CBSE Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes Competency Based Questions Set 01

Multiple Choice Questions (MCQs)

Question 1. \( \text{S}_{\text{N}}1 \) reaction of alkyl halides leads to
(a) Retention of configuration
(b) Racemisation
(c) Inversion of configuration
(d) None of the options
Answer: (b) Racemisation
In simple words: During an \( \text{S}_{\text{N}}1 \) reaction, the intermediate carbocation is flat, meaning the incoming group can join from either side. This results in an equal mix of both spatial arrangements.

Exam Tip: Remember that racemisation occurs because the intermediate is a planar carbocation, allowing equal attack from both the front and back sides.

Question 2. \( p \)-dichlorobenzene has higher melting point than its \( o \)- and \( m \)- isomers because
(a) \( p \)-dichlorobenzene is more polar than \( o \)- and \( m \)- isomer.
(b) \( p \)-isomer has a symmetrical crystalline structure.
(c) boiling point of \( p \)-isomer is more than \( o \)- and \( m \)-isomer.
(d) All of the options are correct reasons.
Answer: (b) \( p \)-isomer has a symmetrical crystalline structure.
In simple words: The para-isomer is highly symmetrical, so it fits very tightly into the solid crystal structure. This makes it require more heat to melt than the other shapes.

Exam Tip: Always associate the higher melting point of para-isomers with symmetrical packing in the crystal lattice.

Question 3. Chloropicrin is formed by the reaction of
(a) steam on carbon tetrachloride.
(b) nitric acid on chlorobenzene.
(c) chlorine on picric acid.
(d) nitric acid on chloroform.
Answer: (d) nitric acid on chloroform.
In simple words: Chloropicrin is made by reacting chloroform with concentrated nitric acid. This process replaces the hydrogen atom of chloroform with a nitro group.

Exam Tip: Chloropicrin is also known as tear gas, and its synthesis from chloroform is a classic organic preparation reaction.

Question 4. Fitting reaction can be used to prepare
(a) Toluene
(b) Acetophenon
(c) Diphenyl
(d) Chlorobenzene
Answer: (c) Diphenyl
In simple words: The Fittig reaction joins two benzene rings together by heating chlorobenzene with sodium metal in dry ether. This creates diphenyl.

Exam Tip: Remember that Fittig is for coupling aryl groups, Wurtz is for alkyl groups, and Wurtz-Fittig joins one of each.

Question 5. Identify the end product (C) in the following sequence:
\( \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{SOCl}_2/\text{Pyridine}} \text{A} \xrightarrow{\text{KCN (alc.)}} \text{B} \xrightarrow{2\text{H}_2\text{O}/\text{H}^+} \text{C} \)
(a) \( \text{C}_2\text{H}_5\text{CH}_2\text{NH}_2 \)
(b) \( \text{C}_2\text{H}_5\text{CONH}_2 \)
(c) \( \text{C}_2\text{H}_5\text{COOH} \)
(d) \( \text{C}_2\text{H}_5\text{NH}_2 + \text{HCOOH} \)
Answer: (c) \( \text{C}_2\text{H}_5\text{COOH} \)
In simple words: Ethanol first turns into ethyl chloride, then into ethyl cyanide, and finally hydrolyzes with acid to become propanoic acid.

Exam Tip: Acid hydrolysis of a nitrile (\( -\text{CN} \)) group always produces a carboxylic acid (\( -\text{COOH} \)) with the same number of carbons.

Question 6. \( \text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} \xrightarrow{\text{alc. KOH}} \text{B} \xrightarrow{\text{HBr}} \text{C} \xrightarrow{\text{Na/ether}} \text{D} \)
In the above reaction, the product D is
(a) Propane
(b) 2, 3-Dimethylbutane
(c) Hexane
(d) Allyl bromide
Answer: (b) 2, 3-Dimethylbutane
In simple words: Propyl chloride undergoes elimination to form propene, which reacts with HBr to become 2-bromopropane. Finally, Wurtz coupling of two 2-bromopropanes yields 2,3-dimethylbutane.

Exam Tip: Be careful with the addition of HBr to asymmetric alkenes; Markovnikov's rule dictates that bromine attaches to the secondary carbon.

