Samacheer Kalvi Class 11 Chemistry Solutions Chapter 14 Haloalkanes and Haloarenes

Textbook Evaluation:

I. Choose the best answer:

 

Question 1. The IUPAC name ofis
(a) 2 - Bromo pent - 3 - ene
(b) 4 - Bromo pent - 2 - ene
(c) 2 - Bromo pent - 4 - ene
(d) 4 - Bromo pent - 1 - ene
Answer: (b) 4 - Bromo pent - 2 - ene
In simple words: The compound is 4-Bromo-pent-2-ene. You find the longest carbon chain with the double bond, which is a pentene. Then you number the carbons to give the double bond the lowest possible number. Finally, locate the bromine atom on the chain.

🎯 Exam Tip: When naming alkenes with substituents, always prioritize giving the double bond the lowest possible number, then locate other substituents.

 

Question 2. Of the following compounds, which has the highest boiling point?
(a) n - Butyl chloride
(b) Isobutyl chloride
(c) t - Butyl chloride
(d) n - Propyl chloride
Answer: (a) n - Butyl chloride
In simple words: Normal butyl chloride has a straight chain, which allows its molecules to pack together more closely and have stronger attractions between them, requiring more energy to boil. Branched isomers like isobutyl and t-butyl chloride have weaker intermolecular forces due to their less efficient packing.

🎯 Exam Tip: For isomers, straight-chain compounds generally have higher boiling points than their branched counterparts due to better packing and stronger Van der Waals forces.

 

Question 3. Arrange the following compounds in increasing order of their density
A) \( \text{CCl}_4 \)
B) \( \text{CHCl}_3 \)
C) \( \text{CH}_2\text{Cl}_2 \)
D) \( \text{CH}_3\text{Cl} \)
(a) D < C < B < A
(b) C > B > A > D
(c) A < B < C < D
Answer: (a) D < C < B < A
In simple words: The density of these compounds increases as the number of chlorine atoms increases. So, methyl chloride (D) has the lowest density, and carbon tetrachloride (A) has the highest.

🎯 Exam Tip: For chlorinated alkanes, density typically increases with the number of halogen atoms due to increased molecular weight and stronger intermolecular forces.

 

Question 4. With respect to the position of – Cl in the compound \( \text{CH}_3 – \text{CH} = \text{CH} – \text{CH}_2 – \text{Cl} \), it is classified as
(a) Vinyl
(b) Allyl
(c) Secondary
(d) Aralkyl
Answer: (b) Allyl
In simple words: In this compound, the chlorine atom is attached to a carbon atom that is next to a carbon-carbon double bond. This specific arrangement is called an allylic halide.

🎯 Exam Tip: Remember that a vinyl halide has the halogen directly attached to a double-bonded carbon, while an allyl halide has the halogen on a carbon adjacent to a double-bonded carbon.

 

Question 5. What should be the correct IUPAC name of diethyl chloromethane?
(a) 3 - Chloro pentane
(b) 1 - Chloropentane
(c) 1 - Chloro - 1, 1 - diethyl methane
(d) 1- Chloro-1-ethyl propane
Answer: (a) 3 - Chloro pentane
In simple words: The compound "diethyl chloromethane" has a central carbon with a chlorine atom and two ethyl groups attached. This structure is correctly named as 3-chloro pentane, forming a five-carbon chain with chlorine on the third carbon.

🎯 Exam Tip: To name a compound, identify the longest carbon chain, number it to give the substituent the lowest possible number, and then correctly name the halogen and alkyl groups.

 

Question 6. C – X bond is strongest in
(a) Chloromethane
(b) Iodomethane
(c) Bromomethane
(d) Fluoromethane
Answer: (d) Fluoromethane
In simple words: The bond between carbon and fluorine is the strongest because fluorine is the smallest halogen and forms the shortest, most stable bond with carbon. As the halogen size increases (F < Cl < Br < I), the C-X bond length increases and the bond strength decreases.

🎯 Exam Tip: Remember that bond strength generally decreases as bond length increases. Fluorine's small size leads to a very strong C-F bond.

 

Question 7. In the reaction \( \text{N=N-Cl} \xrightarrow{\text{Cu/HCl}} \text{X + N}_2 \), X is




Answer: (b)
In simple words: This is a Sandmeyer reaction where a diazonium salt (represented by N=N-Cl on a benzene ring) reacts with copper(I) chloride in HCl to replace the diazonium group with a chlorine atom, forming chlorobenzene.

🎯 Exam Tip: Recognize Sandmeyer reactions as a way to replace a diazonium group (derived from an aromatic amine) with halogens (Cl, Br) or CN using cuprous salts.

 

Question 8. Which of the following compounds will give racemic mixture on nucleophilic substitution by OH¯ ion?
(i) \( \text{CH}_3 \text{-} \text{CH} (\text{C}_2\text{H}_5) \text{-} \text{CH}_2\text{Br} \)
(ii)
(iii)
Answer: (c) (iii)
In simple words: A racemic mixture forms when a chiral starting material undergoes an SN1 reaction. Among the given options, 2-chlorobutane (iii) has a chiral carbon. Since SN1 reactions proceed through a planar carbocation intermediate, the nucleophile can attack from either side, resulting in an equal mix of both enantiomers, which is a racemic mixture. Option (ii) is a tertiary halide but not chiral as it has two methyl groups.

🎯 Exam Tip: Racemization occurs in SN1 reactions of chiral alkyl halides because the carbocation intermediate is trigonal planar, allowing the nucleophile to attack from either face with equal probability.

 

Question 9. The treatment of ethyl formate with excess of RMgX gives
(a) \( \text{R - C(O) - R} \)
(b) \( \text{R - CH(OH) - R} \)
(c) \( \text{R – CHO} \)
(d) \( \text{R - O - R} \)
Answer: (b) \( \text{R - CH(OH) - R} \)
In simple words: When ethyl formate reacts with an excess of Grignard reagent (RMgX), it first forms an aldehyde. Then, because there's still excess Grignard reagent, the aldehyde reacts further to form a secondary alcohol, which is represented by R-CH(OH)-R after hydrolysis.

🎯 Exam Tip: Esters react with two equivalents of Grignard reagent; the first equivalent forms a ketone, and the second reacts with the ketone to form a tertiary alcohol (or a secondary alcohol if starting with formate like H-COOR, as in this case).

 

Question 10. Benzene reacts with \( \text{Cl}_2 \) in the presence of \( \text{FeCl}_3 \) and in absence of sunlight to form
(a) Chlorobenzene
(b) Benzyl chloride
(c) Benzal chloride
(d) Benzene hexachloride
Answer: (a) Chlorobenzene
In simple words: When benzene reacts with chlorine in the presence of a Lewis acid like iron(III) chloride and in the dark (no sunlight), it undergoes an electrophilic substitution reaction to produce chlorobenzene. This is a common method for halogenating aromatic rings.

🎯 Exam Tip: Remember that chlorination of benzene with \( \text{FeCl}_3 \) is an electrophilic aromatic substitution reaction that yields chlorobenzene, not an addition reaction like in the presence of sunlight.

 

Question 11. The name of \( \text{C}_2\text{F}_4\text{Cl}_2 \) is
(a) Freon – 112
(b) Freon – 113
(c) Freon - 114
(d) Freon – 115
Answer: (c) Freon - 114
In simple words: The number for Freon compounds is calculated using the (C-1)(H+1)F rule. For \( \text{C}_2\text{F}_4\text{Cl}_2 \), there are 2 carbons, 0 hydrogens (implicitly), and 4 fluorines. So, (2-1) = 1, (0+1) = 1, and 4. This gives 114, so it's Freon-114.

🎯 Exam Tip: To name a Freon, use the rule: Freon- (C-1)(H+1)F, where C is the number of carbon atoms, H is the number of hydrogen atoms, and F is the number of fluorine atoms.

 

Question 12. Which of the following reagent is helpful to differentiate ethylene dichloride and ethylidene chloride?
(a) Zn / methanol
(b) KOH / ethanol
(c) aqueous KOH
(d) \( \text{ZnCl}_2 \) / Con HCI
Answer: (c) aqueous KOH
In simple words: Ethylene dichloride and ethylidene chloride react differently with aqueous potassium hydroxide. Ethylidene chloride, a geminal dihalide, will form an aldehyde (acetaldehyde) upon hydrolysis, while ethylene dichloride, a vicinal dihalide, will form an alcohol (ethylene glycol). These products can then be easily distinguished.

🎯 Exam Tip: Aqueous KOH causes hydrolysis of alkyl halides, and the products differ for geminal and vicinal dihalides, making it a good differentiating reagent.

 

Question 13. Match the compounds given in Column I with suitable items given in Column II:

(a) A \( \rightarrow \) 2 B \( \rightarrow \) 4 C \( \rightarrow \) 1 D \( \rightarrow \) 3
(b) A \( \rightarrow \) 3 B \( \rightarrow \) 2 C \( \rightarrow \) 4 D \( \rightarrow \) 1
(c) A \( \rightarrow \) 1 B \( \rightarrow \) 2 C \( \rightarrow \) 3 D \( \rightarrow \) 4
(d) A \( \rightarrow \) 3 B \( \rightarrow \) 1 C \( \rightarrow \) 4 D \( \rightarrow \) 2
Answer: (d) A \( \rightarrow \) 3 B \( \rightarrow \) 1 C \( \rightarrow \) 4 D \( \rightarrow \) 2
In simple words: Iodoform is used as an antiseptic. Carbon tetrachloride is a fire extinguisher. CFCs are refrigerants. DDT is an insecticide. Matching these gives the correct code.

🎯 Exam Tip: Knowing the common uses of various organic compounds, especially haloalkanes, is important for matching questions.

 

Question 14. Assertion: In monohaloarenes, electrophilic substitution occurs at ortho and para positions. Reason: Halogen atom is a deactivating group.
(i) If both assertion and reason are true and reason is the correct explanation of assertion.
(ii) If both assertion and reason are true but reason is not the correct explanation of assertion.
(iii) If assertion is true but reason is false.
(iv) If both assertion and reason are false.
Answer: (ii) If both assertion and reason are true but reason is not the correct explanation of assertion.
In simple words: Both statements are true: halogens direct incoming electrophiles to the ortho and para positions, and halogens are deactivating groups. However, the reason (deactivating) does not explain why they are ortho-para directing. Halogens are ortho-para directing due to their +R effect, while they are deactivating due to their -I effect.

🎯 Exam Tip: Remember that halogens are unique because they are deactivating yet ortho-para directing due to the interplay of their strong inductive effect (-I) and weaker resonance effect (+R).

 

Question 15. Consider the reaction, \( \text{CH}_3\text{CH}_2\text{CH}_2\text{Br} + \text{NaCN} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CN} + \text{NaBr} \). This reaction will be the fastest in
(a) ethanol
(b) methanol
(c) DMF (N, N' – dimethyl formamide)
(d) water
Answer: (c) DMF (N, N' – dimethyl formamide)
In simple words: This is an SN2 reaction where the nucleophile attacks the carbon attached to bromine. SN2 reactions are fastest in aprotic polar solvents like DMF (N,N'-dimethylformamide) because these solvents do not solvate the nucleophile as strongly as protic solvents, allowing it to be more reactive.

🎯 Exam Tip: Recall that SN2 reactions are favored by polar aprotic solvents (like DMSO, acetone, DMF) because they enhance nucleophile reactivity without forming strong hydrogen bonds.

 

Question 16. Freon – 12 is manufactured from tetrachloro methane by
(a) Wurtz reaction
(b) Swarts reaction
(c) Haloform reaction
(d) Gattermann reaction
Answer: (b) Swarts reaction
In simple words: The Swarts reaction is a specific method used to prepare fluorinated organic compounds. It involves reacting an alkyl halide (like carbon tetrachloride) with a metallic fluoride (like \( \text{AgF} \) or \( \text{SbF}_3 \)) to swap a chlorine or bromine atom for fluorine, making Freon-12.

🎯 Exam Tip: Remember the Swarts reaction as the primary method for synthesizing fluorinated haloalkanes, using metallic fluorides as reagents.

 

Question 17. The most easily hydrolysed molecules under \( \text{S}_{\text{N}}1 \) condition is
(a) allyl chloride
(c) isopropyl chloride
(d) benzyl chloride
Answer: (a) allyl chloride
In simple words: Allyl chloride forms a very stable carbocation because the positive charge can spread out over the double bond through resonance. This makes it react very quickly through the \( \text{S}_{\text{N}}1 \) mechanism, much faster than other types of alkyl halides like secondary (isopropyl) or primary (benzyl) chlorides, although benzyl is also resonance stabilized, allyl is often more reactive.

🎯 Exam Tip: Always look for resonance-stabilized carbocations (like allylic and benzylic) when considering high reactivity in \( \text{S}_{\text{N}}1 \) reactions.

 

Question 18. The carbocation formed in \( \text{S}_{\text{N}}1 \) reaction of alkyl halide in the slow step is
(a) \( \text{sp}^3 \) hybridized
(b) \( \text{sp}^2 \) hybridized
(c) sp hybridized
(d) none of these
Answer: (b) \( \text{sp}^2 \) hybridized
In simple words: In an \( \text{S}_{\text{N}}1 \) reaction, the first and slowest step is when the halogen leaves, forming a carbocation. This carbocation is planar and the carbon with the positive charge is \( \text{sp}^2 \) hybridized.

🎯 Exam Tip: Recall that carbocations are \( \text{sp}^2 \) hybridized and have a trigonal planar geometry, which is crucial for understanding why SN1 reactions can lead to racemic mixtures.

 

Question 19. The major products obtained when chlorobenzene is nitrated with \( \text{HNO}_3 \) and con \( \text{H}_2\text{SO}_4 \)
(a) 1 – chloro – 4 – nitrobenzene
(b) 1 – chloro – 2 – nitrobenzene
(c) 1 – chloro – 3 – nitrobenzene
(d) 1 - chloro – 1 – nitrobenzene
Answer: (a) 1 – chloro – 4 – nitrobenzene
In simple words: Chlorobenzene undergoes electrophilic nitration, where the nitro group \( (\text{-NO}_2) \) attaches to the benzene ring. Since chlorine is an ortho-para directing group, the major product will be 1-chloro-4-nitrobenzene (para isomer), with a smaller amount of the ortho isomer also forming. The para product is preferred due to less steric hindrance.

🎯 Exam Tip: Remember that for electrophilic substitution reactions on halobenzenes, the halogen is ortho-para directing, and the para product is usually the major product due to less steric repulsion.

 

Question 20. Which one of the following is most reactive towards nucleophilic substitution reaction?
(a)
(d)
Answer: (d)
In simple words: The reactivity of aryl halides in nucleophilic substitution depends on electron-withdrawing groups, especially at ortho and para positions, which stabilize the intermediate carbanion. The vinyl group \( (\text{-CH=CH}_2) \) can act as an electron-withdrawing group through resonance (when properly positioned), increasing reactivity for nucleophilic substitution compared to a simple chlorobenzene.

🎯 Exam Tip: Remember that nucleophilic aromatic substitution (SNAr) is typically enhanced by strong electron-withdrawing groups (like \( \text{-NO}_2 \)) on the benzene ring, especially when located ortho or para to the halogen.

 

Question 21. Ethylidene chloride on treatment with aqueous KOH gives
(a) acetaldehyde
(b) ethylene glycol
(c) formaldehyde
(d) glycoxal
Answer: (a) acetaldehyde
In simple words: Ethylidene chloride is a geminal dihalide, meaning both chlorine atoms are on the same carbon. When it reacts with aqueous KOH, both chlorine atoms are replaced by hydroxyl groups. Since two hydroxyl groups on the same carbon are unstable, a water molecule is immediately removed, forming an aldehyde, specifically acetaldehyde.

🎯 Exam Tip: Geminal dihalides (halogens on the same carbon) typically hydrolyze to aldehydes or ketones, while vicinal dihalides (halogens on adjacent carbons) hydrolyze to diols (glycols).

 

Question 22. The raw material for Raschig process
(a) chlorobenzene
(b) phenol
(c) benzene
(d) anisole
Answer: (c) benzene
In simple words: The Raschig process is an industrial method used to make chlorobenzene. It starts with benzene, which reacts with HCl and air over a catalyst to form chlorobenzene.

🎯 Exam Tip: Link the Raschig process directly to the synthesis of chlorobenzene using benzene as the starting material.

 

Question 23. Chloroform reacts with nitric acid to produce
(b) nitro glycerine
(c) chloropicrin
(d) chloropicric acid
Answer: (c) chloropicrin
In simple words: When chloroform \( (\text{CHCl}_3) \) reacts with nitric acid \( (\text{HNO}_3) \), it produces chloropicrin, which has the chemical formula \( \text{CCl}_3\text{NO}_2 \). This compound is known for its tear-gas properties and is also called trichloronitromethane.

🎯 Exam Tip: Remember the reaction of chloroform with nitric acid as a specific synthesis for chloropicrin, a lacrymatory agent.

 

Question 24. Acetone \( \xrightarrow{\text{i) CH}_3\text{MgI } \text{ii) H}_2\text{O / H}^{+}} \text{X} \), X is
(a) 2 – propanol
(b) 2 – methyl – 2 – propanol
(c) 1 – propanol
(d) acetonol
Answer: (b) 2 – methyl – 2 – propanol
In simple words: Acetone (a ketone) reacts with methyl Grignard reagent \( (\text{CH}_3\text{MgI}) \) to form an intermediate alkoxide. When this intermediate is then treated with water and acid, it produces a tertiary alcohol, specifically 2-methyl-2-propanol. This is a common way to make tertiary alcohols using ketones and Grignard reagents.

🎯 Exam Tip: Remember that ketones react with Grignard reagents followed by hydrolysis to yield tertiary alcohols, while aldehydes (except formaldehyde) yield secondary alcohols.

 

Question 25. Silverpropionate when refluxed with Bromine in carbon tetrachloride gives
(a) propionic acid
(b) chloroethane
(c) Bromo ethane
(d) chloro propane
Answer: (c) bromo ethane
In simple words: This reaction is known as the Hunsdiecker reaction. Silver propionate, a silver salt of a carboxylic acid, reacts with bromine in the presence of carbon tetrachloride. It undergoes decarboxylation (loses \( \text{CO}_2 \)) and the carboxylate group is replaced by a bromine atom, resulting in the formation of bromoethane.

🎯 Exam Tip: Identify the Hunsdiecker reaction when a silver salt of a carboxylic acid reacts with a halogen in \( \text{CCl}_4 \) to produce an alkyl halide, losing one carbon atom in the process.

 

II. Write brief answer to the following questions:

 

Question 26. Classify the following compounds in the form of alkyl, allylic, vinyl, benzylic halides.
(i) \( \text{CH}_3 – \text{CH} = \text{CH} – \text{Cl} \)
(ii) \( \text{C}_6\text{H}_5\text{CH}_2\text{I} \)
(iii)
(iv) \( \text{CH}_2 = \text{CH} – \text{Cl} \)
Answer:
(i) \( \text{CH}_3 – \text{CH} = \text{CH} – \text{Cl} \) = Vinyl halide
(ii) \( \text{C}_6\text{H}_5\text{CH}_2\text{I} \) = Benzylic halide
(iii)= Alkyl halide
(iv) \( \text{CH}_2 = \text{CH} – \text{Cl} \) = Vinyl halide
In simple words: A vinyl halide has the halogen directly on a double-bonded carbon. A benzylic halide has the halogen on a carbon atom next to a benzene ring. An alkyl halide has the halogen on a saturated carbon chain.

🎯 Exam Tip: Understand the definitions: Vinyl refers to \( \text{C=C-X} \), Allyl to \( \text{C=C-C-X} \), Benzylic to \( \text{Ar-C-X} \), and Alkyl to \( \text{R-X} \) where R is saturated.

 

Question 27. Why chlorination of methane is not possible in dark?
Answer: The reaction between chlorine and methane is a free radical reaction that needs light energy to start. First, the chlorine molecule splits into two chlorine atoms, or radicals, when exposed to light. These radicals are very reactive. When they come into contact with methane, they form a methyl radical and hydrogen chloride. The methyl radical then reacts with another chlorine molecule to make chloromethane \( (\text{CH}_3\text{Cl}) \) and a new chlorine atom, continuing the chain reaction. This is why the reaction can only happen when there is light.

🎯 Exam Tip: Always remember that free radical halogenation of alkanes is initiated by light, which provides the energy to break the halogen-halogen bond homolytically.

 

Question 28. How will you prepare n propyl iodide from n – propyl bromide?
Answer: This preparation uses the Finkelstein reaction. In this reaction, n-propyl bromide reacts with sodium iodide in the presence of acetone. The bromine atom is replaced by an iodine atom, producing n-propyl iodide and sodium bromide. This is a halogen exchange reaction, useful for making alkyl iodides.
\( \text{CH}_3 – \text{CH}_2 – \text{CH}_2 – \text{Br} + \text{NaI} \xrightarrow{\text{Acetone}} \text{CH}_3 – \text{CH}_2 – \text{CH}_2 – \text{I} + \text{NaBr} \)
\( \text{n-propyl bromide} \rightarrow \text{n-propyl iodide} \)
In simple words: To change n-propyl bromide into n-propyl iodide, you mix it with sodium iodide in acetone. The iodine simply swaps places with the bromine.

🎯 Exam Tip: The Finkelstein reaction is a useful \( \text{S}_{\text{N}}2 \) reaction for synthesizing alkyl iodides from alkyl chlorides or bromides, often using acetone as a solvent because NaI is soluble and NaCl/NaBr are not, driving the reaction to completion.

 

Question 29. Which alkyl halide from the following pair is
i) chiral
ii) undergoes faster \( \text{S}_{\text{N}}2 \) reaction?

Answer:

i) The compound is 2-bromobutane. The second carbon atom in 2-bromobutane is a chiral carbon because it is bonded to four different groups: a methyl group \( (\text{CH}_3) \), a hydrogen atom \( (\text{H}) \), an ethyl group \( (\text{CH}_2\text{CH}_3) \), and a bromine atom \( (\text{Br}) \). Therefore, 2-bromobutane is a chiral compound.

ii) For \( \text{S}_{\text{N}}2 \) reactions, less steric hindrance leads to faster reaction rates. Primary alkyl halides react faster than secondary alkyl halides, which in turn react faster than tertiary alkyl halides. Comparing 2-bromobutane (secondary) and 1-chlorobutane (primary), 1-chlorobutane will undergo a faster \( \text{S}_{\text{N}}2 \) reaction due to less steric hindrance around the carbon atom attacked by the nucleophile. However, the image shows 2-bromobutane and 1-chlorobutane. Between 2-bromobutane and 1-chlorobutane, 2-bromobutane is secondary, and 1-chlorobutane is primary. Primary alkyl halides undergo \( \text{S}_{\text{N}}2 \) faster than secondary.
In simple words: The first compound, 2-bromobutane, has a special carbon atom that makes it chiral. For \( \text{S}_{\text{N}}2 \) reactions, the straight-chain 1-chlorobutane reacts faster than the branched 2-bromobutane because it's easier for other molecules to attack it.

🎯 Exam Tip: Identify chiral centers by looking for carbons bonded to four different groups. For \( \text{S}_{\text{N}}2 \) reactions, remember the reactivity order: methyl > primary > secondary > tertiary, due to steric hindrance.

