Maharashtra Board Class 12 Chemistry Chapter 10 Halogen Derivatives Solutions

12th Chemistry Chapter 10 Exercise Halogen Derivatives Solutions Maharashtra Board

Halogen Derivatives Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Chemistry Chapter 10 Exercise Solutions Maharashtra Board

Chemistry Class 12 Chapter 10 Exercise Solutions

Exercise 1. Choose The Most Correct Option.

Question i. The correct order of increasing reactivity of C-X bond towards nucleophile in the following compounds is
ℹ️ चित्र व्याख्या (Diagram Explanation): पहले चित्र में एक बेंजीन वलय है जिससे एक X समूह जुड़ा है। दूसरे चित्र में एक बेंजीन वलय है जिससे एक X समूह और एक नाइट्रो समूह (\( NO_2 \)) जुड़ा है। तीसरे चित्र में \( (CH_3)_3C-X \) दिखाया गया है। चौथे चित्र में \( (CH_3)_2CH-X \) दिखाया गया है।
(a) I < II < III < IV
(b) II < I < III < IV
(c) III < IV < II < I
(d) IV < III < I < II
Answer: (d) IV < III < I < II
In simple words: The reactivity of C-X bond towards nucleophiles generally increases with the stability of the carbocation formed, which is hindered by resonance or electron-withdrawing groups on an aromatic ring. Alkyl halides are more reactive than aryl halides.

🎯 Exam Tip: Understanding the stability of carbocations and resonance effects is crucial for determining nucleophilic substitution reactivity in different types of halides.

Question ii.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह अभिक्रिया एक प्रोपीन अणु \( CH_3-CH=CH_2 \) को हाइड्रोजन आयोडाइड (HI) और पेरोक्साइड की उपस्थिति में अभिकृत करने को दर्शाती है। उत्पाद विकल्प हैं: (a) \( I-CH_2-CH=CH_2 \), (b) \( CH_3-CH_2-CH_2I \), (c) \( CH_3-CH-CH_3 \) जिसमें आयोडीन दूसरे कार्बन पर जुड़ा है, (d) \( CH_3-CH-CH_3 \) जिसमें आयोडीन और हाइड्रॉक्सिल (OH) दूसरे और तीसरे कार्बन पर जुड़े हैं। The major product of the above reaction is,
(a) I-CH2-CH=CH2
(b) CH3-CH2-CH2I
(c) CH3-CH-CH3 I
(d) CH3-CH-CH2 I OH
Answer: (c) CH3-CH-CH3 I
In simple words: The reaction of propene with HI in the presence of peroxide typically still follows Markovnikov addition, meaning the iodine adds to the more substituted carbon, yielding 2-iodopropane. The anti-Markovnikov peroxide effect is usually observed specifically with HBr.

🎯 Exam Tip: Remember Markovnikov's and anti-Markovnikov's rules for addition reactions. Peroxide effect (anti-Markovnikov) is generally observed with HBr, not HI, so standard Markovnikov addition often applies to HI even with peroxide unless specified otherwise for HI in a specific context. For HI, the product is 2-iodopropane.

Question iii. Which of the following is likely to undergo racemization during alkaline hydrolysis?
ℹ️ चित्र व्याख्या (Diagram Explanation): पहले चित्र में एक कायरल केंद्र वाला अणु \( CH_3-CH(Cl)-C_2H_5 \) है। दूसरा चित्र एक एलील क्लोराइड को दर्शाता है। तीसरा चित्र एक तृतीयक ब्यूटाइल क्लोराइड को दर्शाता है। चौथा चित्र एक एरील क्लोराइड को दर्शाता है।
(a) Only I
(b) Only II
(c) II and IV
(d) Only IV
Answer: (a) Only I
In simple words: Racemization occurs when an optically active compound with a chiral center undergoes a reaction (like \( S_N1 \) hydrolysis) that produces an equal mixture of enantiomers, leading to an optically inactive product. This typically happens with secondary and tertiary alkyl halides when a carbocation intermediate is formed. Compound (I) has a chiral center and can undergo \( S_N1 \) reaction.

🎯 Exam Tip: Racemization is characteristic of \( S_N1 \) reactions, which proceed through a planar carbocation intermediate. Look for a chiral carbon that can form a stable carbocation.

Question iv. The best method for preparation of alkyl fluorides is
(a) Finkelstein reaction
(b) Swartz reaction
(c) Free radical fluorination
(d) Sandmeyer's reaction
Answer: (b) Swartz reaction
In simple words: Swartz reaction is a specific method used to synthesize alkyl fluorides by reacting alkyl chlorides or bromides with metallic fluorides like \( AgF \), \( Hg_2F_2 \), \( CoF_2 \), or \( SbF_3 \).

🎯 Exam Tip: Finkelstein reaction is for alkyl iodides. Swartz reaction is exclusively for alkyl fluorides. Remember these name reactions for halide interconversions.

Question v. Identify the chiral molecule from the following.
(a) 1-Bromobutane
(b) 1,1- Dibromobutane
(c) 2,3- Dibromobutane
(d) 2-Bromobutane
Answer: (d) 2-Bromobutane
In simple words: A chiral molecule has a non-superimposable mirror image. This usually means it contains a carbon atom bonded to four different groups. In 2-bromobutane, the second carbon atom is bonded to a methyl group, an ethyl group, a hydrogen atom, and a bromine atom, making it a chiral center.

🎯 Exam Tip: To identify a chiral molecule, look for a carbon atom attached to four distinct substituents. This carbon is known as a chiral center or stereocenter.

Question vi. An alkyl chloride on Wurtz reaction gives 2,2,5,5-tetramethylhexane. The same alkyl chloride on reduction with zinc-copper couple in alchol give hydrocarbon with molecular formula \( C_5H_{12} \). What is the structure of alkyl chloride
(a) CH3-C-CH2Cl CH3
(b) CH3-C-CH2CH3 Cl
(c) CH3-CH2-CH-Cl CH3
(d) CH3-CH-CH-CH2 CH3 CH2CH3
Answer: (a) CH3-C-CH2Cl CH3
In simple words: The alkyl chloride that forms 2,2,5,5-tetramethylhexane via Wurtz reaction and 2,2-dimethylpropane (\( C_5H_{12} \)) upon reduction is 1-chloro-2,2-dimethylpropane, whose structure is \( (CH_3)_3C-CH_2Cl \).

🎯 Exam Tip: In Wurtz reaction, R-X combines with R-X to form R-R. For a symmetrical product, halve the product to find R. Then, confirm the structure with the reduction product of R-X to R-H.

Question vii. Butanenitrile may be prepared by heating
(a) propanol with KCN
(b) butanol with KCN
(c) n-butyl chloride with KCN
(d) n-propyl chloride with KCN
Answer: (d) n-propyl chloride with KCN
In simple words: Nitriles can be synthesized by nucleophilic substitution of alkyl halides with cyanide salts (like KCN). To get butanenitrile (4 carbons, \( CH_3CH_2CH_2CN \)), you need an alkyl halide with 3 carbons (n-propyl chloride, \( CH_3CH_2CH_2Cl \)). The cyanide group adds one carbon.

🎯 Exam Tip: Remember that the cyanide group \( (-CN) \) introduces an additional carbon atom into the carbon chain, so for a nitrile with 'n' carbons, the starting alkyl halide should have 'n-1' carbons.

Question viii. Choose the compound from the following that will react fastest by \( S_N1 \) mechanism.
(a) 1-iodobutane
(b) 1-iodopropane
(c) 2-iodo-2 methylbutane
(d) 2-iodo-3-methylbutane
Answer: (c) 2-iodo-2 methylbutane
In simple words: \( S_N1 \) reactions proceed through a carbocation intermediate. The more stable the carbocation, the faster the \( S_N1 \) reaction. Tertiary carbocations are the most stable, followed by secondary, then primary. 2-iodo-2-methylbutane forms a tertiary carbocation, making it the fastest to react by \( S_N1 \).

🎯 Exam Tip: The order of reactivity for \( S_N1 \) reactions is Tertiary > Secondary > Primary. Always identify the type of carbocation formed by the leaving group to predict \( S_N1 \) reactivity.

Question ix.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक क्लोरोबेंजीन जैसा अणु, जिसमें वलय से एक क्लोरीन परमाणु जुड़ा है, को मैग्नीशियम (Mg) के साथ शुष्क ईथर (dry ether) की उपस्थिति में अभिकृत किया जाता है, जिससे उत्पाद A बनता है। फिर उत्पाद A को पानी (\( H_2O \)) के साथ अभिकृत किया जाता है, जिससे उत्पाद B बनता है। उत्पाद B के संभावित संरचनात्मक विकल्प दिए गए हैं। The product 'B' in the above reaction sequence is,
(a) Mg
(b)Mg-Cl
(c) Cl Mg
(d)
Answer: (d)
In simple words: The first step is the formation of a Grignard reagent (A) from chlorobenzene and magnesium in dry ether, resulting in phenylmagnesium chloride. The second step is the hydrolysis of the Grignard reagent (A) with water, which forms benzene (B) and magnesium hydroxychloride.

🎯 Exam Tip: Grignard reagents are strong bases and react readily with any source of protons (like water) to form alkanes (or arenes in this case) and magnesium hydroxyhalides. This reaction is also used to prepare hydrocarbons.

Question x. Which of the following is used as source of dichlorocarbene
(a) tetrachloromethane
(b) chloroform
(c) iodoform
(d) DDT
Answer: (b) chloroform
In simple words: Dichlorocarbene (\( :CCl_2 \)) is a highly reactive intermediate generated by the alpha-elimination of chloroform (\( CHCl_3 \)) with a strong base like potassium tert-butoxide.