Question 7. Identify X and Y in the following sequence
\( \text{C}_2\text{H}_5\text{Br} \xrightarrow{\text{X}} \text{Product} \xrightarrow{\text{Y}} \text{C}_3\text{H}_7\text{NH}_2 \)
(a) X = KCN, Y = \( \text{LiAlH}_4 \)
(b) X = KCN, Y = \( \text{H}_3\text{O}^+ \)
(c) X = \( \text{CH}_3\text{Cl} \), Y = \( \text{AlCl}_3 \text{ HCl} \)
(d) X = \( \text{CH}_3\text{NH}_2 \), Y = \( \text{HNO}_2 \)
Answer: (a) X = KCN, Y = \( \text{LiAlH}_4 \)
In simple words: Treating ethyl bromide with potassium cyanide adds one carbon atom to the chain, and reducing the resulting nitrile with lithium aluminum hydride produces propanamine.

Exam Tip: Nitrile reduction is an excellent synthetic route to step up a carbon chain and obtain a primary amine.

Question 8. In the following sequence of reactions:
\( \text{C}_2\text{H}_5\text{Br} \xrightarrow{\text{AgCN}} \text{X} \xrightarrow{\text{Reduction}} \text{Y} \); Y is
(a) n-propylamine
(b) isopropylamine
(c) ethylamine
(d) ethylmethylamine
Answer: (d) ethylmethylamine
In simple words: Silver cyanide reacts to attach the nitrogen atom directly to the alkyl chain, forming an isocyanide. Reducing this group gives a secondary amine.

Exam Tip: Remember the critical difference: KCN forms cyanides (alkane nitriles), whereas AgCN forms isocyanides (carbylamines).

Question 9. \( \text{X} \xrightarrow{\text{AgNO}_3 / \text{HNO}_3} \text{Yellow or White ppt} \)
Which of the following cannot be X?
(a) \( \text{C}_6\text{H}_5\text{Br} \) (Bromobenzene)
(b) \( (\text{CH}_3)_3\text{Cl} \) (t-Butyl chloride)
(c) \( \text{C}_6\text{H}_5\text{CH}_2\text{Br} \) (Benzyl bromide)
(d) \( \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \) (Benzenediazonium chloride)
Answer: (a) \( \text{C}_6\text{H}_5\text{Br} \) (Bromobenzene)
In simple words: The halogen atom in bromobenzene is directly bound to the aromatic ring, making it extremely unreactive and unable to break away to form a precipitate.

Exam Tip: Aryl halides do not yield halide ions with silver nitrate solution under normal conditions because the carbon-halogen bond has double bond character.

Question 10. Identify Z in the series:
\( \text{CH}_2=\text{CH}_2 \xrightarrow{\text{HBr}} \text{X} \xrightarrow{\text{aq. KOH}} \text{Y} \xrightarrow{\text{Na}_2\text{CO}_3 / \text{I}_2\text{ excess}} \text{Z} \)
(a) \( \text{C}_2\text{H}_5\text{I} \)
(b) \( \text{C}_2\text{H}_5\text{OH} \)
(c) \( \text{CHI}_3 \)
(d) \( \text{CH}_3\text{CHO} \)
Answer: (c) \( \text{CHI}_3 \)
In simple words: Ethene reacts with HBr to form bromoethane, which then hydrolyzes to ethanol. Ethanol reacts with excess iodine and sodium carbonate to give a yellow precipitate of iodoform.

Exam Tip: Any compound containing a \( \text{CH}_3\text{CH(OH)}- \) or \( \text{CH}_3\text{CO}- \) group undergoes the haloform reaction to produce a yellow precipitate of iodoform (\( \text{CHI}_3 \)).

Assertion-Reason Questions

Directions: Mark the option which is most suitable:
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.

Question 1. Assertion: Boiling point of alkyl halides increases with increase in molecular weight.
Reason: Boiling point of alkyl halides is in the order RI > RBr > RCl > RF.
Answer: (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
In simple words: Higher molecular weight raises the boiling point because of stronger intermolecular attractive forces, but the boiling point sequence of halides is due to the size of the halogens.

Exam Tip: Note that the correct reason for the assertion is the increase in the magnitude of van der Waals' forces as molecular mass increases.

Question 2. Assertion: Boiling point of RCl is greater than RF.
Reason: R-Cl is more stable than R-F.
Answer: (c) Assertion is correct statement but reason is wrong statement.
In simple words: RCl boils at a higher temperature than RF because chlorine has a larger surface area and stronger attractions, but the R-F bond is actually more stable than R-Cl.

