nucleophilic substitution faster than chlorobenzene. Explain giving the resonating structures as well.
(ii) Iodoform is obtained by reaction of acetone with hypoiodite ion but not iodide ion.
Answer:
(i) In p-nitrochlorobenzene a carbanion intermediate is formed. This is stabilised by resonance as shown below.

The −I effect of nitro group further stabilises the intermediate. Hence, p-nitrochlorobenzene reacts faster than chlorobenzene.
(ii) Hypoiodite ion can act as an oxidising agent while iodide ion does not.

Teacher's Note:
1. The key resonance form is the one with the negative charge on the oxygen atoms of NO₂; this is possible only for o/p-nitro groups.

Question 26. How will you distinguish between the following pairs of compounds:
(i) Chloroform and carbon tetrachloride.
(ii) Benzyl chloride and chlorobenzene. [CBSE Sample Paper 2014]
Answer:
(i) On heating chloroform and carbon tetrachloride with aniline and ethanolic potassium hydroxide separately chloroform forms pungent smelling isocyanide but carbon tetrachloride does not form this compound.
C₆H₅NH₂ + CHCl₃ + 3KOH (ethanolic) →^Δ C₆H₅NC (Phenyl isocyanide) + 3KCl + 3H₂O
(ii) On adding sodium hydroxide and silver nitrate to both the compounds benzyl chloride forms white precipitate but chlorobenzene does not form white precipitate.
C₆H₅CH₂Cl + NaOH → C₆H₅CH₂OH + NaCl
NaCl + AgNO₃ → AgCl(s) (White ppt.) + NaNO₃

Teacher's Note:
1. The carbylamine test needs a C–H in the halide: CHCl₃ has one, CCl₄ does not.

Question 27. Give the structural formula and IUPAC name of the following compounds: (i) BHC   (ii) DDT
Answer:
(i) BHC: 1,2,3,4,5,6-Hexachlorocyclohexane, C₆H₆Cl₆ (one Cl on each ring carbon).
(ii) DDT: 1,1′-(2,2,2-trichloroethane-1,1-diyl)bis(4-chlorobenzene), (p-ClC₆H₄)₂CH–CCl₃.

Teacher's Note:
1. DDT is often expanded as p,p′-dichlorodiphenyltrichloroethane – the name itself tells you two p-chlorophenyl groups and a CCl₃.

Question 28. Answer the following questions:
(i) What is meant by chirality of a compound? Give an example.
(ii) Which one of the following compound is more easily hydrolyzed by KOH and why?
CH₃CHClCH₂CH₃ or CH₃CH₂CH₂Cl [CBSE OD 2012]
Answer:
(i) A carbon atom surrounded by 4 different groups is termed as chiral carbon and the property of being chiral is chirality.
Eg. 2-Butanol (or any other example).
(ii) CH₃CHClCH₂CH₃ will be easily hydrolysed (towards S_N1) as the cation formed in this case will be secondary, which is more stable.

Teacher's Note:
1. Strictly, chirality is the property of a molecule being non-superimposable on its mirror image; a carbon with four different groups is the usual cause.

Question 29. Which compound in each of the following pairs will react faster in S_N2 reaction with –OH?
(i) CH₃Br or CH₃I
(ii) (CH₃)₃CCl or CH₃Cl [CBSE Comptt. Delhi 2016]
Answer:
(i) CH₃I as I⁻ ion is better leaving group than Br⁻ ion.
(ii) CH₃Cl as 1° alkyl halides are more reactive than tert-alkyl halides in S_N2 reaction with OH⁻.

Teacher's Note:
1. Two separate factors are tested here: the leaving group (I⁻ > Br⁻ > Cl⁻) and steric hindrance at carbon.

