GSEB Class 12 Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes

🎯 Exam Tip: Mastering IUPAC nomenclature and understanding the classifications of halides (alkyl, allyl, benzyl, vinyl, aryl, and their degrees) are fundamental for success in organic chemistry.

Question 2. Give the IUPAC names of the following compounds:
(i) \( \ce{CH3CH(CI)CH(Br)CH3} \)
(ii) \( \ce{CHF2CBrCIF} \)
(iii) \( \ce{CICH2C = CCH2Br} \)
(iv) \( \ce{(CCI3)3CCI} \)
(v) \( \ce{CH3C(p-CIC6H4)2CH(Br)CH3} \)
(vi) \( \ce{(CH3)3CCH=CIC6H4I} \)
Answer:
(i) 2-bromo-3-chlorobutane
(ii) 1-bromo-1-chloro-1,2,2-trifluoroethane
(iii) 1-bromo-4-chlorobut-2-yne
(iv) 1,1,1,2,3,3,3-heptachloro-(2-trichloromethyl) propane
(v) 1-[2-bromo-1-(4-chlorophenyl)-1-methylpropyl]-4-chlorobenzene
(vi) 1-chloro-1-(4-iodophenyl)-3,3-dimethylbut-1-ene
In simple words: The IUPAC names for the given chemical formulas are provided, following systematic nomenclature rules.

🎯 Exam Tip: Practice applying IUPAC rules consistently, focusing on identifying the longest carbon chain, numbering correctly, and assigning substituents in alphabetical order. Halogen names are typically prefixed.

Question 3. Write the structures of the following organic halogen compounds,
(i) 2-Chloro-3-methylpentane
(ii) p-Bromochlorobenzene
(iii) 1-Chloro-4-ethylcyclohexane
(iv) 2-(2-Chlorophenyl)-1-iodooctane
(v) Perfluorobenzene
(vi) 4-tert-Butyl-3-iodoheptane
(vii) 1-Bromo-4-sec-butyl-2-methylbenzene
(viii) 1,4-Dibromobut-2-ene
Answer:
i. ℹ️ चित्र व्याख्या (Diagram Explanation): यह 2-क्लोरो-3-मिथाइलपेंटेन की संरचना है, जिसमें पांच कार्बन की सीधी श्रृंखला है, दूसरे कार्बन पर क्लोरीन और तीसरे पर मिथाइल समूह जुड़ा है।
ii. ℹ️ चित्र व्याख्या (Diagram Explanation): यह पैरा-ब्रोमोक्लोरोबेंजीन की संरचना है, जिसमें बेंजीन रिंग पर ब्रोमीन और क्लोरीन परमाणु एक-दूसरे के पैरा-स्थान पर जुड़े हुए हैं।
iii. ℹ️ चित्र व्याख्या (Diagram Explanation): यह 1-क्लोरो-4-इथाइलसाइक्लोहेक्सेन की संरचना है, जिसमें साइक्लोहेक्सेन रिंग के पहले कार्बन पर क्लोरीन और चौथे कार्बन पर इथाइल समूह जुड़ा है।
iv. ℹ️ चित्र व्याख्या (Diagram Explanation): यह 2-(2-क्लोरोफेनिल)-1-आयोडोऑक्टेन की संरचना है, जिसमें आठ कार्बन की एक लंबी श्रृंखला के पहले कार्बन पर आयोडीन और दूसरे पर 2-क्लोरोफेनिल समूह जुड़ा है।
v. ℹ️ चित्र व्याख्या (Diagram Explanation): यह परफ्लोरोबेंजीन की संरचना है, जिसमें बेंजीन रिंग के सभी छह हाइड्रोजन परमाणुओं को फ्लोरीन परमाणुओं से प्रतिस्थापित किया गया है।
vi. ℹ️ चित्र व्याख्या (Diagram Explanation): यह 4-टर्ट-ब्यूटाइल-3-आयोडोहेप्टेन की संरचना है, जिसमें सात कार्बन की एक श्रृंखला के तीसरे कार्बन पर आयोडीन और चौथे कार्बन पर एक टर्ट-ब्यूटाइल समूह जुड़ा है।
vii. ℹ️ चित्र व्याख्या (Diagram Explanation): यह 1-ब्रोमो-4-सेक-ब्यूटाइल-2-मिथाइलबेंजीन की संरचना है, जिसमें बेंजीन रिंग के पहले कार्बन पर ब्रोमीन, दूसरे पर मिथाइल और चौथे पर सेक-ब्यूटाइल समूह जुड़ा है।
viii. ℹ️ चित्र व्याख्या (Diagram Explanation): यह 1,4-डाइब्रोमोब्यूट-2-ईन की संरचना है, जिसमें चार कार्बन की श्रृंखला के पहले और चौथे कार्बन पर ब्रोमीन परमाणु और दूसरे और तीसरे कार्बन के बीच दोहरा बंधन है।
In simple words: The structures for the given organic halogen compounds are drawn, illustrating the specific arrangement of atoms and functional groups as per their IUPAC names.

🎯 Exam Tip: Accurately translating IUPAC names into chemical structures requires careful attention to the root name (parent chain), suffixes (functional groups), and prefixes (substituents and their positions).

Question 4. Which one of the following has the highest dipole, moment?
(i) \( \ce{CH2Cl2} \)
(ii) \( \ce{CHCl3} \)
(iii) \( \ce{CCl4} \)
Answer: The three-dimensional structures of these compounds are depicted below:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह तीन मेथेन डेरिवेटिव्स - डाइक्लोरोमेथेन (\( \ce{CH2Cl2} \)), क्लोरोफॉर्म (\( \ce{CHCl3} \)) और कार्बन टेट्राक्लोराइड (\( \ce{CCl4} \)) की 3D संरचनाएं दर्शाती है। \( \ce{CH2Cl2} \) में परिणामी द्विध्रुव आघूर्ण 1.62D है, \( \ce{CHCl3} \) में 1.03D है, जबकि \( \ce{CCl4} \) में सममित संरचना के कारण शून्य द्विध्रुव आघूर्ण है। इसमें कार्बन-क्लोरीन और कार्बन-हाइड्रोजन बंधों के कारण उत्पन्न होने वाले द्विध्रुव आघूर्णों के सदिश योग को भी दर्शाया गया है।
In \( \ce{CH2Cl2} \), the net dipole moments from two C-Cl bonds are reinforced by the dipole moments of the two C-H bonds, resulting in a dipole moment of 1.62D. In \( \ce{CHCl3} \), the net dipole moments of two C-Cl bonds are opposed by the combined effects of the C-H and C-Cl bond dipoles, leading to a dipole moment of 1.03D. Since \( \ce{CCl4} \) possesses a symmetrical tetrahedral structure, its individual bond dipoles cancel each other out, resulting in a net dipole moment of zero.
Therefore, \( \ce{CH2Cl2} \) exhibits the highest dipole moment.
In simple words: \( \ce{CH2Cl2} \) has the highest dipole moment because its C-Cl and C-H bond dipoles add up constructively due to its asymmetrical structure, unlike \( \ce{CHCl3} \) (partial cancellation) and \( \ce{CCl4} \) (complete cancellation).

🎯 Exam Tip: Dipole moment depends on both bond polarity and molecular geometry. Symmetrical molecules often have zero net dipole moment even if individual bonds are polar due to cancellation of vector components.