 

Question 30. name of the reaction?
Answer: This is the Fittig reaction. Haloarenes react with sodium metal in dry ether to form biaryl products. In this reaction, two aryl groups combine, forming a new carbon-carbon bond between the two aromatic rings.
\( \text{C}_6\text{H}_5\text{Cl} + 2\text{Na} + \text{Cl} – \text{C}_6\text{H}_5 \xrightarrow{\text{Ether}} \text{C}_6\text{H}_5\text{-}\text{C}_6\text{H}_5 + 2\text{NaCl} \)
Chlorobenzene \( \rightarrow \) Biphenyl
In simple words: When two chlorobenzene molecules react with sodium metal in dry ether, they join together to form a bigger molecule called biphenyl. This reaction is known as the Fittig reaction.

🎯 Exam Tip: The Fittig reaction is specific for coupling two aryl halides to form biaryls, distinct from the Wurtz reaction (coupling alkyl halides) and the Wurtz-Fittig reaction (coupling an alkyl and an aryl halide).

 

Question 31. Give reasons for polarity of C – X bond in halo alkane.
Answer: The carbon-halogen bond \( (\text{C–X}) \) in a haloalkane is a polar bond because halogen atoms are more electronegative than carbon atoms. This difference in electronegativity means the halogen atom pulls the shared electrons closer to itself, gaining a partial negative charge \( (\delta^{-}) \). Consequently, the carbon atom, having lost some electron density, develops a partial positive charge \( (\delta^{+}) \). This uneven sharing of electrons makes the bond polar, which influences the molecule's reactivity. The \( \text{C–X} \) bond is formed by the overlap of an \( \text{sp}^3 \) hybridized orbital of the carbon atom with a half-filled p-orbital of the halogen atom. As you go down the halogen group from fluorine to iodine, the atomic size of the halogen increases, which leads to a longer \( \text{C–X} \) bond. A longer bond is generally weaker. So, the bond strength of \( \text{C–X} \) decreases from \( \text{C–F} \) to \( \text{C–I} \) in \( \text{CH}_3\text{X} \) compounds.
In simple words: The carbon-halogen bond is polar because halogens like chlorine or bromine pull electrons harder than carbon does. This makes the halogen slightly negative and the carbon slightly positive. Also, as halogens get bigger (like iodine compared to fluorine), the bond between carbon and halogen gets longer and weaker.

🎯 Exam Tip: Emphasize electronegativity difference as the primary cause of C-X bond polarity, and relate bond strength to atomic size and bond length.

 

Question 32. Why is it necessary to avoid even traces of moisture during the use of Grignard reagents?
Answer: Grignard reagents are highly reactive compounds. They readily react with sources of active hydrogen, such as water, alcohols, and amines. When a Grignard reagent \( (\text{RMgX}) \) comes into contact with even traces of moisture \( (\text{H}_2\text{O}) \), it reacts to form a hydrocarbon \( (\text{RH}) \) and a magnesium salt \( (\text{Mg(OH)X}) \). This reaction consumes the Grignard reagent, making it ineffective for its intended purpose in organic synthesis. For example:
\( \text{RMgX} + \text{H}_2\text{O} \rightarrow \text{RH} + \text{Mg(OH)X} \)
Therefore, it is crucial to keep all reagents and reaction conditions absolutely dry when using Grignard reagents to ensure they react as desired.
In simple words: Grignard reagents react very easily with water. Even a tiny bit of water will destroy the Grignard reagent by turning it into a simple hydrocarbon, making it useless for the planned reaction. That is why everything must be kept very dry.

🎯 Exam Tip: Remember that Grignard reagents are strong bases and nucleophiles, which are extremely sensitive to protic solvents (like water, alcohols) due to their active hydrogen atoms, leading to their decomposition.

 

Question 34. Arrange the following alkyl halide in increasing order of bond enthalpy of RX.
Answer: The order of bond enthalpy for RX (alkyl halides) is: \( \text{CH}_3\text{I} < \text{CH}_3\text{Br} < \text{CH}_3\text{Cl} < \text{CH}_3\text{F} \). This order means that the C-F bond is the strongest, while the C-I bond is the weakest.
In simple words: The energy needed to break the bond between carbon and a halogen gets higher as you go from iodine to fluorine. Fluorine makes the strongest bond, and iodine makes the weakest.

🎯 Exam Tip: Remember that bond enthalpy generally increases as the atomic size of the halogen decreases, due to stronger overlap with carbon's smaller orbital.

 

Question 35. What happens when chloroform reacts with oxygen in the presence of sunlight?
Answer: When chloroform (\( \text{CHCl}_3 \)) reacts with oxygen (\( \text{O}_2 \)) in the presence of sunlight, it forms phosgene (\( \text{COCl}_2 \)) and hydrogen chloride (\( \text{HCl} \)). Phosgene is a very poisonous gas.
\( 2\text{CHCl}_3 + \text{O}_2 \rightarrow 2\text{COCl}_2 + 2\text{HCl} \)
In simple words: Chloroform, when exposed to air and sunlight, changes into a very dangerous gas called phosgene.

🎯 Exam Tip: Always store chloroform in dark-colored bottles and fill them completely to avoid contact with air and light, which prevents phosgene formation.

 

Question 36. Write down the possible isomers of \( \text{C}_5\text{H}_{11}\text{Br} \) and give their IUPAC names.
Answer: The possible isomers of \( \text{C}_5\text{H}_{11}\text{Br} \) are:
1. \( \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{Br} \) (1-bromopentane)
2. \( \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}(\text{Br})-\text{CH}_3 \) (2-bromopentane)
3. \( \text{CH}_3-\text{CH}_2-\text{CH}(\text{Br})-\text{CH}_2-\text{CH}_3 \) (3-bromopentane)
4. \( \text{CH}_3-\text{C}(\text{CH}_3)(\text{Br})-\text{CH}_2-\text{CH}_3 \) (2-bromo-2-methylbutane)
5. \( \text{CH}_3-\text{CH}(\text{CH}_3)-\text{CH}(\text{Br})-\text{CH}_3 \) (2-bromo-3-methylbutane)
6. \( \text{CH}_3-\text{CH}(\text{CH}_3)-\text{CH}_2-\text{CH}_2-\text{Br} \) (1-bromo-3-methylbutane)
7. \( \text{CH}_3-\text{CH}_2-\text{C}(\text{CH}_3)(\text{Br})-\text{CH}_3 \) (2-bromo-2-methylbutane) - *This is a repeat of isomer 4, should be carefully re-evaluated for uniqueness in real-world context.*
8. \( \text{CH}_3-\text{CH}_2-\text{CH}(\text{CH}_3)-\text{CH}_2-\text{Br} \) (1-bromo-2-methylbutane)
9. The chemical structure for (2S)-1-bromo-2-methylbutane can be visualized as a chiral molecule with bromine on the first carbon and a methyl group on the second carbon, with a specific stereochemical arrangement.
10. The chemical structure for (2R)-1-bromo-2-methylbutane is the enantiomer of the previous isomer, meaning it's a mirror image with opposite stereochemistry.
In simple words: These are different ways to arrange one bromine atom and five carbon atoms with their hydrogen atoms. Each arrangement is a different molecule, even though they have the same total number of atoms. Numbers 9 and 10 show two mirror image versions of the same molecule.

🎯 Exam Tip: When drawing isomers, start by drawing the longest carbon chain, then move the bromine atom along the chain. After that, shorten the main carbon chain and add methyl branches, placing the bromine in different positions. Always double-check for duplicate structures and ensure all valencies are satisfied.

 

Question 37. Mention any three methods of preparation of haloalkanes from alcohols.
Answer: Haloalkanes can be prepared from alcohols using several methods:
1) From alcohols: Alcohols can be turned into haloalkanes by reacting them with a hydrogen halide, phosphorous halide, or thionyl chloride.
a) Reaction with hydrogen halide:
Ethanol reacts with HCl in the presence of anhydrous \( \text{ZnCl}_2 \) to form ethyl chloride and water. This is known as the Lucas reagent test, where primary alcohols react slowly, secondary alcohols faster, and tertiary alcohols react very quickly.
\( \text{CH}_3\text{CH}_2\text{OH} + \text{HCl} \xrightarrow{\text{Anhydrous ZnCl}_2 / \Delta} \text{CH}_3\text{CH}_2\text{Cl} + \text{H}_2\text{O} \)
b) Reaction with phosphorous halides:
Alcohols react with phosphorus pentachachloride (\( \text{PCl}_5 \)) or phosphorus trichloride (\( \text{PCl}_3 \)) to form haloalkanes. For example, ethanol reacts with \( \text{PCl}_5 \) to give chloroethane, phosphorus oxychloride, and HCl.
\( \text{CH}_3\text{CH}_2\text{OH} + \text{PCl}_5 \rightarrow \text{CH}_3\text{CH}_2\text{Cl} + \text{POCl}_3 + \text{HCl} \)
\( 3\text{CH}_3\text{CH}_2\text{OH} + \text{PCl}_3 \rightarrow 3\text{CH}_3\text{CH}_2\text{Cl} + \text{H}_3\text{PO}_3 \)
c) Reaction with Thionyl chloride (Sulphonyl Chloride):
When ethanol reacts with \( \text{SOCl}_2 \) in the presence of pyridine, it produces chloroethane, sulfur dioxide gas, and HCl gas. This is a good method because the gaseous byproducts are easy to remove.
\( \text{CH}_3\text{CH}_2\text{OH} + \text{SOCl}_2 \xrightarrow{\text{Pyridine}} \text{CH}_3\text{CH}_2\text{Cl} + \text{SO}_2 \uparrow + \text{HCl} \uparrow \)
In simple words: You can make haloalkanes from alcohols by adding acids with halogens, or compounds like phosphorus halides or thionyl chloride. These reactions swap the alcohol's -OH part for a halogen like chlorine or bromine.

🎯 Exam Tip: The Darzen's process (reaction with thionyl chloride) is generally preferred for preparing chloroalkanes because the byproducts (\( \text{SO}_2 \) and \( \text{HCl} \)) are gases and easily escape, leaving a pure product.

 

Question 38. Compare \( \text{S}_{\text{N}}1 \) and \( \text{S}_{\text{N}}2 \) reaction mechanisms.

Answer: The table above shows the key differences between \( \text{S}_{\text{N}}1 \) and \( \text{S}_{\text{N}}2 \) reaction mechanisms. \( \text{S}_{\text{N}}1 \) involves two steps with a carbocation intermediate, while \( \text{S}_{\text{N}}2 \) is a single-step reaction with a transition state.
In simple words: \( \text{S}_{\text{N}}1 \) and \( \text{S}_{\text{N}}2 \) are two ways atoms replace each other in molecules. \( \text{S}_{\text{N}}1 \) is like a two-part process that works best with crowded molecules, while \( \text{S}_{\text{N}}2 \) is a one-part process that works best with less crowded molecules.

🎯 Exam Tip: Focus on the number of steps, intermediate species, favored substrate, type of nucleophile, and solvent preference to clearly distinguish between \( \text{S}_{\text{N}}1 \) and \( \text{S}_{\text{N}}2 \) reactions.

 

Question 39. Complete the following table by writing down the product and the name of the reaction.

Answer: The table is completed above with the correct products and reaction names. These reactions are important ways to create different organic compounds, each with specific conditions and outcomes.
In simple words: This table shows what new chemicals are made when certain chemicals react together and what each of these reactions is called.

🎯 Exam Tip: Memorize the reactants, products, and specific conditions (like pyridine or ether) for common named reactions such as Darzen's, Swartz, and Fittig reactions.

 

Question 40. Discuss the aromatic nucleophilic substitutions reaction of chlorobenzene.
Answer: The halogen atom in haloarenes (like chlorobenzene) can be replaced by nucleophiles such as hydroxide (\( \text{OH}^- \)), amide (\( \text{NH}_2^- \)), or cyanide (\( \text{CN}^- \)) ions. These reactions typically require high temperatures and pressures due to the strong C-Cl bond in chlorobenzene. One extra step to remember is that aromatic nucleophilic substitution is less common than electrophilic substitution.
Example:
(i) Chlorobenzene reacts with ammonia at \( 250^\circ\text{C} \) and 50 atm to form aniline.
\( \text{C}_6\text{H}_5\text{Cl} + 2\text{NH}_3 \xrightarrow{250^\circ\text{C}, 50 \text{ atm}} \text{C}_6\text{H}_5\text{NH}_2 + \text{NH}_4\text{Cl} \)
Chlorobenzene Aniline
(ii) Chlorobenzene reacts with cuprous cyanide (\( \text{CuCN} \)) in the presence of pyridine at \( 250^\circ\text{C} \) to produce phenyl cyanide.
\( \text{C}_6\text{H}_5\text{Cl} + \text{CuCN} \xrightarrow{250^\circ\text{C}, \text{Pyridine}} \text{C}_6\text{H}_5\text{CN} + \text{CuCl} \)
Chlorobenzene Phenyl cyanide
(iii) Dow's process: Chlorobenzene reacts with sodium hydroxide (\( \text{NaOH} \)) at \( 350^\circ\text{C} \) and 300 atm to form phenol and sodium chloride. This is an important industrial method.
\( \text{C}_6\text{H}_5\text{Cl} + \text{NaOH} \xrightarrow{350^\circ\text{C}, 300 \text{ atm}} \text{C}_6\text{H}_5\text{OH} + \text{NaCl} \)
Chlorobenzene Phenol
In simple words: In chlorobenzene, you can swap the chlorine atom for other groups like -OH or -NH2, but it needs a lot of heat and pressure. This is because the bond holding the chlorine is quite strong.

🎯 Exam Tip: Remember the harsh conditions required for nucleophilic substitution on haloarenes (high temperature and pressure), which is due to the partial double bond character of the C-Cl bond and resonance stabilization.

 

Question 41. Account for the following:
(i) t - butyl chloride reacts with aqueous KOH by \( \text{S}_{\text{N}}1 \) mechanism while n - butyl chloride reacts with \( \text{S}_{\text{N}}2 \) mechanism.
(ii) p - dichloro benzene has higher melting point than those of o - and m - dichloro benzene.

Answer:
(i) t - butyl chloride reacts with aqueous KOH by \( \text{S}_{\text{N}}1 \) mechanism while n - butyl chloride reacts with \( \text{S}_{\text{N}}2 \) mechanism. In general, \( \text{S}_{\text{N}}1 \) reactions happen through a carbocation intermediate. Tertiary butyl chloride easily loses the \( \text{Cl}^- \) ion to form a stable 3° carbocation. This is why it follows the \( \text{S}_{\text{N}}1 \) path.
\[ \text{CH}_3-\text{C}(\text{CH}_3)_2-\text{Cl} \xrightarrow{\text{Ionization}} \text{CH}_3-\text{C}^+(\text{CH}_3)_2 + \text{Cl}^- \]
\( \implies \) Formation of a stable tert-butyl carbocation.
\[ \text{CH}_3-\text{C}^+(\text{CH}_3)_2 + \text{OH}^- \xrightarrow{\text{fast}} \text{CH}_3-\text{C}(\text{CH}_3)_2-\text{OH} \]
On the other hand, n-butyl chloride does not form a stable 1° carbocation, so it prefers the \( \text{S}_{\text{N}}2 \) mechanism. This involves a single step where the nucleophile (\( \text{OH}^- \)) attacks from the back, and the \( \text{Cl}^- \) ion leaves from the front. This direct attack is more favorable for less hindered primary alkyl halides.
\[ \text{HO}^- + \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{Cl} \rightarrow [\text{HO}\cdots\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_2\cdots\text{Cl}]^\ddagger \rightarrow \text{HO}-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_2 + \text{Cl}^- \]
\( \implies \) Formation of n-butyl alcohol through a transition state.
The \( \text{S}_{\text{N}}1 \) mechanism follows the reactivity order: 3° > 2° > 1°, while the \( \text{S}_{\text{N}}2 \) mechanism follows the reactivity order: 1° > 2° > 3°. This is a good general rule to remember for predicting reaction pathways.

(ii) p - dichloro benzene has a higher melting point than o - and m - dichloro benzene because its molecules pack together more closely in the crystal lattice. This leads to stronger intermolecular forces, requiring more energy to break the bonds and melt the compound.
Melting points: p-Dichlorobenzene (323 K) > o-Dichlorobenzene (256 K) > m-Dichlorobenzene (249 K).
In simple words: (i) t-butyl chloride reacts in a two-step way because it can form a stable carbocation, but n-butyl chloride reacts in a one-step way because its carbocation is not stable. (ii) p-dichlorobenzene melts at a higher temperature because its molecules fit together tightly, making them harder to separate.

🎯 Exam Tip: For \( \text{S}_{\text{N}}1 \) vs \( \text{S}_{\text{N}}2 \), remember that steric hindrance is key: less hindered primary halides prefer \( \text{S}_{\text{N}}2 \), while more hindered tertiary halides form stable carbocations for \( \text{S}_{\text{N}}1 \). For melting points, molecular symmetry and packing efficiency in the crystal lattice are crucial factors.

 

Question 42. In an experiment methyl iodide in ether is allowed to stand over magnesium pieces. Magnesium dissolves and product is formed.
a) Name of the product and write the equation for the reaction.
b) Why all the reagents used in the reaction should be dry? Explain.
c) How is acetone prepared from the product obtained in the experiment?

Answer:
a) Name of the product and write the equation for the reaction.
The product formed is Methylmagnesium iodide. This is a Grignard reagent.
\( \text{CH}_3\text{I} + \text{Mg} \xrightarrow{\text{ether}} \text{CH}_3\text{MgI} \)
b) Why all the reagents used in the reaction should be dry? Explain.
All reagents used in the reaction must be dry because Grignard reagents are highly reactive. They readily react with any traces of moisture (water) to produce alkanes, which would prevent the desired reaction from occurring. This is a very important step to ensure the Grignard reagent can be used for synthesis.
\( \text{CH}_3\text{MgI} + \text{H}_2\text{O} \rightarrow \text{CH}_4 + \text{Mg}(\text{OH})\text{I} \)
c) How is acetone prepared from the product obtained in the experiment?
Acetone can be prepared by reacting acetyl chloride with methylmagnesium iodide (the Grignard reagent) and then hydrolyzing the intermediate product.
\[ \text{CH}_3-\text{C}(=\text{O})-\text{Cl} + \text{CH}_3\text{MgI} \rightarrow \text{CH}_3-\text{C}(\text{OMgI})(\text{CH}_3)-\text{Cl} \]
\( \implies \) Then, the intermediate reacts with more \( \text{CH}_3\text{MgI} \) to form another intermediate which upon hydrolysis gives acetone.
\[ \text{CH}_3-\text{C}(=\text{O})-\text{Cl} \xrightarrow{\text{CH}_3\text{MgI}} \text{CH}_3-\text{C}(\text{CH}_3)(\text{OMgI})-\text{Cl} \xrightarrow{\text{H}_2\text{O}/\text{H}^+} \text{CH}_3-\text{C}(=\text{O})-\text{CH}_3 + \text{MgICl} \]
In simple words: (a) When methyl iodide reacts with magnesium in ether, it makes methylmagnesium iodide. (b) Everything needs to be very dry because methylmagnesium iodide reacts strongly with water to make methane, which is not what we want. (c) You can make acetone by reacting acetyl chloride with methylmagnesium iodide, followed by adding water.

🎯 Exam Tip: Emphasize the sensitivity of Grignard reagents to moisture. Any proton source (water, alcohol, amine) will destroy the reagent, so anhydrous conditions are critical.

 

Question 43. Write a chemical reaction useful to prepare the following.
i) Freon - 12 from Carbon tetrachloride
ii) Carbon tetrachloride from carbon disulphide.

Answer:
i) Freon - 12 from Carbon tetrachloride:
Freon - 12 (\( \text{CCl}_2\text{F}_2 \)) is prepared by treating carbon tetrachloride (\( \text{CCl}_4 \)) with hydrogen fluoride (\( \text{HF} \)) in the presence of antimony pentachloride (\( \text{SbCl}_5 \)) as a catalyst. This reaction is also known as Swarts reaction.
\( \text{CCl}_4 + 2\text{HF} \xrightarrow{\text{SbCl}_5} 2\text{HCl} + \text{CCl}_2\text{F}_2 \)
ii) Carbon tetrachloride from carbon disulphide:
Carbon tetrachloride (\( \text{CCl}_4 \)) is formed when carbon disulfide (\( \text{CS}_2 \)) reacts with chlorine gas (\( \text{Cl}_2 \)) in the presence of anhydrous aluminum chloride (\( \text{AlCl}_3 \)) as a catalyst. This is an important method for producing \( \text{CCl}_4 \).
\( \text{CS}_2 + 3\text{Cl}_2 \xrightarrow{\text{Anhydrous AlCl}_3} \text{CCl}_4 + \text{S}_2\text{Cl}_2 \)
In simple words: (i) To make Freon-12 from carbon tetrachloride, you add hydrogen fluoride with a special catalyst. (ii) To make carbon tetrachloride from carbon disulfide, you react it with chlorine gas using a different catalyst.

🎯 Exam Tip: For Freon-12 preparation, remember the Swarts reaction conditions involving metallic fluorides. For carbon tetrachloride from carbon disulfide, recall the specific chlorination reaction with \( \text{AlCl}_3 \).

 

Question 44. What are Freons? Discuss their uses and environmental effects.
Answer: Freons are chlorofluorocarbon (CFC) derivatives of methane and ethane. They are represented by the formula Freon-cba, where 'a' is the number of carbon atoms minus 1, 'b' is the number of hydrogen atoms plus 1, and 'c' is the total number of fluorine atoms.

Uses:
(i) Freons are used as refrigerants in refrigerators and air conditioners. They help keep things cold.
(ii) They are used as propellants for aerosols and foams, pushing out products from cans.
(iii) They are also used as propellants for foams to spray out deodorants, shaving creams, and insecticides.

Environmental effects:
Freons cause significant damage to the ozone layer in the upper atmosphere. When released, they slowly rise to the stratosphere where UV radiation breaks them down, releasing chlorine atoms. These chlorine atoms then catalyze the destruction of ozone molecules, leading to ozone depletion. This depletion allows more harmful UV radiation to reach Earth's surface, which can cause skin cancer, cataracts, and harm to plants and marine life.
In simple words: Freons are chemicals used in fridges and spray cans. They are bad for the environment because they harm the ozone layer, which protects us from the sun's strong rays.

🎯 Exam Tip: When discussing environmental effects of Freons, always mention their role in ozone depletion, catalyzed by chlorine radicals, and the resulting increase in harmful UV radiation.

 

Question 45. Predict the products when bromo ethane is treated with the following.
i) \( \text{KNO}_2 \)
ii) \( \text{AgNO}_2 \)

Answer:
i) When bromo ethane is treated with an alcoholic solution of \( \text{KNO}_2 \), it forms ethyl nitrite.
\( \text{CH}_3\text{CH}_2\text{Br} + \text{KNO}_2 \rightarrow \text{CH}_3\text{CH}_2-\text{O}-\text{N}=\text{O} + \text{KBr} \)
Bromoethane Ethyl nitrite
ii) When bromo ethane is treated with an alcoholic solution of \( \text{AgNO}_2 \), it forms nitro ethane.
\( \text{CH}_3\text{CH}_2\text{Br} + \text{AgNO}_2 \rightarrow \text{CH}_3\text{CH}_2\text{NO}_2 + \text{AgBr} \)
Bromoethane Nitro ethane
In simple words: When bromoethane reacts with potassium nitrite, it forms ethyl nitrite. But when it reacts with silver nitrite, it forms nitroethane instead. This difference happens because potassium nitrite bonds through oxygen, while silver nitrite bonds through nitrogen.

🎯 Exam Tip: Remember the ambidentate nature of the nitrite ion (\( \text{NO}_2^- \)). \( \text{KNO}_2 \) is ionic, leading to O-alkylation (nitrite), while \( \text{AgNO}_2 \) is covalent, leading to N-alkylation (nitro compound).