Short Answer Questions (3 Marks)

Question 30. Draw the structures of the major monohalo product for each of the following reactions:
(i) p-Cl–C₆H₄–CH₂–CH₃ (1-chloro-4-ethylbenzene) →^{Br₂, heat} ?
(ii) 1-Methylcyclohexene + HBr → ?
(iii) p-HO–CH₂–C₆H₄–OH (4-hydroxybenzyl alcohol) →^{HCl, heat} ? [CBSE (F) 2017]
Answer:
(i) p-Cl–C₆H₄–CH₂–CH₃ →^{+ Br₂, heat} p-Cl–C₆H₄–CH(Br)–CH₃
(ii) 1-Methylcyclohexene + H–Br → 1-Bromo-1-methylcyclohexane (Br and CH₃ on the same ring carbon)
(iii) p-HO–CH₂–C₆H₄–OH →^{HCl, heat} p-Cl–CH₂–C₆H₄–OH

Teacher's Note:
1. In (iii) only the alcoholic OH is replaced; the phenolic C–OH bond has partial double bond character and does not react with HCl.
2. In (i) bromination with heat (no Lewis acid) is a free radical side-chain reaction at the benzylic carbon.

Question 31. Write the major product(s) in the following reactions:
(i) p-HO–C₆H₄–CH₂–OH →^PCl₅ ?   (ii) 1-Methylcyclohexene →^H–Br ?
(iii) CH₃CH₂Cl →^KNO₂ ? [CBSE (F) 2016]
Answer:
(i) p-HO–C₆H₄–CH₂–OH →^PCl₅ p-HO–C₆H₄–CH₂–Cl
(ii) 1-Methylcyclohexene →^H–Br 1-Bromo-1-methylcyclohexane
(iii) CH₃–CH₂–Cl →^KNO₂ CH₃–CH₂–ONO

Teacher's Note:
1. KNO₂ (ionic) attacks through oxygen to give alkyl nitrite, R–ONO; AgNO₂ (covalent) gives nitroalkane, R–NO₂.

Question 32. How do you convert the following:
(i) Prop-1-ene to 1-fluoropropane
(ii) Chlorobenzene to 2-chlorotoluene
(iii) Ethanol to propanenitrile [CBSE Panchkula 2015]
Answer:
(i) CH₃–CH=CH₂ (Prop-1-ene) →^{+ HBr, (C₆H₅COO)₂} CH₃–CH₂–CH₂–Br →^{AgF, Δ} CH₃–CH₂–CH₂–F (1-Fluoropropane)
(ii) C₆H₅Cl (Chlorobenzene) + CH₃–Cl (Chloromethane) →^{Anhyd. AlCl₃} o-ClC₆H₄CH₃ (2-Chlorotoluene)
(iii) CH₃–CH₂–OH (Ethanol) →^{+ PCl₅} CH₃–CH₂–Cl →^{Alc. KCN} CH₃–CH₂–C≡N (Propanenitrile)

Teacher's Note:
1. The peroxide in (i) is essential to put Br on the terminal carbon; AgF then swaps Br for F (Swarts reaction).
2. In (ii) the Friedel-Crafts reaction gives the para isomer too; the ortho isomer is separated from the mixture.

Question 33. Write IUPAC names of the following:
(i) CH₃–CH=CH–CH(Br)–CH₃   (ii) CH₂=C(CH₃)–CH(Br)–CH₃   (iii) CH₃–CH=C(CH₃)–CH(Br)–CH₃ [CBSE (AI) 2011]
Answer:
(i) 4-Bromopent-2-ene
(ii) 3-Bromo-2-methylbut-1-ene
(iii) 4-Bromo-3-methylpent-2-ene

Teacher's Note:
1. The double bond gets priority over the halogen for the lowest locant; halogen is only a prefix.