Question 5. A hydrocarbon \( \ce{C5H10} \) does not react with chlorine in dark but gives a single monochloro compound \( \ce{C5H9Cl} \) in bright sunlight. Identify the hydrocarbon.
Answer:
1. The molecular formula \( \ce{C5H10} \) suggests that the compound could either be a cycloalkane or an alkene.
2. Given that the hydrocarbon does not react with \( \ce{Cl2} \) in the dark, it cannot be an alkene. Alkenes typically undergo addition reactions with \( \ce{Cl2} \) even in the absence of light. Therefore, the compound must be a cycloalkane.
3. The hydrocarbon reacts with \( \ce{Cl2} \) in the presence of bright sunlight to produce only a single monochloro compound \( \ce{C5H9Cl} \). This indicates that all ten hydrogen atoms of the cycloalkane are chemically equivalent, meaning their replacement leads to the identical product.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह साइक्लोपेंटेन की क्लोरीनीकरण अभिक्रिया को दर्शाता है। अंधेरे में क्लोरीन के साथ कोई अभिक्रिया नहीं होती, जो दर्शाता है कि यह एक एल्केन नहीं है। हालांकि, तेज धूप में, क्लोरीन के साथ अभिक्रिया करके 1-क्लोरोसाइक्लोपेंटेन नामक एक एकल मोनोक्लोरो यौगिक बनता है, जिससे यह निष्कर्ष निकलता है कि अभिकारक साइक्लोपेंटेन है, क्योंकि इसके सभी हाइड्रोजन समतुल्य हैं।
The identified hydrocarbon is cyclopentane.
In simple words: The hydrocarbon is cyclopentane, because its \( \ce{C5H10} \) formula, inertness to \( \ce{Cl2} \) in the dark, and single monochloro product in sunlight indicate a symmetrical cycloalkane where all hydrogens are identical.

🎯 Exam Tip: When determining unknown structures, use all given reaction conditions and product information. Lack of dark reaction with \( \ce{Cl2} \) points to saturated structures, while a single monochlorination product indicates high molecular symmetry.

Question 6. Write the isomers of the compound having the formula \( \ce{C4H9Br} \).
Answer:
i. ℹ️ चित्र व्याख्या (Diagram Explanation): यह 1-ब्रोमोब्यूटेन की संरचना है, जिसमें चार कार्बन की एक सीधी श्रृंखला के पहले कार्बन पर ब्रोमीन परमाणु जुड़ा है।
1-bromobutane
ii. ℹ️ चित्र व्याख्या (Diagram Explanation): यह 2-ब्रोमोब्यूटेन की संरचना है, जिसमें चार कार्बन की एक सीधी श्रृंखला के दूसरे कार्बन पर ब्रोमीन परमाणु जुड़ा है।
2-bromobutane
iii. ℹ️ चित्र व्याख्या (Diagram Explanation): यह 2-मिथाइल-1-ब्रोमोप्रोपेन की संरचना है, जिसमें तीन कार्बन की एक श्रृंखला के पहले कार्बन पर ब्रोमीन और दूसरे पर मिथाइल समूह जुड़ा है।
2-methyl-1-bromopropane
iv. ℹ️ चित्र व्याख्या (Diagram Explanation): यह 2-मिथाइल-2-ब्रोमोप्रोपेन की संरचना है, जिसमें तीन कार्बन की एक श्रृंखला के दूसरे कार्बन पर ब्रोमीन और दो मिथाइल समूह जुड़े हैं।
2-methyl-2-bromopropane
In simple words: The compound \( \ce{C4H9Br} \) has four constitutional isomers: 1-bromobutane, 2-bromobutane, 2-methyl-1-bromopropane, and 2-methyl-2-bromopropane, differing in the position of the bromine atom or the carbon skeleton.

🎯 Exam Tip: To identify all isomers, systematically vary the position of the substituent (here, bromine) and the arrangement of the carbon backbone. Don't forget to check for redundancy by naming each proposed structure.

Question 7. Write the equations for the preparation of 1-iodobutane from i. 1-butanol ii. 1-chlorobutane iii. but-1-ene.
Answer:
i. From 1-butanol:
\[\ce{CH3CH2CH2CH2OH (1-butanol) \xrightarrow{Red \ P/I2} CH3CH2CH2CH2I (1-iodobutane)}\]
ii. From 1-chlorobutane:
\[\ce{CH3CH2CH2CH2Cl (1-chlorobutane) + NaI \xrightarrow{acetone} CH3CH2CH2CH2I + NaCl}\]
iii. From but-1-ene:
\[\ce{CH3CH2CH=CH2 (but-1-ene) + HBr \xrightarrow{(CH3COO)2 \ (peroxide)} CH3CH2CH2CH2Br}\]
\[\ce{CH3CH2CH2CH2Br + NaI \xrightarrow{acetone} CH3CH2CH2CH2I + NaBr}\]
In simple words: 1-iodobutane can be synthesized from 1-butanol using red phosphorus and iodine, from 1-chlorobutane via Finkelstein reaction with NaI in acetone, or from but-1-ene through anti-Markovnikov HBr addition followed by Finkelstein reaction.

🎯 Exam Tip: Remember common named reactions like Finkelstein reaction for halogen exchange and Anti-Markovnikov addition (peroxide effect) for regioselective additions to alkenes. Also, know how to convert alcohols to alkyl halides.

Question 8. What are ambident nucleophiles? Explain with an example.
Answer: Ambident nucleophiles are characterized by having two potential nucleophilic centers, meaning they can attack an electrophile from two different atoms. An illustrative example is the cyanide group (\( \ce{CN-} \)), which can initiate an attack through either its carbon atom or its nitrogen atom, as demonstrated by its resonance structures:
\( \ce{:C\equiv N: <-> :\overset{\ominus}{C}=N:} \)
In simple words: Ambident nucleophiles have two different atoms capable of donating electrons in a reaction, like cyanide (\( \ce{CN-} \)), which can bond via carbon or nitrogen.

🎯 Exam Tip: When dealing with ambident nucleophiles, remember that the reaction outcome (which atom attacks) can be influenced by factors such as solvent, nature of the electrophile, and temperature, leading to isomeric products.

Question 9. Which compound in each of the following pairs will react faster in \( \ce{SN2} \) reaction with -OH?
1. \( \ce{CH3Br} \) or \( \ce{CH3I} \)
2. \( \ce{(CH3)3CCI} \) or \( \ce{CH3CI} \)
Answer:
1. \( \ce{CH3I} \): Iodide (\( \ce{I-} \)) is a superior leaving group compared to bromide (\( \ce{Br-} \)) because of its larger size and lower basicity, making it more stable as an anion.
2. \( \ce{CH3Cl} \): Due to steric considerations, primary alkyl halides react significantly faster in \( \ce{SN2} \) mechanisms compared to tertiary alkyl halides, as the nucleophile can approach the reaction center more easily.
In simple words: \( \ce{CH3I} \) reacts faster because iodide is a better leaving group, and \( \ce{CH3Cl} \) reacts faster than \( \ce{(CH3)3CCl} \) because primary halides have less steric hindrance, favoring \( \ce{SN2} \) reactions.

🎯 Exam Tip: For \( \ce{SN2} \) reactions, assess the leaving group ability (I > Br > Cl > F) and steric hindrance at the reaction center (methyl > primary > secondary > tertiary) as key factors for reactivity.