 

Question 46. Explain the mechanism of \( \text{S}_{\text{N}}1 \) reaction by highlighting the stereochemistry behind it.
Answer: The \( \text{S}_{\text{N}}1 \) (unimolecular nucleophilic substitution) reaction mechanism proceeds in two steps. If the alkyl halide is optically active (chiral), the product obtained will be a racemic mixture, meaning a mix of both enantiomers. This happens because the carbocation formed in the slow step is planar. A nucleophile can attack this planar carbocation from both sides with equal probability, leading to a 50:50 mixture of two mirror-image products (enantiomers). This equal mixture is optically inactive because the rotations cancel each other out.

In essence, if the starting alkyl halide is optically active, the \( \text{S}_{\text{N}}1 \) reaction will lead to a loss of optical activity, resulting in a racemic mixture. This inversion of configuration is called Walden inversion, observed for \( \text{S}_{\text{N}}2 \) reactions, where the nucleophile attacks from the back, causing inversion. However, in \( \text{S}_{\text{N}}1 \) reactions, due to the planar carbocation intermediate, attack can occur from both sides, yielding both retention and inversion products, thus resulting in racemization.
In simple words: In \( \text{S}_{\text{N}}1 \) reactions, a flat intermediate molecule is formed. Because of this, the new atom can attach from either side, creating two mirror-image products in equal amounts. This results in a mix that doesn't twist light, even if the starting molecule did.

🎯 Exam Tip: When explaining \( \text{S}_{\text{N}}1 \) stereochemistry, clearly state the formation of a planar carbocation intermediate and how it leads to attack from both faces, resulting in racemization (equal mixture of enantiomers).

 

Question 47. Write short notes on the following.
i) Raschig process
ii) Dows process
iii) Darzen's process

Answer:
i) Raschig process:
The Raschig process is an industrial method for preparing chlorobenzene. It involves passing a mixture of benzene vapor, air, and HCl over a heated cupric chloride catalyst at 525 K. This process is economical and efficient for large-scale production.
\[ \text{C}_6\text{H}_6 + \text{HCl} + 1/2 \text{O}_2 \xrightarrow{\text{CuCl}_2 / 525\text{K}} \text{C}_6\text{H}_5\text{Cl} + \text{H}_2\text{O} \]
ii) Dow's process:
Dow's process is a method to produce phenol from chlorobenzene. In this reaction, chlorobenzene is heated with aqueous sodium hydroxide at very high temperatures (\( 350^\circ\text{C} \)) and pressures (300 atm). It's a significant industrial route for synthesizing phenol.
\[ \text{C}_6\text{H}_5\text{Cl} + \text{NaOH} \xrightarrow{350^\circ\text{C}, 300 \text{ atm}} \text{C}_6\text{H}_5\text{OH} + \text{NaCl} \]
iii) Darzen's process:
Darzen's process (also known as Darzen's halogenation) is used to convert alcohols into alkyl chlorides. The alcohol is reacted with thionyl chloride (\( \text{SOCl}_2 \)) in the presence of pyridine. This method is preferred because the byproducts, sulfur dioxide (\( \text{SO}_2 \)) and hydrogen chloride (\( \text{HCl} \)), are gases and easily escape, leaving a pure alkyl chloride. This is an excellent method for obtaining high yields.
\[ \text{CH}_3\text{CH}_2\text{OH} + \text{SOCl}_2 \xrightarrow{\text{Pyridine}} \text{CH}_3\text{CH}_2\text{Cl} + \text{SO}_2 \uparrow + \text{HCl} \uparrow \]
In simple words: (i) Raschig process makes chlorobenzene from benzene, air, and HCl using copper chloride. (ii) Dow's process makes phenol from chlorobenzene using strong heat and pressure with sodium hydroxide. (iii) Darzen's process makes chloroethane from ethanol using thionyl chloride, where the other products are gases and easily leave, giving a clean product.

🎯 Exam Tip: For each named reaction, remember the specific reactants, products, and key reaction conditions (e.g., catalysts, temperature, pressure) that distinguish it.

 

Question 48. Complete the following reactions.
i) \( \text{CH}_3 - \text{CH} = \text{CH}_2 + \text{HBr} \xrightarrow{\text{Peroxide}} \)
ii) \( \text{CH}_3 - \text{CH}_2 - \text{Br} + \text{NaSH} \xrightarrow{\text{alcohol}/\text{H}_2\text{O}} \)
iii) \( \text{C}_6\text{H}_5\text{Cl} + \text{Mg} \xrightarrow{\text{THF}} \)
iv) \( \text{CHCl}_3 + \text{HNO}_3 \rightarrow \)
v) \( \text{CCl}_4 + \text{H}_2\text{O} \rightarrow \)

Answer:
i) \( \text{CH}_3 - \text{CH} = \text{CH}_2 + \text{HBr} \xrightarrow{\text{Peroxide}} \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{Br} \)
This is an anti-Markovnikov addition, where bromine attaches to the less substituted carbon due to the presence of peroxide.
ii) \( \text{CH}_3 - \text{CH}_2 - \text{Br} + \text{NaSH} \xrightarrow{\text{alcohol}/\text{H}_2\text{O}} \text{CH}_3-\text{CH}_2-\text{SH} + \text{NaBr} \)
This is a nucleophilic substitution reaction, forming ethanethiol.
iii) \( \text{C}_6\text{H}_5\text{Cl} \text{ (Chlorobenzene)} + \text{Mg} \xrightarrow{\text{THF}} \text{C}_6\text{H}_5\text{MgCl} \text{ (Phenyl magnesium chloride)} \)
This reaction forms a Grignard reagent, phenylmagnesium chloride.
iv) \( \text{CHCl}_3 \text{ (Chloroform)} + \text{HNO}_3 \rightarrow \text{CCl}_3\text{NO}_2 + \text{H}_2\text{O} \)
This reaction produces chloropicrin (trichloronitromethane), a tear gas.
v) \( \text{CCl}_4 \text{ (Carbon tetrachloride)} + \text{H}_2\text{O} \rightarrow \text{COCl}_2 \text{ (Carbonyl chloride)} + 2\text{HCl} \)
This reaction shows the hydrolysis of carbon tetrachloride, forming phosgene and hydrogen chloride. This emphasizes why \( \text{CCl}_4 \) should not be mixed with water.
In simple words: (i) Propene reacts with HBr in the presence of peroxide to add bromine to the end carbon. (ii) Ethyl bromide reacts with sodium hydrosulfide to make ethanethiol. (iii) Chlorobenzene reacts with magnesium in THF to form a Grignard reagent. (iv) Chloroform reacts with nitric acid to create chloropicrin. (v) Carbon tetrachloride reacts with water to form phosgene and HCl.

🎯 Exam Tip: Pay close attention to reaction conditions (e.g., peroxide for anti-Markovnikov, solvent for Grignard formation) as they dictate the products. Remember the hazardous nature of products like phosgene and chloropicrin.

 

Question 49. Complete the following reactions.
(i) \( \text{CH}_3 - \text{CH} = \text{CH}_2 + \text{HBr} \xrightarrow{\text{Peroxide}} \)
(ii) \( \text{CH}_3 - \text{CH}_2 - \text{Br} + \text{NaSH} \xrightarrow{\text{alcohol}} \)
(iii) \( \text{C}_6\text{H}_5\text{Cl} + \text{Mg} \xrightarrow{\text{THF}} \)
(iv) \( \text{CHCl}_3 + \text{HNO}_3 \implies \)
(v) \( \text{CCl}_4 + \text{H}_2\text{O} \implies \)
Answer:
(i) \( \text{CH}_3 - \text{CH} = \text{CH}_2 + \text{HBr} \xrightarrow{\text{Peroxide}} \text{CH}_3 - \text{CH}_2 - \text{CH}_2 - \text{Br} \)
This reaction converts propene into n-propyl bromide, showing a specific type of addition where hydrogen adds to the carbon with more hydrogen atoms in the presence of peroxide, following the anti-Markovnikov rule.
(ii) \( \text{CH}_3 - \text{CH}_2 - \text{Br} + \text{NaSH} \xrightarrow{\text{alcohol}} \text{CH}_3 - \text{CH}_2 - \text{SH} + \text{NaBr} \)
Here, propyl bromide reacts with sodium hydrogen sulfide in alcohol to form ethanethiol and sodium bromide.
(iii) \( \text{C}_6\text{H}_5\text{Cl} + \text{Mg} \xrightarrow{\text{THF}} \text{C}_6\text{H}_5\text{MgCl} \)
Chlorobenzene reacts with magnesium in tetrahydrofuran (THF) to produce phenyl magnesium chloride, which is a Grignard reagent.
(iv) \( \text{CHCl}_3 + \text{HNO}_3 \implies \text{CCl}_3\text{NO}_2 + \text{H}_2\text{O} \)
Chloroform reacts with nitric acid to yield chloropicrin, also known as trichloronitromethane, a chemical that was used as a tear gas.
(v) \( \text{CCl}_4 + \text{H}_2\text{O} \implies \text{COCl}_2 + 2\text{HCl} \)
Carbon tetrachloride reacts with water to form carbonyl chloride (phosgene) and hydrochloric acid. Phosgene is a very poisonous gas.
In simple words: For each reaction, we are finding the missing product. These reactions show how different starting chemicals change when they are mixed with other substances under specific conditions like using heat or a catalyst.

🎯 Exam Tip: Remember to balance the chemical equations and identify the reaction conditions (like 'Peroxide', 'alcohol', 'THF') as they are crucial for determining the product.

 

Question 50. Explain the preparation of the following compounds.
(i) DDT
(ii) Chloroform
(iii) Biphenyl
(iv) Chloropicrin
(v) Freon - 12
Answer:
(i) **DDT:** DDT (Dichlorodiphenyltrichloroethane) is made by heating a mixture of chlorobenzene with chloral (trichloroacetaldehyde) in the presence of concentrated sulfuric acid (\( \text{H}_2\text{SO}_4 \)). This reaction creates a compound widely known as an insecticide, although its use is restricted due to environmental concerns.
\( \text{H} \text{C} = \text{O} \text{ (Chloral)} + \text{2 C}_6\text{H}_5\text{Cl (Chlorobenzene)} \xrightarrow{\text{con. H}_2\text{SO}_4} \text{DDT} + \text{H}_2\text{O} \)
(ii) **Chloroform:** Chloroform can be prepared in the lab by reacting ethyl alcohol with bleaching powder. The bleaching powder acts as a source of chlorine and calcium hydroxide. This multi-step process is called the haloform reaction. The reaction proceeds through oxidation, chlorination, and hydrolysis steps to yield chloroform.
Step 1: Oxidation
\( \text{CH}_3\text{CH}_2\text{OH (Ethyl alcohol)} + \text{Cl}_2 \rightarrow \text{CH}_3\text{CHO (Acetaldehyde)} + 2\text{HCl} \)
Step 2: Chlorination
\( \text{CH}_3\text{CHO (Acetaldehyde)} + 3\text{Cl}_2 \rightarrow \text{CCl}_3\text{CHO (Chloral)} + 3\text{HCl} \)
Step 3: Hydrolysis
\( 2\text{CCl}_3\text{CHO (Chloral)} + \text{Ca(OH)}_2 \rightarrow 2\text{CHCl}_3 \text{ (Chloroform)} + \text{(HCOO)}_2\text{Ca} \)
(iii) **Biphenyl:** Biphenyl is formed when chlorobenzene reacts with sodium metal in dry ether. This is known as the Fittig reaction, where two aryl groups combine to produce a biaryl product. The dry ether is important to prevent unwanted side reactions with moisture.
\( \text{C}_6\text{H}_5\text{Cl} + 2\text{Na} + \text{Cl} - \text{C}_6\text{H}_5 \xrightarrow{\text{Ether}} \text{C}_6\text{H}_5 - \text{C}_6\text{H}_5 \text{ (Biphenyl)} + 2\text{NaCl} \)
(iv) **Chloropicrin:** Chloropicrin is formed when chloroform reacts with nitric acid. This reaction produces trichloronitromethane, which is also known as chloropicrin. It is a potent lachrymator and insecticide.
\( \text{CHCl}_3 \text{ (Chloroform)} + \text{HNO}_3 \rightarrow \text{CCl}_3\text{NO}_2 \text{ (Chloropicrin)} + \text{H}_2\text{O} \)
(v) **Freon - 12:** Freon-12 (Dichlorodifluoromethane, \( \text{CCl}_2\text{F}_2 \)) is prepared by reacting carbon tetrachloride with hydrogen fluoride. This reaction happens in the presence of a catalytic amount of antimony pentachloride (\( \text{SbCl}_5 \)). Freons were widely used as refrigerants but are now being phased out due to their ozone-depleting potential.
\( \text{CCl}_4 \text{ (Carbon tetrachloride)} + 2\text{HF} \xrightarrow{\text{SbCl}_5} 2\text{HCl} + \text{CCl}_2\text{F}_2 \text{ (Freon - 12)} \)
In simple words: We explained how to make five different chemical compounds. Each compound has its own special recipe, using different starting materials and sometimes needing heat or other chemicals to help the reaction happen.

🎯 Exam Tip: When explaining preparation methods, always include the reactants, products, and key reaction conditions like catalysts or temperature. Knowing the uses of each compound can also help in remembering the reactions.

 

Question 51. An organic compound (A) with molecular formula \( \text{C}_2\text{H}_5\text{Cl} \) reacts with KOH gives compounds (B) and with alcoholic KOH gives compound (C). Identify (A), (B), (C).
Answer:
(A) \( \text{C}_2\text{H}_5\text{Cl} \) is Ethyl chloride.
When ethyl chloride (A) reacts with aqueous potassium hydroxide (KOH), it undergoes a substitution reaction to form ethyl alcohol (B).
\( \text{C}_2\text{H}_5\text{Cl} \text{ (A)} + \text{KOH (aq)} \rightarrow \text{C}_2\text{H}_5\text{OH (B)} \text{ (Ethyl alcohol)} \)
When ethyl chloride (A) reacts with alcoholic KOH, it undergoes an elimination reaction to form ethylene (C). This is a dehydrohalogenation reaction.
\( \text{C}_2\text{H}_5\text{Cl} \text{ (A)} + \text{KOH (alcoholic)} \rightarrow \text{CH}_2 = \text{CH}_2 \text{ (C)} \text{ (Ethylene)} + \text{KCl} + \text{H}_2\text{O} \)
So, compound (A) is ethyl chloride, (B) is ethyl alcohol, and (C) is ethylene. The type of KOH (aqueous vs. alcoholic) determines whether substitution or elimination occurs.
In simple words: Compound A (ethyl chloride) changes into compound B (ethyl alcohol) if you use KOH in water. But if you use KOH in alcohol, compound A changes into compound C (ethylene). The type of KOH used makes a big difference in the final product.

🎯 Exam Tip: Pay close attention to the conditions (aqueous vs. alcoholic KOH) when dealing with reactions of alkyl halides, as they dictate the reaction pathway (substitution or elimination) and the major product.

 

Question 52. Simplest alkene (A) reacts with HCl to form compound (B). Compound (B) reacts with ammonia to form compound (C) of molecular formula \( \text{C}_2\text{H}_7\text{N} \). Compound (C) undergoes carbylamine test. Identify (A), (B) and (C).
Answer:
(A) The simplest alkene is ethylene, \( \text{CH}_2 = \text{CH}_2 \).
(B) Ethylene reacts with HCl to form ethyl chloride (\( \text{C}_2\text{H}_5\text{Cl} \)).
\( \text{CH}_2 = \text{CH}_2 \text{ (A)} + \text{HCl} \rightarrow \text{C}_2\text{H}_5\text{Cl} \text{ (B)} \)
(C) Ethyl chloride (B) reacts with ammonia (\( \text{NH}_3 \)) to form ethyl amine (\( \text{C}_2\text{H}_5\text{NH}_2 \)). This is a nucleophilic substitution reaction. The molecular formula of ethyl amine is \( \text{C}_2\text{H}_7\text{N} \), which matches the given information.
\( \text{C}_2\text{H}_5\text{Cl} \text{ (B)} + \text{NH}_3 \rightarrow \text{C}_2\text{H}_5\text{NH}_2 \text{ (C)} + \text{HCl} \)
Ethyl amine (C) is a primary amine and therefore undergoes the carbylamine test, producing an isocyanide (foul-smelling compound).
Thus, (A) is Ethylene, (B) is Ethyl chloride, and (C) is Ethyl amine. Knowing common reactions helps identify unknown compounds.
In simple words: We start with a simple alkene (A), which reacts with HCl to make compound B. Then, compound B reacts with ammonia to make compound C. Compound C is ethyl amine, which means A is ethylene and B is ethyl chloride.

🎯 Exam Tip: When identifying unknown compounds, always check if the molecular formula and characteristic reactions (like the carbylamine test) match the proposed structure. This helps confirm your answer.

 

Question 53. A hydrocarbon \( \text{C}_3\text{H}_6(\text{A}) \) reacts with HBr to form compound (B). Compound (B) reacts with aqueous potassium hydroxide to give (C) of molecular formula \( \text{C}_3\text{H}_6\text{O} \). What are the (A), (B) and (C). Explain the reactions.
Answer:
(A) The hydrocarbon with formula \( \text{C}_3\text{H}_6 \) is Propylene (\( \text{CH}_3 - \text{CH} = \text{CH}_2 \)).
(B) Propylene (A) reacts with HBr according to Markovnikov's rule, where the H atom adds to the carbon with more H atoms, and Br adds to the carbon with fewer H atoms. This forms Isopropyl bromide.
\( \text{CH}_3 - \text{CH} = \text{CH}_2 \text{ (A)} + \text{HBr} \rightarrow \text{CH}_3 - \text{CH(Br)} - \text{CH}_3 \text{ (B)} \text{ (Isopropyl bromide)} \)
(C) Isopropyl bromide (B) then reacts with aqueous potassium hydroxide (KOH), undergoing a nucleophilic substitution reaction to produce Isopropyl alcohol (\( \text{C}_3\text{H}_7\text{OH} \)), which has the formula \( \text{C}_3\text{H}_6\text{O} \).
\( \text{CH}_3 - \text{CH(Br)} - \text{CH}_3 \text{ (B)} + \text{KOH (aq)} \rightarrow \text{CH}_3 - \text{CH(OH)} - \text{CH}_3 \text{ (C)} \text{ (Isopropyl alcohol)} + \text{KBr} \)
So, (A) is Propylene, (B) is Isopropyl bromide, and (C) is Isopropyl alcohol. This sequence of reactions illustrates common organic transformations.
In simple words: We started with a hydrocarbon (A) which is propylene. It reacted with HBr to become isopropyl bromide (B). Then, isopropyl bromide reacted with KOH in water to make isopropyl alcohol (C).

🎯 Exam Tip: Remember Markovnikov's rule for addition reactions to unsymmetrical alkenes, and distinguish between aqueous KOH (substitution) and alcoholic KOH (elimination) for haloalkane reactions.

 

Question 54. Two isomers (A) and (B) have the same molecular formula \( \text{C}_2\text{H}_4\text{Cl}_2 \). Compound (A) reacts with aqueous KOH gives compound (C) of molecular formula \( \text{C}_2\text{H}_4\text{O} \). Compound (B) reacts with aqueous KOH gives compound (D) of molecular formula \( \text{C}_2\text{H}_6\text{O}_2 \). Identify (A), (B), (C) and (D).
Answer:
Both compounds (A) and (B) are isomers with the molecular formula \( \text{C}_2\text{H}_4\text{Cl}_2 \).
Compound (A) forms a compound (C) with formula \( \text{C}_2\text{H}_4\text{O} \) when reacted with aqueous KOH. This suggests that (A) is Ethylidene chloride (\( \text{CH}_3\text{CHCl}_2 \)), which on hydrolysis gives acetaldehyde (\( \text{CH}_3\text{CHO} \)).
\( \text{CH}_3\text{CHCl}_2 \text{ (A)} + 2\text{KOH (aq)} \rightarrow \text{CH}_3\text{CHO} \text{ (C)} + 2\text{KCl} + \text{H}_2\text{O} \)
Compound (B) forms a compound (D) with formula \( \text{C}_2\text{H}_6\text{O}_2 \) when reacted with aqueous KOH. This indicates that (B) is Ethylene dichloride (\( \text{ClCH}_2\text{CH}_2\text{Cl} \)), which on hydrolysis gives ethylene glycol (\( \text{HOCH}_2\text{CH}_2\text{OH} \)).
\( \text{ClCH}_2\text{CH}_2\text{Cl} \text{ (B)} + 2\text{KOH (aq)} \rightarrow \text{HOCH}_2\text{CH}_2\text{OH} \text{ (D)} + 2\text{KCl} \)
So, (A) is Ethylidene chloride, (B) is Ethylene dichloride, (C) is Acetaldehyde, and (D) is Ethylene glycol. Knowing the structures of these isomers helps predict their reactions.
In simple words: We have two different chemicals (A and B) that are very similar. When chemical A is mixed with water-based KOH, it turns into C (acetaldehyde). When chemical B is mixed with water-based KOH, it turns into D (ethylene glycol). This helps us figure out what A, B, C, and D are.

🎯 Exam Tip: Differentiate between gem-dihalides (halogens on the same carbon) and vicinal dihalides (halogens on adjacent carbons), as they yield different products upon hydrolysis.

11th Chemistry Guide Haloalkanes and Haloarenes Additional Questions and Answers

I. Choose the best answer:

 

Question 1. Which of the following is 1º alkyl halide
(a) \( \text{R} - \text{CH}_2 - \text{X} \)
(b) \( \text{R}_2\text{CHX} \)
(c) \( \text{R}_3\text{C} - \text{X} \)
(d) \( \text{R} - \text{H} \)
Answer: (a) \( \text{R} - \text{CH}_2 - \text{X} \)
In simple words: A primary (1º) alkyl halide means the carbon atom bonded to the halogen (X) is only attached to one other carbon atom (R).

🎯 Exam Tip: To identify primary, secondary, or tertiary alkyl halides, count the number of carbon atoms directly attached to the carbon bearing the halogen.

 

Question 2. 2º halide among the following
(a) isopropyl chloride
(b) iso-butyl chloride
(c) n-propyl chloride
(d) n-butyl chloride
Answer: (a) isopropyl chloride
In simple words: Isopropyl chloride is a secondary halide because the carbon atom connected to the chlorine is also connected to two other carbon atoms.

🎯 Exam Tip: Draw the structures of the given compounds to accurately determine the type (1º, 2º, or 3º) of halide.

 

Question 3. Ethylidene dibromide is
(a) \( \text{CH}_3\text{CH}_2\text{Br} \)
(b) \( \text{Br} - \text{CH}_2 - \text{CH}_2 - \text{Br} \)
(c) \( \text{CH}_3 - \text{CHBr}_2 \)
(d) \( \text{CH}_2 = \text{CHBr} \)
Answer: (c) \( \text{CH}_3 - \text{CHBr}_2 \)
In simple words: Ethylidene dibromide means both bromine atoms are attached to the same carbon atom in a two-carbon chain.

🎯 Exam Tip: Understand the difference between "ethylidene" (both halogens on the same carbon, a gem-dihalide) and "ethylene" (halogens on adjacent carbons, a vicinal dihalide) when naming dihaloalkanes.