🎯 Exam Tip: Dichlorocarbene is an important intermediate in organic synthesis, notably in Reimer-Tiemann reaction and for cyclopropanation reactions. Chloroform is the most common precursor.

Exercise 2. Do As Directed.

Question i. Write IUPAC name of the following compounds
(a) CH-CH=C-CH-Br С СH,
(b) CH3-CH-CH-CH2-CH CI CH
(c) Cl
(d) Cl CH. 3 CH Cl 25
Answer:

Compound	IUPAC names
(1) \( CH_3-CH=C(CH_3)-CH(Br)-CH_3 \)	4-Bromo-3, 4 dimethyl but-2-ene
(2) \( CH_3-CH(Cl)-CH(CH_3)-CH_2-CH_3 \)	2-Chloro-3-methyl pentane
(3) ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक साइक्लोहेक्सेन वलय है जिससे एक क्लोरीन परमाणु (Cl) और एक एथिल समूह (\( C_2H_5 \)) जुड़े हैं। क्लोरीन पहले कार्बन पर और एथिल समूह चौथे कार्बन पर जुड़ा है।	1-Chloro-4-ethyl cyclohexane
(4) ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक बेंजीन वलय है जिससे दो क्लोरीन परमाणु (Cl) और एक मिथाइल समूह (\( CH_3 \)) जुड़े हैं। एक क्लोरीन पहले कार्बन पर, दूसरा क्लोरीन चौथे कार्बन पर और मिथाइल समूह दूसरे कार्बन पर जुड़ा है।	1,4-Dichloro-2-methyl benzene

In simple words: IUPAC nomenclature systematically names organic compounds based on their structure. For these halogen derivatives, identify the longest carbon chain or ring, number it to give substituents the lowest possible numbers, and then name the substituents and the parent hydrocarbon.

🎯 Exam Tip: Practice identifying the parent chain, functional groups, and substituents. Pay close attention to numbering rules for multiple substituents and double/triple bonds, giving priority to the principal functional group if present.

Question ii. Write structure and IUPAC name of the major product in each of the following reaction.
(a) \( CH_3-CH(CH_3)-CH_2Cl + NaI \xrightarrow{Acetone} \)
(b) \( CH_3-CH_2Br + SbF_3 \)
(c) \( CH_3-CH(CH_3)-CH=CH_2 + HBr \xrightarrow{peroxide} \)
(d)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक साइक्लोहेक्सानोल अणु (Cyclohexanol) को थायोनिल क्लोराइड (\( SOCl_2 \)) के साथ अभिकृत किया जाता है।
(e)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक टॉलूईन अणु (\( C_6H_5CH_3 \)) को क्लोरीन (\( Cl_2 \)) के साथ डार्क (dark) और आयरन (\( Fe \)) की उपस्थिति में अभिकृत किया जाता है।
Answer:
(1) \( CH_3-CH(CH_3)-CH_2Cl + NaI \xrightarrow{Acetone} CH_3-CH(CH_3)-CH_2I + NaCl \)
IUPAC Name: 1-Iodo-2-methyl propane

(2) \( CH_3-CH_2Br + SbF_3 \longrightarrow F_2-CH-CH_2-Br \)
IUPAC Name: 1,1-difluoro-2-bromoethane

(3) \( CH_3-CH(CH_3)-CH=CH_2 + HBr \xrightarrow{Peroxide} CH_3-CH(CH_3)-CH_2-CH_2-Br \)
IUPAC Name: 1-Bromo-3-methyl butane

(4)
ℹ️ चित्र व्याख्या (Diagram Explanation): साइक्लोहेक्सानोल (\( C_6H_{11}OH \)) को थायोनिल क्लोराइड (\( SOCl_2 \)) के साथ अभिकृत करने पर क्लोरोसाइक्लोहेक्सेन (\( C_6H_{11}Cl \)) बनता है, साथ में सल्फर डाइऑक्साइड (\( SO_2 \)) और हाइड्रोजन क्लोराइड (\( HCl \)) गैसें निकलती हैं।
IUPAC Name: chlorocyclohexane

(5)
ℹ️ चित्र व्याख्या (Diagram Explanation): टॉलूईन (\( C_6H_5CH_3 \)) को क्लोरीन (\( Cl_2 \)) के साथ डार्क (dark) और आयरन (\( Fe \)) की उपस्थिति में अभिकृत करने पर o-क्लोरोटॉलूईन और p-क्लोरोटॉलूईन बनते हैं, साथ में हाइड्रोजन क्लोराइड (\( HCl \)) गैस निकलती है।
IUPAC Name: o-chlorotoluene and p-chlorotoluene
In simple words: These reactions involve various halide synthesis and transformation methods. Reaction (a) is a Finkelstein reaction, (b) is a Swartz reaction, (c) is anti-Markovnikov HBr addition, (d) is an alcohol to alkyl chloride conversion using \( SOCl_2 \), and (e) is electrophilic aromatic substitution (chlorination).

🎯 Exam Tip: Pay attention to the reagents and reaction conditions, as they dictate the type of reaction (e.g., \( S_N2 \), electrophilic addition, electrophilic substitution) and the regioselectivity (Markovnikov vs. anti-Markovnikov) or stereoselectivity of the product. Familiarity with name reactions is also helpful.

Question iii. Identify chiral molecule/s from the following.
(a) \( CH_3-CH(OH)-CH_2-CH_3 \)
(b) \( CH_3-CH_2-CH(Br)-CH_2-CH_3 \)
(c) \( CH_3-CH_2-CH_2-CH_2Br \)
(d) \( CH_3-CH(CH_3)-CH_2-CH_3 \)
Answer:Chiral molecule
(a)
ℹ️ चित्र व्याख्या (Diagram Explanation): 2-ब्यूटेनोल की संरचना दिखाई गई है, जहाँ दूसरे कार्बन परमाणु से \( CH_3 \), \( OH \), \( H \), और \( CH_2CH_3 \) जुड़े हैं, जो इसे एक कायरल केंद्र बनाता है।
In simple words: A chiral molecule possesses a non-superimposable mirror image and typically contains a chiral center—a carbon atom bonded to four distinct groups. Out of the given options, 2-butanol (a) has a chiral carbon.

🎯 Exam Tip: To identify a chiral molecule, systematically check each carbon atom to see if it is bonded to four different groups. If such a carbon exists, the molecule is chiral.

Question iv. Which one compound from the following pairs would undergo \( S_N2 \) faster from the?
(a)
ℹ️ चित्र व्याख्या (Diagram Explanation): पहले विकल्प में एक साइक्लोहेक्सेन वलय से एक क्लोरोमिथाइल समूह जुड़ा है (\( -CH_2Cl \)) और दूसरा विकल्प एक साइक्लोहेक्सेन वलय से सीधे एक क्लोरीन परमाणु जुड़ा है (\( -Cl \))।
(b) \( CH_3CH_2CH_2I \) and \( CH_3CH_2CH_2Cl \)
Answer:
(1)
ℹ️ चित्र व्याख्या (Diagram Explanation): पहले विकल्प में एक साइक्लोहेक्सेन वलय से एक क्लोरोमिथाइल समूह जुड़ा है (\( -CH_2Cl \)) और दूसरा विकल्प एक साइक्लोहेक्सेन वलय से सीधे एक क्लोरीन परमाणु जुड़ा है (\( -Cl \))। पहले विकल्प में \( CH_2Cl \) एक प्राथमिक हैलाइड है जबकि दूसरा एक एरील हैलाइड है। Since \( CH_2Cl \) is a primary halide it undergoes \( S_N2 \) reaction faster than
ℹ️ चित्र व्याख्या (Diagram Explanation): साइक्लोहेक्सेन वलय से सीधे एक क्लोरीन परमाणु जुड़ा है (\( -Cl \))।
(2) Since iodine is a better leaving group than chloride, 1-iodo propane (\( CH_3CH_2CH_2I \)) undergoes \( S_N2 \) reaction faster than 1-chloropropane (\( CH_3CH_2CH_2Cl \)).
In simple words: \( S_N2 \) reactions are favored by less steric hindrance around the carbon bearing the leaving group (primary > secondary > tertiary) and by good leaving groups (I > Br > Cl > F). Therefore, primary halides and those with iodide as a leaving group react faster.

🎯 Exam Tip: For \( S_N2 \) reactivity, consider two main factors: steric hindrance (least steric hindrance at the reaction center is faster) and the leaving group ability (a weaker base is a better leaving group). Primary alkyl halides and iodides are generally most reactive in \( S_N2 \).

Question v. Complete the following reactions giving major product.
(a) \( CH_3-CH=CH_2 \xrightarrow{HBr} A \xrightarrow{alc. KOH} B \)
(b) \( CH_3-CH=CH_2 \xrightarrow{Red P/Br_2} A \xrightarrow{Ag_2O/H_2O} B \)
(c) \( CH_3-C(CH_3)_2-CH_2-Cl \xrightarrow{Na/dry ether} A \)
(d)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक क्लोरोबेंजीन जैसा अणु (\( C_6H_5Cl \)) को मैग्नीशियम (Mg) के साथ शुष्क ईथर (dry ether) की उपस्थिति में अभिकृत किया जाता है, जिससे उत्पाद A बनता है।
Answer:
(1) \( CH_3-CH=CH_2 \xrightarrow{HBr \atop Peroxide} CH_3-CH_2-CH_2-Br \) (A: 1-Bromopropane (major product))
\( CH_3-CH_2-CH_2-Br \xrightarrow{alc.KOH} CH_3-CH=CH_2 + KBr + H_2O \) (B: Propene)
In simple words: Reaction (a) shows anti-Markovnikov addition of HBr to propene (due to peroxide), forming 1-bromopropane. Subsequent reaction with alcoholic KOH is an elimination reaction, regenerating propene.