Exam Tip: Bond stability depends on bond dissociation energy, which is higher for the smaller C-F bond compared to the C-Cl bond.

Question 3. Assertion: n-Butyl bromide has higher boiling point than isobutyl bromide.
Reason: The branching of the chain makes the molecule compact and therefore decreases the surface area.
Answer: (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
In simple words: Branched structures are more spherical and have less outer area than straight chains. This decreases their attractive forces, making them easier to boil.

Exam Tip: For isomeric alkyl halides, boiling points decrease as branching increases because of a reduction in surface area and van der Waals' forces.

Question 4. Assertion: p-Dichlorobenzene has higher melting point than o-dichlorobenzene.
Reason: Stronger the van der Waals’ forces of attraction, higher is the melting point.
Answer: (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
In simple words: The para-isomer is very symmetrical, allowing it to pack exceptionally well into a crystal structure. This tight fit requires more energy to disrupt, causing its melting point to be much higher.

Exam Tip: High melting points in para-isomers are always attributed to crystal lattice symmetry and tight packing, not simply to overall polar forces.

Question 5. Assertion: Vinyl chloride is less reactive than alkyl chloride.
Reason: Stability of alkyl halide decreases as the strength of C-X bond decreases.
Answer: (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
In simple words: Vinyl chloride has resonance that gives the C-Cl bond double bond character, making it difficult to split. This is why it is less reactive than normal alkyl chlorides.

Exam Tip: Remember that resonance in vinyl halides creates partial double-bond character in the C-X bond, rendering it inert to nucleophilic attacks.

Question 6. Assertion: Nucleophilic substitution reaction on an optically active alkyl halide gives a mixture of enantiomers.
Reason: The reaction occurs by SN2 mechanism.
Answer: (c) Assertion is correct statement but reason is a wrong statement.
In simple words: Obtaining an equal mixture of enantiomers (racemisation) occurs during an SN1 reaction, where a planar, symmetric carbocation is attacked from either side.

Exam Tip: Recall that SN2 reactions proceed with complete inversion of configuration (Walden inversion), whereas SN1 leads to racemisation.

Question 7. Assertion: p-Dichlorobenzene is less soluble in organic solvents than the corresponding o-isomer.
Reason: o-Dichlorobenzene is polar while p-dichlorobenzene is not.
Answer: (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
In simple words: The para-isomer packs tightly in the solid state due to its symmetry, so it takes a lot of energy to break its lattice and dissolve it compared to the ortho-isomer.

Exam Tip: Solubility of organic solids is strongly influenced by lattice energy; a more symmetrical crystal lattice is harder to dissolve.

Question 8. Assertion: Tertiary haloalkanes are more reactive than primary haloalkanes towards elimination reactions.
Reason: The +I-effect of the alkyl groups weakens the C—X bond.
Answer: (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
In simple words: The multiple alkyl groups in tertiary halides push electrons toward the carbon, weakening the bond to the halogen and easily yielding highly stable, substituted alkenes.

Exam Tip: Elimination ease follows the order 3° > 2° > 1° because tertiary carbocations/alkenes are more stable.

Question 9. Assertion: Benzyl chloride is more reactive than p-chlorotoluene towards aqueous NaOH.
Reason: The C—Cl bond in benzyl chloride is more polar than C—Cl bond in p-chlorotoluene.
Answer: (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
In simple words: In benzyl chloride, the chlorine is easy to replace because its bond is highly polar and it forms a stable carbocation, whereas chlorotoluene has a very strong ring-attached bond.

Exam Tip: Benzyl carbocations are resonance stabilized, allowing benzyl halides to undergo rapid nucleophilic substitutions.

Question 10. Assertion: Lower members of alkyl halides are colourless gases.
Reason: Alkyl iodides in general turn black on exposure to air and light.
Answer: (c) Assertion is correct statement but reason is a wrong statement.
In simple words: Alkyl iodides slowly decompose when exposed to light and oxygen, turning brown (rather than black) due to the formation of free iodine.

Exam Tip: The decomposition of alkyl iodides in light yields iodine molecule (\( \text{I}_2 \)), which causes the solution to take on a brown tint.