 

Question 4. Which of the following is gemdihalide?
(a) \( \text{CH}_3\text{CHBrCH}_2\text{Br} \)
(b) \( \text{CH}_3\text{CHBr}_2 \)
(c) \( \text{CH}_3\text{CHBrCH}_2\text{CH}_2\text{Br} \)
(d) \( \text{BrCH}_2\text{CH}_2\text{Br} \)
Answer: (b) \( \text{CH}_3\text{CHBr}_2 \)
In simple words: A gem-dihalide is a molecule where two halogen atoms are attached to the exact same carbon atom. Here, both bromine atoms are on the middle carbon.

🎯 Exam Tip: Geminal dihalides (gem-dihalides) have both halogen atoms on the same carbon, while vicinal dihalides (vic-dihalides) have them on adjacent carbons.

 

Question 5. Vicinal dihalide is
(a) \( \text{CH}_3\text{CH}_2\text{Br} \)
(b) \( \text{Br} - \text{CH}_2 - \text{CH}_2 - \text{Br} \)
(c) \( \text{CH}_3 - \text{CHBr}_2 \)
(d) \( \text{CH}_2 = \text{CHBr} \)
Answer: (b) \( \text{Br} - \text{CH}_2 - \text{CH}_2 - \text{Br} \)
In simple words: A vicinal dihalide means that the two halogen atoms are on carbon atoms that are right next to each other.

🎯 Exam Tip: Always visualize the structure to correctly identify geminal versus vicinal dihalides; "vicinal" means neighbors.

 

Question 6. The reagent used to get alkyl halide from alcohol
(a) \( \text{PCl}_5 \)
(b) \( \text{SOCl}_2 \)
(c) Both (a) and (b)
(d) \( \text{Cl}_2 \)
Answer: (c) Both (a) and (b)
In simple words: You can turn an alcohol into an alkyl halide by using either \( \text{PCl}_5 \) (phosphorus pentachloride) or \( \text{SOCl}_2 \) (thionyl chloride).

🎯 Exam Tip: Thionyl chloride (\( \text{SOCl}_2 \)) is often preferred because the byproducts (\( \text{SO}_2 \) and \( \text{HCl} \)) are gases, making the purification of the alkyl halide easier.

 

Question 7. For the preparation of alkyl halides from alcohols which among the following cannot be used
(a) \( \text{PCl}_5 \)
(b) \( \text{SOCl}_2 \)
(c) \( \text{PCl}_3 \)
(d) \( \text{NaCl} \)
Answer: (d) \( \text{NaCl} \)
In simple words: To make an alkyl halide from an alcohol, you can use phosphorus pentachloride, thionyl chloride, or phosphorus trichloride. But you cannot use sodium chloride for this reaction.

🎯 Exam Tip: \( \text{NaCl} \) is a stable ionic compound and lacks the reactivity required to replace the hydroxyl group of an alcohol with a chlorine atom.

 

Question 8. In the preparation of alkyl halide from alkane and halogen which of the following reaction involved
(a) Electrophilic addition
(b) Nucleophilic addition
(c) Electrophilic substitution
(d) Nucleophilic substitution
Answer: (a) Electrophilic addition
In simple words: When you make an alkyl halide from an alkane and a halogen, the process usually involves a step where an electrophile (an electron-loving particle) adds to the molecule.

🎯 Exam Tip: Alkane halogenation often proceeds via free radical substitution, while halogenation of alkenes/alkynes is typically electrophilic addition. The question phrasing here points to the latter, or potentially a free radical pathway if 'alkane' implies saturated hydrocarbons reacting with halogens under UV light.

 

Question 9. In the preparation of alkyl halide from alkane and halogen which of the following reaction involved
(a) Free radical substitution
(b) Nucleophilic addition
(c) Electrophilic substitution
(d) Nucleophilic substitution
Answer: (a) Free radical substitution
In simple words: When you make an alkyl halide by combining an alkane and a halogen, the reaction typically involves free radicals. In this type of reaction, a hydrogen atom from the alkane is replaced by a halogen atom.

🎯 Exam Tip: Alkane halogenation (e.g., methane with chlorine in UV light) is a classic example of a free radical substitution reaction, proceeding through initiation, propagation, and termination steps.

 

Question 10. Grignard reagent is formed when alkyl halide reacts with which one of the following
(a) Mg in alcohol
(b) Mg in acid
(c) Mg in dry ether
(d) MgO
Answer: (c) Mg in dry ether
In simple words: To make a Grignard reagent, you need to mix an alkyl halide with magnesium metal in a dry ether solvent. The dry conditions are very important.

🎯 Exam Tip: Grignard reagents are highly reactive and react with even traces of moisture or acidic protons; hence, dry conditions are essential for their synthesis.

 

Question 11. When alkyl halide reacts with moist \( \text{Ag}_2\text{O} \) gives
(a) alcohol
(b) ether
(c) alkane
(d) Alkene
Answer: (a) alcohol
In simple words: When an alkyl halide is treated with moist silver oxide, it forms an alcohol. The water present with the silver oxide acts as a nucleophile.

🎯 Exam Tip: Moist silver oxide (\( \text{Ag}_2\text{O/H}_2\text{O} \)) behaves like aqueous \( \text{KOH} \) and promotes nucleophilic substitution to form alcohols, while dry \( \text{Ag}_2\text{O} \) forms ethers.

 

Question 12. Alkyl halide on reduction with Zn + HCl gives
(a) alcohol
(b) alkene
(c) alkane
(d) ether
Answer: (c) alkane
In simple words: When an alkyl halide is reacted with zinc and hydrochloric acid, the halogen atom is removed, and a hydrogen atom takes its place, turning it into an alkane.

🎯 Exam Tip: Reduction reactions typically add hydrogen or remove oxygen/halogens. \( \text{Zn} + \text{HCl} \) is a common reducing agent system used for dehalogenation.

 

Question 13. Cyanide is formed as the major product of the reaction when alkyl halide is treated with one of the following :
(a) \( \text{AgNO}_2 \)
(b) \( \text{KNO}_2 \)
(c) \( \text{AgCN} \)
(d) \( \text{KCN} \)
Answer: (c) \( \text{AgCN} \)
In simple words: When an alkyl halide is reacted with silver cyanide (\( \text{AgCN} \)), the main product formed is an isocyanide, not a simple cyanide. The carbon atom of the cyanide group attaches to the alkyl group.

🎯 Exam Tip: The ambidentate nature of the cyanide ion (\( \text{CN}^- \)) results in different products: \( \text{KCN} \) gives alkyl cyanides (due to ionic nature), while \( \text{AgCN} \) gives alkyl isocyanides (due to covalent nature).

 

Question 14. Which of the reactions are most common in alkyl halides
(a) Nucleophilic addition
(b) Electrophilic addition
(c) Nucleophilic substitution
(d) Electrophilic substitution
Answer: (c) Nucleophilic substitution
In simple words: Alkyl halides frequently take part in reactions where a nucleophile (an electron-rich group) replaces the halogen atom. This is a very common type of reaction for them.

🎯 Exam Tip: Alkyl halides have a polar C-X bond, making the carbon atom electrophilic and prone to attack by nucleophiles. Elimination reactions are also common, especially with strong bases.

 

Question 15. Treatment of ammonia with excess of ethyl chloride will yield
(a) Diethyl amine
(b) Ethane
(c) Tetra ethyl ammonium chloride
(d) methyl amine
Answer: (c) Tetra ethyl ammonium chloride
In simple words: If you keep adding ethyl chloride to ammonia, you don't just get one product. Ammonia reacts multiple times to form primary, secondary, tertiary amines, and finally, a quaternary ammonium salt like tetraethylammonium chloride.

🎯 Exam Tip: Amination of alkyl halides is usually not practical for preparing single primary amines because the product primary amine itself is a better nucleophile than ammonia, leading to over-alkylation and a mixture of products.

 

Question 16. In chloro ethane the carbon bearing halogen is bonded to ____.
(a) three, primary
(b) two, secondary
(c) one, tertiary
(d) two, primary
Answer: (d) two, primary
In simple words: In chloroethane, the carbon holding the chlorine atom is also connected to two hydrogen atoms and one other carbon atom. This means it is a primary carbon.

🎯 Exam Tip: Draw the structure of chloroethane (\( \text{CH}_3\text{CH}_2\text{Cl} \)). The carbon attached to the Cl atom is \( \text{CH}_2 \), which is directly bonded to one other carbon and two hydrogen atoms, making it primary.

 

Question 17. Which of the following is used as refrigerant?
(a) \( \text{CH}_3\text{COCH}_3 \)
(b) \( \text{CCl}_4 \)
(c) \( \text{C}_2\text{H}_5\text{Cl} \)
(d) \( \text{CF}_4 \)
Answer: (c) \( \text{C}_2\text{H}_5\text{Cl} \)
In simple words: Ethyl chloride is one of the substances that can be used as a refrigerant, helping things get cold. However, chlorofluorocarbons (like \( \text{CF}_4 \) as a general category) are also refrigerants, but ethyl chloride is a specific option given.

🎯 Exam Tip: While \( \text{CF}_4 \) (tetrafluoromethane) is a refrigerant (a type of Freon), \( \text{C}_2\text{H}_5\text{Cl} \) (ethyl chloride) also has applications as a refrigerant and local anesthetic, making it a valid answer among the options.

 

Question 18. \( \text{S}_N\text{1} \) reaction occurs through the intermediate formation of
(a) carbocation
(b) carbanion
(c) free radicals
(d) transition
Answer: (a) carbocation
In simple words: The \( \text{S}_N\text{1} \) reaction happens in two steps, and in the first step, a carbocation is formed. This is a positively charged carbon atom which acts as a temporary intermediate.

🎯 Exam Tip: The stability of the carbocation formed in the \( \text{S}_N\text{1} \) reaction (tertiary > secondary > primary) determines the reaction rate. Rearrangements can also occur to form a more stable carbocation.

 

Question 19. The rate of \( \text{S}_N\text{2} \) reaction is maximum when the solvent is
(a) Dimethyl sulphoxide
(b) Water
(c) Dimethyl sulphoxide
(d) Benzene
Answer: (c) Dimethyl sulphoxide
In simple words: The \( \text{S}_N\text{2} \) reaction goes fastest when a special kind of solvent called an aprotic polar solvent is used. Dimethyl sulphoxide (DMSO) is one such solvent that helps the reaction happen quickly.

🎯 Exam Tip: \( \text{S}_N\text{2} \) reactions are favored by polar aprotic solvents (like DMSO, acetone, DMF) because they solvate cations but leave anions (nucleophiles) relatively bare and highly reactive.

 

Question 20. The most reactive nucleophile among the following is
(a) \( \text{CH}_3\text{O}^- \)
(b) \( \text{C}_6\text{H}_5\text{O}^- \)
(c) \( (\text{CH}_3)_2\text{CHO}^- \)
(d) \( (\text{CH}_3)_3\text{CO}^- \)
Answer: (a) \( \text{CH}_3\text{O}^- \)
In simple words: Among these options, the methoxide ion (\( \text{CH}_3\text{O}^- \)) is the most reactive nucleophile. This is because it is less bulky and more concentrated in negative charge, making it better at attacking other molecules.

🎯 Exam Tip: For nucleophiles with the same attacking atom (e.g., oxygen), nucleophilicity generally decreases with increasing steric hindrance (bulkiness) around the attacking site. Smaller nucleophiles can approach the electrophilic carbon more easily.

 

Question 21. The correct order of reactivity towards nucleophilic substitution reaction is
(a) \( \text{CH}_3\text{F} > \text{CH}_3\text{Cl} > \text{CH}_3\text{Br} > \text{CH}_3\text{I} \)
(b) \( \text{CH}_3\text{I} > \text{CH}_3\text{Br} > \text{CH}_3\text{Cl} > \text{CH}_3\text{F} \)
(c) \( \text{CH}_3\text{I} > \text{CH}_3\text{Cl} > \text{CH}_3\text{Br} > \text{CH}_3\text{F} \)
(d) \( \text{CH}_3\text{I} > \text{CH}_3\text{Br} > \text{CH}_3\text{F} > \text{CH}_3\text{Cl} \)
Answer: (b) \( \text{CH}_3\text{I} > \text{CH}_3\text{Br} > \text{CH}_3\text{Cl} > \text{CH}_3\text{F} \)
In simple words: For nucleophilic substitution reactions, alkyl iodides react the fastest, followed by bromides, then chlorides, and finally fluorides. This is because the carbon-halogen bond becomes weaker as the halogen atom gets larger, making it easier to break.

🎯 Exam Tip: The leaving group ability is the most important factor in determining the rate of nucleophilic substitution reactions. Iodide is the best leaving group because of its large size and weak C-I bond, while fluoride is the poorest.

 

Question 22. In \( \text{S}_N\text{2} \) reactions the order of reactivity of the halides.
\( \text{CH}_3\text{X}, \text{C}_2\text{H}_5\text{X}, \text{n} - \text{C}_3\text{H}_7\text{X}, \text{n}- \text{C}_4\text{H}_9\text{X} \) is
(a) \( \text{CH}_3\text{X} > \text{C}_2\text{H}_5\text{X} > \text{n} - \text{C}_3\text{H}_7\text{X} > \text{n} - \text{C}_4\text{H}_9\text{X} \)
(b) \( \text{CH}_3\text{X} > \text{n} - \text{C}_3\text{H}_7\text{X} > \text{C}_2\text{H}_5\text{X} > \text{n} - \text{C}_4\text{H}_9\text{X} \)
(c) \( \text{C}_2\text{H}_5\text{X} > \text{n} - \text{C}_3\text{H}_7\text{X} > \text{n} - \text{C}_4\text{H}_9\text{X} < \text{CH}_3\text{X} \)
(d) \( \text{n} - \text{C}_4\text{H}_9\text{X} > \text{n} - \text{C}_3\text{H}_7\text{X} > \text{C}_2\text{H}_5\text{X} > \text{CH}_3\text{X} \)
Answer: (a) \( \text{CH}_3\text{X} > \text{C}_2\text{H}_5\text{X} > \text{n} - \text{C}_3\text{H}_7\text{X} > \text{n} - \text{C}_4\text{H}_9\text{X} \)
In simple words: For \( \text{S}_N\text{2} \) reactions, smaller alkyl halides react faster than larger ones. This is because the attacking molecule finds it easier to reach the carbon atom with less crowding around it.

🎯 Exam Tip: \( \text{S}_N\text{2} \) reactions are very sensitive to steric hindrance. Methyl halides are the most reactive, followed by primary, then secondary, with tertiary halides being essentially unreactive in \( \text{S}_N\text{2} \) due to high steric hindrance.

 

Question 23. \( \text{S}_N\text{2} \) mechanism proceeds through the formation of a
(a) carbocation
(b) transition state
(c) free radical
(d) carbanion
Answer: (b) transition state
In simple words: The \( \text{S}_N\text{2} \) reaction happens in just one step, where the nucleophile attacks and the leaving group leaves at the same time. During this, a temporary, unstable structure called a transition state is formed.

🎯 Exam Tip: Unlike \( \text{S}_N\text{1} \) reactions which involve a carbocation intermediate, \( \text{S}_N\text{2} \) reactions are concerted, meaning bond breaking and bond forming occur simultaneously through a single transition state without any true intermediate.

 

Question 24. In Dow's process the starting raw material is
(a) Phenol
(b) Chloro benzene
(c) Aniline
(d) Diazobenzene
Answer: (b) Chloro benzene
In simple words: Dow's process is a method used to make phenol. It starts by taking chlorobenzene as the main ingredient.

🎯 Exam Tip: Dow's process is a significant industrial method for producing phenol, which involves the hydrolysis of chlorobenzene under harsh conditions (high temperature and pressure).

 

Question 25. Chloro benzene is prepared commercially by
(a) Dow's process
(b) Decon's process
(c) Raschig process
(d) Etard's process
Answer: (c) Raschig process
In simple words: Chloro benzene is made on a large scale in factories using a method called the Raschig process. This method involves using benzene, air, and HCl to create chlorobenzene.

🎯 Exam Tip: The Raschig process is an older but significant industrial method for synthesizing chlorobenzene, contrasting with other named reactions like Dow's process which uses chlorobenzene as a reactant to produce phenol.

 

Question 26. Chloro benzene is reactive than benzene towards electrophilic substitution and directs incoming electrophile to the position.
(a) more, ortho & para
(b) less, ortho & para
(c) more, meta
(d) less, meta
Answer: (b) less, ortho & para
In simple words: Chlorobenzene is less reactive than benzene when it comes to attracting electrons, but it guides new electron-loving groups to attach at the ortho and para spots on its ring.

🎯 Exam Tip: Remember that halogens are deactivating but ortho-para directing due to resonance effects overriding inductive effects.

 

Question 27. The raw material for raschig; process is
(a) chloro benzene
(b) phenol
(c) benzene
(d) anisol
Answer: (c) benzene
In simple words: To start the Raschig process, which makes chlorobenzene, you need benzene as the main ingredient.

🎯 Exam Tip: Knowing the starting materials for named reactions helps you understand the reaction pathway and products.

 

Question 28. Chloro benzene on treatment with sodium in dry ether gives diphenyl. The name of the reaction is
(a) Fitting reaction
(b) Wurtz fittig reaction
(c) Wurtz reaction
(d) Sandmeyer reaction
Answer: (a) Fitting reaction
In simple words: When chlorobenzene reacts with sodium metal in dry ether to form diphenyl, it is called the Fittig reaction.

🎯 Exam Tip: Distinguish between Wurtz reaction (alkyl halides), Fittig reaction (aryl halides), and Wurtz-Fittig reaction (alkyl and aryl halides) based on the reactants.

 

Question 29. An organic compound which produces a bluish green coloured flame on heating in the presence of copper is,
(a) chloro benzene
(b) benzaldehyde
(c) aniline
(d) benzoic acid
Answer: (a) chloro benzene
In simple words: When you heat chlorobenzene with copper, it makes a bluish-green flame. This is a special test that shows chlorine is present in the compound.

🎯 Exam Tip: This is a classic test for halogens in organic compounds, often called Beilstein test.

 

Question 30. The raw materials for the commercial manufacture of DDT are
(a) chloro benzene and chloroform
(b) chloro benzene and chloro methane
(c) chloro benzene and chloral
(d) chloro benzene and iodoform
Answer: (c) chloro benzene and chloral
In simple words: To make DDT on a large scale, you need to combine chlorobenzene and chloral.

🎯 Exam Tip: Always remember the specific reactants for the synthesis of important compounds like DDT, as they are often tested.

 

Question 31. Iodoform is used as
(a) anaesthetic
(b) antiseptic
(c) analgesic
(d) anti febrin
Answer: (b) antiseptic
In simple words: Iodoform is used to clean wounds and prevent infections, acting as an antiseptic.

🎯 Exam Tip: Understand the common uses of halogenated organic compounds, especially those with medicinal or industrial applications.

 

Question 32. The following is used in paint removing
(a) \( \text{CHCl}_3 \)
(b) \( \text{CH}_2\text{Cl}_2 \)
(c) \( \text{CCl}_4 \)
(d) \( \text{CH}_3\text{Cl} \)
Answer: (b) CH2Cl2
In simple words: Dichloromethane, or methylene chloride, is often used to strip paint because it can dissolve many paint chemicals.

🎯 Exam Tip: Learn the practical applications of common organic solvents; methylene chloride is a key one for paint removal.

 

Question 33. In fire extinguishers, following is used
(a) \( \text{CHCl}_3 \)
(b) \( \text{CS}_2 \)
(c) \( \text{CCl}_4 \)
(d) \( \text{CH}_2\text{Cl}_2 \)
Answer: (c) CCl4
In simple words: Carbon tetrachloride, also known as pyrene, was historically used in fire extinguishers because its vapors are heavy and non-flammable.

🎯 Exam Tip: Note that while \( \text{CCl}_4 \) was used, it's now known to be toxic and damaging to the ozone layer, so its use is restricted.

 

Question 34. The following is used for metal cleaning and finishing
(a) \( \text{CHCl}_3 \)
(b) \( \text{CHI}_3 \)
(c) \( \text{CH}_2\text{Cl}_2 \)
(d) \( \text{C}_6\text{H}_6 \)
Answer: (c) CH2Cl2
In simple words: Dichloromethane is effective for cleaning metals and giving them a good finish because it can remove grease and oils.

🎯 Exam Tip: Dichloromethane's versatility as a solvent makes it useful for various cleaning applications, including metal surfaces.

 

Question 35. First chlorinated insecticide
(a) DDT
(b) Gammaxene
Answer: (a) DDT
In simple words: DDT was the first widely used chemical to kill insects, especially those that spread diseases.

🎯 Exam Tip: Remember DDT as a historical benchmark in pesticide use, understanding its initial impact and later environmental concerns.

 

Question 36. The following is used as anaesthetic
(a) \( \text{C}_2\text{H}_4 \)
(b) \( \text{CHCl}_3 \)
(c) \( \text{CH}_2\text{Cl}_2 \)
(d) DDT
Answer: (b) CHCl3
In simple words: Chloroform, with the chemical formula \( \text{CHCl}_3 \), was once a commonly used substance to make patients unconscious during operations.

🎯 Exam Tip: Chloroform is a historically significant anesthetic, but its use has largely been replaced due to safety concerns.

 

Question 37. Freon – 12 is
(a) \( \text{CF}_3\text{Cl} \)
(b) \( \text{CHCl}_2\text{F} \)
(c) \( \text{CF}_2\text{Cl}_2 \)
(d) DDT
Answer: (c) CF2Cl2
In simple words: Freon-12 is the common name for dichlorodifluoromethane, a compound used in refrigeration.

🎯 Exam Tip: Freons are chlorofluorocarbons (CFCs); learn their common names and formulas, especially the numerical designation system.

 

Question 38. The name of DDT
(a) p, p' – dichloro diphenyl trichloro ethane
(b) p, p' – dichloro diphenyl trichloro ethene
(d) p, p' - tetra chloro ethane
Answer: (a) p, p' – dichloro diphenyl trichloro ethane
In simple words: The full chemical name for DDT is p,p'-dichlorodiphenyltrichloroethane, which describes its structure.

🎯 Exam Tip: The systematic IUPAC names for complex compounds like DDT are crucial for identifying them unambiguously.

 

Question 39. Freon R - 22 is
(a) \( \text{CHCIF}_2 \)
(b) \( \text{CCl}_2\text{F}_2 \)
(c) \( \text{CH}_3\text{Cl} \)
(d) \( \text{CH}_2\text{Cl}_2 \)
Answer: (a) CHCIF2
In simple words: Freon R-22 is the common name for chlorodifluoromethane, another type of refrigerant.

🎯 Exam Tip: Understand the 'R-number' system for refrigerants: R-22 means (C-1) = 0, (H+1) = 2, F = 2, so it's \( \text{CHClF}_2 \).

 

Question 40. Molecular formula of DDT has
(a) 5 Cl atoms
(b) 4 Cl atoms
(c) 3 Cl atoms
(d) 2 Cl atoms
Answer: (a) 5 Cl atoms
In simple words: The DDT molecule contains a total of five chlorine atoms in its structure.

🎯 Exam Tip: Visualize the structure of DDT to count the chlorine atoms accurately (3 on the trichloroethane part, 2 on the dichlorophenyl parts).

 

Question 41. What is DDT among the following
(a) Green house gas
(b) A fertilizer
(c) Bio degradable pollutant
(d) Non – Bio degradable pollutant
Answer: (d) Non – Bio degradable pollutant
In simple words: DDT is considered a non-biodegradable pollutant because it does not break down easily in nature and stays in the environment for a long time.

🎯 Exam Tip: The persistence of DDT in the environment and its accumulation in the food chain make it a significant environmental concern.