Question 10. Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene:
(i) 1-Bromo-l-methylcyclohexane
(ii) 2-Chloro-2-methylbutane
(iii) 2,2,3-Trimethyl-3-bromopentane.
Answer:
i. ℹ️ चित्र व्याख्या (Diagram Explanation): 1-ब्रोमो-1-मिथाइलसाइक्लोहेक्सेन को सोडियम एथोक्साइड और इथेनॉल के साथ डीहाइड्रोहैलोजेनीकरण पर, दो एल्कीन बनते हैं: मिथाइलेनेसाइक्लोहेक्सेन और 1-मिथाइलसाइक्लोहेक्सीन। 1-मिथाइलसाइक्लोहेक्सीन मुख्य उत्पाद है, क्योंकि यह अधिक प्रतिस्थापित एल्कीन है।
Formation of methylenecyclohexane and 1-methylcyclohexene. The major product is 1-methylcyclohexene (more substituted alkene).
ii. ℹ️ चित्र व्याख्या (Diagram Explanation): 2-क्लोरो-2-मिथाइलब्यूटेन को सोडियम एथोक्साइड और इथेनॉल के साथ डीहाइड्रोहैलोजेनीकरण पर, दो एल्कीन बनते हैं: 2-मिथाइलप्रोपिन और 2-मिथाइल-2-ब्यूटेन। 2-मिथाइल-2-ब्यूटेन मुख्य उत्पाद है, क्योंकि यह अधिक प्रतिस्थापित एल्कीन है।
Formation of 2-methylpropene and 2-methyl-2-butene. The major product is 2-methyl-2-butene (more substituted alkene).
iii. ℹ️ चित्र व्याख्या (Diagram Explanation): 2,2,3-ट्राइमिथाइल-3-ब्रोमोपेंटेन को सोडियम एथोक्साइड और इथेनॉल के साथ डीहाइड्रोहैलोजेनीकरण पर, 2,2,3-ट्राइमिथाइलपेंट-2-ईन मुख्य उत्पाद के रूप में बनता है, जो अधिक प्रतिस्थापित एल्कीन है।
Formation of 2,2,3-trimethylpent-2-ene. This is the major product as it is the more substituted alkene.
More substituted alkene is the major product (Saytzeff rule).
In simple words: Dehydrohalogenation of alkyl halides with sodium ethoxide in ethanol produces alkenes, with the more substituted alkene being the major product according to Saytzeff's rule.

🎯 Exam Tip: Always apply Saytzeff's rule when predicting the major product of elimination (dehydrohalogenation) reactions: the most substituted alkene is generally the most stable and thus the major product.

Question 11. How will you bring about the following conversions?
(i) Ethanol to but-1-yne
(ii) Ethane to bromoethene
(iii) Propene to 1-nitropropane
(iv) Toluene to benzyl alcohol
(v) Propene to propyne
(vi) Ethanol to ethyl fluoride
(vii) Bromomethane to propanone
(viii) But-1-ene to but-2-ene
(ix) 1-Chlorobutane to n-octane
(x) Benzene to biphenyl
Answer:
i. Ethanol to but-1-yne:
\[\ce{CH3CH2OH (Ethanol) \xrightarrow{HCl/anhyd.ZnCl2, -H2O} CH3CH2Cl \xrightarrow{CH \equiv CNa} CH3-CH2-C \equiv CH (but-1-yne)}\]
ii. Ethane to bromoethene:
\[\ce{CH3-CH3 (ethane) \xrightarrow{Cl2/UV \ light} CH3-CH2Cl \xrightarrow{alc. \ KOH} CH2=CH2 \xrightarrow{Br2/CCl4} BrCH2-CH2Br \xrightarrow{alc. \ KOH} CH2=CHBr (bromoethene)}\]
iii. Propene to 1-nitropropane:
\[\ce{CH3-CH=CH2 (propene) + HBr \xrightarrow{(CH3COO)2 \ (peroxide)} CH3-CH2-CH2Br \xrightarrow{AgNO2} CH3CH2CH2NO2 (1-nitropropane)}\]
iv. Toluene to benzyl alcohol:
\[\ce{CH3-C6H5 (Toluene) \xrightarrow{Cl2/hv} C6H5CH2Cl (Benzyl \ chloride) \xrightarrow{KOH(aq)} C6H5CH2OH (Benzyl \ alcohol)}\]
v. Propene to propyne:
\[\ce{CH3-CH=CH2 (Propene) \xrightarrow{Cl2/Ether} CH3-CH(Cl)CH2Cl \xrightarrow{2NaNH2} CH3-C \equiv CH (propyne) + 2NaCl + 2NH3}\]
vi. Ethanol to ethyl fluoride:
\[\ce{CH3CH2OH (Ethanol) \xrightarrow{HBr, -H2O} CH3CH2Br \xrightarrow{AgF \ or \ AgBr} CH3CH2F (Ethyl \ fluoride)}\]
vii. Bromomethane to propanone:
\[\ce{CH3Br (Bromomethane) \xrightarrow{NaC \equiv CH, -NaBr} CH3-C \equiv CH \xrightarrow{40% \ H2SO4, 1% \ HgSO4} CH3COCH3 (propanone)}\]
viii. But-1-ene to but-2-ene:
\[\ce{CH3CH2CH=CH2 (but-1-ene) + HBr \xrightarrow{KOH(alc)} CH3-CH=CH-CH3 (but-2-ene)}\]
ix. 1-Chlorobutane to n-octane:
\[\ce{2CH3CH2CH2CH2Cl (1-chlorobutane) + 2Na \xrightarrow{ether} CH3-(CH2)6-CH3 (n-octane) + 2NaCl}\]
x. Benzene to biphenyl:
\[\ce{2C6H6 (benzene) + Cl2 \xrightarrow{FeCl3} 2C6H5Cl (chlorobenzene) \xrightarrow{2Na/dry \ ether} C6H5-C6H5 (biphenyl) + 2NaCl}\]
In simple words: This section outlines multi-step synthesis pathways for various organic conversions, utilizing a sequence of reactions such as halogenation, dehydrohalogenation, nucleophilic substitution, Wurtz reaction, and addition reactions to transform starting materials into desired products.

🎯 Exam Tip: For conversions, break down the process into smaller, manageable steps. Identify the change in functional groups or carbon skeleton and select appropriate reagents for each step. Consider named reactions like Wurtz and Finkelstein.

Question 12. Explain why
(i) the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.
(ii) alkyl halides, though polar, are immiscible with water.
(iii) Grignard reagents should be prepared under anhydrous conditions.
Answer:
(i) The carbon atom directly bonded to chlorine in chlorobenzene is \( \ce{sp^2} \) hybridized, while in cyclohexyl chloride, it is \( \ce{sp^3} \) hybridized. \( \ce{sp^2} \) hybridized carbons are more electronegative than \( \ce{sp^3} \) hybridized carbons. This makes the C-Cl bond in chlorobenzene less polar than in cyclohexyl chloride, resulting in a lower dipole moment.
(ii) Although alkyl halides are polar molecules, they are generally immiscible with water. This is because they cannot form hydrogen bonds with water molecules. To dissolve in water, the energy required to break existing hydrogen bonds between water molecules and form new attractions with the alkyl halide would be greater than the energy released, making the process energetically unfavorable.
(iii) Grignard reagents are highly reactive and readily undergo hydrolysis when exposed to water or any protic solvent, forming alkanes. The reaction is:
\[\ce{R-X + Mg \xrightarrow{ether} R-Mg-X}\]
\[\ce{RMgX + H2O -> R-H + Mg(OH)X}\]
To prevent this hydrolysis and ensure the Grignard reagent is formed and maintained for further reactions, it must be prepared and handled under strictly anhydrous (water-free) conditions.
In simple words: Chlorobenzene has a lower dipole moment due to \( \ce{sp^2} \) carbon making the C-Cl bond less polar; alkyl halides are immiscible with water because they can't hydrogen bond; Grignard reagents need anhydrous conditions to prevent hydrolysis to alkanes.