🎯 Exam Tip: Peroxides direct HBr addition to alkenes to follow the anti-Markovnikov rule. Alcoholic KOH is a strong base and favors E2 elimination reactions over \( S_N2 \) substitution, especially with primary or secondary alkyl halides.

(2) \( CH_3-CH=CH_2 \xrightarrow{Red P/Br_2} A \xrightarrow{Ag_2O/H_2O} B \)
ℹ️ चित्र व्याख्या (Diagram Explanation): पहले चरण में प्रोपीन को रेड फॉस्फोरस और ब्रोमीन (\( Red P/Br_2 \)) के साथ अभिकृत करने पर \( CH_3-CH(OH)-CH_2-Br \) (A) बनता है। दूसरे चरण में A को सिल्वर ऑक्साइड और पानी (\( Ag_2O/H_2O \)) के साथ अभिकृत करने पर \( CH_3-CH(OH)-CH_2-OH \) (B) बनता है।
A = \( CH_3-CH(OH)-CH_2-Br \)
B = \( CH_3-CH(OH)-CH_2-OH \)
In simple words: Reaction (2) first involves bromination of propene with Red P/Br2, followed by hydrolysis with moist silver oxide. The product (A) is 2-bromo-1-propanol, which upon reaction with silver oxide and water yields 1,2-propanediol (B).

🎯 Exam Tip: \( Red P/Br_2 \) is often used to brominate alcohols. \( Ag_2O/H_2O \) (moist silver oxide) acts as a source of \( OH^- \) and promotes hydrolysis of alkyl halides, converting them to alcohols.

(3) \( CH_3-C(CH_3)_2-CH_2-Cl \xrightarrow{Na/dry ether} A \)
ℹ️ चित्र व्याख्या (Diagram Explanation): 1-क्लोरो-2,2-डाइमिथाइलप्रोपेन (\( (CH_3)_3CCH_2Cl \)) को सोडियम (Na) के साथ शुष्क ईथर (dry ether) की उपस्थिति में अभिकृत करने पर 2,2,5,5-टेट्रामिथाइलहेक्सेन (A) बनता है।
A = \( CH_3-C(CH_3)_2-CH_2-CH_2-C(CH_3)_2-CH_3 \) (2,2,5,5-tetramethylhexane)
In simple words: This is a Wurtz reaction, where two molecules of 1-chloro-2,2-dimethylpropane couple in the presence of sodium and dry ether to form 2,2,5,5-tetramethylhexane.

🎯 Exam Tip: The Wurtz reaction is used for synthesizing symmetrical alkanes by reacting an alkyl halide with sodium metal in dry ether. The carbon chain of the product is double that of the alkyl group from the starting material.

(4)
ℹ️ चित्र व्याख्या (Diagram Explanation): क्लोरोबेंजीन (\( C_6H_5Cl \)) को मैग्नीशियम (Mg) के साथ शुष्क ईथर (dry ether) की उपस्थिति में अभिकृत करने पर फिनाइलमैटिनिअम क्लोराइड (A) बनता है।
A =
ℹ️ चित्र व्याख्या (Diagram Explanation): यह फिनाइलमैटिनिअम क्लोराइड की संरचना है, जिसमें एक बेंजीन वलय से \( -MgCl \) समूह जुड़ा है।
In simple words: This reaction forms a Grignard reagent (phenylmagnesium chloride) from chlorobenzene and magnesium in dry ether. Grignard reagents are important organometallic compounds.

🎯 Exam Tip: Grignard reagents are prepared in dry ether to prevent them from reacting with moisture. They are highly reactive and serve as versatile nucleophiles in organic synthesis.

Question vi. Name the reagent used to bring about the following conversions.
(a) Bromoethane to ethoxyethane
(b) 1-Chloropropane to 1 nitropropane
(c) Ethyl bromide to ethyl isocyanide
(d) Chlorobenzene to biphenyl
Answer:

Reactant	Ans.	Reagent	Product
(1) Bromoethane		Sodium alkoxide	ethoxy ethane
(2) 1-Chloropropane		Silver nitrite	1-nitropropane
(3) Ethyl bromide		AgCN	Ethyl isocyanide
(4) Chlorobenzene		Na	Biphenyl
(5) Ethyl iodide		Mg	Ethyl magnesium iodide
(6) Ethyl chloride		Nal	Ethyl iodide
(7) Toluene + \( Br_2 \)		Fe	ortho and para bromo toluene

In simple words: This question tests knowledge of various synthetic reactions involving halogen derivatives. Each conversion requires a specific reagent to achieve the desired product, such as sodium alkoxide for Williamson ether synthesis, silver nitrite for nitroalkanes, or sodium for Wurtz/Fittig reactions.

🎯 Exam Tip: Master common organic reagents and their specific applications. Pay attention to the distinction between \( AgNO_2 \) (nitroalkanes) and \( NaNO_2 \) (alkyl nitrites), and between KCN (nitriles) and AgCN (isocyanides), as these are frequent trick questions.

Question vii. Arrange the following in the increase order of boiling points
(a) 1-Bromopropane
(b) 2- Bromopropane
(c) 1- Bromobutane
(d) 1-Bromo-2-methylpropane
Answer: 1-Bromo-2-methylpropane, 2-Bromopropane, 1-Bromopropane, 1-Bromo butane
In simple words: Boiling points of alkyl halides are influenced by molecular size, branching, and intermolecular forces. Generally, boiling points increase with molecular weight (more halogens or longer carbon chain) and decrease with increased branching (smaller surface area for Van der Waals forces).

🎯 Exam Tip: For isomeric alkyl halides, the straight-chain isomer will have a higher boiling point than branched isomers due to better packing and stronger Van der Waals forces. For different halogens with the same alkyl group, boiling points increase with the atomic mass of the halogen (I > Br > Cl > F).

Question viii. Match the pairs.
Column I
a. CH2CH-CH2X
b. CH2=CH-CH2X
c. CH2=CH-X
Column II
i. vinyl halide
ii. alkyl halide
iii. allyl halide
iv. benzyl halide
v. aryl halide
Answer:
(1) CH3-CH-CH3-(b) alkyl halide X
(2) CH2=CH-CH2X-(c) allyl halide
(3) CH2=CH-X-(a) vinyl halide
In simple words: This question matches common classifications of organic halides based on the position of the halogen atom relative to other functional groups or the hybridization of the carbon atom it's attached to. An alkyl halide has the halogen on an \( sp^3 \) carbon, an allyl halide has the halogen on an \( sp^3 \) carbon adjacent to a C=C double bond, and a vinyl halide has the halogen directly attached to an \( sp^2 \) carbon of a C=C double bond.

🎯 Exam Tip: Clearly differentiate between alkyl, allyl, vinyl, and aryl halides by focusing on the carbon atom directly bonded to the halogen. This understanding is fundamental to predicting their chemical reactivity.

