CBSE Class 12 Chemistry Haloalkanes and Haloarenes Case Study Questions Set 01

Case Study 1

The polarity of C–X bond is responsible for the nucleophilic substitution reactions of alkyl halides which mostly occur by S_N1 and S_N2 mechanisms. The rates of S_N1 reaction are governed by the stability of intermediate carbocations while that of S_N2 reactions are governed by steric factors. Chirality has a great role in understanding the mechanism of S_N1 and S_N2 reactions. S_N1 reactions of chiral alkyl halides are accompanied by racemisation. Whereas S_N2 reactions are characterised by inversion of configuration.

Question 1. What do you mean by chirality?
Answer:
The objects which are non-superimposable on their mirror images are called chiral and the property is known as chirality.

Teacher's Note:
1. For molecules, a carbon atom carrying four different groups is the usual source of chirality.

Question 2. Which of the following compounds would undergo S_N1 reaction faster and why?
C₆H₁₁–CH₂–Cl (cyclohexylmethyl chloride)   or   C₆H₅–CH₂–Cl (benzyl chloride) [CBSE 2020]
Answer:
Benzyl chloride undergoes S_N1 reaction faster than cyclohexyl methyl chloride because in case of benzyl chloride, the carbocation formed after the loss of Cl⁻ is stabilised by resonance, whereas, no such stabilisation is possible in the carbocation obtained from cyclohexyl methyl chloride.

Teacher's Note:
1. Both are primary halides; the difference comes only from resonance of the benzyl cation with the ring.

Question 3. In the pair, (CH₃)₃C–Br and CH₃–CH₂–Br, which one is more reactive towards S_N2 reaction and why?
Answer:
CH₃–CH₂–Br is more reactive.
In CH₃–CH₂–Br, the carbon carrying Br atom is attached to one methyl group: less steric hindrance, more reactive.
In (CH₃)₃C–Br, the carbon carrying Br atom is attached to three bigger methyl groups: more steric hindrance, less reactive.

Teacher's Note:
1. For S_N2 the order is 1° > 2° > 3° – exactly the reverse of S_N1.

Question 4. What is meant by the term racemic mixture?
Answer:
An equimolar mixture of a pair of enantiomers is known as racemic mixture.

Teacher's Note:
1. A racemic mixture shows zero rotation because the rotations of the two enantiomers cancel each other.

Question 5. Predict the order of reactivity of the four isomeric bromobutanes in S_N1 and S_N2 reactions.
Answer:
CH₃CH₂CH₂CH₂Br < (CH₃)₂CHCH₂Br < CH₃CH₂CH(Br)CH₃ < (CH₃)₃CBr (S_N1)
CH₃CH₂CH₂CH₂Br > (CH₃)₂CHCH₂Br > CH₃CH₂CH(Br)CH₃ > (CH₃)₃CBr (S_N2)
Of the two primary bromides, the carbocation intermediate derived from (CH₃)₂CHCH₂Br is more stable than that derived from CH₃CH₂CH₂CH₂Br because of greater electron donating inductive effect of (CH₃)₂CH group. So, (CH₃)₂CHCH₂Br is more reactive than CH₃CH₂CH₂CH₂Br in S_N1 reactions. CH₃CH₂CH(Br)CH₃ is a secondary bromide and (CH₃)₃CBr is a tertiary bromide. Thus, the above order is followed in S_N1. The reactivity in S_N2 reactions follows the reverse order as the steric hindrance around the electrophilic carbon increases in that order.

Teacher's Note:
1. Write the S_N2 order as the exact reverse of the S_N1 order – examiners look for both orders with one line of reason each.

Case Study 2

Grignard reagents are alkyl magnesium halides. Due to large electronegativity difference between carbon and magnesium, the carbon-magnesium bond has significant ionic character. The hydrocarbon part of the Grignard reagents acts as a source of carbanions. Therefore, Grignard reagents readily undergo nucleophilic addition reactions to aldehydes and ketones forming the addition products which upon hydrolysis yield alcohols.