🎯 Exam Tip: Dipole moments are affected by hybridization and electronegativity. Immiscibility often relates to hydrogen bonding capability. The high reactivity of Grignard reagents with protic compounds is a critical concept for their synthesis and use.

Question 13. Give the uses of freon 12, DDT, carbon tetrachloride and iodoform.
Answer:
- Freon 12: Primarily utilized as a refrigerant in air conditioning systems and as a propellant in aerosol sprays.
- DDT (Dichlorodiphenyltrichloroethane): Historically used as a potent insecticide for agricultural purposes and mosquito control, though its use is now restricted due to environmental concerns.
- Carbon Tetrachloride (\( \ce{CCl4} \)): Formerly employed as a fire extinguisher, a cleaning solvent, and a precursor in the production of refrigerants, but its use is largely phased out due to its toxicity and ozone-depleting properties.
- Iodoform (\( \ce{CHI3} \)): Traditionally used as an antiseptic for dressing wounds due to its mild antibacterial properties.
In simple words: Freon 12 is for refrigeration and aerosols, DDT was an insecticide, carbon tetrachloride was a fire extinguisher and solvent (now restricted), and iodoform served as an antiseptic.

🎯 Exam Tip: Knowing the common applications and environmental impacts of important organic compounds, especially halomethanes and haloarenes, is valuable for both theoretical and practical understanding.

Question 14. Write the structure of the major organic product in each of the following reactions:
Answer:
i. ℹ️ चित्र व्याख्या (Diagram Explanation): 1-क्लोरोप्रोपेन की सोडियम आयोडाइड के साथ एसीटोन और गर्मी की उपस्थिति में अभिक्रिया से 1-आयोडोप्रोपेन बनता है, जो फिंकलस्टीन अभिक्रिया का एक उदाहरण है।
Major product: \( \ce{CH3CH2CH2CH2I} \)
ii. ℹ️ चित्र व्याख्या (Diagram Explanation): टर्ट-ब्यूटाइल ब्रोमाइड की पोटेशियम हाइड्रॉक्साइड के साथ इथेनॉल और गर्मी की उपस्थिति में अभिक्रिया से 2-मिथाइलप्रोपिन बनता है, जो एलिमिनेशन अभिक्रिया का एक उदाहरण है।
Major product: \( \ce{CH3-C(=CH2)CH3} \)
iii. ℹ️ चित्र व्याख्या (Diagram Explanation): 1-ब्रोमो-2-मिथाइलब्यूटेन की सोडियम हाइड्रॉक्साइड के साथ जल की उपस्थिति में अभिक्रिया से 2-मिथाइलब्यूटेन-1-ol बनता है, जो न्यूक्लियोफिलिक प्रतिस्थापन अभिक्रिया का एक उदाहरण है।
Major product: \( \ce{CH3CH(Br)CH2CH3} \)
iv. ℹ️ चित्र व्याख्या (Diagram Explanation): 1-ब्रोमोप्रोपेन की पोटेशियम साइनाइड के साथ अभिक्रिया से ब्यूटाइलोनाइट्राइल बनता है, जो न्यूक्लियोफिलिक प्रतिस्थापन अभिक्रिया का एक उदाहरण है।
Major product: \( \ce{CH3CH2CH2CN} \)
v. ℹ️ चित्र व्याख्या (Diagram Explanation): 1-ब्रोमो-2-मिथाइलप्रोपेन की सोडियम एथोक्साइड के साथ इथेनॉल की उपस्थिति में अभिक्रिया से 2-मिथाइलप्रोपेन बनता है, जो एलिमिनेशन अभिक्रिया का एक उदाहरण है।
Major product: \( \ce{CH2=CH2} \)
vi. ℹ️ चित्र व्याख्या (Diagram Explanation): प्रोपेन-1-ol की थायोनिल क्लोराइड ( \( \ce{SOCl2} \) ) के साथ अभिक्रिया से 1-क्लोरोप्रोपेन बनता है।
Major product: \( \ce{CH3CH2CH2Cl} \)
vii. ℹ️ चित्र व्याख्या (Diagram Explanation): प्रोपीन की HBr और पेरोक्साइड की उपस्थिति में अभिक्रिया से 1-ब्रोमोप्रोपेन बनता है, जो एंटी-मार्कोवनिकोव योग का एक उदाहरण है।
Major product: \( \ce{CH3CH2CH2Br} \)
viii. ℹ️ चित्र व्याख्या (Diagram Explanation): 2-मिथाइलप्रोपिन की HBr के साथ अभिक्रिया से 2-ब्रोमो-2-मिथाइलप्रोपेन बनता है, जो मार्कोवनिकोव योग का एक उदाहरण है।
Major product: \( \ce{CH3CH2-C(CH3)2 + HBr} \)
In simple words: This section provides the major organic products for various reactions involving haloalkanes and related compounds, including nucleophilic substitutions, eliminations, and additions, under specified conditions.

🎯 Exam Tip: Carefully identify the type of reaction (substitution, elimination, addition) based on the reagents and conditions. Apply rules like Markovnikov's, anti-Markovnikov's, and Saytzeff's to predict regioselectivity and major products accurately.

Question 15. Write the mechanism of the following reaction.
nBuBr + KCN \( \xrightarrow{\text{EtOH-H₂O}} \) nBuCN
Answer: Cyanide ion (\( \text{CN}^- \)) functions as an ambident nucleophile, meaning it can attack the carbon atom of the C-Br bond in n-butyl bromide through either its carbon (C) or nitrogen (N) atom. The reaction preferentially proceeds via the carbon atom because the carbon-carbon bond formed is stronger than the carbon-nitrogen bond. Consequently, the attack occurs through the carbon atom to yield n-butyl cyanide.
In simple words: Cyanide can attack a molecule from two places, carbon or nitrogen. For n-butyl bromide, it attacks through carbon because the bond formed (C-C) is stronger than a C-N bond, leading to n-butyl cyanide.

🎯 Exam Tip: Understanding ambident nucleophiles and bond strength is crucial for predicting reaction products and mechanisms in organic chemistry.