Question 34. (i) Explain why the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.
(ii) An optically active compound having molecular formula C₇H₁₅Br reacts with aqueous KOH to give a racemic mixture of products. Write the mechanism involved in this reaction. [CBSE Sample Paper 2016]
Answer:
(i) Due to greater s-character, sp² hybrid carbon atom is more electronegative than sp³ hybrid carbon atom. Therefore, the sp² hybrid carbon of C–Cl bond in chlorobenzene has less tendency to release electrons to Cl than an sp³ hybrid carbon of cyclohexyl chloride. As a result, the magnitude of negative charge is less on Cl atom of chlorobenzene than in cyclohexyl chloride.
Further, due to resonance the C–Cl bond in chlorobenzene acquires partial double bond character whereas the bond in cyclohexyl chloride is a pure single bond. In other words, C–Cl bond in chlorobenzene is shorter than in cyclohexyl chloride.
As dipole moment is a product of charge and distance. Therefore, dipole moment of chlorobenzene is less than that of cyclohexyl chloride.
(ii) Since the optically active compound, C₇H₁₅Br reacts with KOH forms a racemic mixture, therefore it must be tertiary alkyl halide and the reaction will follow S_N1 mechanism.
CH₃–CH₂–CH₂–C(CH₃)(Br)–CH₂–CH₃ + OH⁻ → CH₃–CH₂–CH₂–C(CH₃)(OH)–CH₂–CH₃ + Br⁻   [(±) 3-Methylhexan-3-ol]
Mechanism:
(a) The C–Br bond ionises in the slow step: 3-bromo-3-methylhexane ⇄^Slow planar 3° carbocation, (CH₃)(C₂H₅)(CH₃CH₂CH₂)C⁺ + Br⁻
(b) The planar carbocation is attacked by ⁻OH from either side (front or back) with equal ease, giving (+) 3-methylhexan-3-ol and (−) 3-methylhexan-3-ol in equal amounts, i.e., a racemic mixture.

Teacher's Note:
1. The tertiary carbon in 3-bromo-3-methylhexane carries CH₃, C₂H₅ and C₃H₇ groups, so it is chiral – that is why the starting compound is optically active.

Question 35. Compound ‘A’ with molecular formula C₄H₉Br is treated with aq. KOH solution. The rate of this reaction depends upon the concentration of the compound ‘A’ only. When another optically active isomer ‘B’ of this compound was treated with aq. KOH solution, the rate of reaction was found to be dependent on concentration of compound and KOH both.
(i) Write down the structural formula of both compounds ‘A’ and ‘B’.
(ii) Out of these two compounds, which one will be converted to the product with inverted configuration?
Answer:
(i) Compound A: CH₃–C(CH₃)(Br)–CH₃, i.e., (CH₃)₃C–Br;
Compound B: CH₃–CH₂–CH(Br)–CH₃
(ii) Compound ‘B’.

Teacher's Note:
1. Rate depending on [RX] only means S_N1 (best for 3°); rate depending on both means S_N2, which gives inversion.

Question 36. Predict the major product formed when HCl is added to isobutylene. Explain the mechanism involved.
Answer:
CH₃–C(CH₃)=CH₂ (Isobutylene) + HCl → CH₃–C(CH₃)(Cl)–CH₃ (2-Chloro-2-methylpropane)
The mechanism involved in this reaction is:
Step I CH₃–C(CH₃)=CH₂ →^H+ CH₃–C⁺(CH₃)–CH₃ (3° carbocation, more stable) + CH₃–CH(CH₃)–C⁺H₂ (1° carbocation, less stable)
Step II CH₃–C⁺(CH₃)–CH₃ →^Cl− CH₃–C(CH₃)(Cl)–CH₃

Teacher's Note:
1. This is Markovnikov addition: H⁺ adds to the carbon with more H atoms so that the more stable carbocation forms.

Question 37. Consider the three types of replacement of group X by group Y as shown here.
This can result in giving compound (A) or (B) or both. What is the process called if
(i) (A) is the only compound obtained?
(ii) (B) is the only compound obtained?
(iii) (A) and (B) are formed in equal proportions? [CBSE (F) 2013]

Answer:
(i) Retention
(ii) Inversion
(iii) Racemisation

Teacher's Note:
1. Compare where Y ends up: in the same position as X means retention, on the opposite side (groups flipped like an umbrella) means inversion.

Question 38. Following compounds are given to you:
2-Bromopentane, 2-Bromo-2-methylbutane, 1-Bromopentane
(i) Write the compound which is most reactive towards S_N2 reaction.
(ii) Write the compound which is optically active.
(iii) Write the compound which is most reactive towards β-elimination reaction. [CBSE Delhi 2017]
Answer:
(i) 1-Bromopentane, it is primary halide therefore undergoes S_N2 reaction faster.
(ii) 2-Bromopentane as carbon number two is asymmetric carbon.
(iii) 2-Bromo-2-methyl butane, because tertiary alkyl halides on dehydrohalogenation form most substituted alkene which is more stable.