Exercise 3. Give Reasons

Question i. Haloarenes are less reactive than haloalkanes.
Answer: Haloarenes (Aryl halides) are less reactive than (alkyl halides) haloalkanes due to the following reasons :
(1) Resonance effect : In haloarenes, the electron pairs on halogen atom are in conjugation with \( \pi \)-electrons of the benzene ring. The delocalization of these electrons C-Cl bond acquires partial double bond character.
ℹ️ चित्र व्याख्या (Diagram Explanation): इसमें क्लोरोबेंजीन की पांच अनुनादी संरचनाएँ दिखाई गई हैं। पहली संरचना में बेंजीन वलय से एक क्लोरीन परमाणु जुड़ा है। अन्य संरचनाओं में, क्लोरीन परमाणु पर एकाकी इलेक्ट्रॉन युग्म वलय के \( \pi \)-इलेक्ट्रॉनों के साथ संयुग्मन में भाग लेते हैं, जिससे C-Cl बंध में आंशिक द्विबंध गुण आ जाता है और वलय पर ऑर्थो और पैरा स्थितियों पर नकारात्मक आवेश आता है।
(I) \( \iff \) (II) \( \iff \) (III) \( \iff \) (IV) \( \iff \) (V)
Due to partial double bond character of C-Cl bond in aryl halides, the bond cleavage in haloarene is difficult and are less reactive. On the other hand, in alkyl halides, carbon is attached to chlorine by a single bond and it can be easily broken.
(2) Aryl halides are stabilized by resonance but alkyl halides are not. Hence, the energy of activation for the displacement of halogen from aryl halides is much greater than that of alkyl halides.
(3) Different hybridization state of carbon atom in C-X bond :
(i) In alkyl halides, the carbon of C-X bond is \( sp^3 \)-hybridized with less S-character and greater bond length of 178 pm, which requires less energy to break the C-X bond.
(ii) In aryl halides, the carbon of C-X bond is \( sp^2 \)-hybridized with more S-character and shorter bond length which requires more energy to break C-X bond. Therefore, aryl halides are less reactive than alkyl halides.
(iii) Polarity of the C-X bond : In aryl halide C-X bond is less polar than in alkyl halides. Because \( sp^2 \)-hybrid carbon of C-X bond has less tendency to release electrons to the halogen than a \( sp^3 \)-hybrid carbon in alkyl halides. Thus halogen atom in aryl halides cannot be easily displaced by nucleophile.
(2) Aryl halides are extremely less reactive towards nucleophilic substitution reactions.
Answer: Aryl halides are extremely less reactive towards nucleophilic substitution reaction due to the following reasons : (1) Resonance effect : In haloarenes, the electron pairs on halogen atom are in conjugation with \( \pi \)-electrons of the benzene ring. The delocalization of these electrons C-Cl bond acquires partial double bond character.
ℹ️ चित्र व्याख्या (Diagram Explanation): इसमें क्लोरोबेंजीन की पांच अनुनादी संरचनाएँ दिखाई गई हैं। पहली संरचना में बेंजीन वलय से एक क्लोरीन परमाणु जुड़ा है। अन्य संरचनाओं में, क्लोरीन परमाणु पर एकाकी इलेक्ट्रॉन युग्म वलय के \( \pi \)-इलेक्ट्रॉनों के साथ संयुग्मन में भाग लेते हैं, जिससे C-Cl बंध में आंशिक द्विबंध गुण आ जाता है और वलय पर ऑर्थो और पैरा स्थितियों पर नकारात्मक आवेश आता है।
(I) \( \iff \) (II) \( \iff \) (III) \( \iff \) (IV) \( \iff \) (V)
Due to partial double bond character of C-Cl bond in aryl halides, the bond cleavage in haloarene is difficult and are less reactive towards nucleophilic substitution.
(2) \( Sp^2 \) hybrid state of C : Different hybridization state of carbon atom in C-X bond : In aryl halides, the carbon of C-X bond is \( sp^2 \)-hybridized with more S-character and shorter bond length of 169 pm which requires more energy to break C-X bond. It is difficult to break a shorter bond than a longer bond, in alkyl chloride (bond length 178 pm) therefore, aryl halides are less reactive towards nucleophilic substitution reaction.
(3) Instability of phenyl cation : In aryl halides, the phenyl cation formed due to self ionisation will not be stabilized by resonance which rules out possibility of \( S_N1 \) mechanism. Also backside attack of nucleophile is blocked by the aromatic ring which rules out \( S_N2 \) mechanism. Thus cations are not formed and hence aryl halides do not undergo nucleophilic substitution reaction easily.
(4) As any halides are electron rich molecules due to the presence of \( \pi \)-bond, they repel electron rich nucleophilic, attack. Hence, aryl halides are less reactive toward nucleophilic substitution reactions. However, the presence of electron withdrawing groups at o/p position activates the halogen of aryl halides towards substitution.
ℹ️ चित्र व्याख्या (Diagram Explanation): क्लोरोबेंजीन जिसमें पैरा स्थिति पर नाइट्रो समूह है (\( Cl-C_6H_4-NO_2 \)), उसे जलीय \( NaOH \) के साथ 433 K पर और फिर \( H_2O^+ \) के साथ अभिकृत करने पर पैरा-नाइट्रोफेनोल (\( HO-C_6H_4-NO_2 \)) बनता है।
(3) Aryl halides undergo electrophilic substitution reactions slowly.
Answer: Aryl halides undergo electrophilic substitution reactions slowly and it can be explained as follows :
(1) Inductive effect : The strongly electronegative halogen atom withdraws the electrons from carbon, atom of the ring, hence aryl halides show reactivity towards electrophilic attack.
(2) Resonance effect : The resonating structures of aryl halides show increase in electron density at ortho and para position, hence it is o, p directing.
ℹ️ चित्र व्याख्या (Diagram Explanation): इसमें क्लोरोबेंजीन की अनुनादी संरचनाएँ दिखाई गई हैं, जो दर्शाती हैं कि क्लोरीन परमाणु पर एकाकी इलेक्ट्रॉन युग्म वलय के \( \pi \)-इलेक्ट्रॉनों के साथ संयुग्मन में भाग लेते हैं, जिससे ऑर्थो और पैरा स्थितियों पर इलेक्ट्रॉन घनत्व बढ़ जाता है।
(I) \( \iff \) (II) \( \iff \) (III) \( \iff \) (IV)
The inductive effect and resonance effect compete with each other. The inductive effect is stronger than resonance effect. The reactivity of aryl halides is controlled by stronger inductive effect and o, p orientation is controlled by weaker resonating effect.
ℹ️ चित्र व्याख्या (Diagram Explanation): इसमें क्लोरोनियम आयन की संरचना दिखाई गई है, जो बेंजीन वलय पर एक क्लोरीन परमाणु और एक इलेक्ट्रोफाइल Y के हमले के बाद बनती है, जहाँ क्लोरीन पर धनात्मक आवेश होता है और यह ऑर्थो या पैरा स्थिति पर Y से जुड़ा होता है। The attack of electrophile (Y) on haloarenes at ortho and para positions are more stable due to formation of chloronium ion. The chloronium ion formed is comparatively more stable than other hybrid structures of carbonium ion.
In simple words: Haloarenes are less reactive than haloalkanes due to the partial double bond character of the C-X bond from resonance, the greater S-character of the \( sp^2 \) hybridized carbon, and the instability of the phenyl carbocation. However, for electrophilic substitution, halogens are deactivating but ortho/para directing due to their lone pair resonance donation.

🎯 Exam Tip: Remember that \( sp^2 \) hybridized carbons form shorter and stronger bonds than \( sp^3 \) carbons. This, combined with resonance stabilization of the C-X bond, significantly reduces the reactivity of aryl halides in nucleophilic substitution. For electrophilic reactions, the inductive effect deactivates, but resonance directs to o/p positions.

Question iv. Reactions involving Grignard reagent must be carried out under anhydrous condition.
Answer:
(1) Grignard reagent (R Mg X) is an organometallic compound. The carbon-magnesium bond is highly polar and magnesium halogen bond is in ionic in nature. Grignard reagent is highly reactive.
(2) The reactions of Grignard reagent are carried out in dry conditions because traces of moisture may spoil the reaction and Grignard reagent reacts with water to produce alkane. Hence, reactions involving Grignard reagent must be carried out under anhydrous condition.
In simple words: Grignard reagents are extremely reactive and strong bases. They readily react with even trace amounts of water (an acid) to form alkanes, thereby destroying the reagent. Therefore, all reactions involving Grignard reagents must be performed under strictly anhydrous conditions.

🎯 Exam Tip: Always remember the sensitivity of Grignard reagents to protic solvents (water, alcohols, etc.). Dry conditions are essential to prevent their premature decomposition and ensure the desired reaction proceeds.

Question v. Alkyl halides are generally not prepared by free radical halogenation of alkane.
Answer:
(1) Free radical halogenation of alkane gives a mixture of all different possible Monohaloalkanes as well as polyhalogen alkanes.
(2) In this method, by changing the quantity of halogen the desired product can be
In simple words: Free radical halogenation of alkanes is generally not a good method for preparing specific alkyl halides because it produces a complex mixture of mono- and poly-halogenated products and isomers, making isolation of a single desired product difficult and inefficient.

🎯 Exam Tip: Free radical reactions are often non-selective. For precise synthesis of a single alkyl halide, other more selective methods (like addition to alkenes or alcohol conversions) are preferred.

 

Question. Alkyl halides though polar are immiscible with water.
Answer:(1) In alkyl halide, the halogen atom is more electronegative than carbon atom, the C - X bond is polar. (2) Though alkyl halide is polar, it is insoluble in water because alkyl halide is not able to form hydrogen bonds with water. Attraction between alkyl halide molecule is stronger than attraction between alkyl halide and water.
In simple words: Alkyl halides are polar due to the electronegativity difference between carbon and halogen, but they cannot form hydrogen bonds with water, which is necessary for dissolution. The intermolecular forces within alkyl halides and within water are stronger than the attractions between alkyl halide and water molecules.

🎯 Exam Tip: Understanding the role of hydrogen bonding and intermolecular forces is crucial for explaining solubility trends. Focus on why alkyl halides cannot establish these bonds with water.

 

Question. C-F bond in CH3F is the strongest bond and C-I bond in CH3l is the weakest bond. Explain.
Answer:(1) Methyl fluoride (CH3F) is highly polar molecule and has the shortest C-F bond length (139 pm) and the strongest C-F bond due to greater overlap of orbitals of the same principal quantum number i.e., overlap of 2sp³ orbital of carbon with 2pz orbital of fluorine. (2) Methyl iodide (CH3l) is much less polar and has the longest (C-I) bond length (214 pm) and the weakest C-I bond due to poor overlap of 2sp³ orbital carbon with 5pz orbital of iodine i.e., 2sp³ orbital of carbon cannot penetrate into larger p-orbitals.
In simple words: The C-F bond is strong and short because fluorine's small size allows for excellent orbital overlap with carbon. Conversely, the C-I bond is weak and long due to iodine's large size, leading to poor orbital overlap and reduced bond strength.

🎯 Exam Tip: Remember that bond strength is inversely related to bond length, and both are influenced by the effectiveness of orbital overlap, which depends on the atomic size of the bonded atoms.

 

Question. The boiling point of alkyl iodide is higher than that of alkyl fluoride.
Answer:For a given alkyl group, the boiling point increases with increasing atomic mass of the halogen, because magnitude of van der Waals force increases with increase in size and mass of halogen. Therefore, boiling point of alkyl iodide is higher than that of alkyl fluoride.
In simple words: Alkyl iodides have higher boiling points than alkyl fluorides because iodine is much larger and heavier than fluorine. This leads to stronger van der Waals forces of attraction between alkyl iodide molecules, requiring more energy to overcome them.

🎯 Exam Tip: Boiling points of alkyl halides are primarily determined by van der Waals forces. Larger and heavier halogens result in stronger van der Waals forces and thus higher boiling points.