Question 6. Give the mechanism of addition of Grignard reagent to carbonyl compound forming an alcohol.
Answer:
Step I: Nucleophilic addition of Grignard reagent to carbonyl group.
>C^δ+=O^δ− + R^δ−–Mg^δ+–X → [>C(R)–O⁻Mg⁺X] (Adduct)
Step II: Hydrolysis
[>C(R)–O⁻Mg⁺X] + H₂O → >C(R)–OH + Mg(OH)X

Teacher's Note:
1. The carbanion-like R group of RMgX attacks the electron-deficient carbonyl carbon; the new C–C bond is the whole point of the reaction.

Question 7. How will you convert propanone to 2-methylpropan-2-ol?
Answer:
CH₃–CO–CH₃ (Propanone) + CH₃MgBr (Methyl magnesium bromide) → [(CH₃)₃C–O⁻Mg⁺Br] →^+H₂O (CH₃)₃C–OH (2-methylpropan-2-ol)

Teacher's Note:
1. Ketones give tertiary alcohols with Grignard reagents, aldehydes (other than HCHO) give secondary alcohols.

Question 8. Write the structures of A and B in the following reactions:
CH₃–CH₂–OH →^{Cu, 573 K} A →^{(i) CH₃MgBr, (ii) H₂O} B
Answer:
A = CH₃–CHO (Ethanal); B = CH₃–CH(OH)–CH₃ (Propan-2-ol)

Teacher's Note:
1. Cu at 573 K dehydrogenates a primary alcohol to an aldehyde; adding CH₃MgBr then adds one carbon and gives a 2° alcohol.

Question 9. Complete the following:
Bromobenzene →^{Mg, dry ether} .......... →^{(i) CO₂, (ii) H₂O/H+} ..........
Answer:
C₆H₅MgBr (Phenyl magnesium bromide) and C₆H₅COOH (Benzoic acid)

Teacher's Note:
1. Grignard reagent + CO₂ followed by hydrolysis is a standard way to make a carboxylic acid with one extra carbon.

Question 10. What will be the product formed when chlorobenzene reacts with magnesium in presence of dry ether?
Answer:
Phenyl magnesium chloride (Grignard reagent), C₆H₅MgCl.

Teacher's Note:
1. Dry ether is essential: any moisture would destroy the Grignard reagent and give benzene.

Case Study 3

In the laboratory, haloalkanes (except fluoroalkanes) can be prepared either by the action of phosphorus halides on alcohols or halogen acids on alcohols, alkenes and alkynes. However, addition of HBr to alkenes and alkynes in presence and absence of peroxides often gives isomeric products. Chloroalkanes are, however best prepared by the action of thionyl chloride on alcohols.
Although direct halogenation (except fluorination) of alkanes usually gives a mixture of isomeric haloalkanes yet it may be a useful method in some cases. In addition, certain specific reactions such as Finkelstein and Hunsdiecker reactions are also useful methods for preparation of haloalkanes.

Question 11. Explain why thionyl chloride method is preferred for preparing alkyl chlorides from alcohols ?
Answer:
Because the by products of the reaction, i.e., SO₂ and HCl being gases escape into the atmosphere leaving behind alkyl chlorides in almost pure state.
R–OH + SOCl₂ \( \xrightarrow{\text{Pyridine}} \) R–Cl + SO₂↑ + HCl↑

Teacher's Note:
1. No separation step is needed because both by-products are gases; this is the key word in the answer.

Question 12. Under what conditions, 2-methylpropene can be converted into isobutyl bromide by hydrogen bromide ?
Answer:
In presence of peroxides, anti-Markovnikov's addition of HBr to 2-methylpropene gives isobutyl bromide.
CH₃–C(CH₃)=CH₂ (2-Methylpropene) + HBr \( \xrightarrow{\text{ROOR}\ (\text{Anti-Mark addition})} \) CH₃–CH(CH₃)–CH₂Br (Isobutyl bromide)

Teacher's Note:
1. Without peroxide the same reaction gives tert-butyl bromide through the 3° carbocation.

