Class 12 Chemistry MCQs | Chapter 6: Haloalkanes and Haloarenes – Part 4

Explanation: Ammonia acts as a nucleophile and replaces the halogen atom of the haloalkane. The first substitution product is a primary amine. Excess ammonia is used so that the newly formed amine is less likely to react further with more haloalkane. This helps increase the yield of the primary amine.

Explanation: The primary amine formed in the first step is itself nucleophilic and can react further with more haloalkane. This can produce secondary, tertiary, or even quaternary ammonium salts. Using excess ammonia reduces the chance that the amine product will be alkylated further. So excess \( NH_3 \) helps favor the primary amine as the main product.

Explanation: In the Williamson ether synthesis, an alkoxide ion acts as a nucleophile and displaces the halide from a suitable haloalkane. Here, ethoxide ion attacks methyl bromide and forms an ether. The product therefore contains one methyl group and one ethyl group attached to oxygen. That gives methoxyethane, \( CH_3OC_2H_5 \).

Explanation: Williamson ether synthesis proceeds through nucleophilic substitution and works best when the substrate allows easy backside attack. Primary haloalkanes are least hindered and therefore undergo substitution more readily. Secondary substrates may give more competition from elimination, while tertiary ones strongly favor elimination. Aryl halides do not usually react by ordinary nucleophilic substitution under these conditions.

Explanation: Sulfur nucleophiles can replace the halogen atom in a haloalkane by nucleophilic substitution. When \( RS^- \) attacks the carbon attached to halogen, the product formed is \( R-S-R' \), which is a thioether. This is the sulfur analogue of ether formation from alkoxides. It is another example showing how the product of a haloalkane depends on the nucleophile used.

Explanation: Aqueous \( KOH \) supplies \( OH^- \), which acts as a nucleophile toward the haloalkane. The bromine atom leaves and hydroxide takes its place on carbon. This is a hydrolysis reaction of the haloalkane. The product formed is therefore an alcohol.

Explanation: In hydrolysis of a haloalkane with aqueous \( KOH \), the halogen atom is replaced by the hydroxyl group. Since the substrate is \( CH_3CH_2CH_2Cl \), the product keeps the same three-carbon chain. The chlorine atom is replaced by \( OH \), giving \( CH_3CH_2CH_2OH \). That compound is propan-1-ol.

Explanation: \( \textbf{Given:} \) Substrate: alkyl halide, \( R-X \) Reagent: aqueous \( KOH \) \( \textbf{Required:} \) Correct equation for hydrolysis \( \textbf{Relevant Principle:} \) Aqueous \( KOH \) supplies \( OH^- \), which replaces the halogen atom in nucleophilic substitution. \( \textbf{Why this principle applies:} \) The reaction medium is aqueous, so substitution to form alcohol is favored here. \( \textbf{Identify the organic product:} \) Replacing \( X \) by \( OH \) gives \( R-OH \). \( \textbf{Identify the inorganic product:} \) The potassium ion combines with the leaving halide to give \( KX \). \( \textbf{Balanced equation:} \) \[ R-X + KOH \xrightarrow{\text{aq.}} R-OH + KX \] \( \textbf{Final simplification:} \) This equation represents hydrolysis of the alkyl halide. All formulas, symbols, and reaction conditions are written correctly. \( \textbf{Final Answer:} \) \[ R-X + KOH \xrightarrow{\text{aq.}} R-OH + KX \]

Explanation: In chlorobenzene, chlorine is attached directly to an aromatic \( sp^2 \)-hybridised carbon. The lone pair on chlorine can interact with the benzene ring, giving the \( C-Cl \) bond partial double-bond character. This makes the bond shorter, stronger, and much harder to break than the \( C-Cl \) bond in an ordinary alkyl chloride. Backside attack at the aryl carbon is also difficult. Therefore chlorobenzene needs drastic conditions for nucleophilic substitution, while ethyl chloride is hydrolysed much more readily.

Explanation: When a haloalkane reacts with \( KCN \), the nucleophile enters through the carbon end of the cyanide ion and forms a nitrile. The nitrile carbon becomes part of the product chain. Because of that added carbon, the total number of carbon atoms in the product is one more than in the starting haloalkane. This is why cyanide substitution is useful in chain extension.

Explanation: \( KCN \) gives nitriles because the cyanide ion attacks through carbon. Starting from \( CH_3CH_2CH_2Br \), the bromine atom is replaced by \( CN \). This produces \( CH_3CH_2CH_2CN \), which has four carbon atoms in total. The product is therefore butanenitrile.