 

Question. The boiling point of isopropyl bromide is lower than that of n-propyl bromide.
Answer:For isomeric alkyl halides (isopropyl bromide and n-propyl bromide), the boiling point decreases as the branching increases, surface area decreases on branching and van der Waals forces decrease, therefore, the boiling point of isopropyl bromide is lower than that of n-propyl bromide.
In simple words: Isopropyl bromide has a branched structure, which reduces its surface area compared to the straight-chain n-propyl bromide. A smaller surface area leads to weaker van der Waals forces between molecules, resulting in a lower boiling point.

🎯 Exam Tip: For isomers, increased branching leads to a more spherical shape, reducing surface area and weakening van der Waals forces, thus lowering the boiling point. Remember this trend for comparing boiling points of isomers.

 

Question. p-Dichlorobenzene has mp. higher than those of o-and rn-isomers.
Answer:p-Dichlorobenzene has higher melting point than those of o-and m-isomers. This is because of its symmetrical structure which can easily fits in crystal lattice. As a result intermolecular forces of attraction are stronger and therefore greater energy is required to overcome its lattice energy.
In simple words: Para-dichlorobenzene has a more symmetrical structure than its ortho and meta isomers, allowing it to pack more efficiently and tightly in a crystal lattice. This tight packing leads to stronger intermolecular forces and a higher melting point, as more energy is needed to break the lattice.

🎯 Exam Tip: Symmetry in molecules can significantly impact melting points due to better packing in the crystal lattice, leading to stronger intermolecular forces. This effect is often more pronounced for melting points than boiling points.

 

Question iii. Reactions involving Grignard reagent must be carried out under anhydrous conditions.
Answer:(1) Grignard reagent (R Mg X) is an organometallic compound. The carbon-magnesium bond is highly polar and magnesium halogen bond is in ionic in nature. Grignard reagent is highly reactive. (2) The reactions of Grignard reagent are carried out in dry conditions because traces of moisture may spoil the reaction and Grignard reagent reacts with water to produce alkane. Hence, reactions involving Grignard reagent must be carried out under anhydrous condition.
In simple words: Grignard reagents are extremely reactive organometallic compounds. They react vigorously with water, destroying the reagent and forming an alkane, so all reactions must be done in a completely water-free (anhydrous) environment to ensure the desired reaction proceeds.

🎯 Exam Tip: Always remember the extreme sensitivity of Grignard reagents to protic solvents like water. Anhydrous conditions are a critical experimental requirement for their successful use in synthesis.

 

Question iv. Alkyl halides are generally not prepared by free radical halogenation of alkanes.
Answer:(1) Direct fluorination of alkanes is highly exothermic, explosive and invariably leads to polyfluorination and decomposition of the alkanes. It is difficult to control the reaction. (2) Direct iodination of alkanes is highly reversible and difficult to carry out. (3) In direct chlorination and bromination, the reaction is not selective. It can lead to different isomeric monohalogenated alkanes (alkyl halides) as well as polyhalogenated alkanes. Hence, halogenation of alkanes is not a good method of preparation of alkyl halides.
In simple words: Free radical halogenation of alkanes is generally not preferred for preparing alkyl halides because fluorination is too violent, iodination is reversible, and chlorination/bromination are non-selective, producing a mixture of different mono- and polyhalogenated products that are hard to separate.

🎯 Exam Tip: Recognize the limitations of free radical halogenation for alkyl halide synthesis. Focus on why selectivity and control are poor for chlorine and bromine, and the extreme reactivity or reversibility for fluorine and iodine, respectively.

 

4. Distinguish between - S№1 and S№2 mechanism of substitution reaction ?
Answer:

Factor	SN1	SN2
(1) Number of steps	Two steps	One step
(2) Molecularity/Order	Unimolecular/1st order	Bimolecular/2nd order
(3) Reaction rate	Depends upon concentration of one reacting species	Depends upon concentration of two reacting species
(4) Attack of a nucleophile	Back side attack and front side attack on a substrate with equal probability	Only back side attack on a substrate
(5) Transition state	Two steps, two transition states	One step, one transition state
(6) Type of substrate	Mainly tertiary (3°) substrate	Mainly primary (1°) substrate
(7) Stereochemistry	50% inversion and 50% retention of configuration	100% inversion of configuration
(8) Enantiomer	Forms racemic mixture	Forms opposite enantiomer
(9) Solvent	Polar solvent favourable	Nonpolar solvent favourable
(10) Energy of activation	Two values of energies of activation	One value of energy activation
(11) Intermediate	Carbocation intermediates	No intermediate
(12) Nucleophile	Weak nucleophile favourable	Strong nucleophile favourable
(13) Order of reactivity in alkyl halides	Tertiary > Secondary > Primary	Primary > Secondary > Tertiary

In simple words: S_N1 reactions occur in two steps via a carbocation intermediate, leading to racemization and favoring tertiary substrates and polar solvents. S_N2 reactions happen in one concerted step with a single transition state, causing inversion of configuration and favoring primary substrates and strong nucleophiles.

🎯 Exam Tip: Focus on the key differences: number of steps, intermediate formation, stereochemistry (racemization vs. inversion), and substrate/solvent preferences. These are core distinctions for SN1 vs. SN2 mechanisms.

 

5. Explain - Optical isomerism in 2-chlorobutane.
Answer:Cl \( \text{CH}_3\text{-CH-CH}_2\text{-CH}_3 \) (1) 2-Chlorobutane contains an asymmetric carbon atom (the starred carbon atom) which is attached to four different groups, i.e., ethyl (-CH2-CH3), methyl (CH3), chloro (Cl) and hydrogen (H) groups.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र 2-क्लोरोब्यूटेन के दो एनेंटियोमर को दर्शाता है. बाईं ओर d-आइसोमर ( (+) 2-क्लोरोब्यूटेन) का एक त्रिविमीय मॉडल है जहाँ H, Cl, CH3 और C2H5 समूह एक असममित कार्बन से जुड़े हैं. दाहिनी ओर l-आइसोमर ((-) 2-क्लोरोब्यूटेन) का एक दर्पण प्रतिबिंब है, जिसमें समान समूह विपरीत स्थानिक विन्यास में जुड़े हैं. यह दर्शाता है कि ये दोनों संरचनाएँ एक दूसरे पर अध्यारोपित नहीं हो सकती हैं.

(2) Two different arrangements of these groups around the carbon atom are possible as shown in the figure. Hence, it exists as a pair of enanti-omers. The two enantiomers are mirror images of each other and are not superimposable. (3) One of the enantiomers will rotate the plane of plane-polarized light to the left hand side and is called the laevorotatory isomer (l-isomer). The other enantiomer will rotate the plane of plane-polarized light to the right hand side and is called the dextrorotatory isomer (d-isomer). (4) Equimolar mixture of the d- and the l-isomers is optically inactive and is called the racemic mixture or the racemate (dl-mixture). The optical inactivity of the racemic mixture is due to external compensation.
In simple words: 2-Chlorobutane has a chiral carbon bonded to four different groups, meaning it exists as two non-superimposable mirror images called enantiomers. These enantiomers rotate plane-polarized light in opposite directions (d- and l-forms), and an equal mixture of both is optically inactive, known as a racemic mixture.

🎯 Exam Tip: To identify optical isomerism, look for chiral centers (carbon atoms bonded to four different groups). Understand that enantiomers are non-superimposable mirror images and a racemic mixture is optically inactive due to compensation.

 

6. Convert the following.

 

Question i. Propene to propan-1-ol
Answer:\( \text{CH}_3\text{-CH=CH}_2 \xrightarrow{\text{HBr, peroxide}} \text{CH}_3\text{-CH}_2\text{-CH}_2\text{-Br} \) Propene n-propyl bromide \( \text{CH}_3\text{-CH}_2\text{-CH}_2\text{-Br} + \text{aq. KOH} \xrightarrow{\Delta} \text{CH}_3\text{-CH}_2\text{-CH}_2\text{-OH} + \text{KBr} \) Propan-1-ol
In simple words: First, propene reacts with HBr in the presence of peroxide to form n-propyl bromide (anti-Markovnikov addition). Then, n-propyl bromide undergoes hydrolysis with aqueous KOH to yield propan-1-ol.

🎯 Exam Tip: This conversion involves anti-Markovnikov addition (HBr/peroxide) followed by nucleophilic substitution. Remember to use appropriate reagents for regioselectivity and functional group transformation.

 

Question ii. Benzyl alcohol to benzyl cyanide
Answer:\( \text{C}_6\text{H}_5\text{-CH}_2\text{OH} \xrightarrow{\text{NaBr/H}_2\text{SO}_4} \text{C}_6\text{H}_5\text{-CH}_2\text{Br} + \text{H}_2\text{O} \) Benzyl alcohol Benzyl bromide \( \text{C}_6\text{H}_5\text{-CH}_2\text{Br} \xrightarrow{\text{KCN, alc.}} \text{C}_6\text{H}_5\text{-CH}_2\text{CN} + \text{KBr} \) Benzyl bromide Benzyl cyanide
In simple words: Benzyl alcohol is first converted to benzyl bromide using NaBr and H2SO4. Then, benzyl bromide reacts with alcoholic potassium cyanide (KCN) to replace the bromine with a cyano group, forming benzyl cyanide.

🎯 Exam Tip: This is a two-step conversion involving the formation of an alkyl halide intermediate (benzyl bromide) and then a nucleophilic substitution reaction with KCN to introduce the nitrile group. Remember that KCN is a common reagent for converting halides to nitriles.