 

Question 42. The IUPAC name of \( (\text{CH}_3)_3\text{CHCH}_2\text{Br} \) is
(a) 1 – bromo – 2 – methyl propane
(b) 2 – bromo – 2 -methyl propane
(c) 1 – bromo – 1 – methyl propane
(d) 2 – bromo – 1 -methylpropane
Answer: (a) 1 – bromo – 2 – methyl propane
In simple words: When naming the compound \( (\text{CH}_3)_3\text{CHCH}_2\text{Br} \), it's called 1-bromo-2-methylpropane following the rules for chemical names.

🎯 Exam Tip: For IUPAC naming, identify the longest carbon chain, number it correctly to give the lowest number to the principal functional group (here, bromine), and then name the substituents.

 

Question 43. IUPAC name of allyl chloride is
(a) 1 – chloro ethane
(b) 3 – chloro- 1 – propyne
(c) 3 – chloro – 1 – propene
(d) 1 - chloro propane
Answer: (c) 3 – chloro – 1 – propene
In simple words: Allyl chloride, which has a double bond and a chlorine atom, is correctly named 3-chloro-1-propene according to IUPAC rules.

🎯 Exam Tip: Remember that in IUPAC nomenclature, the double bond gets preference over the halogen substituent for numbering the parent chain.

 

Question 44. The number of structural isomers possible with the formula \( \text{C}_4\text{H}_9\text{Cl} \) are
(a) 5
(b) 4
(c) 3
(d) 2
Answer: (b) 4
In simple words: For the chemical formula \( \text{C}_4\text{H}_9\text{Cl} \), there are four different ways to arrange the atoms, creating four structural isomers. These include 1-chlorobutane, 2-chlorobutane, 1-chloro-2-methylpropane, and 2-chloro-2-methylpropane.

🎯 Exam Tip: Systematically draw all possible carbon skeletons first, then place the chlorine atom at different non-equivalent positions to find all isomers.

 

Question 45. Density is highest for
(a) \( \text{CHCl}_3 \)
(b) \( \text{CH}_2\text{Cl}_2 \)
(c) \( \text{CCl}_4 \)
(d) \( \text{C}_6\text{H}_6 \)
Answer: (c) CCl4
In simple words: Among the options, carbon tetrachloride (\( \text{CCl}_4 \)) has the highest density because it has the most heavy chlorine atoms in its molecule.

🎯 Exam Tip: Generally, for halogenated organic compounds, density increases with the number of halogen atoms and the atomic mass of the halogen.

 

Question 46. \( \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{PCl}_5} \text{X} \). In this reaction 'X' is
(a) Ethanol
(b) Ethylene chloride
(c) ethylidene chloride
(d) ethyl chloride
Answer: (d) ethyl chloride
In simple words: When ethanol reacts with phosphorus pentachloride (\( \text{PCl}_5 \)), it changes into ethyl chloride.

🎯 Exam Tip: This is a standard method for converting alcohols into alkyl halides; \( \text{PCl}_5 \) replaces the -OH group with -Cl.

 

Question 47. Thionyl chloride is preferred in the preparation of chloro compound from alcohol since
(a) Both the byproducts are gases and they escape out leaving product in pure state
(b) It is a chlorinating agent
(c) It is an oxidising agent
(d) All other reagents are unstable
Answer: (a) Both the byproducts are gases and they escape out leaving product in pure state
In simple words: Thionyl chloride is a good choice for making chloro compounds from alcohols because the other substances made in the reaction are gases, which easily leave, leaving behind a very clean product.

🎯 Exam Tip: This reaction, known as Darzen's process, is particularly useful for synthesizing pure alkyl chlorides because the gaseous byproducts \( \text{SO}_2 \) and \( \text{HCl} \) can be easily removed.

 

Question 48.
(a) \( \text{C}_2\text{H}_4 \)
(b) \( \text{C}_3\text{H}_6 \)
(c) \( \text{C}_4\text{H}_8 \)
(d) \( \text{C}_5\text{H}_{10} \)
Answer: (a) C2H4
In simple words: Ethylene (\( \text{C}_2\text{H}_4 \)) is a simple alkene molecule.

🎯 Exam Tip: Recognize the basic chemical formulas for simple alkanes, alkenes, and alkynes, which are fundamental in organic chemistry.

 

Question 49. – OH cannot be replaced by – Cl if we use
(a) \( \text{PCl}_5 \)
(b) \( \text{PCl}_3 \)
(c) \( \text{S}_2\text{Cl}_2 \)
(d) \( \text{SOCl}_2 \)
Answer: (c) S2Cl2
In simple words: You cannot replace a hydroxyl group (\( \text{-OH} \)) with a chlorine atom (\( \text{-Cl} \)) if you use \( \text{S}_2\text{Cl}_2 \). The other options are common reagents for this conversion.

🎯 Exam Tip: Be familiar with common reagents like \( \text{PCl}_5 \), \( \text{PCl}_3 \), and \( \text{SOCl}_2 \) (thionyl chloride) that facilitate the substitution of an -OH group with a halogen, particularly -Cl.

 

Question 50. In the hydrohalogenation of ethylene for adding HCl, the catalyst used is
(a) Anhydrous \( \text{AlCl}_3 \)
(b) Conc. Sulphuric acid
(c) Dilute Sulphuric acid
(d) Anhydrous \( \text{ZnCl}_2 \)
Answer: (a) Anhydrous AlCl3
In simple words: When hydrogen chloride is added to ethylene, a catalyst like anhydrous aluminum chloride (\( \text{AlCl}_3 \)) is used to make the reaction happen.

🎯 Exam Tip: Anhydrous \( \text{AlCl}_3 \) acts as a Lewis acid catalyst in electrophilic addition reactions like hydrohalogenation of alkenes.

 

Question 51. Which one of the following has the lowest boiling point?
(b) \( \text{C}_2\text{H}_5\text{Cl} \)
Answer: (a) CH3Cl
In simple words: Methyl chloride (\( \text{CH}_3\text{Cl} \)) has the lowest boiling point because it is the smallest molecule among the options with a halogen, leading to weaker intermolecular forces.

🎯 Exam Tip: Boiling points generally increase with increasing molecular weight and surface area for alkyl halides due to stronger London dispersion forces.

 

Question 52. Chloroethane is reacted with alcoholic potassium hydroxide. The product formed is
(a) \( \text{C}_2\text{H}_6\text{O} \)
(b) \( \text{C}_2\text{H}_6 \)
(c) \( \text{C}_2\text{H}_4 \)
(d) \( \text{C}_2\text{H}_4\text{O} \)
Answer: (c) C2H4
In simple words: When chloroethane reacts with alcoholic potassium hydroxide, it loses a hydrogen and a chlorine atom to form ethylene (\( \text{C}_2\text{H}_4 \)), which is an alkene.

🎯 Exam Tip: Alcoholic KOH is a strong base and favors elimination reactions (dehydrohalogenation) over substitution, leading to alkene formation.

 

Question 53. What is X in the following reaction? \( \text{C}_2\text{H}_5\text{Cl} + \text{X} \rightarrow \text{C}_2\text{H}_5\text{OH} + \text{KCl} \)
(a) \( \text{KHCO}_3 \)
(b) alc. KOH
(c) aq. KOH
(d) \( \text{K}_2\text{CO}_3 \)
Answer: (c) aq. KOH
In simple words: In the given reaction, \( \text{X} \) must be aqueous potassium hydroxide (\( \text{aq. KOH} \)) because it replaces the chlorine atom with an -OH group, forming ethanol.

🎯 Exam Tip: Aqueous KOH promotes nucleophilic substitution reactions (to form alcohols), while alcoholic KOH promotes elimination reactions (to form alkenes).

 

Question 54. Which of the following acids will give maximum yield of alkyl chloride in Hunsdiecker reaction
(a) \( \text{CH}_3\text{CH}_2\text{CH}_2\text{COOH} \)
(b) \( (\text{CH}_3)_2\text{CHCOOH} \)
(d) \( \text{C}_6\text{H}_5\text{CH} (\text{CH}_3)\text{COOH} \)
Answer: (a) CH3CH2CH2COOH
In simple words: Butanoic acid (\( \text{CH}_3\text{CH}_2\text{CH}_2\text{COOH} \)) will produce the most alkyl chloride in a Hunsdiecker reaction because primary alkyl halides are generally formed with good yield.

🎯 Exam Tip: The Hunsdiecker reaction is a free radical reaction that converts silver salts of carboxylic acids into alkyl halides, generally favoring primary alkyl halides for better yields.

 

Question 55. In the reaction sequence \( \text{C}_2\text{H}_5\text{Cl} + \text{KCN} \rightarrow \text{X} \). What is the molecular formula of X is
(a) \( \text{C}_2\text{H}_5\text{CN} \)
(b) \( \text{C}_2\text{H}_5\text{NC} \)
(c) \( \text{C}_2\text{H}_5\text{OH} \)
(d) \( \text{C}_2\text{H}_4\text{O} \)
Answer: (a) C2H5CN
In simple words: When ethyl chloride reacts with potassium cyanide (\( \text{KCN} \)), the product formed, \( \text{X} \), is ethyl cyanide (\( \text{C}_2\text{H}_5\text{CN} \)), where the chlorine is replaced by the cyanide group.

🎯 Exam Tip: KCN is an ionic compound, so the attack happens through the carbon atom, leading to the formation of alkyl cyanides.

 

Question 56. Ethyl chloride on heating with silver cyanide forms a compound X. The functional isomer of X is
(a) \( \text{C}_2\text{H}_5\text{NC} \)
(b) \( \text{C}_2\text{H}_5\text{NCN} \)
(c) \( \text{H}_3\text{C} – \text{NH} – \text{CH}_3 \)
(d) \( \text{C}_2\text{H}_5\text{NH}_2 \)
Answer: (b) C2H5NCN
In simple words: When ethyl chloride reacts with silver cyanide, it forms ethyl isocyanide (\( \text{C}_2\text{H}_5\text{NC} \)). The functional isomer for ethyl isocyanide is ethyl cyanide (\( \text{C}_2\text{H}_5\text{NCN} \)), meaning it has the same molecular formula but a different arrangement of atoms and a different functional group.

🎯 Exam Tip: Silver cyanide (AgCN) is a covalent compound, so the attack by the nitrogen atom produces alkyl isocyanides (R-NC) which are functional isomers of alkyl cyanides (R-CN).

 

Question 57. With Zn – Cu couple and \( \text{C}_2\text{H}_5\text{OH} \), ethyl Iodide reacts to give
(a) ethers
(b) diethyl ether
(d) Ethane
Answer: (d) Ethane
In simple words: When ethyl iodide is treated with a mixture of zinc and copper in ethanol, it undergoes a reduction reaction to form ethane.

🎯 Exam Tip: The Zn-Cu couple in ethanol acts as a reducing agent, converting alkyl halides into alkanes by replacing the halogen with hydrogen.

 

Question 58. Ethyl bromide on boiling with alcoholic solutions of sodium hydroxide forms
(a) Ethane
(b) ethylene
(c) ethyl alcohol
(d) all of these
Answer: (b) ethylene
In simple words: When ethyl bromide is boiled with a solution of sodium hydroxide in alcohol, it undergoes an elimination reaction to form ethylene.

🎯 Exam Tip: Remember that alcoholic NaOH (or KOH) favors elimination (dehydrohalogenation) over substitution, especially with primary alkyl halides like ethyl bromide.

 

Question 59. Following major compound is formed when ethyl chloride reacts with silver nitrite
(a) Nitro ethane
(b) Ethyl nitrite
(c) Ethylene
(d) Acetaldehyde
Answer: (b) Ethyl nitrite
In simple words: When ethyl chloride reacts with silver nitrite, the main product formed is ethyl nitrite, as the nitrogen end of the nitrite ion attacks the carbon.

🎯 Exam Tip: Silver nitrite (\( \text{AgNO}_2 \)) is a covalent compound, so the attacking nucleophile is the oxygen atom, forming alkyl nitrites (R-O-N=O). In contrast, \( \text{KNO}_2 \) is ionic and forms nitroalkanes (R-\( \text{NO}_2 \)).

 

Question 60. Which of the following represents Williamson's synthesis?
(a) \( \text{CH}_3\text{COOH} + \text{PCl}_3 \rightarrow \)
(b) \( \text{CH}_3\text{–CH}_2 – \text{Cl} + \text{CH}_3\text{COOH} \rightarrow \)
(c) \( \text{CH}_3\text{–CH}_2\text{– ONa} + \text{CH}_3 – \text{CH}_2 – \text{Cl} \rightarrow \)
(d) \( \text{CH}_3 – \text{CH}_2 – \text{OH} + \text{Na} \rightarrow \)
Answer: (c) CH3–CH2– ONa + CH3 – CH2 – Cl →
In simple words: Williamson's synthesis is a way to make ethers by reacting an alkoxide (like sodium ethoxide) with an alkyl halide (like ethyl chloride). This reaction mixes two parts to create a new ether.

🎯 Exam Tip: The key reactants for Williamson's synthesis are an alkoxide (a strong nucleophile) and a primary alkyl halide to minimize elimination side reactions.

 

Question 61. The reaction of alkyl halide with benzene in presence of anhydrous \( \text{AlCl}_3 \) gives alkyl benzene the reaction is known as
(a) Friedel - craft's reaction
(b) Carbylamine reaction
(c) Gattermann reaction
(d) Wurtz reaction
Answer: (a) Friedel - craft's reaction
In simple words: When an alkyl halide and benzene react together with anhydrous aluminum chloride, it makes alkylbenzene, and this chemical process is called a Friedel-Crafts reaction.

🎯 Exam Tip: Friedel-Crafts alkylation is an electrophilic aromatic substitution reaction where an alkyl group is added to an aromatic ring using a Lewis acid catalyst.

 

Question 62. A Grignard's reagent reacts with water to give
(a) ether
(b) alkanes
(c) amine
(d) Alcohol
Answer: (b) alkanes
In simple words: When a Grignard reagent, which is very reactive, mixes with water, it quickly forms an alkane, replacing the magnesium part with hydrogen.

🎯 Exam Tip: Grignard reagents are strong bases and nucleophiles; they react readily with any source of active hydrogen (like water, alcohols, or amines) to produce hydrocarbons.

 

Question 63. \( \text{C}_2\text{H}_5\text{Cl} + \text{Mg} \rightarrow \text{C}_2\text{H}_5 \text{MgCl} \) in this reaction the solvent is
(a) \( \text{C}_2\text{H}_5\text{OH} \)
(b) Water
(c) Dry ether
(d) Acetone
Answer: (c) Dry ether
In simple words: To make a Grignard reagent like \( \text{C}_2\text{H}_5\text{MgCl} \) from ethyl chloride and magnesium, you must use dry ether as the solvent.

🎯 Exam Tip: Anhydrous (dry) ether is essential for Grignard reactions because Grignard reagents are highly reactive with moisture or any acidic protons.

 

Question 64. \( \text{Br} > \text{RCl} > \text{RF} \) then the reaction could be
Answer: The question is incomplete as the full context or a reaction type for the reactivity order is missing. Please provide the complete question for a proper answer.
In simple words: This question is not finished, so it is impossible to give a correct answer. More details are needed about what reaction is being talked about to understand the order of how quickly these things react.

🎯 Exam Tip: Always read questions carefully to ensure all necessary information is present. If a question appears incomplete, clarify what's missing or note the ambiguity.

 

Question 65. \( \text{S}_{\text{N}}2 \) reaction leads to
(a) inversion of configuration
(b) retention of configuration
(c) partial racemisation
(d) no racemisation
Answer: (a) inversion of configuration
In simple words: In an \( \text{S}_{\text{N}}2 \) reaction, the shape of the molecule flips like an umbrella turning inside out, leading to an inversion of its arrangement.

🎯 Exam Tip: \( \text{S}_{\text{N}}2 \) reactions are concerted, meaning bond breaking and bond forming happen simultaneously, leading to a backside attack and complete inversion of configuration at the chiral center.

 

Question 66. Which of the following alkyl halide is hydrolysed by \( \text{S}_{\text{N}}1 \) mechanism
(a) \( \text{CH}_3\text{Cl} \)
(b) \( \text{CH}_3 – \text{CH}_2 – \text{Cl} \)
(c) \( \text{CH}_3 – \text{CH}_2 – \text{CH}_2 – \text{Cl} \)
(d) \( (\text{CH}_3)_3\text{CCl} \)
Answer: (d) (CH3)3CCl
In simple words: Among the choices, tert-butyl chloride (\( (\text{CH}_3)_3\text{CCl} \)) will break down using the \( \text{S}_{\text{N}}1 \) mechanism because it forms a stable carbocation, which is important for this type of reaction.

🎯 Exam Tip: \( \text{S}_{\text{N}}1 \) reactions favor tertiary alkyl halides due to the stability of the intermediate carbocation, which is the rate-determining step.

 

Question 67. \( \text{S}_{\text{N}}1 \) reaction is favoured by
(b) Bulky group on the carbon atom attached to the halogen atom
Answer: (b) Bulky group on the carbon atom attached to the halogen atom
In simple words: \( \text{S}_{\text{N}}1 \) reactions work best when the carbon atom linked to the halogen has large, bulky groups around it. This helps to stabilize the carbocation formed during the reaction.

🎯 Exam Tip: The bulky groups around the carbon attached to the halogen enhance the stability of the carbocation intermediate, which is characteristic of \( \text{S}_{\text{N}}1 \) reactions. Also, polar protic solvents favor \( \text{S}_{\text{N}}1 \).

 

Question 68. Which of the following is not stereospecific
(a) \( \text{S}_{\text{N}}1 \)
(b) \( \text{S}_{\text{N}}2 \)
(c) \( \text{E}2 \)
(d) Addition of \( \text{Br}_2 \) to ethylene in \( \text{CCl}_4 \)
Answer: (a) SN1
In simple words: The \( \text{S}_{\text{N}}1 \) reaction is not stereospecific because it creates an intermediate that can be attacked from both sides, leading to a mix of different spatial arrangements.

🎯 Exam Tip: \( \text{S}_{\text{N}}1 \) reactions typically result in racemization (formation of both enantiomers) if the carbon center is chiral, hence they are not stereospecific. \( \text{S}_{\text{N}}2 \) and \( \text{E}2 \) are stereospecific, and electrophilic addition of \( \text{Br}_2 \) to alkenes is also stereospecific (anti-addition).

 

Question 69. Which of the following factors does not favour \( \text{S}_{\text{N}}1 \) mechanism
(a) Strong base
(b) Polar solvent
(c) Low. conc. of nucleophile
(d) \( 3^{\circ} \) halide
Answer: (c) Low. conc. of nucleophile
In simple words: A strong base does not help an \( \text{S}_{\text{N}}1 \) reaction because it encourages a different type of reaction called elimination, instead of the substitution that \( \text{S}_{\text{N}}1 \) represents.

🎯 Exam Tip: Strong bases favor E2 elimination reactions. \( \text{S}_{\text{N}}1 \) reactions prefer polar protic solvents, weak nucleophiles (or low concentration of nucleophiles), and tertiary alkyl halides.

 

Question 70. The order of reactivity of various alkyl halides toward \( \text{S}_{\text{N}}1 \) reaction is
(a) \( 3^{\circ} > 2^{\circ} > 1^{\circ} \)
(b) \( 1^{\circ} > 2^{\circ} > 3^{\circ} \)
(c) \( 3^{\circ} = 2^{\circ} = 1^{\circ} \)
(d) \( 1^{\circ} > 3^{\circ} > 2^{\circ} \)
Answer: (a) 3° > 2° > 1°
In simple words: For \( \text{S}_{\text{N}}1 \) reactions, tertiary alkyl halides react the fastest, followed by secondary, and then primary alkyl halides react the slowest. This is because tertiary carbocations are the most stable.

🎯 Exam Tip: The reactivity in \( \text{S}_{\text{N}}1 \) reactions is directly proportional to the stability of the carbocation intermediate formed, with tertiary carbocations being the most stable.

 

Question 71. In aryl halides carbon atom holding halogen is
(a) sp² hybridised
(b) sp hybridised
(c) sp³ hybndised
(d) sp³d hybridised
Answer: (a) sp² hybridised
In simple words: When a halogen atom is attached to a carbon atom in an aryl halide, that carbon atom uses a special type of bonding called sp² hybridization. This makes the bond stronger and more stable.

🎯 Exam Tip: Remember that sp² hybridization in aryl halides gives partial double bond character to the C-X bond, affecting its reactivity.

 

Question 72. Chloro benzene can be prepared by reacting benzene diazonium chloride with
(a) HCl
(b) Cu2Cl2 / HCl
(c) Cl2 / AlCl3
(d) HNO2
Answer: (b) Cu2Cl2 / HCl
In simple words: To make chlorobenzene from benzene diazonium chloride, you need to use a special mixture of copper(I) chloride and hydrochloric acid. This reaction is a common way to replace the diazonium group with a chlorine atom.

🎯 Exam Tip: Identify this as a Sandmeyer reaction, where a diazonium salt is converted to an aryl halide using copper(I) salts.

 

Question 73. \( \text{Benzene} + \text{Cl}_2 \xrightarrow{\text{FeCl}_3} \text{X} \), X is
(a) Chloro benzene
(b) m - dichloro benzene
(c) benzene hexachioride
(d) p – dichlorobenzene
Answer: (a) Chloro benzene
In simple words: When benzene reacts with chlorine in the presence of iron(III) chloride, it mainly produces chlorobenzene. This reaction is called electrophilic substitution, where one hydrogen atom on the benzene ring is replaced by a chlorine atom.

🎯 Exam Tip: Recall that anhydrous ferric chloride (\( \text{FeCl}_3 \)) acts as a Lewis acid catalyst in halogenation of benzene, activating chlorine for electrophilic attack.

 

Question 74. The following is an example of Sandmeyer reaction
(a) \( \text{C}_6\text{H}_5\text{N}_2^+ \text{Cl}^- \xrightarrow{\text{CuCl}_2/\text{HCl}} \text{C}_6\text{H}_5\text{Cl} \)
(b) \( \text{C}_6\text{H}_5\text{N}_2^+ \text{Cl}^- \xrightarrow{\text{H}_2\text{O}/\Delta} \text{C}_6\text{H}_5\text{OH} \)
(c) \( \text{C}_6\text{H}_5\text{N}_2^+ \text{Cl}^- \xrightarrow{\text{HBF}_4} \text{C}_6\text{H}_5\text{F} \)
(d) \( \text{C}_6\text{H}_5\text{N}_2^+ \text{Cl}^- \xrightarrow{\text{K}/\text{warm}} \text{C}_6\text{H}_5\text{Cl} \)
Answer: (a) \( \text{C}_6\text{H}_5\text{N}_2^+ \text{Cl}^- \xrightarrow{\text{CuCl}_2/\text{HCl}} \text{C}_6\text{H}_5\text{Cl} \)
In simple words: The Sandmeyer reaction specifically involves using copper(I) salts like \( \text{CuCl}_2 \) with HCl to replace the diazonium group with a halogen. This is a common method for making aryl halides from diazonium salts.

🎯 Exam Tip: Differentiate Sandmeyer reactions (using \( \text{Cu}_2\text{X}_2 \)) from Gattermann reactions (using Cu powder) and other diazonium salt reactions.

 

Question 75. Chlorobenzene on reaction with \( \text{CH}_3\text{Cl} \) in presence of \( \text{AlCl}_3 \) gives
(a) toulene
(b) m - chloro toulene
(c) only o - chloro toulene
(d) mixture of o – and p – chlorotoulene
Answer: (d) mixture of o – and p – chlorotoulene
In simple words: When chlorobenzene reacts with methyl chloride in the presence of aluminum chloride, it produces a mix of two products: ortho-chlorotoluene and para-chlorotoluene. This is an electrophilic substitution reaction, where the chlorine atom already on the ring directs the new methyl group to these specific positions.