Case Based Questions

Case 1
Nucleophilic substitution reactions are of two types; substitution nucleophilic bimolecular (\( \text{S}_{\text{N}}2 \)) and substitution nucleophilic unimolecular (\( \text{S}_{\text{N}}1 \)) depending on molecules taking part in determining the rate of reaction. Reactivity of alkyl halide towards \( \text{S}_{\text{N}}1 \) and \( \text{S}_{\text{N}}2 \) reactions depends on various factors such as steric hindrance, stability of intermediate or transition state and polarity of solvent. \( \text{S}_{\text{N}}2 \) reaction mechanism is favoured mostly by primary alkyl halide followed by secondary and tertiary. This order is reversed in case of \( \text{S}_{\text{N}}1 \) reactions.

Question (i). Which of the following is most reactive towards nucleophilic substitution reactions?
(a) \( \text{C}_6\text{H}_5\text{Cl} \)
(b) \( \text{CH}_2=\text{CHCl} \)
(c) \( \text{ClCH}_2\text{CH}=\text{CH}_2 \)
(d) \( \text{CH}_3\text{CH}=\text{CHCl} \)
Answer: (c) \( \text{ClCH}_2\text{CH}=\text{CH}_2 \)
In simple words: Allylic halides are extremely reactive because the intermediate carbocation formed after losing the halogen is highly stabilized by resonance with the double bond.

Exam Tip: Allyl and benzyl halides are highly reactive towards both SN1 and SN2 mechanisms due to resonance stabilization of the intermediates/transition states.

Question (ii). Isopropyl chloride undergoes hydrolysis by
(a) \( \text{S}_{\text{N}}1 \) mechanism
(b) \( \text{S}_{\text{N}}2 \) mechanism
(c) \( \text{S}_{\text{N}}1 \) and \( \text{S}_{\text{N}}2 \) mechanism
(d) neither \( \text{S}_{\text{N}}1 \) nor \( \text{S}_{\text{N}}2 \) mechanism
Answer: (c) \( \text{S}_{\text{N}}1 \) and \( \text{S}_{\text{N}}2 \) mechanism
In simple words: Isopropyl chloride is a secondary alkyl halide, so it can react through either pathway depending on the strength of the nucleophile and the type of solvent used.

Exam Tip: Secondary alkyl halides are on the borderline and can undergo substitution via either SN1 or SN2 depending on experimental conditions.

Question (iii). The most reactive nucleophile among the following is
(a) \( \text{CH}_3\text{O}^- \)
(b) \( \text{C}_6\text{H}_5\text{O}^- \)
(c) \( (\text{CH}_3)_2\text{CHO}^- \)
(d) \( (\text{CH}_3)_2\text{CO}^- \)
Answer: (a) \( \text{CH}_3\text{O}^- \)
In simple words: Smaller, less bulky nucleophiles can easily approach and attack the electrophilic carbon center, making them much more reactive than crowded ones.

Exam Tip: Bulky nucleophiles act as strong bases but poor nucleophiles due to steric hindrance preventing attack on carbon.

Question (iv). Tertiary alkyl halides are practically inert to substitution by \( \text{S}_{\text{N}}2 \) mechanism because of
(a) insolubility
(b) instability
(c) inductive effect
(d) steric hindrance
Answer: (d) steric hindrance
In simple words: The three bulky alkyl groups surrounding the central carbon block the path of the incoming nucleophile, preventing the back-side attack required for an \( \text{S}_{\text{N}}2 \) reaction.

Exam Tip: Always associate the complete inertness of tertiary halides towards SN2 with steric crowding of the alpha carbon.

Question (v). Which of the following is the correct order of decreasing \( \text{S}_{\text{N}}2 \) reactivity?
(a) \( \text{RCH}_2\text{X} > \text{R}_2\text{CHX} > \text{R}_3\text{CX} \)
(b) \( \text{R}_3\text{CX} > \text{R}_2\text{CHX} > \text{RCH}_2\text{X} \)
(c) \( \text{R}_2\text{CHX} > \text{R}_2\text{CX} > \text{RCH}_2\text{X} \)
(d) \( \text{RCH}_2\text{X} > \text{R}_3\text{CX} > \text{RCHX} \)
Answer: (a) \( \text{RCH}_2\text{X} > \text{R}_2\text{CHX} > \text{R}_3\text{CX} \)
In simple words: Primary alkyl halides are the least crowded and react the fastest, while tertiary alkyl halides are highly crowded and react the slowest.

Exam Tip: The rate of SN2 substitution is strictly governed by steric hindrance, hence primary is always faster than secondary and tertiary.