Explanation: In reactions with \( AgCN \), the cyanide group behaves differently from \( KCN \). Because \( AgCN \) is more covalent, substitution occurs mainly through the nitrogen end. This leads to formation of an isocyanide rather than a nitrile. So 1-bromopropane gives propyl isocyanide, \( CH_3CH_2CH_2NC \).

Explanation: \( \textbf{Given:} \) Substrate: haloalkane, \( R-X \) Reagent: ammonia, \( NH_3 \) \( \textbf{Required:} \) Correct equation for formation of a primary amine \( \textbf{Relevant Principle:} \) Ammonia acts as a nucleophile and replaces the halogen atom in the haloalkane. \( \textbf{Why this principle applies:} \) One molecule of ammonia forms the amine, while another accepts the hydrogen halide equivalent formed during reaction. \( \textbf{Identify the organic product:} \) Replacement of \( X \) by \( NH_2 \) gives \( R-NH_2 \). \( \textbf{Identify the inorganic product:} \) The ammonium halide formed is \( NH_4X \). \( \textbf{Balanced equation:} \) \[ R-X + 2NH_3 \rightarrow R-NH_2 + NH_4X \] \( \textbf{Final simplification:} \) This is the standard equation for primary amine formation from a haloalkane. All formulas and coefficients are correctly written. \( \textbf{Final Answer:} \) \[ R-X + 2NH_3 \rightarrow R-NH_2 + NH_4X \]

Explanation: Williamson ether synthesis proceeds mainly through an \( S_N2 \) pathway when a suitable alkyl halide reacts with an alkoxide ion. Methyl bromide is highly suitable because the carbon bearing bromine is very unhindered. Ethoxide can attack this carbon from the backside and displace bromide efficiently. A carbocation pathway is not needed, and elimination is not the expected route for methyl bromide. This is why methyl and primary halides are preferred substrates for clean Williamson ether synthesis.

Explanation: Methanethiolate ion, \( CH_3S^- \), is a sulfur nucleophile. It attacks methyl bromide by nucleophilic substitution and replaces bromine. The product formed has sulfur between two methyl groups. This compound is dimethyl sulfide, \( CH_3SCH_3 \), which is a thioether.

Explanation: Alcoholic \( KOH \) usually favors elimination rather than substitution. In this reaction, a hydrogen atom is removed from the carbon adjacent to the one bearing bromine, and \( HBr \) is eliminated. This process forms a double bond between the two carbon atoms. Therefore \( 2 \)-bromopropane gives propene as the main product.

Explanation: Formation of an alkene from a haloalkane requires dehydrohalogenation. Alcoholic \( KOH \) acts mainly as a base and removes a \( \beta \)-hydrogen, while the halogen leaves from the adjacent carbon. Heating helps favor elimination over substitution. So bromoethane gives ethene under alcoholic \( KOH \) and heat.

Explanation: In dehydrohalogenation, the haloalkane loses one hydrogen atom and one halogen atom as hydrogen halide. The hydrogen is removed from the \( \beta \)-carbon, while the halogen leaves from the \( \alpha \)-carbon. This creates a carbon-carbon double bond in the product. So the reaction is an elimination process, not a substitution one.

Explanation: The carbon attached directly to halogen is called the \( \alpha \)-carbon. The next carbon is called the \( \beta \)-carbon, and the hydrogen removed in elimination is usually a \( \beta \)-hydrogen. Removal of this hydrogen and the halogen from neighboring carbons produces the double bond. This is why dehydrohalogenation is often described as \( \beta \)-elimination.

Explanation: \( \textbf{Given:} \) Substrate: haloalkane, \( R-CH_2-CH_2-X \) Reagent: alcoholic \( KOH \) \( \textbf{Required:} \) Correct equation for dehydrohalogenation \( \textbf{Relevant Principle:} \) Alcoholic \( KOH \) favors elimination of \( HX \) from adjacent carbon atoms to form an alkene. \( \textbf{Why this principle applies:} \) The base removes a \( \beta \)-hydrogen while the halogen leaves from the \( \alpha \)-carbon. \( \textbf{Identify the organic product:} \) Loss of \( H \) and \( X \) forms a double bond, giving \( R-CH=CH_2 \). \( \textbf{Identify the inorganic products:} \) The remaining ions form \( KX \), and water is also produced. \( \textbf{Balanced equation:} \) \[ R-CH_2-CH_2-X + KOH \xrightarrow{\text{alc.}} R-CH=CH_2 + KX + H_2O \] \( \textbf{Final simplification:} \) This is the standard dehydrohalogenation equation. All symbols, formulas, and reaction conditions are correctly written. \( \textbf{Final Answer:} \) \[ R-CH_2-CH_2-X + KOH \xrightarrow{\text{alc.}} R-CH=CH_2 + KX + H_2O \]