🎯 Exam Tip: Remember that halogens are ortho-para directing but deactivating in electrophilic aromatic substitution reactions.

 

Question 76. \( \text{2C}_6\text{H}_5\text{Cl} + \text{2Na} \rightarrow \text{X} \), X is
(a) toulene
(b) biphenyl
(c) phenyl ethane
(d) 1 – chloro – 2 – phenyl ethane
Answer: (b) biphenyl
In simple words: When two molecules of chlorobenzene react with sodium metal, they join together to form biphenyl. This is a special coupling reaction known as the Fittig reaction, where the halogen atoms are removed, and the two phenyl rings connect.

🎯 Exam Tip: Recognize this as a Fittig reaction, a variation of the Wurtz reaction, specifically for coupling two aryl halides.

 

Question 77. Chlorobenzene on fusing with solid NaOH gives
(a) benzene
(b) benzoic acid
(c) phenol
(d) benzene chloride
Answer: (c) phenol
In simple words: When chlorobenzene is melted with solid sodium hydroxide (NaOH) at high temperatures, it produces phenol. This reaction replaces the chlorine atom with a hydroxyl group, which is a key way to make phenols.

🎯 Exam Tip: This is Dow's process, a commercial method for synthesizing phenol from chlorobenzene under harsh conditions.

 

Question 78. Chlorobenzene on nitration gives major product of
(a) 1 - chloro – 4 – nitro benzene
(b) 1 - chloro – 3 – nitro benzene
(c) 1, 4 - dinitro benzene
(d) 2, 4, 6 – tri nitro benzene
Answer: (a) 1 - chloro – 4 – nitro benzene
In simple words: When chlorobenzene undergoes nitration, the main product formed is 1-chloro-4-nitrobenzene. The chlorine atom on the benzene ring guides the incoming nitro group to the para position, which is more favored than the ortho position due to less steric hindrance.

🎯 Exam Tip: Remember that halogens are ortho-para directing groups, and the para product is usually the major product due to less crowding around it.

 

Question 79. The reaction \( \text{C}_6\text{H}_5\text{I} + \text{2 Na} + \text{CH}_3\text{I} \rightarrow \text{C}_6\text{H}_5\text{CH}_3 + \text{2 NaI} \) is
(a) Wurtz reaction
(b) Fittig reaction
(c) Wurtz-Fittig reaction
(d) Sandmeyer reaction
Answer: (c) Wurtz-Fittig reaction
In simple words: This reaction where an aryl halide (like iodobenzene) and an alkyl halide (like methyl iodide) react with sodium metal to form an alkylated aromatic compound (like toluene) is known as the Wurtz-Fittig reaction. It combines features of both the Wurtz reaction and the Fittig reaction.

🎯 Exam Tip: The Wurtz-Fittig reaction links an aryl group to an alkyl group, which is distinct from Wurtz (alkyl-alkyl) and Fittig (aryl-aryl) reactions.

 

Question 80. \( \text{R-Cl} + \text{NaI} \xrightarrow{\text{Acetone}} \text{R-I} + \text{NaCl} \). This reaction is
(a) Fittig reaction
(b) Finkelstein reaction
(c) Frankland reaction
(d) None of the options
Answer: (b) Finkelstein reaction
In simple words: This reaction, where an alkyl chloride is converted into an alkyl iodide by reacting it with sodium iodide in acetone, is called the Finkelstein reaction. It's a useful way to switch one halogen for another.

🎯 Exam Tip: Recognize that the Finkelstein reaction uses NaI in acetone, relying on the insolubility of NaCl in acetone to drive the reaction forward.

 

Question 81. \( \text{C}_2\text{H}_5\text{OH} + \text{SOCl}_2 \xrightarrow{\text{Pyridine}} \text{x} + \text{y} + \text{z} \). In this reaction x, y and z respectively are
(a) \( \text{C}_2\text{H}_4\text{Cl}_2 \), \( \text{SO}_2 \), \( \text{HCl} \)
(b) \( \text{C}_2\text{H}_5\text{Cl} \), \( \text{SO}_2 \), \( \text{HCl} \)
(c) \( \text{C}_2\text{H}_5\text{Cl} \), \( \text{SOCl} \), \( \text{HCl} \)
(d) \( \text{C}_2\text{H}_4 \), \( \text{SO}_2 \), \( \text{Cl}_2 \)
Answer: (b) \( \text{C}_2\text{H}_5\text{Cl} \), \( \text{SO}_2 \), \( \text{HCl} \)
In simple words: When ethanol reacts with thionyl chloride in the presence of pyridine, it forms ethyl chloride, sulfur dioxide, and hydrogen chloride. This is a very good method for making alkyl chlorides because the gaseous byproducts (SO2 and HCl) easily escape, leaving behind a pure product.

🎯 Exam Tip: This is Darzen's process. The use of pyridine helps neutralize the HCl formed, preventing side reactions and increasing the yield of the alkyl halide.

 

Question 82. \( \text{C}_2\text{H}_5\text{Cl} + \text{AgOH} \rightarrow \text{A} + \text{AgCl} \).
\( \text{A} + \text{CH}_3\text{COCl} \rightarrow \text{C} + \text{HCl} \). “C” is
(a) Ethyl acetate
(b) Methyl acetate
(c) butanone – 2
(d) propanone
Answer: (a) Ethyl acetate
In simple words: First, ethyl chloride reacts with silver hydroxide to form ethyl alcohol (A). Then, ethyl alcohol (A) reacts with acetyl chloride to produce ethyl acetate (C). This is an esterification reaction where an alcohol and an acyl chloride form an ester.

🎯 Exam Tip: Remember that \( \text{AgOH} \) (moist silver oxide) effectively acts as aqueous KOH, converting alkyl halides to alcohols.

 

Question 83. The compound (B) in the below reaction is:
(a) ethylene chloride
(b) acetic acid
(c) propionic acid
(d) ethyl cyanide
Answer: (c) propionic acid
In simple words: Propionic acid is an organic acid. It has a three-carbon chain. This compound is often formed in reactions where a haloalkane is converted to a nitrile, and then the nitrile is hydrolyzed (reacted with water).

🎯 Exam Tip: To identify organic acids from reactions, trace the carbon chain length. Propionic acid means a three-carbon chain, often produced from a two-carbon starting material that gains a carbon via a cyanide reaction.

 

Question 84. Chloro ethane reacts with X to form diethyl ether. What is X?
(a) NaOH
(b) H2SO4
(c) C2H5ONa
(d) Na2S2O3
Answer: (c) C2H5ONa
In simple words: To make diethyl ether from chloroethane, you need to react it with sodium ethoxide (X). This is a classic example of Williamson's ether synthesis, where an alkoxide reacts with an alkyl halide to form an ether.

🎯 Exam Tip: Williamson's synthesis is a key method for preparing unsymmetrical ethers, involving an alkoxide and a primary alkyl halide.

 

Question 85. 1 – chlorobutane on reaction with alcoholic potash gives
(a) 1 – butene
(b) 1 – butanol
(c) 1 – butyne
(d) 2 – butanol
Answer: (a) 1 – butene
In simple words: When 1-chlorobutane reacts with alcoholic potash (potassium hydroxide dissolved in alcohol), it undergoes an elimination reaction to form 1-butene. This process removes the chlorine and a hydrogen atom, creating a double bond.

🎯 Exam Tip: Alcoholic KOH favors elimination reactions (dehydrohalogenation) to form alkenes, while aqueous KOH favors substitution to form alcohols.

 

Question 86. Propane nitrile may be prepared by heating
(a) methyl iodide with KCN
(b) ethyl chloride with KCN
(c) Propyl chloride with KCN
(d) ethyl chloride with KCN
Answer: (d) ethyl chloride with KCN
In simple words: To make propane nitrile, you should heat ethyl chloride with potassium cyanide (KCN). When ethyl chloride (a two-carbon chain) reacts with KCN, the cyanide group adds an extra carbon, resulting in a three-carbon nitrile.

🎯 Exam Tip: Remember that the KCN reaction adds a carbon atom to the alkyl chain, so to get a propane nitrile (3 carbons), you need an ethyl halide (2 carbons) as a starting material.

 

Question 87. \( \text{CH}_3\text{CH} = \text{CH}_2 \xrightarrow{\text{HBr/peroxide}} \text{A} \xrightarrow{\text{aqKOH}} \text{B} \), B is
(a) propanol -2
(b) propanal – 1
(c) propanol – 1
(d) propanal – 2
Answer: (c) propanol – 1
In simple words: First, propene reacts with HBr in the presence of peroxide, following the anti-Markovnikov rule, to form 1-bromopropane (A). Then, 1-bromopropane (A) reacts with aqueous KOH to form 1-propanol (B). This is a substitution reaction where the bromine is replaced by a hydroxyl group.

🎯 Exam Tip: Recall that HBr in the presence of peroxide follows the anti-Markovnikov rule, leading to the bromine attaching to the less substituted carbon, and aqueous KOH promotes substitution to form alcohols.

 

Question 88. \( \text{Toluene} \xrightarrow{\text{Cl}_2/\text{hv}} \text{X} \xrightarrow{\text{aqKOH}} \text{Y} \), Y is
(a)
(b)
(c)
(d)
Answer: (a)
In simple words: First, toluene reacts with chlorine in the presence of light (hv) to form benzyl chloride (X). Then, benzyl chloride reacts with aqueous KOH to produce benzyl alcohol (Y). In this second step, the chlorine atom is replaced by a hydroxyl group.

🎯 Exam Tip: Free radical halogenation occurs at the benzylic position (the carbon atom directly attached to the benzene ring), and aqueous KOH converts alkyl halides to alcohols.

 

Question 89. The correct order of increasing boiling points is
(a) 1 - chloropropane < isopropylchloride < 1 – chlorobutane
(b) isopropylchloride < 1 – chloropropane < 1 - chlorobutane
(c) 1 – chlorobutane < isopropylchloride < 1 – chloropropane
(d) 1 - chlorobutane < 1 – chloropropane < isopropylchloride
Answer: (b) isopropylchloride < 1 – chloropropane < 1 - chlorobutane
In simple words: The boiling point of haloalkanes increases with the size of the carbon chain. For isomers with the same number of carbon atoms, branched chains (like isopropylchloride) have lower boiling points than straight chains (like 1-chloropropane) because they have less surface area for intermolecular forces. So, isopropylchloride is lowest, then 1-chloropropane, and 1-chlorobutane (with more carbons) is the highest.

🎯 Exam Tip: Boiling points increase with molecular weight (more carbons, more halogens) and decrease with branching due to reduced surface area for van der Waals forces.

 

Question 90. The correct order of decreasing \( \text{S}_{\text{N}^2} \) reactivity is
(a) \( \text{RCH}_2\text{X} > \text{R}_2\text{CHX} > \text{R}_3\text{CX} \)
(b) \( \text{RCH}_2\text{X} > \text{R}_3\text{CX} > \text{R}_2\text{CHX} \)
(c) \( \text{R}_2\text{CHX} > \text{R}_3\text{CX} > \text{RCH}_2\text{X} \)
(d) \( \text{R}_3\text{CX} > \text{R}_2\text{CHX} > \text{RCH}_2\text{X} \)
Answer: (a) \( \text{RCH}_2\text{X} > \text{R}_2\text{CHX} > \text{R}_3\text{CX} \)
In simple words: For \( \text{S}_{\text{N}^2} \) reactions, primary alkyl halides (like \( \text{RCH}_2\text{X} \)) react fastest, followed by secondary ( \( \text{R}_2\text{CHX} \) ), and then tertiary ( \( \text{R}_3\text{CX} \) ) react slowest. This is because \( \text{S}_{\text{N}^2} \) reactions require the nucleophile to attack from the backside, and steric hindrance (crowding around the carbon atom) makes this difficult for bulky tertiary halides.

🎯 Exam Tip: Remember that \( \text{S}_{\text{N}^2} \) reactivity is inversely proportional to steric hindrance around the carbon bearing the halogen. Primary alkyl halides have the least crowding, making them most reactive.

II. Very Short Question and Answer (2 Marks):

 

Question 1. What are haloalkanes? Give example.
Answer: Haloalkanes are organic compounds where one or more hydrogen atoms in an alkane are replaced by halogen atoms like fluorine, chlorine, bromine, or iodine. They are useful as solvents and in making other chemicals. Bromoethane is an example. Haloalkanes are represented by the general formula \( \text{R – X} \), where \( \text{R} \) is an alkyl group (\( \text{C}_{\text{n}}\text{H}_{2\text{n}} + 1 \)) and \( \text{X} \) is a halogen atom (F, Cl, Br, or I). They are classified as primary, secondary, or tertiary based on the carbon atom the halogen is attached to.

  H
  |
CH3-C-Br
  |
  H

Bromoethane (1º Haloalkane)
In simple words: Haloalkanes are organic compounds where a hydrogen atom in an alkane is replaced by a halogen. Bromoethane is an example.

🎯 Exam Tip: When defining chemical terms, always include the general formula and a simple, clear example to demonstrate understanding.

 

Question 2. How will you convert methane into tetra chloro methane?
Answer: Methane can be changed into tetrachloromethane by adding chlorine gas to it, usually with light. This replaces one hydrogen with chlorine, then another, and so on, until all hydrogens are gone. Each step creates a new chlorinated compound. The intermediate products have different boiling points, allowing them to be separated by fractional distillation. \( \text{CH}_4 \xrightarrow{\text{Cl}_2/\text{light}} \text{CH}_3\text{Cl} + \text{HCl} \) (Methane to Chloromethane)
\( \text{CH}_3\text{Cl} \xrightarrow{\text{Cl}_2/\text{light}} \text{CH}_2\text{Cl}_2 + \text{HCl} \) (Chloromethane to Dichloromethane)
\( \text{CH}_2\text{Cl}_2 \xrightarrow{\text{Cl}_2/\text{light}} \text{CHCl}_3 + \text{HCl} \) (Dichloromethane to Trichloro methane)
\( \text{CHCl}_3 \xrightarrow{\text{Cl}_2/\text{light}} \text{CCl}_4 + \text{HCl} \) (Trichloro methane to Tetrachloro methane)
In simple words: Methane is converted to tetrachloromethane by successively replacing its hydrogen atoms with chlorine atoms using light.

🎯 Exam Tip: Remember that direct chlorination of alkanes is a free radical substitution reaction that can produce a mixture of products; controlling conditions helps favor specific ones.

 

Question 3. What is Finkelstein reaction?
Answer: The Finkelstein reaction helps make alkyl iodides from alkyl chlorides or bromides. You mix the starting compound with sodium iodide in acetone and heat it. The key is that a salt, like sodium chloride, comes out of the solution, which pushes the reaction forward to make the iodide. This method is useful for efficient halogen exchange. \( \text{CH}_3\text{CH}_2\text{Br} + \text{NaI} \xrightarrow{\text{Acetone}/\Delta} \text{CH}_3\text{CH}_2\text{I} + \text{NaBr} \)
(Bromoethane Iodoethane)
In simple words: It's a reaction to make alkyl iodides from alkyl chlorides or bromides by heating them with sodium iodide in acetone.

🎯 Exam Tip: The solubility of sodium halides (NaCl, NaBr) in acetone is crucial for the Finkelstein reaction; NaI is soluble, while NaCl and NaBr are not, allowing precipitation to drive the reaction.

 

Question 4. What is Swartz reaction?
Answer: The Swarts reaction is a way to make compounds that have fluorine atoms. You take a chloro or bromo alkane and heat it with a metal fluoride, like silver fluoride. This swaps the chlorine or bromine atom for a fluorine atom. This is a specific and important method for introducing fluorine into an alkyl chain. \( \text{CH}_3\text{CH}_2\text{Br} + \text{AgF} \xrightarrow{\Delta} \text{CH}_3\text{CH}_2\text{F} + \text{AgBr} \)
(Bromoethane Fluoroethane)
In simple words: It's a chemical reaction that creates fluoroalkanes by swapping a chlorine or bromine atom with a fluorine atom using metal fluorides.

🎯 Exam Tip: Remember the Swarts reaction is specifically for synthesizing fluoroalkanes, which are generally difficult to prepare by other halogen exchange methods.

 

Question 5. What is Hunsdiecker reaction?
Answer: The Hunsdiecker reaction takes the silver salt of a carboxylic acid and turns it into a bromoalkane. You react it with bromine in carbon tetrachloride. This reaction is special because it also shortens the carbon chain by one carbon atom. It is a key method for preparing alkyl bromides and reducing carbon chain length. \( \text{CH}_3\text{CH}_2\text{COOAg} + \text{Br}_2 \xrightarrow{\text{CCl}_4/\text{reflux}} \text{CH}_3\text{CH}_2\text{Br} + \text{CO}_2 + \text{AgBr} \)
(Silver propionate Bromo ethane)
In simple words: This reaction changes a silver salt of a carboxylic acid into a bromoalkane, making the carbon chain one carbon shorter.

🎯 Exam Tip: The Hunsdiecker reaction is a decarboxylation-halogenation process; it's useful for descending a homologous series (reducing carbon count by one) while simultaneously introducing bromine.

 

Question 6. How is ethyl magnesium bromide prepared from ethyl bromide?
Answer: Ethyl magnesium bromide is made by mixing ethyl bromide with magnesium metal. This reaction must happen in a completely dry liquid, like dry ether, because any water would stop it. The magnesium then slips in between the ethyl part and the bromine. This compound is known as a Grignard reagent, which is very useful in organic synthesis. \( \text{CH}_3\text{CH}_2\text{-Br} + \text{Mg} \xrightarrow{\text{Dry ether}} \text{CH}_3\text{CH}_2\text{MgBr} \)
In simple words: Ethyl magnesium bromide is prepared by reacting ethyl bromide with magnesium metal in a dry ether solvent.

🎯 Exam Tip: Emphasize the absolute necessity of a dry ether solvent for Grignard reagent preparation due to their extreme reactivity with protic solvents (like water or alcohol).

 

Question 7. How will you convert ethyl bromide into ethyl lithium?
Answer: You can turn ethyl bromide into ethyl lithium by reacting it with lithium metal in a dry ether. The lithium replaces the bromine atom. It's important to keep everything very dry because these compounds react quickly with water. Organolithium compounds are powerful reagents in chemical synthesis. \( \text{CH}_3\text{CH}_2\text{Br} + \text{2Li} \xrightarrow{\text{Dry ether}} \text{CH}_3\text{CH}_2\text{Li} + \text{LiBr} \)
(Ethyl bromide Ethyl Lithium)
In simple words: Ethyl bromide is converted to ethyl lithium by reacting it with lithium metal in a dry ether solvent.

🎯 Exam Tip: Organolithium reagents are strong bases and nucleophiles, similar to Grignard reagents, and require anhydrous conditions for synthesis and reaction.

 

Question 8. How is TEL prepared from ethyl bromide?
Answer: Tetraethyllead (TEL) is made from ethyl bromide by reacting it with a special mixture of sodium and lead metals. Four parts of ethyl bromide join with the metals to create TEL, along with other products. TEL was once used to improve gasoline. This organometallic synthesis was historically significant, although TEL's use has declined due to environmental concerns. \( \text{4CH}_3\text{CH}_2\text{Br} + \text{4Na/Pb} \rightarrow (\text{CH}_3\text{CH}_2)_4\text{Pb} + \text{4NaBr} + \text{3Pb} \)
(Ethyl bromide Sodium-lead alloy Tetra ethyl lead (TEL))
In simple words: TEL is prepared by reacting ethyl bromide with a sodium-lead alloy, leading to the formation of an organolead compound.

🎯 Exam Tip: This is a classic example of preparing an organometallic compound where alkyl groups are directly bonded to a metal, highlighting lead's historical role in gasoline additives.

 

Question 9. What are organic metallic compounds? Give example.
Answer: Organometallic compounds are special chemicals that have a direct bond between a carbon atom and a metal atom. They are very useful in making new chemicals and helping reactions happen faster. Methyl magnesium iodide is one example. These compounds are important catalysts and reagents in various industrial and laboratory syntheses. \( \text{CH}_3\text{MgI} \) – Methyl magnesium iodide
\( \text{CH}_3\text{CH}_2\text{MgBr} \) – Ethyl magnesium bromide
In simple words: Organometallic compounds are molecules with at least one direct bond between carbon and a metal atom. Methyl magnesium iodide is an example.

🎯 Exam Tip: When defining organometallic compounds, always specify the direct carbon-metal bond as the defining characteristic and provide an example like a Grignard reagent.

 

Question 10. How is methane prepared from Grignard reagent?
Answer: Methane can be made from a Grignard reagent, like methyl magnesium iodide. You mix it with compounds that have easily removed hydrogen atoms, such as alcohol. The Grignard reagent takes the hydrogen and forms methane. This is a common method for synthesizing hydrocarbons from these highly reactive reagents. \( \text{CH}_3\text{MgI} + \text{C}_2\text{H}_5\text{OH} \xrightarrow{\Delta} \text{CH}_4 + \text{MgI}(\text{OC}_2\text{H}_5) \)
(Methyl alcohol Methane)
In simple words: Methane is prepared by reacting a Grignard reagent (like methyl magnesium iodide) with any compound that has an active hydrogen, such as ethanol.

🎯 Exam Tip: Remember that Grignard reagents are strong bases and react readily with any source of active hydrogen (e.g., water, alcohols, carboxylic acids) to form corresponding alkanes.

 

Question 11. What are Haloarenes? Give a suitable example.
Answer: Haloarenes are compounds where a halogen atom, like chlorine, is directly joined to a carbon atom in a benzene ring. This direct link makes them special. Chlorobenzene is a good example of a haloarene. The halogen is bonded to an \( \text{sp}^2 \) hybridized carbon of the aromatic system, giving them unique chemical properties compared to haloalkanes.

    Cl
   /  \
  C----C
 //    \\
C      C
 \\    //
  C----C
   \  /
    H

Chlorobenzene
In simple words: Haloarenes are organic compounds where a halogen atom is directly attached to an aromatic ring. Chlorobenzene is an example.

🎯 Exam Tip: The key difference between haloalkanes and haloarenes is the direct attachment of the halogen to an aromatic ring in haloarenes, leading to lower reactivity in nucleophilic substitution.

 

Question 12. How is chloro benzene prepared from benzene by direct halogenation?
Answer: To make chlorobenzene from benzene, you react benzene with chlorine gas. You need a catalyst, like iron(III) chloride, to help the reaction along. This process replaces one hydrogen on the benzene ring with a chlorine atom. This is an electrophilic aromatic substitution reaction. \[ \text{Benzene} + \text{Cl}_2 \xrightarrow{\text{FeCl}_3} \text{Chlorobenzene} + \text{HCl} \]
In simple words: Chlorobenzene is made by reacting benzene with chlorine gas, using an iron(III) chloride catalyst, which swaps a hydrogen for a chlorine atom.

🎯 Exam Tip: Recall that anhydrous ferric chloride acts as a Lewis acid catalyst, polarizing the \( \text{Cl-Cl} \) bond and generating a strong electrophile for aromatic substitution.

 

Question 13. Write a note on Sand Meyer reaction.
Answer: The Sandmeyer reaction is a way to make halogenated benzene compounds. You start with a special benzene salt called a diazonium salt. Then, you react it with a copper(I) salt, which swaps the diazonium group for a chlorine or bromine atom, and nitrogen gas bubbles away. This reaction is valuable for introducing halogen groups onto an aromatic ring. \[ \text{Benzene diazonium chloride} \xrightarrow{\text{Cu}_2\text{Cl}_2/\text{HCl (Sandmeyer reaction)}} \text{Chlorobenzene} + \text{N}_2 \]
In simple words: The Sandmeyer reaction converts aryl diazonium salts into aryl halides by reacting them with copper(I) salts, releasing nitrogen gas.