Case 2
A chlorocompound (A) on reduction with Zn-Cu ethanol gives the hydrocarbon (B) with five carbon atoms. When (A) is dissolved in dry ether and treated with sodium metal it gave 2,2,5,5-tetramethylhexane. The treatment of (A) with alcoholic KCN gives compound (C).

Question (i). The compound (A) is
(a) 1-chloro-2,2-dimethylpropane
(b) 1-chloro-2,2-dimethyl butane
(c) 1-chloro-2 methyl butane
(d) 2-chloro-2-methyl butane.
Answer: (a) 1-chloro-2,2-dimethylpropane
In simple words: Linking two 1-chloro-2,2-dimethylpropane units together through a Wurtz reaction yields a symmetrical ten-carbon alkane, which is 2,2,5,5-tetramethylhexane.

Exam Tip: Work backward from the symmetric dimer to identify the monomer unit in a Wurtz coupling reaction.

Question (ii). The reaction of (C) with Na, \( \text{C}_2\text{H}_5\text{OH} \) gives
(a) \( (\text{CH}_3)_3\text{C} \text{CH}_2\text{CONH}_2 \)
(b) \( (\text{CH}_3)_3\text{C} \text{NH}_2 \)
(c) \( (\text{CH}_3)_3\text{C} \text{CH}_2\text{CH}_2\text{NH}_2 \)
(d) \( (\text{CH}_3)_2\text{CHCH}_2\text{NH}_2 \)
Answer: (c) \( (\text{CH}_3)_3\text{C} \text{CH}_2\text{CH}_2\text{NH}_2 \)
In simple words: Reaction of 1-chloro-2,2-dimethylpropane with cyanide yields a nitrile. Reducing this nitrile group with sodium in ethanol yields a primary amine containing an additional carbon.

Exam Tip: Nitrile reduction converts the \( -\text{CN} \) group into a primary amine \( -\text{CH}_2\text{NH}_2 \).

Question (iii). The reaction of (C) with Na, \( \text{C}_2\text{H}_5\text{OH} \) is called
(a) Gilman reaction
(b) Mendius reaction
(c) Grooves process
(d) Swart’s reaction
Answer: (b) Mendius reaction
In simple words: The reduction of organic nitriles to primary amines using sodium in an alcohol solvent is specifically known as the Mendius reaction.

Exam Tip: Be sure to memorize named reduction reactions like Mendius reduction for converting cyanides to amines.

Question (iv). The reaction of (A) with aq. KOH will preferably favour
(a) \( \text{S}_{\text{N}}1 \) mechanism
(b) \( \text{S}_{\text{N}}2 \) mechanism
(c) \( \text{E}1 \) mechanism
(d) \( \text{E}2 \) mechanism
Answer: (a) \( \text{S}_{\text{N}}1 \) mechanism
In simple words: The bulky neopentyl structure of compound (A) hinders direct attack by a nucleophile, forcing the substitution to take place via a carbocation intermediate.

Exam Tip: Neopentyl-type systems are highly hindered toward back-side attack, making them undergo substitution predominantly via SN1.

Question (v). Compound (B) is
(a) n-pentane
(b) 2, 2-dimethylpropane
(c) 2-methylbutane
(d) None of the options
Answer: (b) 2, 2-dimethylpropane
In simple words: Reducing 1-chloro-2,2-dimethylpropane replaces its chlorine atom with a hydrogen atom, yielding the branched five-carbon alkane 2,2-dimethylpropane.

Exam Tip: Reduction of an alkyl halide with Zn-Cu/ethanol replaces the halogen atom with a hydrogen atom to form the corresponding alkane.

Case 3
When haloalkanes with \( \beta \)-hydrogen atom are boiled with alcoholic solution of KOH, they undergo elimination of hydrogen halide resulting in the formation of alkenes. These reactions are called \( \beta \)-elimination reactions or dehydrohalogenation reactions. These reactions follow Saytzeff’s rule. Substitution and elimination reactions often compete with each other. Mostly bases behave as nucleophiles and therefore can engage in substitution or elimination reactions depending upon the alkyl halide and the reaction conditions.

Question (i). Among the following the most reactive towards alcoholic KOH is
(a) \( \text{CH}_2=\text{CHBr} \)
(b) \( \text{CH}_3\text{COCH}_2\text{CH}_2\text{Br} \)
(c) \( \text{CH}_3\text{CH}_2\text{Br} \)
(d) \( \text{CH}_3\text{CH}_2\text{CH}_2\text{Br} \)
Answer: (d) \( \text{CH}_3\text{CH}_2\text{CH}_2\text{Br} \)
In simple words: A longer alkyl chain polarizes the carbon-halogen bond more effectively, which makes the molecule more reactive during elimination.