Question 13. Among the isomers of pentane (C₅H₁₂), write the one which on photochemical chlorination yields a single monohalide ?
Answer:
Pentane (C₅H₁₂) has the following three isomers : CH₃CH₂CH₂CH₂CH₃ (n-Pentane), CH₃–CH(CH₃)–CH₂CH₃ (Isopentane) and CH₃–C(CH₃)₂–CH₃ (Neopentane).
All the twelve hydrogens of neopentane are equivalent, therefore, replacement of any one of them by Cl gives the same product, i.e., neopentyl chloride.
C(CH₃)₄ (Neopentane) \( \xrightarrow{\text{Cl}_2,\ h\nu,\ -\text{HCl}} \) (CH₃)₃C–CH₂Cl (Neopentyl chloride)
Thus, neopentane on photochemical chlorination yields a single monohalide.

Teacher's Note:
1. The number of monochloro products equals the number of sets of equivalent hydrogens: 3 for n-pentane, 4 for isopentane, 1 for neopentane.

Question 14. Write equation for the preparation of 1-iodobutane from 1-chlorobutane.
Answer:
It can be easily prepared by Finkelstein reaction
CH₃CH₂CH₂CH₂–Cl (1-Chlorobutane) + NaI \( \xrightarrow{\text{Acetone},\ \Delta\ (\text{Finkelstein reaction})} \) CH₃CH₂CH₂CH₂–I (1-Iodobutane) + NaCl

Teacher's Note:
1. Dry acetone is essential: NaCl is insoluble in it and precipitates, driving the equilibrium forward.

Case Study 4

The characteristic reactions of haloalkanes are nucleophilic substitution reactions which occur by two different mechanisms : S_N1 and S_N2. S_N1 reactions are two-step reactions and occur through carbocation intermediates. Their reactivity depends upon the relative stability of carbocations.
In contrast, S_N2 reactions are concerted reactions which occur through a transition state. Their relative rates depend upon steric factors. An important difference between S_N1 and S_N2 reactions is that whereas S_N1 reactions of optically active haloalkanes usually give racemic products with some degree of inversion but S_N2 reactions are always accompanied by only inversion of configuration.

Question 15. Write the mechanism of S_N1 reactions involving formation of ion pair and its conversion to carbocations. Which entity gives inverted product and which entity gives racemization?
Answer:
As shown below, Ion pair gives inversion product while carbocation intermediate gives racemic mixture.
R*–X (Optically active alkyl halide) ⇌ [R*⁺][X⁻] (Ion pair) ⇌ R*⁺ (Carbocation)
Ion pair + OH⁻ → inverted product (OH⁻ attacks from the side opposite to X⁻, which still shields the front side)
R*⁺ + OH⁻ (attack from both faces) → HO–R* + R*–OH (Racemic mixture)

Teacher's Note:
1. This is why S_N1 reactions of optically active halides give racemic product with some excess of the inverted isomer.

Question 16. Taking (–)-2-bromooctane as an example, write the mechanism of S_N2 reaction.
Answer:
HO⁻ attacks the carbon carrying Br from the side opposite to Br (back side attack). In the transition state, C is partly bonded to both OH and Br; as the C–O bond forms, the C–Br bond breaks and the configuration is inverted :
HO⁻ + (–)-2-Bromooctane [CH₃, H and n-C₆H₁₃ on C-2] \( \xrightarrow{\text{Heat},\ \text{S}_\text{N}2} \) [HO^δ−···C···Br^δ−] → (+)-2-Octanol (inverted configuration) + Br⁻

Teacher's Note:
1. The three groups on carbon flip over like an umbrella in the wind (Walden inversion); the product has the opposite configuration.

Question 17. With the help of a suitable optically active alkyl halide explain the terms : (i) retention and (ii) inversion.

Answer:
Consider an optically active alkyl halide CH₃–C*H(C₂H₅)–X reacting with a nucleophile Y⁻ (+Y, –X) :
(i) Retention : If the nucleophile Y takes the same position as that of X, so that the spatial arrangement of groups in the product is the same as in the reactant, the configuration is said to be retained (retention).
(ii) Inversion : If the nucleophile Y takes a position opposite to that of X, so that the product has the opposite (mirror image) arrangement of groups, the configuration is said to be inverted (inversion).

Teacher's Note:
1. Retention and inversion describe the arrangement in space; they do not tell you whether the sign of rotation (+ or −) changes.