 

Question iii. Ethanol to propane nitrile
Answer:\( \text{CH}_3\text{-CH}_2\text{-OH} + \text{HBr} \longrightarrow \text{CH}_3\text{-CH}_2\text{-Br} + \text{H}_2\text{O} \) Ethanol Ethyl bromide \( \text{CH}_3\text{-CH}_2\text{-Br} + \text{KCN} \xrightarrow{\text{alc.}} \text{CH}_3\text{-CH}_2\text{-CN} + \text{KBr} \) Ethyl bromide Propane nitrile
In simple words: Ethanol is first converted into ethyl bromide by reacting it with HBr. Subsequently, ethyl bromide undergoes a nucleophilic substitution reaction with alcoholic potassium cyanide (KCN) to form propane nitrile, extending the carbon chain by one atom.

🎯 Exam Tip: This synthesis involves two common transformations: converting an alcohol to an alkyl halide, then using KCN for a chain-extending nucleophilic substitution to form a nitrile. This is a crucial method for increasing carbon chain length.

 

Question iv. But-1-ene to n-butyl iodide
Answer:\( \text{6CH}_3\text{-CH}_2\text{-CH=CH}_2 + \text{B}_2\text{H}_6 \longrightarrow \text{2 (CH}_3\text{-CH}_2\text{-CH}_2\text{-CH}_2\text{)}_3\text{B} \) But-1-ene Tributyl borane \( \text{(CH}_3\text{-CH}_2\text{-CH}_2\text{-CH}_2\text{)}_3\text{B} + \text{3H}_2\text{O}_2 \xrightarrow{\text{OH}^-, \text{H}_2\text{O}} \text{3CH}_3\text{-CH}_2\text{-CH}_2\text{-CH}_2\text{-OH} + \text{B(OH)}_3 \) Butan-1-ol \( \text{3CH}_3\text{-CH}_2\text{-CH}_2\text{-CH}_2\text{-OH} + \text{PI}_3 \xrightarrow{\text{red P/I}_2, \Delta} \text{3CH}_3\text{-CH}_2\text{-CH}_2\text{-CH}_2\text{I} + \text{H}_3\text{PO}_3 \) Butan-1-ol n-butyl iodide
In simple words: But-1-ene is first converted to butan-1-ol via hydroboration-oxidation, which is an anti-Markovnikov hydration. Then, butan-1-ol reacts with phosphorus triiodide (PI3) to replace the hydroxyl group with iodine, yielding n-butyl iodide.

🎯 Exam Tip: This is a multi-step synthesis demonstrating hydroboration-oxidation for anti-Markovnikov hydration of an alkene, followed by conversion of an alcohol to an alkyl iodide using PI3. Pay attention to the regioselectivity of the first step.

 

Question v. 2-Chloropropane to propan-1-ol
Answer:\( \text{CH}_3\text{-CH(Cl)-CH}_3 \xrightarrow{\text{alc.KOH}, \Delta} \text{CH}_3\text{-CH=CH}_2 + \text{KCl} + \text{H}_2\text{O} \) 2-Chloropropane Propene \( \text{6CH}_3\text{-CH=CH}_2 + \text{B}_2\text{H}_6 \longrightarrow \text{2(CH}_3\text{-CH}_2\text{-CH}_2\text{)}_3\text{B} \) Propene tripropyl borane \( \text{(CH}_3\text{-CH}_2\text{-CH}_2\text{)}_3\text{B} + \text{3H}_2\text{O}_2 \xrightarrow{\text{OH}^-, \text{H}_2\text{O}} \text{3CH}_3\text{-CH}_2\text{-CH}_2\text{-OH} + \text{B(OH)}_3 \) Propan-1-ol
In simple words: 2-Chloropropane is first dehydrohalogenated using alcoholic KOH to form propene. Propene is then converted to propan-1-ol via hydroboration-oxidation, which ensures the anti-Markovnikov addition of water.

🎯 Exam Tip: This conversion involves an elimination reaction (dehydrohalogenation) to form an alkene, followed by a hydroboration-oxidation sequence to achieve anti-Markovnikov hydration. Note the change in functional groups and regioselectivity.

 

Question vi. tert-Butyl bromide to isobutyl bromide
Answer:CH3 | \( \text{CH}_3\text{-C(Br)-CH}_3 \xrightarrow{\text{alc.KOH}, \Delta} \text{CH}_3\text{-C=CH}_2 + \text{KBr} + \text{H}_2\text{O} \) tert-Butyl bromide 2-Methyl prop-1-ene CH3 | \( \text{CH}_3\text{-C=CH}_2 \xrightarrow{\text{HBr, Peroxide}} \text{CH}_3\text{-C(CH}_3\text{)-CH}_2\text{Br} \) 2-Methyl prop-1-ene isobutyl bromide
In simple words: tert-Butyl bromide undergoes elimination with alcoholic KOH to form 2-methylprop-1-ene. This alkene then reacts with HBr in the presence of peroxide, leading to anti-Markovnikov addition of HBr and yielding isobutyl bromide.

🎯 Exam Tip: This conversion requires an elimination reaction to remove the tertiary halogen and create an alkene, followed by an anti-Markovnikov addition of HBr to place the bromine at the less substituted carbon, thus forming the primary halide (isobutyl bromide).

 

Question vii. Aniline to chlorobenzene
Answer:\( \text{C}_6\text{H}_5\text{-NH}_2 \xrightarrow{\text{NaNO}_2/\text{HCl}, \Delta} \text{C}_6\text{H}_5\text{-N}_2^+\text{Cl}^- \) Aniline Benzene diazonium chloride \( \text{C}_6\text{H}_5\text{-N}_2^+\text{Cl}^- \xrightarrow{\text{Cu}_2\text{Cl}_2} \text{C}_6\text{H}_5\text{-Cl} + \text{N}_2 \) Benzene diazonium chloride Chlorobenzene
In simple words: Aniline is first converted to benzene diazonium chloride through diazotization using sodium nitrite and HCl. This diazonium salt then reacts with cuprous chloride (Cu2Cl2) in a Sandmeyer reaction to replace the diazonium group with chlorine, forming chlorobenzene.

🎯 Exam Tip: This is a classic Sandmeyer reaction. Remember the two key steps: diazotization of an aromatic amine and subsequent reaction of the diazonium salt with cuprous halide to form an aryl halide. Diazonium salts are versatile intermediates.

 

Question viii. Propene to 1-nitropropane
Answer:\( \text{CH}_3\text{-CH=CH}_2 \xrightarrow{\text{HBr, Peroxide}} \text{CH}_3\text{-CH}_2\text{-CH}_2\text{-Br} \) Propene 1-bromopropane \( \text{CH}_3\text{-CH}_2\text{-CH}_2\text{-Br} + \text{AgNO}_2 \xrightarrow{-\text{NaBr}, \Delta} \text{CH}_3\text{-CH}_2\text{-CH}_2\text{-NO}_2 + \text{AgBr(s)} \) 1-bromopropane 1-nitropropane
In simple words: Propene reacts with HBr in the presence of peroxide to yield 1-bromopropane via anti-Markovnikov addition. Subsequently, 1-bromopropane reacts with silver nitrite (AgNO2) to form 1-nitropropane, as AgNO2's covalent nature favors N-attack.

🎯 Exam Tip: Note the crucial role of peroxide for anti-Markovnikov addition of HBr. Also, remember that AgNO2 (covalent) yields nitroalkanes, while KNO2/NaNO2 (ionic) yields alkyl nitrites, due to ambident nucleophile behavior.

 

7. Answer the following

 

Question i. HCl is added to a hydrocarbon 'A' (C4H8) to give a compound 'B' which on hydrolysis with aqueous alkali forms tertiary alcohol 'C' (C4H10O). Identify 'A', 'B' and 'C'.
Answer:CH3 | \( \text{CH}_3\text{-C=CH}_2 + \text{HCl} \longrightarrow \text{CH}_3\text{-C(Cl)-CH}_3 \) (A) 2-methyl propene (B) tert.butyl chloride CH3 CH3 | | \( \text{CH}_3\text{-C(Cl)-CH}_3 + \text{aq.KOH} \longrightarrow \text{H}_3\text{C-C(OH)-CH}_3 \) (B) tert.butyl chloride (C) tert. butyl alcohol A = CH3-C=CH2 CH3 B = CH3-C-CH3 | Cl CH3 C = CH3-C-OH | CH3
In simple words: Hydrocarbon A is 2-methyl propene. Adding HCl to A gives compound B, which is tert-butyl chloride (Markovnikov's rule). Hydrolysis of B with aqueous alkali yields compound C, which is tert-butyl alcohol.

🎯 Exam Tip: This question tests your knowledge of Markovnikov's rule for HX addition to alkenes and nucleophilic substitution for converting alkyl halides to alcohols. Identifying the tertiary alcohol is key to deducing the structures.