Question 16. Arrange the compounds of each set in order of reactivity towards \( \text{S}_{\text{N}}2 \) displacement:
(i) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane
(ii) 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 3-Bromo-2-methylbutane
(iii) 1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methylbutane, 1-Bromo-3-methylbutane.
Answer: The order of reactivity for \( \text{S}_{\text{N}}2 \) displacement reactions is primarily determined by steric hindrance. Primary halides react fastest, followed by secondary halides, and tertiary halides react slowest due to the increasing bulkiness around the electrophilic carbon center, which hinders the nucleophile's approach.
(i) 1-Bromopentane > 2-Bromopentane > 2-Bromo-2-methylbutane
(Due to steric effect, the order of reactivity in \( \text{S}_{\text{N}}2 \) reaction is 1° > 2° > 3°)
(ii) 1-Bromo-3-methylbutane > 3-Bromo-2-methylbutane > 2-Bromo-2-methylbutane
(This is due to steric effect)
(iii) 1-Bromobutane > 1-Bromo-3-methylbutane > 1-Bromo-2-methylbutane > 1-Bromo-2,2-dimethylpropane
In simple words: \( \text{S}_{\text{N}}2 \) reactions are fastest when the attacking site is less crowded. So, less branched molecules react quicker than highly branched ones.

🎯 Exam Tip: Remember that steric hindrance significantly impacts the rate of \( \text{S}_{\text{N}}2 \) reactions. Rank compounds by their primary, secondary, or tertiary nature to determine relative reactivity. For isomers of the same type (e.g., all primary), compare the bulkiness of substituents near the reaction center.

Question 17. p-Dichlorobenzene has higher m.p. and solubility than those of o- and m-isomers. Discuss.
Answer: Para-dichlorobenzene exhibits a higher melting point and greater solubility compared to its ortho- and meta-isomers. This difference is attributed to the symmetrical structure of the p-isomer, which allows it to pack more efficiently into the crystal lattice. This close packing results in stronger intermolecular attractive forces, requiring more energy to overcome during melting and increasing its stability in the solid state. While solubility generally correlates with polarity, the enhanced lattice energy of p-dichlorobenzene plays a more dominant role here.
In simple words: The para-form of dichlorobenzene melts at a higher temperature and is more soluble because its symmetrical shape allows molecules to fit together tightly in a solid crystal, making it harder to break apart.

🎯 Exam Tip: When comparing physical properties of isomers, always consider symmetry and its impact on crystal packing (for melting point) and intermolecular forces (for both melting point and solubility).