Exam Tip: Elimination reaction rates increase with the accessibility and stability of the forming double bond.

Question (ii). The general reaction, R—X \( \xrightarrow{\text{aq. OH}^-} \) ROH + X–, is expected to follow decreasing order of reactivity as in
(a) t-BuI > t-BuBr > t-BuCl > t-BuF
(b) t-BuF > t-BuCl > t-BuBr > t-BuI
(c) t-BuBr > t-BuCl > t-BuI > t-BuF
(d) t-BuF > t-BuCl > t-BuI > t-BuBr
Answer: (a) t-BuI > t-BuBr > t-BuCl > t-BuF
In simple words: Iodide is the best leaving group because the carbon-iodine bond is the weakest and easiest to break, while carbon-fluorine is the strongest.

Exam Tip: Reactivity order for different halogens with the same alkyl group is always iodide > bromide > chloride > fluoride.

Question (iii). Reaction of t-butyl bromide with sodium methoxide produces
(a) sodium t-butoxide
(b) t-butyl methyl ether
(c) iso-butane
(d) iso-butylene
Answer: (d) iso-butylene
In simple words: Because sodium methoxide is a strong base and t-butyl bromide is a highly crowded tertiary halide, the system undergoes elimination instead of substitution, forming iso-butylene.

Exam Tip: Strong bases react with tertiary halides to yield alkenes via elimination rather than ethers via Williamson synthesis.

Question (iv). In the elimination reactions the reactivity of alkyl halides follows the sequence
(a) R – F > R – Cl > R – Br > R – I
(b) R – I > R – Br > R – Cl > R – F
(c) R – I > R – F > R – Br > R – Cl
(d) R – F > R – I > R – Br > R – Cl
Answer: (b) R – I > R – Br > R – Cl > R – F
In simple words: The energy needed to break a carbon-iodine bond is much lower than for a carbon-fluorine bond, making iodides react the fastest in elimination.

Exam Tip: Bond dissociation energy dictates leaving group ability: C-I is easiest to cleave, making it highly reactive.

Question (v). The ease of dehydrohalogenation of alkyl halide with alcoholic KOH is
(a) 3º < 2º < 1º
(b) 3º > 2º > 1º
(c) 3º < 2º > 1º
(d) 3º > 2º < 1º
Answer: (b) 3º > 2º > 1º
In simple words: Tertiary alkyl halides lose hydrogen halide fastest because they yield highly substituted alkenes, which are thermodynamically more stable.

Exam Tip: Ease of dehydrohalogenation is directly related to the stability of the alkene product formed, following Saytzeff's rule.

Case 4
Consider the given sequence of reactions:
Benzene + \( \text{Cl}_2/\text{FeCl}_3 \rightarrow \text{X} \)
\( \text{X} + \text{Na, Dry ether} \rightarrow \text{W} \)
\( \text{X} + \text{conc. HNO}_3 + \text{conc. H}_2\text{SO}_4 \rightarrow \text{Y} \)
\( \text{Y} + \text{aq. NaOH (15%), 433 K, dil. HCl} \rightarrow \text{Z} \)

Question (iii). When X reacts with \( \text{CH}_3\text{COCl} \) in the presence of anhy. \( \text{AlCl}_3 \), the reaction is known as
(a) Fitting reaction
(b) Ullmann reaction
(c) Wurtz-Fittig reaction
(d) Friedel-Crafts acylation reaction.
Answer: (d) Friedel-Crafts acylation reaction.
In simple words: Introducing an acetyl group (\( \text{CH}_3\text{CO}- \)) into the benzene ring using an acid chloride and anhydrous aluminum chloride is a Friedel-Crafts acylation.

Exam Tip: Friedel-Crafts acylation is a key electrophilic aromatic substitution reaction used to prepare aromatic ketones.

Question (iv). When X is treated Ni-Al/NaOH the product obtained is
(a) benzene
(b) phenol
(c) p-chlorophenol
(d) triphenyl
Answer: (a) benzene
In simple words: Treating chlorobenzene with a nickel-aluminum alloy in sodium hydroxide reduces it by replacing the chlorine atom with hydrogen, forming benzene.