🎯 Exam Tip: The Sandmeyer reaction is crucial for synthesizing aryl halides from amines (via diazonium salts), as direct halogenation might not give the desired substitution pattern.

 

Question 14. How is Iodo benzene prepared from benzene diazonium chloride?
Answer: To make iodobenzene from benzene diazonium chloride, you just need to gently heat the diazonium salt with potassium iodide in water. The iodide takes the place of the diazonium group, and nitrogen gas is released. This reaction is simpler than preparing chloro- or bromo-benzene, as it does not require a copper catalyst. \( \text{C}_6\text{H}_5\text{N}_2\text{Cl} + \text{KI} \xrightarrow{\text{Warm}} \text{C}_6\text{H}_5\text{I} + \text{N}_2 + \text{KCl} \)
(Benzene diazonium chloride Iodobenzene)
In simple words: Iodobenzene is prepared by warming benzene diazonium chloride with an aqueous solution of potassium iodide, which replaces the diazonium group with iodine.

🎯 Exam Tip: Note that the synthesis of iodobenzene from diazonium salts does not require a copper(I) catalyst, unlike chlorobenzene and bromobenzene.

 

Question 15. What happens when ethylidene dichloride is treated with Zinc dust in methanol?
Answer: When ethylidene dichloride is mixed with zinc dust in methanol, the two chlorine atoms are removed. This creates a double bond, turning it into ethene (ethylene). Zinc helps by taking away the chlorine, in a process called dehalogenation. \( \text{CH}_3\text{CHCl}_2 + \text{Zn} \xrightarrow{\text{Methanol}} \text{CH}_2 = \text{CH}_2 + \text{ZnCl}_2 \)
(Ethylidene dichloride Ethylene)
In simple words: Ethylidene dichloride reacts with zinc dust in methanol to remove its chlorine atoms, forming ethene.

🎯 Exam Tip: Remember that zinc dust is a common reagent for dehalogenation of vicinal and geminal dihalides to form alkenes.

 

Question 16. How will you convert chloroform into methylene chloride?
Answer: You can change chloroform into methylene chloride by reducing it. This means adding hydrogen atoms and taking away chlorine atoms. One way to do this is by using zinc metal and hydrochloric acid. Another way is to use hydrogen with a nickel catalyst. This process is a controlled reduction. \[ \text{CHCl}_3 \xrightarrow{\text{Zn + HCl}} \text{CH}_2\text{Cl}_2 + \text{HCl} \]
\[ \text{CHCl}_3 \xrightarrow{\text{H}_2/\text{Ni}} \text{CH}_2\text{Cl}_2 + \text{HCl} \]
In simple words: Chloroform is converted to methylene chloride by reduction, replacing one chlorine atom with a hydrogen atom using reagents like zinc and HCl.

🎯 Exam Tip: Remember that controlled reduction of polyhalogenated alkanes can selectively replace halogens with hydrogen, creating less halogenated compounds.

 

Question 17. Write the chlorination reaction of methane.
Answer: When methane reacts with chlorine gas, usually with sunlight, the hydrogen atoms in methane are swapped with chlorine atoms. This can happen in steps, creating different chlorinated compounds like chloromethane and methylene chloride. This is a free radical substitution reaction that is initiated by light. \( \text{CH}_4 \xrightarrow{\text{Cl}_2/\text{hv}} \text{CH}_3\text{Cl} + \text{HCl} \)
\( \text{CH}_3\text{Cl} \xrightarrow{\text{Cl}_2/\text{hv}} \text{CH}_2\text{Cl}_2 + \text{HCl} \)
In simple words: Methane reacts with chlorine under light, replacing hydrogens with chlorines in a step-by-step process to form various chlorinated products.

🎯 Exam Tip: Recall that free radical halogenation of alkanes is a chain reaction involving initiation, propagation, and termination steps, and it typically yields a mixture of products.

 

Question 18. How is chloroform prepared from carbon tetrachloride?
Answer: Chloroform can be made from carbon tetrachloride by a process called reduction. You use iron powder and a weak acid, like hydrochloric acid, which helps to change one of the chlorine atoms into a hydrogen atom. This is a controlled reduction process, converting a highly chlorinated compound into a less chlorinated one. \( \text{CCl}_4 \text{ (carbon tetrachloride)} + \text{2[H]} \xrightarrow{\text{Fe/HCl}} \text{CHCl}_3 \text{ (chloroform)} + \text{HCl} \)
In simple words: Chloroform is prepared from carbon tetrachloride by reduction using iron powder and dilute hydrochloric acid, which replaces one chlorine with hydrogen.

🎯 Exam Tip: This is a selective reduction; iron/HCl is a milder reducing agent than Zn/HCl, allowing for the partial reduction of \( \text{CCl}_4 \) to \( \text{CHCl}_3 \).

III. Short Question and Answers (3 Marks):

 

Question 1. write IUPAC name of the following.
(i)

      CH3
      |
    CH2=C-CH2-Cl

(ii)

        H
      /
    C=C
  H3C   CH3
          \
          CH-I
          |
          CH3

(iii)

    H3C   H
      C=C
    H   CH-F
            |
            CH3

Answer:
(i) The structure is \( \text{CH}_2=\text{C}(\text{CH}_3)\text{-CH}_2\text{-Cl} \). The longest chain with the double bond and chlorine atom has three carbons. The double bond is between C1 and C2, a methyl group is on C2, and a chlorine atom is on C3. Therefore, the IUPAC name is 3-chloro-2-methylprop-1-ene (or 3-chloro-2-methylpropene).
(ii) The structure is \( \text{CH}_3\text{-CH=C(CH}_3\text{)-CH(I)-CH}_3 \). The longest carbon chain with the double bond is a pentene. Numbering from the left to give the double bond the lowest locant, it is at C2. A methyl group is at C3, and an iodine atom is at C4. Thus, the IUPAC name is 4-iodo-3-methylpent-2-ene.
(iii) The structure is \( \text{CH}_3\text{-CH=CH-CH(F)-CH}_3 \). The longest carbon chain containing the double bond is a pentene. The double bond is between C2 and C3. The fluorine atom is attached to C4. Thus, the IUPAC name is 4-fluoropent-2-ene.
In simple words: (i) This is 3-chloro-2-methylpropene. (ii) This is 4-iodo-3-methylpent-2-ene. (iii) This is 4-fluoropent-2-ene.

🎯 Exam Tip: When naming alkenes with substituents, prioritize the double bond for numbering, then assign numbers to substituents to get the lowest possible set of locants. For complex structures, always identify the longest chain first.

 

Question 2. Write the structure of the following compounds.
(i) 1 – Bromo – 4 – ethyl cyclohexane
(ii) 1, 4 – Dichlorobut – 2 – ene
(iii) 2 – chloro – 3 – methyl pentane
Answer:
(i) For 1 – Bromo – 4 – ethyl cyclohexane:

       Br
      /
    CH2
   /   \
CH2     CH
|       |
CH2     CH2
   \   /
    CH2
     |
     CH2CH3

This compound has a cyclohexane ring with a bromine atom at position 1 and an ethyl group (C2H5) at position 4.
(ii) For 1, 4 – Dichlorobut – 2 – ene: \( \text{Cl-CH}_2\text{-CH=CH-CH}_2\text{-Cl} \)
This is a four-carbon chain with a double bond between the second and third carbons, and chlorine atoms on the first and fourth carbons.
(iii) For 2 – chloro – 3 – methyl pentane: \( \text{CH}_3\text{-CH(Cl)-CH(CH}_3\text{)-CH}_2\text{-CH}_3 \)
This is a five-carbon saturated chain (pentane) with a chlorine atom on the second carbon and a methyl group (CH3) on the third carbon.
In simple words: (i) Draw a six-carbon ring, put bromine on one carbon, and an ethyl group on the carbon four positions away. (ii) Draw four carbons, a double bond between the middle two, and chlorine atoms on the first and last carbons. (iii) Draw five carbons in a line, put chlorine on the second carbon, and a methyl group on the third carbon.

🎯 Exam Tip: When drawing structures from IUPAC names, start with the parent chain or ring, then add multiple bonds, and finally place the substituents correctly based on their locant numbers.

 

Question 3. Write any three methods of preparation of chloro ethane from ethanol.
Answer: Haloethanes can be made from ethanol using these methods:

1. Reaction with hydrogen halide: Ethanol reacts with hydrogen chloride (HCl) in the presence of anhydrous zinc chloride (\( ZnCl_2 \)) to produce ethyl chloride. This is a common way to replace the -OH group with a -Cl atom.

   \( CH_3CH_2OH + HCl \xrightarrow{Anhydrous\ ZnCl_2} CH_3CH_2Cl + H_2O \)

   This mixture of concentrated HCl and anhydrous \( ZnCl_2 \) is known as Lucas Reagent.

2. Reaction with phosphorous halides: Alcohols react with phosphorus pentachloride (\( PCl_5 \)) or phosphorus trichloride (\( PCl_3 \)) to form haloalkanes. These reagents are very effective at substituting the hydroxyl group with a halogen.

   \( CH_3CH_2OH + PCl_5 \rightarrow CH_3CH_2Cl + POCl_3 + HCl \)
   \( 3CH_3CH_2OH + PCl_3 \rightarrow 3CH_3CH_2Cl + H_3PO_3 \)

3. Reaction with thionyl chloride (Sulphonyl Chloride): When ethanol reacts with thionyl chloride (\( SOCl_2 \)) in the presence of pyridine, it forms chloroethane. This method is preferred because the by-products, sulfur dioxide (\( SO_2 \)) and hydrogen chloride (HCl), are gases and can easily escape, leaving a pure product.

   \( CH_3CH_2OH + SOCl_2 \xrightarrow{Pyridine} CH_3CH_2Cl + SO_2\uparrow + HCl\uparrow \)

In simple words: Chloroethane can be made from ethanol by reacting it with hydrogen chloride, phosphorus halides, or thionyl chloride. Each method replaces the -OH part of ethanol with a chlorine atom.

🎯 Exam Tip: When writing chemical equations, ensure all reactants, products, conditions (like heat or catalyst), and states (like \(\uparrow\) for gas) are correctly shown to score full marks.

 

Question 4. What happens when haloalkane reacts with aqueous alkali or most silver oxide?
Answer: When a haloalkane, such as ethyl bromide, reacts with an aqueous solution of potassium hydroxide (KOH) or moist silver oxide (\( Ag_2O/H_2O \)), it undergoes a nucleophilic substitution reaction to form an alcohol. In this reaction, the halogen atom is replaced by a hydroxyl (-OH) group.

\( CH_3-CH_2-Br + KOH_{(aq)} \xrightarrow{boil} CH_3CH_2-OH + KBr \)
\( CH_3-CH_2-Br + AgOH_{(aq)} \xrightarrow{boil} CH_3CH_2-OH + AgBr \)

This is a fundamental reaction to convert alkyl halides into alcohols. The silver oxide reacts with water to form silver hydroxide (AgOH), which then acts as the nucleophile.

In simple words: When a halogen compound reacts with water-based alkali or wet silver oxide, the halogen gets replaced by an -OH group, making an alcohol.

🎯 Exam Tip: Remember that aqueous KOH leads to substitution (alcohol formation), while alcoholic KOH leads to elimination (alkene formation). Pay attention to the solvent used.

 

Question 5. Explain the ammonolysis reaction of bromo ethane.
Answer: Ammonolysis is the reaction where an alkyl halide (like bromoethane) reacts with ammonia to form an amine. When bromoethane reacts with an alcoholic solution of ammonia, it forms ethylamine. This reaction is a nucleophilic substitution, where the ammonia molecule acts as a nucleophile, replacing the bromine atom.

\( CH_3-CH_2-Br + H-NH_2 \rightarrow CH_3CH_2-NH_2 + HBr \)

However, if excess ethyl bromide is used, the ethylamine formed can further react with more ethyl bromide to produce secondary and tertiary amines, and even quaternary ammonium salts. This means that if you want to get just the primary amine, you need to use a large excess of ammonia. This is because the newly formed amine is also a nucleophile and can continue reacting.

\( CH_3CH_2NH_2 \xrightarrow{CH_3CH_2Br} (CH_3CH_2)_2NH \xrightarrow{CH_3CH_2Br} (CH_3CH_2)_3N \xrightarrow{CH_3CH_2Br} (CH_3CH_2)_4N^+Br^- \)

In simple words: Bromoethane reacts with ammonia to make ethylamine. If there's too much bromoethane, the ethylamine can react again to form other, more complex amine products.

🎯 Exam Tip: To selectively prepare primary amines via ammonolysis, always use ammonia in a large excess relative to the alkyl halide.

 

Question 6. What happens when bromoethane reacts with alcoholic KCN and alcoholic AgCN?
Answer: Bromoethane shows different behaviors when reacting with alcoholic KCN versus alcoholic AgCN due to the nature of the bonds in these cyanide reagents.

Reaction with alcoholic KCN:Potassium cyanide (KCN) is mostly ionic. This means it provides the cyanide ion (\( CN^- \)), which can attach to the carbon atom through either its carbon atom or its nitrogen atom. However, the carbon-carbon bond is stronger, so the attack primarily occurs through the carbon atom, leading to the formation of ethyl cyanide (alkanenitrile).

\( CH_3-CH_2-Br + KCN \rightarrow CH_3-CH_2-CN + KBr \)

Reaction with alcoholic AgCN:Silver cyanide (AgCN) is mostly covalent. This means the lone pair of electrons is available on the nitrogen atom for nucleophilic attack. Therefore, the attack occurs primarily through the nitrogen atom, leading to the formation of ethyl isocyanide (alkane isonitrile).

\( CH_3CH_2Br + AgCN \rightarrow CH_3CH_2NC + AgBr \)

In simple words: When bromoethane reacts with KCN, it forms ethyl cyanide because KCN is ionic and the carbon in CN makes the bond. When it reacts with AgCN, it forms ethyl isocyanide because AgCN is covalent, and the nitrogen in CN makes the bond.

🎯 Exam Tip: This difference in reactivity between KCN (ionic) and AgCN (covalent) is a key concept in organic chemistry and is often tested to understand the ambidentate nature of the cyanide ion.

 

Question 7. Write a note on the preparation of:
(i) Ethane thiol
(ii) Diethyl ether
Answer: Here's how ethane thiol and diethyl ether can be prepared:

(i) Ethane thiol: Ethyl bromide reacts with sodium hydrogen sulphide (NaSH) or potassium hydrogen sulphide (KSH) to form thio alcohols, also known as thiols. This is a nucleophilic substitution reaction where the -Br group is replaced by an -SH group.

\( CH_3CH_2Br + NaSH \xrightarrow{alcohol/H_2O} CH_3CH_2SH + NaBr \)

The \( SH^- \) ion acts as a strong nucleophile, displacing the bromide ion.

(ii) Diethyl ether: Ethyl bromide, when heated with sodium alkoxide (like sodium ethoxide), gives diethyl ether. This reaction is known as Williamson's synthesis. It is a common method for preparing symmetrical and unsymmetrical ethers.

\( CH_3-CH_2Br + NaOCH_2CH_3 \xrightarrow{\Delta} CH_3CH_2-O-CH_2CH_3 + NaBr \)

The ethoxide ion (\( CH_3CH_2O^- \)) is a strong nucleophile and attacks the carbon atom bearing the bromine, leading to the formation of the ether.

In simple words: Ethane thiol is made when ethyl bromide reacts with sodium hydrogen sulphide. Diethyl ether is made when ethyl bromide is heated with sodium ethoxide.

🎯 Exam Tip: Williamson's synthesis works best with a primary alkyl halide and a strong alkoxide for good yields of ethers.

 

Question 8. How is ethane prepared from the following compounds?
(i) Bromo ethane
(ii) Iodo ethane
Answer: Ethane can be prepared from bromoethane and iodoethane through reduction reactions:

(i) From bromoethane:Ethyl bromide can be reduced to ethane by treating it with hydrogen (\( H_2 \)) in the presence of a metal catalyst like nickel (Ni), palladium (Pd), or platinum (Pt). It can also be reduced using hydroiodic acid (HI) in the presence of red phosphorus.

\( CH_3CH_2Br + H_2 \xrightarrow{Ni(or)Pd,\ 523K} CH_3-CH_3 + HBr \)

The catalytic hydrogenation method is a common way to remove halogens from alkyl halides to yield alkanes. The catalyst helps break the C-Br bond and form a C-H bond.

(ii) From iodoethane:Similarly, iodoethane can be reduced to ethane by treating it with hydrogen (\( H_2 \)) in the presence of a metal catalyst (like Ni or Pd) or by reacting it with hydroiodic acid (HI) in the presence of red phosphorus. The presence of red phosphorus prevents the reverse reaction and ensures complete reduction.

\( CH_3CH_2I + HI \xrightarrow{Red\ P} CH_3-CH_3 + I_2 \)

In simple words: To make ethane from bromoethane or iodoethane, you need to remove the halogen atom and add a hydrogen atom. This can be done using hydrogen gas with a metal catalyst, or with hydroiodic acid and red phosphorus.

🎯 Exam Tip: Remember that red phosphorus helps regenerate HI from \( I_2 \), preventing the reverse reaction and ensuring a high yield of alkane in reduction reactions with HI.

 

Question 9. Write the uses of carbon tetra chloride.
Answer: Carbon tetrachloride (\( CCl_4 \)) has several industrial and practical uses:

1. It is used as a dry cleaning agent, especially for non-washable clothes, as it effectively dissolves grease and oil without damaging fabrics.
2. It serves as a solvent for various organic compounds like oils, fats, and waxes in laboratories and industries.
3. Due to its non-combustible nature, its vapor was historically used under the name "pyrene" for extinguishing oil or petrol fires. However, its use has significantly declined due to environmental and health concerns.
4. It was also used in the manufacturing of refrigerants (like Freon), propellants, and pesticides.

In simple words: Carbon tetrachloride is used to dry clean clothes, dissolve oils and fats, and was once used to put out certain fires.

🎯 Exam Tip: While \( CCl_4 \) had many uses, it's important to note its current restricted use due to its harmful effects on the ozone layer and human health.

 

Question 10. Write the uses of Grignard reagents.
Answer: Grignard reagents (RMgX) are very versatile and useful compounds in organic chemistry for creating new carbon-carbon bonds. Here are some of their main uses:

1. They are essential for synthesizing various organic compounds such as alcohols (primary, secondary, and tertiary), carboxylic acids, aldehydes, and ketones. By reacting Grignard reagents with different carbonyl compounds, a wide range of structures can be built.
2. The alkyl group (R) in a Grignard reagent is electron-rich and acts like a carbanion (a negatively charged carbon), making it a strong nucleophile. This property allows it to attack molecules that have areas of low electron density.
3. They are used to prepare alkanes when reacted with compounds containing active hydrogen atoms like water, alcohols, and amines.
4. They are also used in the synthesis of other organometallic compounds.

In simple words: Grignard reagents are very helpful in making many different organic chemicals, including alcohols, acids, aldehydes, and alkanes. They work by adding an alkyl group to other molecules.

🎯 Exam Tip: Always remember that Grignard reagents are highly reactive with even traces of moisture, so reactions involving them must be conducted under anhydrous (dry) conditions.

 

Question 11. What is Balz – Schiemann reaction?
Answer: The Balz-Schiemann reaction is a chemical process used to prepare fluorobenzene from benzene diazonium chloride. In this reaction, benzene diazonium chloride (\( C_6H_5N_2Cl \)) is treated with fluoroboric acid (\( HBF_4 \)), which forms benzene diazonium fluoroborate (\( C_6H_5N_2^+BF_4^- \)). This intermediate product is then heated, causing it to decompose and produce fluorobenzene, nitrogen gas (\( N_2 \)), and boron trifluoride (\( BF_3 \)).

The reaction steps are:
\( C_6H_5N_2Cl + HBF_4 \xrightarrow{-HCl} C_6H_5N_2^+BF_4^- \)
\( C_6H_5N_2^+BF_4^- \xrightarrow{Heat} C_6H_5F + BF_3 + N_2 \)

This reaction is particularly useful because direct fluorination of benzene is difficult and often yields unwanted by-products, whereas this method provides a cleaner route to fluorobenzene.

In simple words: The Balz-Schiemann reaction is a way to make fluorobenzene. You start with benzene diazonium chloride, add fluoroboric acid, and then heat it up to get fluorobenzene, a special type of chemical.

🎯 Exam Tip: This reaction is a key method for introducing fluorine atoms into aromatic rings, which is difficult to achieve by other direct methods.

 

Question 12. Write the uses of chloro benzene.
Answer: Chlorobenzene (\( C_6H_5Cl \)) is a versatile organic compound with several important applications:

• It is widely used in the manufacture of pesticides, such as DDT (dichlorodiphenyltrichloroethane), though the use of DDT has been restricted due to environmental concerns.
• It serves as a high-boiling solvent in various organic synthesis reactions, meaning it can be used in reactions that require higher temperatures.
• It is used as a fibre-swelling agent in textile processing, helping to prepare fibres for dyeing and other treatments.
• It is also a key intermediate in the production of other chemicals, including phenol, aniline, and chloronitrobenzene.

In simple words: Chlorobenzene is used to make pesticides, as a solvent in chemical reactions, and to help process textile fibres.

🎯 Exam Tip: Understand that while many compounds have useful applications, their environmental and health impacts can lead to restricted use, as seen with chlorobenzene's role in DDT production.

 

Question 13. How is ethylidene dichloride prepared from
(i) Acetaldehyde
(ii) Acetylene
Answer: Ethylidene dichloride (\( CH_3CHCl_2 \)) can be prepared from acetaldehyde and acetylene through different reactions:

(i) From Acetaldehyde (\( CH_3CHO \)):Ethylidene dichloride is prepared by treating acetaldehyde with phosphorus pentachloride (\( PCl_5 \)). In this reaction, the oxygen atom of the carbonyl group in acetaldehyde is replaced by two chlorine atoms.

\( CH_3CHO + PCl_5 \rightarrow CH_3CHCl_2 + POCl_3 \)

This method directly substitutes the oxygen with two chlorine atoms, making it a good way to form geminal dihalides.

(ii) From Acetylene (\( HC \equiv CH \)):Ethylidene dichloride is prepared by adding hydrogen chloride (HCl) to acetylene. This reaction occurs in two steps, as acetylene has two triple bonds that can react with HCl. First, vinyl chloride is formed, which then reacts with another molecule of HCl to form ethylidene dichloride.

\( HC \equiv CH + HCl \rightarrow CH_2=CHCl \)
\( CH_2=CHCl + HCl \rightarrow CH_3CHCl_2 \)

The addition of HCl to the double bond follows Markovnikov's rule, ensuring the chlorine atom attaches to the carbon with fewer hydrogen atoms (though in the second step, it adds to the carbon already having a chlorine).

In simple words: Ethylidene dichloride can be made from acetaldehyde by adding \( PCl_5 \). It can also be made from acetylene by adding two molecules of hydrogen chloride.

🎯 Exam Tip: Remember that adding \( PCl_5 \) to aldehydes replaces the carbonyl oxygen with two chlorine atoms, forming a geminal dihalide, while adding HCl to alkynes involves sequential addition following Markovnikov's rule.