Question 18. How the following conversions can be carried out?
(i) Propene to propan-1-ol
(ii) Ethane to bromoethene
(iii) 1-Bromopropane to 2-bromopropane
(iv) Toluene to benzyl alcohol
(v) Benzene to 4-bromonitrobenzene
(vi) Benzyl alcohol to 2-phenylethanoic acid
(vii) Ethanol to propanenitrile
(viii) Aniline to chlorobenzene
(ix) 2-Chlorobutane to 3,4-dimethylhexane
(x) 2-Methyl-1-propene to 2-chloro-2-methylpropane
(xi) Ethyl chloride to propanoic acid
(xii) But-1-ene to n-butyliodide
(xiii) 2-Chloropropane to 1-propanol
(xiv) Chlorobenzene to p-nitrophenol
(xv) 2-Bromopropane to 1-bromopropane
(xvi) Chloroethane to butane
(xvii) Benzene to biphenyl
(xviii) tert-Butyl bromide to isobutyl bromide
(xix) Aniline to phenylisocyanide
Answer:
(i) Propene to propan-1-ol
\( \text{CH}_3\text{CH}=\text{CH}_2 \xrightarrow{\text{HBr/peroxide}} \text{CH}_3\text{CH}_2\text{CH}_2\text{Br} \xrightarrow{\text{KOH (aq)}} \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \)
(ii) Ethane to bromoethene
\( \text{CH}_3\text{CH}_3 \xrightarrow{\text{Cl}_2/\text{UV light}} \text{CH}_3\text{CH}_2\text{Cl} \xrightarrow{\text{alc. KOH}} \text{CH}_2=\text{CH}_2 \xrightarrow{\text{Br}_2/\text{CCl}_4} \text{BrCH}_2-\text{CH}_2\text{Br} \xrightarrow{\text{KOH (alc)}} \text{CH}_2=\text{CHBr} \)
(iii) 1-Bromopropane to 2-bromopropane
\( \text{CH}_3\text{CH}_2\text{CH}_2\text{Br} \xrightarrow{\text{KOH (alc)}} \text{CH}_3\text{CH}=\text{CH}_2 \xrightarrow{\text{HBr}} \text{CH}_3\text{CH(Br)}\text{CH}_3 \)
(iv) Toluene to benzyl alcohol
ℹ️ चित्र व्याख्या (Diagram Explanation): इस अभिक्रिया में टोल्यून को क्लोरीन के साथ पराबैंगनी प्रकाश की उपस्थिति में अभिकृत करने पर बेंजाइल क्लोराइड बनता है, जिसे जलीय KOH से अभिकृत करने पर बेंजाइल अल्कोहल प्राप्त होता है।
\( \text{CH}_3\text{C}_6\text{H}_5 \xrightarrow{\text{Cl}_2/\text{hv}} \text{ClCH}_2\text{C}_6\text{H}_5 \xrightarrow{\text{KOH(aq)}} \text{HOCH}_2\text{C}_6\text{H}_5 \)
(v) Benzene to 4-bromonitrobenzene
ℹ️ चित्र व्याख्या (Diagram Explanation): इस अभिक्रिया में बेंजीन को Br2 और FeBr3 से अभिकृत करने पर ब्रोमोबेंजीन बनता है, जिसे सान्द्र HNO3 और सान्द्र H2SO4 से नाइट्रेशन करने पर 4-ब्रोमोनाइट्रोबेंजीन (पैरा उत्पाद) प्राप्त होता है।
\( \text{C}_6\text{H}_6 \xrightarrow{\text{Br}_2/\text{FeBr}_3} \text{C}_6\text{H}_5\text{Br} \xrightarrow{\text{conc.HNO}_3/\text{conc.H}_2\text{SO}_4} \text{BrC}_6\text{H}_4\text{NO}_2 \)
(vi) Benzyl alcohol to 2-phenylethanoic acid
ℹ️ चित्र व्याख्या (Diagram Explanation): इस अभिक्रिया में बेंजाइल अल्कोहल को SOCl2 से अभिकृत करने पर बेंजाइल क्लोराइड बनता है, जिसे KCN से अभिक्रिया कराने पर बेंजाइल साइनाइड बनता है। बेंजाइल साइनाइड का जलीय अम्लीय हाइड्रोलिसिस करने पर 2-फिनाइलइथेनोइक एसिड प्राप्त होता है।
\( \text{C}_6\text{H}_5\text{CH}_2\text{OH} \xrightarrow{\text{SOCl}_2} \text{C}_6\text{H}_5\text{CH}_2\text{Cl} \xrightarrow{\text{KCN}} \text{C}_6\text{H}_5\text{CH}_2\text{CN} \xrightarrow{\text{H}_2\text{O/H}^+} \text{C}_6\text{H}_5\text{CH}_2\text{COOH} \)
(vii) Ethanol to propanenitrile
\( \text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{I}_2/\text{P}} \text{CH}_3\text{CH}_2\text{I} \xrightarrow{\text{KCN}} \text{CH}_3\text{CH}_2\text{CN} \)
(viii) Aniline to chlorobenzene
ℹ️ चित्र व्याख्या (Diagram Explanation): इस अभिक्रिया में एनिलीन को NaNO2 और HCl के साथ 0-5°C पर डाइएज़ोटाइज़ेशन करने पर बेंजीन डाइएज़ोनियम क्लोराइड बनता है, जिसे CuCl और HCl के साथ अभिकृत करने पर क्लोरोबेंजीन प्राप्त होता है।
\( \text{C}_6\text{H}_5\text{NH}_2 \xrightarrow{\text{NaNO}_2, \text{HCl}/0-5^{\circ}\text{C}} \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \xrightarrow{\text{CuCl/HCl}} \text{C}_6\text{H}_5\text{Cl} \)
(ix) 2-Chlorobutane to 3,4-dimethylhexane
\( 2\text{CH}_3-\text{CH(Cl)}-\text{CH}_2\text{CH}_3 + 2\text{Na} \longrightarrow \text{CH}_3\text{CH}_2\text{CH}(\text{CH}_3)-\text{CH}(\text{CH}_3)\text{CH}_2\text{CH}_3 + 2\text{NaCl} \)
(x) 2-Methyl-1-propene to 2-chloro-2-methylpropane
\( \text{CH}_3-\text{C(CH}_3)=\text{CH}_2 \xrightarrow{\text{HCl}} \text{CH}_3-\text{C(Cl)(CH}_3)-\text{CH}_3 \)
(xi) Ethyl chloride to propanoic acid
\( \text{CH}_3\text{CH}_2\text{Cl} \xrightarrow{\text{KCN}} \text{CH}_3\text{CH}_2\text{CN} \xrightarrow{\text{H}_2\text{O/H}^+} \text{CH}_3\text{CH}_2\text{COOH} \)
(xii) But-1-ene to n-butyliodide
\( \text{CH}_3\text{CH}_2\text{CH}=\text{CH}_2 \xrightarrow{\text{HBr/peroxide}} \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br} \xrightarrow{\text{NaI/acetone}} \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{I} \)
(xiii) 2-Chloropropane to 1-propanol
\( \text{CH}_3-\text{CH(Cl)}-\text{CH}_3 \xrightarrow{\text{KOH (alc)}} \text{CH}_3-\text{CH}=\text{CH}_2 \xrightarrow{\text{HBr/peroxide}} \text{CH}_3\text{CH}_2\text{CH}_2\text{Br} \xrightarrow{\text{KOH(aq)}} \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \)
(xiv) Chlorobenzene to p-nitrophenol
ℹ️ चित्र व्याख्या (Diagram Explanation): इस अभिक्रिया में क्लोरोबेंजीन का सान्द्र HNO3 और सान्द्र H2SO4 से नाइट्रेशन करने पर p-नाइट्रोक्लोरोबेंजीन बनता है, जिसे NaOH से 433K पर और फिर H2O/H+ से अभिकृत करने पर p-नाइट्रोफीनोल प्राप्त होता है।
\( \text{ClC}_6\text{H}_5 \xrightarrow{\text{conc.HNO}_3/\text{conc.H}_2\text{SO}_4} \text{ClC}_6\text{H}_4\text{NO}_2 \xrightarrow{\text{(i) NaOH, 433K; (ii) H}_2\text{O/H}^+} \text{HOC}_6\text{H}_4\text{NO}_2 \)
(xv) 2-Bromopropane to 1-bromopropane
\( \text{CH}_3\text{CH(Br)}\text{CH}_3 \xrightarrow{\text{KOH (alc)}} \text{CH}_3-\text{CH}=\text{CH}_2 \xrightarrow{\text{HBr/peroxide}} \text{CH}_3\text{CH}_2\text{CH}_2\text{Br} \)
(xvi) Chloroethane to butane
\( 2\text{CH}_3\text{CH}_2\text{Cl} + 2\text{Na} \xrightarrow{\text{Dry ether}} \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3 + 2\text{NaCl} \)
(xvii) Benzene to biphenyl
ℹ️ चित्र व्याख्या (Diagram Explanation): इस अभिक्रिया में बेंजीन को Br2 और FeBr3 से अभिकृत करने पर ब्रोमोबेंजीन बनता है, जिसे 2Na के साथ शुष्क ईथर की उपस्थिति में वुर्ट्ज़-फिटिग अभिक्रिया करने पर बाइफिनाइल प्राप्त होता है।
\( 2\text{C}_6\text{H}_6 \xrightarrow{\text{Br}_2/\text{FeBr}_3} 2\text{C}_6\text{H}_5\text{Br} \xrightarrow{2\text{Na}/\text{Dry ether}} \text{C}_6\text{H}_5-\text{C}_6\text{H}_5 + 2\text{NaBr} \)
(xviii) tert-Butyl bromide to isobutyl bromide
\( \text{CH}_3\text{C(Br)(CH}_3)\text{CH}_3 \xrightarrow{\text{KOH (alc)}} \text{CH}_3\text{C(CH}_3)=\text{CH}_2 \xrightarrow{\text{HBr/peroxide}} \text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{Br} \)
(xix) Aniline to phenylisocyanide
ℹ️ चित्र व्याख्या (Diagram Explanation): इस अभिक्रिया में एनिलीन को CHCl3 और 3KOH के साथ गरम करने पर फेनिल आइसोसायनाइड बनता है, साथ ही 3KCl और 3H2O भी प्राप्त होते हैं।
\( \text{C}_6\text{H}_5\text{NH}_2 + \text{CHCl}_3 + 3\text{KOH} \longrightarrow \text{C}_6\text{H}_5\text{NC} + 3\text{KCl} + 3\text{H}_2\text{O} \)
In simple words: This section outlines various chemical transformations, converting one organic compound into another using a sequence of specific reactions like substitution, elimination, and addition, often involving catalysts or specific reagents.

🎯 Exam Tip: For conversion questions, always identify the starting material and the desired product. Break down complex conversions into simpler, known reactions. Pay close attention to reagents, reaction conditions (like presence of peroxide, alcoholic/aqueous KOH), and regioselectivity (Markovnikov's vs. anti-Markovnikov's rule).

GSEB Class 12 Chemistry Haloalkanes and Haloarenes Additional Important Questions and Answers

Question 1. Give the structure and IUPAC names of any five haloalkanes having the molecular formula \( \text{C}_5\text{H}_{11}\text{Cl} \).
Answer: Here are five isomeric haloalkanes with the molecular formula \( \text{C}_5\text{H}_{11}\text{Cl} \), along with their structures and IUPAC names:
1. \( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{Cl} \) - 1-chloropentane
2. \( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH(Cl)}\text{CH}_3 \) - 2-chloropentane
3. \( \text{CH}_3\text{CH}_2\text{CH(Cl)}\text{CH}_2\text{CH}_3 \) - 3-chloropentane
4. \( \text{CH}_3\text{CH(CH}_3)\text{CH}_2\text{CH}_2\text{Cl} \) - 1-chloro-3-methylbutane
5. \( \text{CH}_3\text{CH(Cl)}\text{CH(CH}_3)\text{CH}_3 \) - 2-chloro-3-methylbutane
In simple words: Five different ways to arrange one chlorine atom and five carbon atoms with eleven hydrogens, giving different structures and names for the same chemical formula.

🎯 Exam Tip: When drawing isomers, start with the longest possible carbon chain, then systematically move the halogen and methyl groups to generate all unique structures. Always check for duplicates by naming them using IUPAC rules.