 

Question ii. Complete the following reaction sequences by writing the structural formulae of the organic compounds 'A', 'B' and 'C'.
Answer:(1) \( \text{CH}_3\text{-CH(Br)-CH}_2\text{-CH}_3 \xrightarrow{\text{alc.KOH}, \Delta} \text{CH}_3\text{-CH=CH-CH}_3 + \text{CH}_3\text{-CH}_2\text{-CH=CH}_2 + \text{2KBr} + \text{2H}_2\text{O} \) 2-Bromobutane 80% But-2-ene (A) 20% But-1-ene \( \text{CH}_3\text{-CH=CH-CH}_3 + \text{Br}_2 \longrightarrow \text{CH}_3\text{-CH(Br)-CH(Br)-CH}_3 \) (A) But-2-ene (B) 2,3-Dibromobutane \( \text{CH}_3\text{-CH(Br)-CH(Br)-CH}_3 + \text{2NaNH}_2 \longrightarrow \text{CH}_3\text{-CH(NH}_2\text{)-CH(NH}_2\text{)-CH}_3 + \text{2NaBr} \) (B) 2,3-Dibromobutane (C) 2,3-diaminobutane A = CH3-CH=CH-CH3 B = CH3-CH(Br)-CH(Br)-CH3 C = CH3-CH(NH2)-CH(NH2)-CH3
(2) \( \text{CH}_3\text{-CH(OH)-CH}_3 \xrightarrow{\text{PBr}_3} \text{CH}_3\text{-CH(Br)-CH}_3 + \text{H}_3\text{PO}_3 \) isopropyl alcohol (A) isopropyl bromide \( \text{CH}_3\text{-CH(Br)-CH}_3 + \text{NH}_3 \text{ (excess)} \longrightarrow \text{CH}_3\text{-CH(NH}_2\text{)-CH}_3 + \text{HBr} \) (A) isopropyl bromide (B) isopropyl amine A = CH3-CH(Br)-CH3 isopropyl bromide B = CH3-CH(NH2)-CH3 isopropyl amine
In simple words: For sequence (1), 2-bromobutane eliminates to form but-2-ene (A), which then reacts with Br2 to give 2,3-dibromobutane (B). B then reacts with NaNH2 to form 2,3-diaminobutane (C). For sequence (2), isopropyl alcohol reacts with PBr3 to form isopropyl bromide (A), which then reacts with excess ammonia to yield isopropyl amine (B).

🎯 Exam Tip: Analyze each step carefully: (1) involves elimination (Saytzeff's rule for major product), addition of bromine, and substitution with amide. (2) involves converting an alcohol to an alkyl halide and then to an amine via nucleophilic substitution. Pay attention to reagents and products in multi-step syntheses.

 

Question iii. Observe the following and answer the questions given below.
\( \text{CH}_2\text{=CH-X} \longleftrightarrow \text{CH}_2\text{-CH=X}^- \) a. Name the type of halogen derivative b. Comment on the bond length of C-X bond in it c. Can react by S№1 mechanism? Justify your answer.
Answer:a. Vinyl halide b. C - X bond length shorter in vinyl halide than alkyl halide. Vinyl halide has partial double bond character due to resonance. c. Yes, It reacts by S№1 mechanism. S№1 mechanism involves formation of carbocation intermediate. The vinylic carbocation intermediate formed is resonance stabilized, hence S№1 mechanism is favoured.
In simple words: This derivative is a vinyl halide. The C-X bond is shorter than in alkyl halides due to partial double bond character from resonance. While generally less reactive, S_N1 is possible if the vinylic carbocation formed is resonance stabilized.

🎯 Exam Tip: For vinyl halides, resonance affects bond length and reactivity. The partial double bond character makes the C-X bond shorter. Though traditionally considered unreactive, stabilized vinylic carbocations can undergo S_N1 reactions.

 

12th Chemistry Digest Chapter 9 Halogen Derivatives Intext Questions And Answers

 

Use Your Brain Power..... (Textbook Page 212)

 

Question 1. Write IUPAC name of the following:
Answer:(i) CH3-CH(Br)-CH3: 2-Bromopropane (ii) CH3-CH(CH3)-CH2I: 1-Iodo-2-methyl propane (iii) CH3-CH=CH-CH2-Cl: 1-Chloro-but-2-ene (iv) CH3-C≡C-CH2-Br: 1-Bromo-but-2-yne (v) C6H5-Br: Bromobenzene (vi) C6H3(Br)2(CH3): 1, 4-Dibromobenzene (p-dibromobenzene)
In simple words: The IUPAC names describe the position of halogens and any alkyl groups on the parent hydrocarbon chain or ring. For alkenes and alkynes, the double or triple bond position is also indicated, ensuring the lowest possible numbering for substituents and functional groups.

🎯 Exam Tip: Master IUPAC nomenclature rules for halogen derivatives. Prioritize functional groups, use lowest locants for substituents, and correctly identify the parent chain or ring. Practice with diverse structures to ensure accuracy.

 

Question 10.1 : (Textbook Page 213)

 

Question. How will you obtain 1.bromo.1-methylcyclohexane from alkene? Write possible structures of alkene and the reaction involved.
Answer:(1) Cyclohexene with a methyl group at position 1 (1-methylcyclohexene) reacts with HBr. \( \text{CH}_3\text{-C}_6\text{H}_9 + \text{HBr} \longrightarrow \text{CH}_3\text{-CH(Br)-C}_6\text{H}_9 \) (2) Cyclohexene with a methyl group at position 2 (2-methylcyclohexene) reacts with HBr. \( \text{CH}_2\text{=C}_6\text{H}_9 + \text{HBr} \longrightarrow \text{CH}_3\text{-CH(Br)-C}_6\text{H}_9 \)
In simple words: 1-bromo-1-methylcyclohexane can be obtained by reacting either 1-methylcyclohexene or 2-methylcyclohexene with HBr. In both cases, the bromine atom adds to the carbon already bearing the methyl group according to Markovnikov's rule, because that carbon can form a more stable tertiary carbocation intermediate.

🎯 Exam Tip: For addition of HBr to unsymmetrical alkenes, always consider Markovnikov's rule, which states that the hydrogen adds to the carbon with more hydrogens, and the halogen adds to the more substituted carbon, via the most stable carbocation intermediate.

 

Use Your Brain Power .... (Textbook Page 213)

 

Question 1. Rewrite the following reaction by filling the blanks:
Answer:(1) \( \text{CH}_3\text{-CH = CH}_2 + \text{HBr} \longrightarrow \text{CH}_3\text{-CH(Br)-CH}_3 \text{ (major)} + \text{CH}_3\text{-CH}_2\text{-CH}_2\text{-Br} \text{ (minor)} \) Propene isopropyl bromide n-propyl bromide (2) \( \text{(CH}_3\text{)}_2\text{C=CHCH}_3 + \text{HBr} \xrightarrow{\text{peroxide}} \text{(CH}_3\text{)}_2\text{CH-CH(Br)-CH}_3 \text{ (major)} + \text{(CH}_3\text{)}_2\text{C(Br)-CH}_2\text{-CH}_3 \text{ (minor)} \) 2-methyl but-2-ene 2-bromo-3-methyl butane 2-bromo-2-methyl butane (3) \( \text{CH}_3\text{-CH = CH}_2 + \text{HBr} \xrightarrow{\text{peroxide}} \text{CH}_3\text{-CH}_2\text{-CH}_2\text{-Br} \text{ (major)} + \text{CH}_3\text{-CH(Br)-CH}_3 \text{ (minor)} \) Propene n-propyl bromide isopropyl bromide
In simple words: For (1), HBr adds to propene following Markovnikov's rule, forming major 2-bromopropane and minor 1-bromopropane. For (2), HBr with peroxide adds to 2-methylbut-2-ene in an anti-Markovnikov fashion, favoring 2-bromo-3-methylbutane. For (3), HBr with peroxide adds to propene, also anti-Markovnikov, producing major 1-bromopropane.

🎯 Exam Tip: Remember that HBr addition to alkenes follows Markovnikov's rule unless peroxide is present, in which case it follows anti-Markovnikov's rule. For unsymmetrical alkenes, identifying major and minor products correctly is essential.

 

Question 10.2 : (Textbook Page 216)

 

Question. Arrange the following compounds in order of increasing boiling points : bromoform, chloromethane, dibromomethane, bromomethane.
Answer:The comparative boiling points of halogen derivatives are mainly related with van der Waals forces of attraction which depend upon the molecular size. In the present case all the compounds contain only one carbon. Thus the molecular size depends upon the size of halogen and number of halogen atoms present. Thus increasing order of boiling point is, CH3Cl < CH3Br < CH2Br2 < CHBr3
In simple words: The boiling point of these compounds increases with increasing molecular mass and the number of halogen atoms, due to stronger van der Waals forces. Therefore, chloromethane has the lowest boiling point, followed by bromomethane, then dibromomethane, and bromoform has the highest.

🎯 Exam Tip: Boiling points of haloalkanes are directly proportional to their molecular weight and the strength of van der Waals forces. For compounds with the same carbon skeleton, more halogen atoms or heavier halogens lead to higher boiling points.

 

Try This (Textbook Page 2016)

 

Question 1. (1) Make a three-dimensional model of 2-chlorobutane. (2) Make another model which is a mirror image of the first model. (3) Try to superimpose the two models on each other. (4) Do they superimpose on each other exactly? (5) Comment on whether the two models are identical or not.
Answer:(1) (2) and (3)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र 2-क्लोरोब्यूटेन के गैर-अध्यारोपण योग्य दर्पण प्रतिबिंबों को दर्शाता है. इसमें एक केंद्रीय असममित कार्बन परमाणु होता है जिससे हाइड्रोजन (H), क्लोरीन (Cl), मिथाइल (CH3), और एथिल (C2H5) समूह जुड़े होते हैं. बाईं ओर एक त्रिविमीय संरचना (d-आइसोमर) है, और दाहिनी ओर उसका दर्पण प्रतिबिंब (l-आइसोमर) है. यह स्पष्ट रूप से दिखाता है कि दोनों संरचनाएँ एक दूसरे पर अध्यारोपित नहीं हो सकतीं, जिससे उनके एनेंटियोमेरिक संबंध की पुष्टि होती है.