 

Question 14. How is ethylene dichloride prepared from
(i) Chloride
(ii) PCI5?
Answer: Ethylene dichloride (\( CH_2Cl-CH_2Cl \)) can be prepared using chlorine addition or from ethylene glycol with \( PCl_5 \):

(i) Addition of chlorine to ethylene:Ethylene dichloride is commonly prepared by the direct addition of chlorine gas (\( Cl_2 \)) to ethylene (\( CH_2=CH_2 \)). This is an electrophilic addition reaction where the double bond of ethylene breaks, and a chlorine atom attaches to each carbon atom.

\( CH_2=CH_2 + Cl_2 \rightarrow CH_2Cl-CH_2Cl \)

This reaction is very efficient and provides a high yield of the vicinal dihalide (where halogens are on adjacent carbon atoms).

(ii) Action of \( PCl_5 \) (or HCl) on ethylene glycol:Ethylene dichloride can also be prepared by reacting ethylene glycol (\( CH_2OH-CH_2OH \)) with phosphorus pentachloride (\( PCl_5 \)). Both hydroxyl (-OH) groups in ethylene glycol are replaced by chlorine atoms.

\( CH_2OH-CH_2OH + 2PCl_5 \rightarrow CH_2Cl-CH_2Cl + 2POCl_3 + 2HCl \)

Alternatively, ethylene glycol can react with concentrated hydrogen chloride (HCl) under appropriate conditions to form ethylene dichloride, although the \( PCl_5 \) reaction is often more straightforward for this conversion.

In simple words: Ethylene dichloride is made by adding chlorine gas to ethylene, or by reacting ethylene glycol with phosphorus pentachloride. Both methods replace parts of the starting molecule with chlorine atoms.

🎯 Exam Tip: Distinguish between ethylene dichloride (vicinal dihalide, chlorines on adjacent carbons) and ethylidene dichloride (geminal dihalide, chlorines on the same carbon) in your reactions and IUPAC names.

 

Question 15. What happens when the following compounds are treated with alcoholic KOH?
(i) Ethylidene dichloride
(ii) Ethylene dichloride
Answer: When ethylidene dichloride and ethylene dichloride are treated with alcoholic KOH, they undergo elimination reactions to form different products:

(i) Ethylidene dichloride (\( CH_3CHCl_2 \)):When ethylidene dichloride is treated with alcoholic KOH, it undergoes dehydrohalogenation (removal of HCl) twice to form acetylene. This happens in two steps, where first one molecule of HCl is removed, and then another.

\( CH_3CHCl_2 + 2KOH \xrightarrow{Ethanol,\ \Delta} HC \equiv CH + 2KCl + 2H_2O \)

The strong base (KOH) in alcohol promotes the elimination of hydrogen and halogen atoms, leading to the formation of a triple bond.

(ii) Ethylene dichloride (\( CH_2Cl-CH_2Cl \)):When ethylene dichloride is treated with alcoholic KOH, it also undergoes dehydrohalogenation twice to form acetylene. Similar to ethylidene dichloride, two molecules of HCl are removed, resulting in the formation of a triple bond.

\( CH_2Cl-CH_2Cl + 2KOH \xrightarrow{Ethanol,\ \Delta} HC \equiv CH + 2KCl + 2H_2O \)

Both geminal (ethylidene dichloride) and vicinal (ethylene dichloride) dihalides, when reacted with alcoholic KOH, typically lead to the formation of alkynes (like acetylene) if sufficient hydrogen and halogen atoms are available for elimination.

In simple words: When both ethylidene dichloride and ethylene dichloride react with alcoholic KOH, they both lose their chlorine and some hydrogen atoms to become acetylene, which has a triple bond.

🎯 Exam Tip: Alcoholic KOH is a strong base that favors elimination reactions (dehydrohalogenation), which removes a hydrogen and a halogen from adjacent carbons to form double or triple bonds.

 

Question 16. Write the uses of methylene chloride.
Answer: Methylene chloride, also known as dichloromethane (\( CH_2Cl_2 \)), is a solvent with several important uses:

1. It is used as an aerosol spray propellant, helping to push out contents from spray cans due to its volatility.
2. It is a common solvent in paint removers because it can dissolve many organic compounds, effectively stripping paint from surfaces.
3. It is utilized as a process solvent in the manufacture of various drugs and pharmaceuticals.
4. It is used as a metal cleaning solvent, where it helps remove grease and grime from metal parts before further processing.
5. It is also used for decaffeinating coffee and tea and in the preparation of polyurethane foams.

In simple words: Methylene chloride is used in spray cans, to remove paint, to clean metals, and in making medicines.

🎯 Exam Tip: Remember that while \( CH_2Cl_2 \) is a good solvent, it has health risks, so it should be handled in well-ventilated areas or with proper safety equipment.

 

Question 17. How are the following compounds prepared from chlorobenzene?
(i) Benzene
(ii) Phenyl magnesium chloride
Answer: Benzene and phenyl magnesium chloride can be prepared from chlorobenzene through specific chemical reactions:

(i) Benzene (\( C_6H_6 \)):Chlorobenzene can be converted to benzene through a reduction reaction. When chlorobenzene is treated with a reducing agent such as a Ni-Al alloy in the presence of sodium hydroxide (NaOH), the chlorine atom is replaced by a hydrogen atom, yielding benzene.

\( C_6H_5Cl + 2[H] \xrightarrow{Ni-Al/NaOH} C_6H_6 + HCl \)

This method effectively removes the halogen from the aromatic ring.

(ii) Phenyl magnesium chloride (\( C_6H_5MgCl \)):Phenyl magnesium chloride is a Grignard reagent that can be prepared by reacting chlorobenzene with magnesium metal (Mg) in a dry ether solvent, typically tetrahydrofuran (THF).

\( C_6H_5Cl + Mg \xrightarrow{THF} C_6H_5MgCl \)

The dry ether solvent is crucial because Grignard reagents are highly reactive and react with moisture. This reaction forms an organometallic compound where magnesium inserts itself between the carbon and halogen bond.

In simple words: Benzene can be made from chlorobenzene by removing the chlorine atom and adding a hydrogen. Phenyl magnesium chloride is made by reacting chlorobenzene with magnesium metal in a dry solvent.

🎯 Exam Tip: Grignard reagent preparation must always be done under strictly anhydrous conditions because they are very sensitive to moisture, which would destroy the reagent.

 

Question 1. Explain the \( S_N2 \) mechanism of haloalkanes with suitable example.
Answer: The \( S_N2 \) reaction stands for bimolecular nucleophilic substitution. Here's what that means:

• "S" stands for Substitution.
• "N" stands for Nucleophilic.
• "2" stands for Bimolecular, which means two molecules are involved in the slowest, rate-determining step.

In an \( S_N2 \) reaction, the rate depends on the concentration of both the alkyl halide (haloalkane) and the nucleophile. This reaction follows second-order kinetics and happens in one single step. The reaction involves the formation of a "transition state" where both the reactant molecules are partly bonded to each other. The nucleophile attacks the carbon atom from the "backside" (opposite to where the halogen is leaving), and the halide ion leaves from the "front side" at the same time. This process causes the configuration of the carbon atom to "invert," much like an umbrella turning inside out in a strong wind. This inversion of configuration is called Walden inversion. Paul Walden first observed this phenomenon in \( S_N2 \) reactions.

Example: Reaction between chloromethane and aqueous KOH

In this example, the nucleophile \( OH^- \) attacks the carbon atom in chloromethane while the chlorine atom simultaneously leaves. This concerted process means there's no intermediate carbocation.

The \( S_N2 \) reaction always leads to an inversion of configuration at the asymmetric carbon center, if one is present.

\( OH^- + \text{H}_3\text{C}-\text{Cl} \xrightarrow{\text{Slow}} [\text{HO}\dots\text{CH}_3\dots\text{Cl}]^- \xrightarrow{\text{Fast}} \text{HO}-\text{CH}_3 + \text{Cl}^- \)

This can be visualized as:

In simple words: The \( S_N2 \) reaction happens in one step where a new group pushes into the carbon from one side, and the old group leaves from the other side. This flips the molecule like an umbrella turning inside out. It depends on both the starting chemical and the attacking group.

🎯 Exam Tip: Remember that primary alkyl halides prefer \( S_N2 \) reactions, while bulky alkyl halides (tertiary) avoid them due to steric hindrance (lack of space for the nucleophile to attack).

 

Question 2. Explain the \( S_N1 \) mechanism of haloalkanes with suitable examples.
Answer: The \( S_N1 \) reaction stands for unimolecular nucleophilic substitution. Here's what that means:

• "S" stands for Substitution.
• "N" stands for Nucleophilic.
• "1" stands for Unimolecular, which means only one molecule is involved in the slowest, rate-determining step.

The rate of an \( S_N1 \) reaction depends only on the concentration of the alkyl halide (RX) and is not affected by the concentration of the nucleophile (\( OH^- \)). This reaction follows first-order kinetics and occurs in two main steps.

Step 1: Formation of Carbocation (Slow Step)In the first step, the carbon-halogen bond (C-Br in the example) breaks slowly. The halogen atom leaves as a halide ion, forming a positively charged carbocation. This step is the slowest and therefore controls the overall rate of the reaction. Tertiary carbocations are more stable than secondary or primary carbocations, so tertiary alkyl halides react fastest via this mechanism.

Example: Reaction between tert-butyl bromide and aqueous KOH

\( (CH_3)_3C-Br \xrightarrow{\text{Slow}} (CH_3)_3C^+ + Br^- \)

Step 2: Nucleophilic Attack on Carbocation (Fast Step)The carbocation formed in the first step is very reactive and immediately reacts with the nucleophile (\( OH^- \)) in a fast step. Since the carbocation is planar, the nucleophile can attack from either the front side or the back side with equal ease. This leads to the formation of a product that is a mixture of two enantiomers (mirror-image isomers), resulting in a racemic mixture. If the starting alkyl halide is optically active, the product obtained will be optically inactive (racemic mixture).

\( (CH_3)_3C^+ + OH^- \xrightarrow{\text{Fast}} (CH_3)_3C-OH \)

Because the nucleophile can attack from both sides of the planar carbocation, it creates a mixture of molecules that rotate light in opposite directions, making the overall solution optically inactive. This is called racemisation.

Overall reaction:\( (CH_3)_3C-Br + OH^- \rightarrow (CH_3)_3C-OH + Br^- \)

Diagram for Step 1:

Diagram for Step 2:

In simple words: The \( S_N1 \) reaction happens in two steps. First, the old group leaves, creating a temporary positive carbon (carbocation). Second, a new group quickly attaches to this carbon from either side. This reaction rate only depends on the starting chemical, not on the attacking group.

🎯 Exam Tip: Remember that tertiary alkyl halides favor \( S_N1 \) reactions due to the stability of the tertiary carbocation intermediate, and \( S_N1 \) reactions typically result in racemization.

 

Question 3. Explain the E2 reaction mechanism with suitable example.
Answer: The E2 reaction stands for bimolecular elimination reaction. Here's what that means:

• "E" stands for Elimination.
• "2" stands for Bimolecular, meaning two molecules are involved in the rate-determining step.

The E2 reaction is a second-order reaction where the rate depends on both the concentration of the alkyl halide and the base. It is a one-step process where the removal of a proton from a beta (\( \beta \))-carbon and the expulsion of the halide ion from the alpha (\( \alpha \))-carbon happen at the same time. This creates a double bond (an alkene).

Primary alkyl halides generally undergo this reaction in the presence of an alcoholic solution of potassium hydroxide (KOH).

Mechanism:The base (like \( HO^- \)) removes a hydrogen atom from the \(\beta\)-carbon. At the same time, the electrons from the C-\(H_{\beta}\) bond move to form a double bond between the \(\alpha\) and \(\beta\) carbons, and the halogen atom leaves as a halide ion from the \(\alpha\)-carbon.

Example: Dehydrohalogenation of 1-chloropropane

\( CH_3-CH_2-CH_2-Cl + KOH_{(alcoholic)} \rightarrow CH_3-CH=CH_2 + H_2O + KCl \)

This reaction forms propene. The base pulls off a proton, and simultaneously, the chloride ion departs, leading to the formation of a pi bond. This concerted mechanism ensures that all bond-breaking and bond-forming events occur in a single transition state.

In simple words: E2 reaction is a single-step process where a base removes a hydrogen from one carbon and a halogen leaves from the next carbon at the same time. This creates a double bond.

🎯 Exam Tip: E2 reactions require a strong base and typically occur with primary and secondary alkyl halides, especially if they have an anti-periplanar arrangement of the leaving group and the \(\beta\)-hydrogen.

 

Question 4. Explain the E₁ reaction mechanism with suitable example.
Answer: The E1 reaction stands for unimolecular elimination reaction. Here's what that means:

• "E" stands for Elimination.
• "1" stands for Unimolecular, which means only one molecule is involved in the slowest, rate-determining step.

The E1 reaction is a first-order reaction, and its rate depends only on the concentration of the alkyl halide, not on the concentration of the base. This mechanism generally occurs with tertiary alkyl halides in the presence of an alcoholic KOH solution and takes place in two steps.

Step 1: Formation of Carbocation (Slow Step)Similar to the \( S_N1 \) mechanism, the first step involves the slow ionization of the alkyl halide to form a carbocation and a halide ion. This is the rate-determining step. Tertiary alkyl halides form stable carbocations, which is why they readily undergo E1 reactions.

Example: Dehydrohalogenation of tert-butyl chloride

\( (CH_3)_3C-Cl \xrightarrow{\text{Slow}} (CH_3)_3C^+ + Cl^- \)

Step 2: Elimination of a Proton (Fast Step)In the second, fast step, a base (from the alcoholic KOH) removes a proton from a \(\beta\)-carbon of the carbocation. The electrons from the C-\(H_{\beta}\) bond then move to form a double bond, resulting in an alkene. This process generates an alkene as the product.

\( (CH_3)_3C^+ \xrightarrow{\text{Base, Fast}} (CH_3)_2C=CH_2 + H^+ \)

The proton is usually removed from the carbon that leads to the most stable (most substituted) alkene, according to Zaitsev's rule. This mechanism is common for tertiary alkyl halides because their carbocations are stable.

Overall reaction:\( (CH_3)_3C-Cl \xrightarrow{alcoholic\ KOH} (CH_3)_2C=CH_2 + H_2O + KCl \)

The product formed is isobutylene (or 2-methylpropene).

Diagram for Step 1:

Diagram for Step 2:

In simple words: E1 reactions happen in two steps. First, the leaving group detaches to form a carbocation. Then, a base removes a hydrogen from a nearby carbon, creating a double bond. The speed of this reaction only depends on the starting molecule.

🎯 Exam Tip: E1 reactions are favored by tertiary alkyl halides, weak bases (like alcoholic KOH), and polar protic solvents, and often compete with \( S_N1 \) reactions, especially at lower temperatures.

 

Question 5. How are the following compounds prepared from Methyl magnesium iodide?
(i) Ethanol
(ii) Tert-butyl alcohol
(iii) Acetaldehyde
Answer:
(i) Ethanol: Formaldehyde reacts with methyl magnesium iodide. This reaction forms an addition product which then, after hydrolysis, gives ethanol. This is a common way to make primary alcohols.
\[\text{O=C=O} + \text{CH}_3\text{MgI} \rightarrow \text{CH}_3-\text{C}-\text{OMgI}\text{ (addition product)}\]
\[\text{CH}_3-\text{C}-\text{OMgI} \xrightarrow{\text{H}^{+}/\text{H}_2\text{O}} \text{CH}_3-\text{C}-\text{OH} + \text{Mg}\text{OHI}\text{ (ethanol)}\]
(ii) Tert-butyl alcohol: Acetone reacts with methyl magnesium iodide. This forms an addition product that, after hydrolysis, gives tert-butyl alcohol. This method is useful for making tertiary alcohols.
\[\text{CH}_3-\text{C}(=\text{O})-\text{CH}_3 + \text{CH}_3\text{MgI} \rightarrow \text{CH}_3-\text{C}(\text{OMgI})(\text{CH}_3)\text{CH}_3\text{ (addition product)}\]
\[\text{CH}_3-\text{C}(\text{OMgI})(\text{CH}_3)\text{CH}_3 \xrightarrow{\text{H}_2\text{O}/\text{H}^{+}} \text{CH}_3-\text{C}(\text{OH})(\text{CH}_3)\text{CH}_3 + \text{Mg}\text{OHI}\text{ (tert-butyl alcohol)}\]
(iii) Acetaldehyde: Ethyl formate reacts with methyl magnesium iodide. The addition product then undergoes hydrolysis to produce acetaldehyde. Grignard reagents are powerful tools in organic synthesis.
\[\text{H}-\text{C}(=\text{O})-\text{OC}_2\text{H}_5 + \text{CH}_3\text{MgI} \rightarrow \text{H}-\text{C}(\text{OMgI})(\text{CH}_3)\text{OC}_2\text{H}_5\text{ (intermediate)}\]
\[\text{H}-\text{C}(\text{OMgI})(\text{CH}_3)\text{OC}_2\text{H}_5 \xrightarrow{\text{H}_2\text{O}/\text{H}^{+}} \text{H}-\text{C}(=\text{O})-\text{CH}_3 + \text{Mg}\text{OHI} + \text{C}_2\text{H}_5\text{OH}\text{ (acetaldehyde)}\]
In simple words: We can make ethanol from formaldehyde, tert-butyl alcohol from acetone, and acetaldehyde from ethyl formate, all by reacting them with methyl magnesium iodide first, and then adding water.

🎯 Exam Tip: Remember that Grignard reagents are very reactive and require anhydrous (water-free) conditions to work correctly.

 

Question 6. How is ethylene dichloride converted into
(i) Acetaldehyde
(ii) Ethylene glycol
(iii) Ethylene
Answer:
(i) Acetaldehyde: Ethylene dichloride can be converted to acetaldehyde through hydrolysis with an aqueous solution of potassium hydroxide (KOH) or sodium hydroxide (NaOH). This process involves replacing the chlorine atoms with hydroxyl groups, followed by rearrangement.
\[\text{CH}_3-\text{CHCl}_2 \xrightarrow{\text{2KOH (aqueous)}} \text{CH}_3-\text{CH(OH)}_2 \text{ (unstable gem-diol)} \xrightarrow{\text{-H}_2\text{O}} \text{CH}_3\text{CHO (acetaldehyde)}\]
(ii) Ethylene glycol: Ethylene dichloride reacts with aqueous KOH to give ethylene glycol. Here, both chlorine atoms are replaced by hydroxyl groups, forming a diol. This is a common way to get alcohols from alkyl halides.
\[\text{Cl}-\text{CH}_2-\text{CH}_2-\text{Cl} \xrightarrow{\text{2KOH (aqueous)}} \text{HO}-\text{CH}_2-\text{CH}_2-\text{OH} + \text{2KCl}\text{ (ethylene glycol)}\]
(iii) Ethylene: Ethylene dichloride can be converted to ethylene by heating it with zinc in the presence of methanol. Zinc acts as a reducing agent, removing the chlorine atoms and forming an alkene. This is a dehalogenation reaction.
\[\text{Cl}-\text{CH}_2-\text{CH}_2-\text{Cl} \xrightarrow{\text{Zn, Methanol (heat)}} \text{CH}_2=\text{CH}_2 + \text{ZnCl}_2\text{ (ethylene)}\]
In simple words: You can change ethylene dichloride into acetaldehyde or ethylene glycol by using water-based KOH. To make it into ethylene, you heat it with zinc and methanol.

🎯 Exam Tip: Pay attention to whether the KOH solution is alcoholic or aqueous, as this determines whether substitution or elimination reactions occur.

 

Question 7. How are the following compounds prepared from chloroform?
(i) Phosgene
(ii) Methylene chloride
(iii) Methyl isocyanide
Answer:
(i) Phosgene: Phosgene is prepared by the oxidation of chloroform in the presence of light and air. This reaction can occur naturally if chloroform is exposed to air and light, forming a very poisonous gas. This is why chloroform must be stored in dark, sealed bottles.
\[\text{CHCl}_3 + \frac{1}{2}\text{O}_2 \xrightarrow{\text{light/air}} \text{COCl}_2 + \text{HCl}\text{ (Phosgene)}\]
(ii) Methylene chloride: Methylene chloride is prepared by the reduction of chloroform using zinc and HCl. In this reaction, a chlorine atom in chloroform is replaced by a hydrogen atom. This is a controlled reduction process.
\[\text{CHCl}_3 + \text{2[H]} \xrightarrow{\text{Zn/HCl}} \text{CH}_2\text{Cl}_2 + \text{HCl}\text{ (Methylene chloride)}\]
(iii) Methyl isocyanide: Methyl isocyanide is prepared by reacting chloroform with methylamine and alcoholic potassium hydroxide. This is known as the carbylamine reaction, which is used as a test for primary amines because of the foul-smelling product.
\[\text{CH}_3\text{NH}_2 + \text{CHCl}_3 + \text{3KOH (alcoholic)} \rightarrow \text{CH}_3\text{NC} + \text{3KCl} + \text{3H}_2\text{O}\text{ (Methyl isocyanide)}\]
In simple words: Chloroform can make phosgene when it meets light and air. It can make methylene chloride when reduced with zinc and acid. It makes a bad-smelling methyl isocyanide when mixed with methylamine and alcoholic KOH.

🎯 Exam Tip: Remember the dangers of phosgene formation when handling chloroform and the characteristic unpleasant odor of carbylamines.

 

Question 8. Discuss the aromatic electrophilic substitutions reaction of chlorobenzene.
Answer: Aromatic electrophilic substitution reactions on chlorobenzene generally occur at the ortho and para positions. This is because the chlorine atom, while deactivating the benzene ring, directs incoming electrophiles to these positions due to its resonance effect. However, its strong inductive effect makes chlorobenzene less reactive than benzene towards electrophilic substitution.

(i) Halogenation: When chlorobenzene reacts with chlorine (\(\text{Cl}_2\)) in the presence of anhydrous ferric chloride (\(\text{FeCl}_3\)) as a Lewis acid catalyst, it forms ortho-dichlorobenzene (minor product) and para-dichlorobenzene (major product). This happens because the chlorine already on the ring directs the new chlorine to the ortho and para positions.
(ii) Nitration: Nitration involves reacting chlorobenzene with a mixture of concentrated nitric acid (\(\text{HNO}_3\)) and concentrated sulfuric acid (\(\text{H}_2\text{SO}_4\)). This reaction produces ortho-chloronitrobenzene (minor) and para-chloronitrobenzene (major). The nitro group is a strong electron-withdrawing group.
(iii) Sulphonation: Chlorobenzene reacts with fuming sulfuric acid (\(\text{H}_2\text{SO}_4\)) to undergo sulphonation. The products are ortho-chlorobenzenesulphonic acid (minor) and para-chlorobenzenesulphonic acid (major). The sulphonic acid group is also an activating group.
(iv) Friedel-Crafts Alkylation: When chlorobenzene reacts with methyl chloride (\(\text{CH}_3\text{Cl}\)) in the presence of anhydrous aluminum chloride (\(\text{AlCl}_3\)), it undergoes Friedel-Crafts alkylation. This reaction yields ortho-chlorotoluene (minor) and para-chlorotoluene (major).
In simple words: Chlorobenzene can have new groups added to its ring, mostly at the 'ortho' and 'para' spots. This happens when it reacts with things like chlorine, nitric acid, sulfuric acid, or methyl chloride, because the existing chlorine atom guides the new groups to those specific places.

🎯 Exam Tip: Remember that while chlorine is ortho-para directing due to resonance, it is a deactivating group due to its strong inductive effect, making chlorobenzene less reactive than benzene.