Question 2. How will you prepare
(a) methane from methyl chloride
(b) ethane from methyl chloride
(c) toluene from methyl chloride
Answer:
(a) Methane from methyl chloride
\( \text{CH}_3\text{Cl} + \text{H}_2 \xrightarrow{\text{Ni}} \text{CH}_4 + \text{HCl} \)
(b) Ethane from methyl chloride
\( 2\text{CH}_3\text{Cl} + 2\text{Na} \longrightarrow \text{CH}_3\text{CH}_3 + 2\text{NaCl} \)
(c) Toluene from methyl chloride
ℹ️ चित्र व्याख्या (Diagram Explanation): इस अभिक्रिया में क्लोरोबेंजीन को शुष्क ईथर में सोडियम और मेथिल क्लोराइड के साथ वुर्ट्ज़-फिटिग अभिक्रिया करने पर टोल्यून बनता है।
\( \text{C}_6\text{H}_5\text{Cl} + \text{CH}_3\text{Cl} \xrightarrow{\text{Anhy. AlCl}_3} \text{C}_6\text{H}_5\text{CH}_3 \)
In simple words: This shows how to make methane, ethane, and toluene using methyl chloride through reactions like reduction, Wurtz reaction, and Friedel-Crafts alkylation respectively.

🎯 Exam Tip: Learn common named reactions like Wurtz reaction and Friedel-Crafts alkylation, as they are frequently used for carbon-carbon bond formation and changing the carbon skeleton.

Question 3. Grignard reagents are widely used in synthetic organic chemistry.
(i) What are Grignard reagents?
(ii) Give an example.
(iii) How are they prepared?
(iv) What happens when Grignard reagent is treated with water?
Answer:
(i) Grignard reagents are organometallic compounds with the general formula \( \text{R-Mg-X} \), where R is an alkyl or aryl group and X is a halogen (Cl, Br, I). They are highly reactive and nucleophilic.
(ii) An example is methyl magnesium chloride (\( \text{CH}_3\text{MgCl} \)).
(iii) Grignard reagents are typically prepared by reacting an alkyl or aryl halide with magnesium metal in the presence of anhydrous ether.
\( \text{CH}_3\text{Cl} + \text{Mg} \xrightarrow{\text{Ether}} \text{CH}_3\text{MgCl} \)
(iv) When a Grignard reagent is treated with water, it readily undergoes hydrolysis to form an alkane, owing to its highly basic nature.
\( \text{CH}_3\text{MgCl} + \text{H}_2\text{O} \longrightarrow \text{CH}_4 + \text{Mg(OH)Cl} \)
In simple words: Grignard reagents are special chemicals containing carbon, magnesium, and a halogen; they are made by reacting organic halides with magnesium in dry ether. They react strongly with water to produce alkanes.

🎯 Exam Tip: Remember the critical anhydrous conditions for preparing Grignard reagents to prevent their immediate hydrolysis, which would yield unwanted alkanes instead of allowing them to function as powerful nucleophiles.

Question 4. Match the following

🎯 Exam Tip: Memorize the key reagents and conditions associated with important named reactions in organic chemistry, as they are frequently tested in match-the-column and direct recall questions.

Question 5. An organic compound 'A' reacts with chlorine gas in presence of \( \text{FeCl}_3 \) to give B. B on reaction with a mixture of conc.\(\text{HNO}_3\) and conc.\(\text{H}_2\text{SO}_4\) gives a mixture of ortho and para nitrochlorobenzene. Identify A and B. Write their IUPAC names.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): इस अभिक्रिया में बेंजीन (A) क्लोरीन गैस और FeCl3 के साथ अभिकृत करने पर क्लोरोबेंजीन (B) बनाता है। क्लोरोबेंजीन का सान्द्र HNO3 और सान्द्र H2SO4 के साथ नाइट्रेशन करने पर o- और p-नाइट्रोक्लोरोबेंजीन का मिश्रण प्राप्त होता है।
\[ \text{A (Benzene)} + \text{Cl}_2 \xrightarrow{\text{FeCl}_3} \text{B (Chlorobenzene)} \] \[ \text{B (Chlorobenzene)} + \text{conc.HNO}_3 + \text{conc.H}_2\text{SO}_4 \longrightarrow \text{o-Nitrotoluene} + \text{p-Nitrotoluene} \] A: Benzene
B: Chlorobenzene
In simple words: Compound A is benzene, which reacts with chlorine and iron chloride to form chlorobenzene (Compound B). Chlorobenzene then undergoes nitration to produce a mix of ortho and para nitrochlorobenzenes.

🎯 Exam Tip: Recognize electrophilic aromatic substitution reactions. Benzene undergoes halogenation to form halobenzenes (e.g., chlorobenzene), which then undergo further substitution, like nitration, at ortho and para positions due to the activating/deactivating nature of the existing substituent.

Question 6. The teacher told the students to treat KOH solution with 1-chloropropane. One student used alcoholic KOH solution while others used aqueous KOH solution.
(i) What would be the products obtained in each case?
(ii) Justify your answer with chemical equations.
Answer:
(i) If alcoholic KOH is used, the product will be propene (an alkene) via elimination. If aqueous KOH is used, the product will be 1-propanol (an alcohol) via substitution.
(ii) Justification with chemical equations:
When alcoholic KOH is used, it acts as a strong base, promoting a dehydrohalogenation (elimination) reaction, leading to the formation of an alkene.
\( \text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} + \text{KOH (alc)} \longrightarrow \text{CH}_3\text{CH}=\text{CH}_2 + \text{KCl} + \text{H}_2\text{O} \)
When aqueous KOH is used, the hydroxide ion acts as a nucleophile, leading to a nucleophilic substitution reaction, forming an alcohol.
\( \text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} + \text{KOH (aq)} \longrightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} + \text{KCl} \)
In simple words: Alcoholic KOH removes HCl from 1-chloropropane to make propene (an alkene), while aqueous KOH replaces chlorine with an -OH group to make 1-propanol (an alcohol).

🎯 Exam Tip: Distinguish between the roles of alcoholic and aqueous KOH. Alcoholic KOH favors elimination (E2 reaction) to form alkenes, whereas aqueous KOH favors nucleophilic substitution (S_N2 reaction) to form alcohols. This is a common trap in organic chemistry questions.

Question 7. A hydrocarbon \( \text{C}_5\text{H}_{12} \) does not react with chlorine in dark but gives a single monochloro compound \( \text{C}_5\text{H}_9\text{Cl} \) in bright sunlight. Identify the hydrocarbon. Write the structure and IUPAC name of the reactant and product in this reaction.
Answer:
1. The molecular formula \( \text{C}_5\text{H}_{12} \) indicates that the hydrocarbon is either a cycloalkane or an alkane. It has the general formula \( \text{C}_n\text{H}_{2n+2} \), which means it is an alkane. (The initial text says cycloalkane or alkene, but the formula explicitly states alkane).
2. The hydrocarbon does not react with \( \text{Cl}_2 \) in the dark, which is characteristic of saturated hydrocarbons (alkanes) and rules out alkenes that would undergo addition in the dark.
3. It reacts with \( \text{Cl}_2 \) in the presence of bright sunlight to give a single monochloro compound \( \text{C}_5\text{H}_9\text{Cl} \). This implies that all ten hydrogen atoms of the alkane must be equivalent, so that substitution of any one hydrogen yields the same product.
The hydrocarbon with molecular formula \( \text{C}_5\text{H}_{12} \) having all equivalent hydrogens is 2,2-dimethylpropane.
ℹ️ चित्र व्याख्या (Diagram Explanation): 2,2-डाइमिथाइलप्रोपेन (जिसके सभी हाइड्रोजन समतुल्य हैं) क्लोरीन गैस के साथ सूर्य के प्रकाश की उपस्थिति में अभिकृत होने पर 1-क्लोरो-2,2-डाइमिथाइलप्रोपेन का एकमात्र उत्पाद देता है।
\[ (\text{CH}_3)_4\text{C} + \text{Cl}_2 \xrightarrow{\text{Sunlight}} (\text{CH}_3)_3\text{CCH}_2\text{Cl} \] Reactant: 2,2-dimethylpropane
Product: 1-chloro-2,2-dimethylpropane
In simple words: The hydrocarbon is 2,2-dimethylpropane because its formula fits an alkane, it doesn't react with chlorine in the dark, and its structure ensures only one monochloro product forms when reacted in sunlight due to all hydrogens being identical.