(4) Two models are non-superimposable mirror images of each other called enantiomers. (5) Two enantiomers are identical. They have the same physical properties (such as melting points, boiling points, densities refractive index). They also have identical chemical properties. The magnitude of their optical rotation is equal but the sign of optical rotation is opposite.
In simple words: When you make models of 2-chlorobutane and its mirror image, you'll find they cannot be perfectly aligned (non-superimposable), meaning they are enantiomers. These enantiomers are chemically identical and share most physical properties, but they rotate plane-polarized light in opposite directions.

🎯 Exam Tip: The concept of non-superimposable mirror images (enantiomers) is fundamental to optical isomerism. Remember that enantiomers share most properties but differ in optical rotation direction, while an equimolar mixture (racemic mixture) is optically inactive.

 

Try This (Textbook Page 219)

 

Question 1. 1. Draw structures of enantiomers of lactic acid using Fischer projection formulae. 2. Draw structures of enantiomers of 2-bromobutane using wedge formula.
Answer:(1) Fischer projection formula: Lactic acid: CH3-CH(OH)-COOH To draw the enantiomers: \( \text{COOH} \) \( | \) \( \text{H - C - OH} \) \( | \) \( \text{CH}_3 \) (This is one enantiomer, e.g., (R)-Lactic acid) Its mirror image (S)-Lactic acid: \( \text{COOH} \) \( | \) \( \text{OH - C - H} \) \( | \) \( \text{CH}_3 \) (2) Wedge formula: 2-bromobutane CH3-CH(Br)-C2H5 To draw the enantiomers: One enantiomer (e.g., (R)-2-bromobutane): CH3 / H ---C--- Br \ C2H5 (Here, H is on a dashed wedge (going away) and Br is on a solid wedge (coming out), or vice-versa, with CH3 and C2H5 in the plane) Its mirror image (S)-2-bromobutane: CH3 / Br ---C--- H \ C2H5 (Here, Br is on a dashed wedge (going away) and H is on a solid wedge (coming out), or vice-versa)
In simple words: For lactic acid, Fischer projections show the chiral carbon at the intersection, with vertical lines representing bonds going away and horizontal lines coming out. For 2-bromobutane, wedge-dash formulas illustrate the 3D arrangement, with solid wedges for bonds coming towards you, dashed wedges for bonds going away, and straight lines for bonds in the plane.

🎯 Exam Tip: Familiarize yourself with both Fischer projection and wedge-dash notation for depicting stereochemistry. Practice drawing enantiomers in both formats to clearly show the non-superimposable mirror image relationship and understand the conventions for each representation.

 

Can You Tell? (Textbook Page 220)

 

Question 1. Alkyl halides, when treated with alcoholic solution of silver nitrite, give nitroalkanes whereas with sodium nitrite they give alkyl nitrites. Explain.
Answer:Nitrite ion is an ambident nucleophile, which can attack through 'O' or 'N'. \[ \text{X-R} + \text{:O=N=O} \longrightarrow \text{R-NO}_2 \] \[ \text{R}^+ + \text{:O=N-O}^- \longrightarrow \text{R-O-N=O} \] Nitroalkane Alkyl nitrite Both nitrogen and oxygen are capable of donating electron pair. C - N bond, being stronger than N - O bond, attack occurs through C atom from alkyl halide forming nitroalkane. However, sodium nitrite (NaNO2) is an ionic compound and oxygen is free to donate pair of electrons. Hence, attack occurs through oxygen resulting in the formation of alkyl nitrite.
In simple words: The nitrite ion can attack via either its nitrogen or oxygen atom. Silver nitrite (AgNO2) is mostly covalent, so it attacks via nitrogen to form a stronger C-N bond, yielding nitroalkanes. Sodium nitrite (NaNO2) is ionic, allowing its oxygen atom to readily attack the alkyl halide, leading to alkyl nitrites.

🎯 Exam Tip: The ambident nature of the nitrite ion is key. Remember that the preference for N-attack (nitroalkane) versus O-attack (alkyl nitrite) depends on the nature of the attacking reagent (covalent AgNO2 vs. ionic NaNO2/KNO2).

 

Use Your Brain Power! (Textbook Page 222)

 

Question 1. Draw the Fischer projection formulae of two products obtained when compound (A) reacts with OHe by S№1 mechanis.
Answer:Compound (A): C2H5-C(Br)(CH3)-nC3H7 (3-Bromo-3-methylhexane) Reaction with OH- by S_N1 mechanism: \( \text{C}_2\text{H}_5\text{-C(Br)(CH}_3\text{)-nC}_3\text{H}_7 + \text{OH}^- \longrightarrow \text{C}_2\text{H}_5\text{-C(OH)(CH}_3\text{)-nC}_3\text{H}_7 + \text{HO-C(CH}_3\text{)(C}_2\text{H}_5\text{)-nC}_3\text{H}_7 \) (Racemic mixture) Fischer projections for the two enantiomeric products (alcohols): For the (R) enantiomer (e.g., if Br was (R) and OH- attacks from one side): \( \text{CH}_3 \) \( | \) \( \text{C}_2\text{H}_5 \text{-C-OH} \) \( | \) \( \text{n-C}_3\text{H}_7 \) For the (S) enantiomer (OH- attacks from the other side): \( \text{CH}_3 \) \( | \) \( \text{HO-C-C}_2\text{H}_5 \) \( | \) \( \text{n-C}_3\text{H}_7 \)
In simple words: When a chiral alkyl halide like compound (A) undergoes an S_N1 reaction, it forms a planar carbocation intermediate. The nucleophile (OH-) can attack from either side with equal probability, leading to a racemic mixture, meaning equal amounts of both enantiomeric alcohols.

🎯 Exam Tip: Remember that S_N1 reactions involving chiral centers proceed through a planar carbocation, allowing for attack from both sides and thus producing a racemic mixture of enantiomers. This loss of stereochemical purity is a hallmark of S_N1.

 

Question 2. Draw the Fischer projection formula of the product formed when compound (B) reacts with OHO by SN2 mechanism.
Answer:Compound (B): CH3-CH(Cl)-C2H5 (2-Chlorobutane) Reaction with OH- by S_N2 mechanism: \( \text{CH}_3\text{-CH(Cl)-C}_2\text{H}_5 + \text{OH}^- \longrightarrow \text{HO-CH(CH}_3\text{)-C}_2\text{H}_5 + \text{Cl}^- \) (Product with inverted configuration) If the reactant (B) has a specific Fischer projection (e.g., (R)-2-chlorobutane): \( \text{CH}_3 \) \( | \) \( \text{H-C-Cl} \) \( | \) \( \text{C}_2\text{H}_5 \) The product formed by S_N2 mechanism will have inverted configuration ((S)-butan-2-ol): \( \text{CH}_3 \) \( | \) \( \text{HO-C-H} \) \( | \) \( \text{C}_2\text{H}_5 \)
In simple words: In an S_N2 reaction, the nucleophile (OH-) attacks the chiral carbon from the side opposite to the leaving group (Cl-). This back-side attack leads to a complete inversion of configuration at the chiral center, producing an alcohol with the opposite stereochemistry compared to the starting alkyl halide.

🎯 Exam Tip: The S_N2 mechanism is characterized by concerted bond breaking and formation with backside attack, resulting in 100% inversion of configuration at a chiral center. Practice drawing inverted Fischer projections to demonstrate this outcome.

 

Question 10.4 : (Textbook Page 223)

 

Question. Allylic and benzylic halides show high reactivity towards the Sn1 mechanism than other primary alkyl halides. Explain.
Answer:In allylic and benzylic halide, the carbocation formed undergoes stabilization through the resonance. Hence, allylic and benzylic halides show high reactivity towards the Sn1 reaction. The resonating structures are Resonance stabilization of allylic carbocation Resonance stabilization of benzylic carbocation
In simple words: Allylic and benzylic halides are highly reactive in S_N1 reactions because the carbocations formed after the leaving group departs are stabilized by resonance. This delocalization of positive charge across multiple atoms lowers the activation energy, making S_N1 favored compared to other primary alkyl halides.

🎯 Exam Tip: The stability of the carbocation intermediate is the primary driving force for S_N1 reactions. Always look for resonance stabilization in allylic and benzylic systems as a key factor for their enhanced reactivity in S_N1 processes.

 

Question 1. Conversion of chlorobenzene to phenol by aqueous sodium hydroxide requires a high temperature of about 623K and high pressure. Explain.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र क्लोरोबेंजीन को फिनोल में परिवर्तित करने की प्रक्रिया को दर्शाता है। क्लोरोबेंजीन, जिसमें एक क्लोरीन परमाणु एक बेंजीन वलय से जुड़ा होता है, कठोर परिस्थितियों (623K तापमान और 300 वायुमंडलीय दबाव पर जलीय NaOH, उसके बाद H2O+ के साथ अम्लीय उपचार) में अभिक्रिया करके फिनोल बनाता है, जिसमें एक -OH समूह बेंजीन वलय से जुड़ा होता है।
Answer: Due to the partial double bond character in chlorobenzene, the bond cleavage in chlorobenzene is difficult and is less reactive. Hence, during the conversion of chlorobenzene to phenol by a question NaOH requires high temperature & high pressure.
In simple words: Chlorobenzene has a strong carbon-chlorine bond due to resonance, which gives it partial double bond character, making it difficult to break. Therefore, harsh conditions like high temperature and pressure are needed to achieve its conversion to phenol.

🎯 Exam Tip: Understanding the resonance effect and its impact on bond strength is crucial for explaining the reactivity of aryl halides in nucleophilic substitution reactions.