Exam Tip: Ni-Al alloy in aqueous alkaline medium acts as a reducing agent capable of removing halogens from aromatic rings.

Question (v). Compound Z is
(a) phenol
(b) p-chlorophenol
(c) p-nitrophenol
(d) nitrobenzene
Answer: (c) p-nitrophenol
In simple words: In p-chloronitrobenzene, the electron-withdrawing nitro group activates the chlorine, allowing it to be easily replaced by a hydroxyl group to yield p-nitrophenol.

Exam Tip: Nucleophilic aromatic substitution (\( \text{S}_{\text{N}}\text{Ar} \)) is strongly activated by electron-withdrawing groups at the ortho and para positions relative to the leaving halogen.

Case 5
Haloarenes are less reactive than haloalkanes. The low reactivity of haloarenes can be attributed to:
— resonance effect
— \( sp^2 \) hybridisation of C—X bond
— polarity of C—X bond
— instability of phenyl cation (formed by self-ionisation of haloarene)
— repulsion between the electron rich attacking nucleophiles and electron rich arenes.
Reactivity of haloarenes can be increased or decreased by the presence of certain groups at certain positions. For example, nitro (\( -\text{NO}_2 \)) group at o/p positions increases the reactivity of haloarenes towards nucleophilic substitution reactions.

Question (i). Aryl halides are less reactive towards nucleophilic substitution reaction as compared to alkyl halides due to
(a) the formation of less stable carbonium ion
(b) resonance stabilisation
(c) larger carbon-halogen bond
(d) inductive effect
Answer: (b) resonance stabilisation
In simple words: Resonance in aryl halides allows the halogen's lone pair to be shared with the ring. This gives the carbon-halogen bond partial double-bond character, making it very hard to break.

Exam Tip: Resonance and the \( sp^2 \) hybridization of the carbon bonded to the halogen are the primary reasons for the low reactivity of aryl halides.

Question (v). The reactivity of the compounds (i) MeBr, (ii) \( \text{PhCH}_2\text{Br} \), (iii) MeCl, (iv) \( p\text{-MeOC}_6\text{H}_4\text{Br} \) decreases as
(a) (i) > (ii) > (iii) > (iv)
(b) (iv) > (ii) > (i) > (iii)
(c) (iv) > (iii) > (i) > (ii)
(d) (ii) > (i) > (iii) > (iv)
Answer: (d) (ii) > (i) > (iii) > (iv)
In simple words: Benzyl bromide is highly reactive because its carbocation/transition state is stabilized. Bromides are more reactive than chlorides, and aryl halides are extremely unreactive.

Exam Tip: The overall order of nucleophilic substitution reactivity is: benzyl halides > alkyl bromides > alkyl chlorides > aryl halides.

Case 6
Nucleophilic substitution reaction of haloalkane can be conducted according to both \( \text{S}_{\text{N}}1 \) and \( \text{S}_{\text{N}}2 \) mechanisms. However, which mechanism it is based on is related to such factors as the structure of haloalkane, and properties of leaving group, nucleophilic reagent and solvent.
Influences of halogen: No matter which mechanism the nucleophilic substitution reaction is based on, the leaving group always leaves the central carbon atom with electron pair. This is just the opposite of the situation that nucleophilic reagent attacks the central carbon atom with electron pair. Therefore, the weaker the alkalinity of leaving group is, the more stable the anion formed is and it will be more easier for the leaving group to leave the central carbon atom; that is to say, the reactant is more easier to be substituted. The alkalinity order of halogen ion is \( \text{I}^- < \text{Br}^- < \text{Cl}^- < \text{F}^- \) and the order of their leaving tendency should be \( \text{I}^- > \text{Br}^- > \text{Cl}^- > \text{F}^- \). Therefore, in four halides with the same alkyl and different halogens, the order of substitution reaction rate is RI > RBr > RCl > RF. In addition, if the leaving group is very easy to leave, many carbocation intermediates are generated in the reaction and the reaction is based on \( \text{S}_{\text{N}}1 \) mechanism. If the leaving group is not easy to leave, the reaction is based on \( \text{S}_{\text{N}}2 \) mechanism.
Influences of solvent polarity: In \( \text{S}_{\text{N}}1 \) reaction, the polarity of the system increases from the reactant to the transition state, because polar solvent has a greater stabilizing effect on the transition state than the reactant, thereby reduce activation energy and accelerate the reaction. In \( \text{S}_{\text{N}}2 \) reaction, the polarity of the system generally does not change from the reactant to the transition state and only charge dispersion occurs. At this time, polar solvent has a great stabilizing effect on Nu than the transition state, thereby increasing activation energy and slow down the reaction rate. For example, the decomposition rate (\( \text{S}_{\text{N}}1 \)) of tertiary chlorobutane in 25ºC water (dielectric constant 79) is 3,00,000 times faster than in ethanol (dielectric constant 24). The reaction rate (\( \text{S}_{\text{N}}2 \)) of 2-bromopropane and NaOH in ethanol containing 40% water is twice slower than in absolute ethanol. In a word, the level of solvent polarity has influence on both \( \text{S}_{\text{N}}1 \) and \( \text{S}_{\text{N}}2 \) reactions, but with different results. Generally speaking, weak polar solvent is favourable for \( \text{S}_{\text{N}}2 \) reaction, while strong polar solvent is favourable for \( \text{S}_{\text{N}}1 \) reaction, because only under the action of polar solvent can halogenated hydrocarbon dissociate into carbocation and halogen ion and solvents with a strong polarity is favourable for solvation of carbocation, increasing its stability. Generally speaking, the substitution reaction of tertiary haloalkane is based on \( \text{S}_{\text{N}}1 \) mechanism in solvents with a strong polarity (for example, ethanol containing water).