Question 18. Write the order of reactivity of the four isomeric bromobutanes towards S_N1 reactions.
Answer:
The order of reactivity depends upon the stability of the intermediate carbocation, i.e.,
CH₃CH₂CH₂CH₂–Br < (CH₃)₂CHCH₂Br < CH₃CH₂CH(Br)CH₃ < (CH₃)₃CBr

Teacher's Note:
1. Classify first: two primary, one secondary and one tertiary bromide; the tertiary one is fastest in S_N1.

Case Study 5

Unlike haloalkanes, haloarenes cannot be prepared from phenols. However, they can be prepared by nuclear and side chain halogenation of arenes. Aryl bromides have also been prepared by Hunsdiecker reaction.
Haloarenes and vinyl halides are much less reactive than haloalkanes towards nucleophilic substitution reactions. But the presence of electron-withdrawing groups at o- and p-positions enhances their reactivity considerably.

Question 19. Why does phenol not react with PCl₅ to give chlorobenzene ?
Answer:
Due to resonance, C–OH has some double bond character and hence difficult to break. In contrast, in alcohols, R–OH is a single bond and hence easy to break.
(In phenol, the lone pair of O is delocalised into the ring : C₆H₅–OH ↔ ring carbanion with C=O⁺H, so the C–O bond is partly a double bond.)

Teacher's Note:
1. The C–O bond of phenol is also attached to an sp² carbon, which makes it shorter and stronger than in alcohols.

Question 20. What is Hunsdiecker reaction ? Give one example.
Answer:
When the silver salt of a carboxylic acid is refluxed with bromine in CCl₄, an alkyl or aryl bromide having one carbon atom less than the acid is formed, with the evolution of CO₂. This is called Hunsdiecker reaction. For example,
C₆H₅COOAg (Silver benzoate) + Br₂ \( \xrightarrow{\text{CCl}_4,\ \text{Reflux}\ (\text{Hunsdiecker reaction})} \) C₆H₅Br (Bromobenzene) + CO₂ + AgBr

Teacher's Note:
1. Hunsdiecker reaction is a decarboxylative bromination: the product always has one carbon fewer than the acid.

Question 21. With a chemical equation, explain what happens when m-iodobenzyl iodide is treated with AgNO₂ ?
Answer:
m-IC₆H₄CH₂I (m-Iodobenzyl iodide) + AgNO₂ → m-IC₆H₄CH₂NO₂ (m-Iodophenylnitromethane) + AgI
Substitution occurs in the side chain since side chain aryl halides are more reactive than aryl halides.

Teacher's Note:
1. Only the iodine on the CH₂ group is replaced; the iodine on the ring stays because of its partial double bond character.

Question 22. Why are haloarenes and vinyl halides much less reactive than haloalkanes towards nucleophilic substitution reactions ? Explain why presence of –NO₂ groups at o- and p-positions makes substituted aryl halides much more reactive than the parent aryl halide towards nucleophilic substitution reactions.
Answer:
Due to resonance C–X bond in aryl halides has some double bond character and hence difficult to break. On the other hand, C–X in alkyl halides has only single bond and hence easy to break. (The same is true of vinyl halides, CH₂=CH–X ↔ ⁻CH₂–CH=X⁺.)
o- and p-Substituted nitro aryl halides are more reactive than parent aryl halides since in the former case carbanions are stabilized by resonance over the nitro group (the negative charge is delocalised onto the oxygen atoms of NO₂, more reactive), whereas in the parent aryl halide the carbanion formed by attack of the nucleophile is not stabilized in this way (less reactive).

Teacher's Note:
1. Only o- and p-NO₂ can take the negative charge directly by resonance; m-NO₂ acts only through its −I effect.

Case Study 6

One of the important methods of preparation of alkyl halides is the addition of halogen acids to alkenes. The order of reactivity being : HI > HBr > HCl > HF. Depending upon the structure of the alkene, sometimes rearranged products are formed. The addition of HBr to alkenes is of special interest since addition of HBr to alkenes occurs in the presence or absence of peroxides giving isomeric products. Since the addition of HF to alkenes is very slow, therefore, alkyl fluorides are prepared by Swarts reaction.