🎯 Exam Tip: When given a molecular formula and reaction conditions, consider isomerism. The formation of a "single" monochloro product strongly suggests a highly symmetrical alkane where all hydrogen atoms are equivalent, such as 2,2-dimethylpropane or methane.

Question 8. When benzene is treated with chlorine in the presence of sunlight a compound which is commonly used as an insecticide is formed.
(i) Identify the compound obtained.
(ii) Write the chemical equation.
(iii) If the reaction is carried out in the presence of \( \text{FeCl}_3 \) and in the absence of sunlight, what would be the product obtained? Write the chemical equation.
Answer:
(i) The compound obtained is Benzene Hexachloride (BHC), also known as Lindane ( \( \text{C}_6\text{H}_6\text{Cl}_6 \) ).
(ii) Chemical equation for formation of BHC:
ℹ️ चित्र व्याख्या (Diagram Explanation): बेंजीन क्लोरीन के साथ सूर्य के प्रकाश की उपस्थिति में योगात्मक अभिक्रिया करके बेंजीन हेक्साक्लोराइड (BHC) बनाता है।
\[ \text{C}_6\text{H}_6 + 3\text{Cl}_2 \xrightarrow{\text{Sunlight}} \text{C}_6\text{H}_6\text{Cl}_6 \] (iii) If the reaction is carried out in the presence of \( \text{FeCl}_3 \) and in the absence of sunlight, the product obtained would be chlorobenzene. This is an electrophilic aromatic substitution reaction where a hydrogen atom on the benzene ring is replaced by a chlorine atom, catalyzed by \( \text{FeCl}_3 \).
ℹ️ चित्र व्याख्या (Diagram Explanation): बेंजीन को \( \text{FeCl}_3 \) उत्प्रेरक की उपस्थिति में क्लोरीन के साथ अभिकृत करने पर क्लोरोबेंजीन बनता है, जो एक प्रतिस्थापन अभिक्रिया है।
\[ \text{C}_6\text{H}_6 + \text{Cl}_2 \xrightarrow{\text{FeCl}_3} \text{C}_6\text{H}_5\text{Cl} + \text{HCl} \]
In simple words: Reacting benzene with chlorine under sunlight gives BHC (an insecticide). If the same reaction is done with iron chloride and no sunlight, it produces chlorobenzene instead.

🎯 Exam Tip: Differentiate between the reaction conditions for addition and substitution reactions on benzene. Sunlight promotes free radical addition (forming BHC), while a Lewis acid catalyst (like \( \text{FeCl}_3 \)) promotes electrophilic substitution (forming chlorobenzene).

Question 9. Complete the equation.
\( \text{CH}_3-\text{CH}=\text{CH}_2 + \text{HBr} \longrightarrow ...... \)
(i) Identify the product and why do we get this product?
(ii) If the reaction is carried out in presence of benzyl peroxide what would be the product? Explain.
Answer:
(i) The product obtained when propene reacts with HBr is 2-bromopropane.
\[ \text{CH}_3-\text{CH}=\text{CH}_2 + \text{HBr} \longrightarrow \text{CH}_3-\text{CH(Br)}-\text{CH}_3 \] This product is formed according to Markovnikov's rule, which states that in the addition of a protic acid (HX) to an unsymmetrical alkene, the hydrogen atom adds to the carbon atom of the double bond that already has a greater number of hydrogen atoms, and the halogen atom adds to the carbon atom with fewer hydrogen atoms. This happens because the more stable carbocation (secondary carbocation in this case) is formed as an intermediate.
(ii) If the reaction is carried out in the presence of benzyl peroxide, the product obtained would be 1-bromopropane.
\[ \text{CH}_3-\text{CH}=\text{CH}_2 + \text{HBr} \xrightarrow{\text{Benzyl peroxide}} \text{CH}_3-\text{CH}_2-\text{CH}_2\text{Br} \] This occurs due to the anti-Markovnikov's addition (peroxide effect or Kharasch effect). Peroxides initiate a free radical mechanism, causing the HBr to add such that the bromine atom attaches to the carbon with more hydrogen atoms, resulting in the less substituted alkyl halide. This effect is specific to HBr and not observed with HCl or HI.
In simple words: Without peroxide, propene and HBr make 2-bromopropane (Markovnikov's rule). With peroxide, they make 1-bromopropane (anti-Markovnikov's rule), changing where the bromine attaches.

🎯 Exam Tip: Clearly distinguish between Markovnikov's and anti-Markovnikov's (peroxide effect) additions. Remember that the peroxide effect is only observed with HBr and involves a free radical mechanism, leading to the less substituted product.

Question 10. How will you prepare 2-bromopropane from propene?
Answer: 2-bromopropane can be prepared from propene by reacting it with hydrogen bromide (HBr). This reaction follows Markovnikov's rule, where the bromine atom adds to the more substituted carbon of the double bond.
\[ \text{CH}_3-\text{CH}=\text{CH}_2 + \text{HBr} \longrightarrow \text{CH}_3-\text{CH(Br)}-\text{CH}_3 \]In simple words: To make 2-bromopropane from propene, just add HBr directly to propene; the bromine will attach to the middle carbon.

🎯 Exam Tip: Direct addition of HBr to propene without peroxides leads to the Markovnikov product (2-bromopropane). This is a straightforward application of the rule.

Question 11. How will you prepare 1-bromopropane from propene?
Answer: 1-bromopropane can be prepared from propene by reacting it with hydrogen bromide (HBr) in the presence of a peroxide. This reaction proceeds via the anti-Markovnikov's addition, also known as the peroxide effect, where the bromine atom adds to the less substituted carbon of the double bond.
\[ \text{CH}_3-\text{CH}=\text{CH}_2 + \text{HBr} \xrightarrow{\text{Peroxide}} \text{CH}_3-\text{CH}_2-\text{CH}_2\text{Br} \]In simple words: To make 1-bromopropane from propene, add HBr in the presence of a peroxide, which makes the bromine attach to the end carbon instead of the middle one.

🎯 Exam Tip: The key to forming the anti-Markovnikov product (1-bromopropane) from propene and HBr is the presence of a peroxide, which steers the reaction through a free radical mechanism.

Question 12. Match the following

🎯 Exam Tip: Familiarize yourself with common chemicals mentioned in environmental chemistry or daily applications, as their uses and properties are frequently tested.