Question (i). \( \text{S}_{\text{N}}1 \) mechanism is favoured in which of the following solvents:
(a) benzene
(b) carbon tetrachloride
(c) acetic acid
(d) carbon disulphide
Answer: (c) acetic acid
In simple words: Acetic acid is a polar protic solvent, which excels at surrounding and stabilizing the carbocation intermediate formed in \( \text{S}_{\text{N}}1 \) reactions.

Exam Tip: Polar protic solvents stabilize the carbocation intermediate through hydrogen bonding, strongly favoring the SN1 pathway.

Question (ii). Nucleophilic substitution will be fastest in case of:
(a) 1-Chloro-2,2-dimethyl propane
(b) 1-Iodo-2,2-dimethyl propane
(c) 1-Bromo-2,2-dimethyl propane
(d) 1-Fluoro-2,2-dimethyl propane
Answer: (b) 1-Iodo-2,2-dimethyl propane
In simple words: The carbon-iodine bond is the longest and weakest among the options, making iodide the best leaving group and allowing it to react the fastest.

Exam Tip: For any alkyl halide series, the iodide derivative always reacts fastest because iodide is the best leaving group.

Question (iii). \( \text{S}_{\text{N}}1 \) reaction will be fastest in which of the following solvents?
(a) Acetone (dielectric constant 21)
(b) Ethanol (dielectric constant 24)
(c) Methanol (dielectric constant 32)
(d) Chloroform (dielectric constant 5)
Answer: (c) Methanol (dielectric constant 32)
In simple words: Methanol has the highest dielectric constant here. Higher solvent polarity stabilizes the transition state, lowering the activation energy for the reaction.

Exam Tip: Solvents with a high dielectric constant are more polar and thus stabilize the ionic transition states of SN1 reactions more effectively.

Question (iv). Polar solvents make the reaction faster as they:
(a) destabilize transition state and decrease the activation energy
(b) destabilize transition state and increase the activation energy
(c) stabilize transition state and increase the activation energy
(d) stabilize transition state and decrease the activation energy
Answer: (d) stabilize transition state and decrease the activation energy
In simple words: Polar solvents interact strongly with and stabilize the charged transition state. This lowers the energy barrier, causing the reaction to run faster.

Exam Tip: Stabilization of the transition state always lowers the activation energy, which increases the rate of the reaction.

Question (v). \( \text{S}_{\text{N}}1 \) reaction will be fastest in case of:
(a) 1-Chloro-2-methyl propane
(b) 1-Iodo-2-methyl propane
(c) 1-Chlorobutane
(d) 1-Iodobutane
Answer: (b) 1-Iodo-2-methyl propane
In simple words: 1-Iodo-2-methyl propane has more steric hindrance and a branched structure that can undergo rearrangement, and it contains iodide, which is a highly reactive leaving group.

Exam Tip: A combination of a branched carbon frame and an excellent leaving group like iodide maximizes the rate of SN1 reactions.