Teacher's Note:
1. 2-Bromo-2-methylbutane is not optically active: its C-2 carries two identical CH₃ groups.

Question 39. Give reasons for the following:
(i) Benzyl chloride is highly reactive towards the S_N1 reaction.
(ii) 2-bromobutane is optically active but 1-bromobutane is optically inactive.
(iii) Electrophilic reactions in haloarenes occur slowly. [CBSE Bhubaneshwar 2015]
Answer:
(i) Benzyl chloride is highly reactive towards the S_N1 reaction because the intermediate benzyl carbocation formed in slowest step is stabilized through resonance.

(ii) 2-bromobutane is a chiral molecule as it contains an asymmetric carbon atom therefore, it is optically active whereas 1-bromobutane is an achiral molecule as it does not contain asymmetric carbon atom therefore it is optically inactive.
CH₃–CH₂–C*H(Br)–CH₃ (2-Bromobutane; C* is the asymmetric carbon)
(iii) Halogen in haloarenes withdraws electrons through –I effect and release electrons through +R effect. The inductive effect is stronger than resonance effect and causes net electron withdrawal. As a result, the electrophilic substitution reactions in haloarenes occur slowly.

Teacher's Note:
1. In the benzyl cation the positive charge spreads to the two ortho and the para carbons – draw all four structures for full marks.

Long Answer Questions (5 Marks)

Question 40. (i) Among all the isomers of molecular formula C₄H₉Br, identify
(a) the one isomer which is optically active.
(b) the one isomer which is highly reactive towards S_N2.
(c) the two isomers which give same product on dehydrohalogenation with alcoholic KOH.
(ii) Give IUPAC name of the following organic compounds:
(a) the compound shown below   (b) (CH₃)₃CCH₂Br [CBSE 2019 (56/2/1)]

Answer:
(i) (a) CH₃–CH₂–CH(Br)–CH₃ (2-Bromobutane)
(b) CH₃–CH₂–CH₂–CH₂–Br (1-Bromobutane)
(c) CH₃–CH(CH₃)–CH₂–Br (1-Bromo-2-methylpropane) and CH₃–C(CH₃)(Br)–CH₃ (2-Bromo-2-methylpropane)
(ii) (a) 2-chlorocyclopent-1-ene-1-carboxylic acid
(b) 1-bromo-2,2-dimethylpropane

Teacher's Note:
1. For (i)(c), both isomers lose HBr to give the same alkene, 2-methylpropene, (CH₃)₂C=CH₂.
2. In (ii)(a) the carbon bearing –COOH is C-1 and the double bond is numbered to pass through it.

Question 41. (i) (a) Write equation for preparation of 1-iodobutane from 1-chlorobutane.
(b) Out of 2-bromopentane, 2-bromo-2-methylbutane and 1-bromopentane, which compound is most reactive towards elimination reaction and why?
(c) Give IUPAC name of CH₃–CH=CH–C(CH₃)(Br)–CH₃ [CBSE 2019 (56/4/1)]
(ii) What are enantiomers? Draw the structures of the possible enantiomers of 3-methylpent-1-ene.
Answer:
(i) (a) CH₃CH₂CH₂CH₂Cl + NaI →^{Dry acetone} CH₃CH₂CH₂CH₂–I + NaCl
(b) 2-Bromo-2-methylbutane is more reactive as it gives more substituted alkene on elimination.
(c) 4-Bromo-4-methylpent-2-ene
(ii) Stereoisomers which are non-superimposable mirror images of each other are called enantiomers.

Teacher's Note:
1. The Finkelstein reaction in (i)(a) goes forward because NaCl is insoluble in dry acetone and precipitates out.
2. In 3-methylpent-1-ene, C-3 carries H, CH₃, C₂H₅ and CH=CH₂ – four different groups.