Haloalkanes and halo arenes practice · Mixed difficulty

Option D ($R\mathrm{I}\; > \;R\mathrm{Br\;} > \;R\mathrm{Cl}\; > \;R\mathrm{F}$) is correct because the reactivity of alkyl halides in $S_N2$ reactions increases as the size of the halogen increases, due to the weakening of the C-X bond.

Option C (Nucleophilic substitution) is correct because the reaction involves the substitution of a leaving group (Br) by a nucleophile (H2O), which is a characteristic of nucleophilic substitution reactions.

Option B (Antiseptic) is correct because iodoform is known for its antiseptic properties due to the slow liberation of iodine.

Option B (1.5 times) is correct because the rate of an $SN_2$ reaction is directly proportional to the concentration of the alkyl halide and the nucleophile. Tripling the alkyl halide concentration increases the rate by a factor of 3, while halving the hydroxide concentration decreases the rate by a factor of 2, resulting in a net increase of 3/2 = 1.5 times.

Option A (Benzene carboxylic acid) is correct because the Grignard reagent formed from bromobenzene reacts with carbon dioxide to form benzoic acid after hydrolysis.

Option B (Polarity of $\mathrm{C}-\mathrm{Cl}$ bond in chlorobenzene is less than that in cyclohexyl chloride) is correct because the C-Cl bond in chlorobenzene has partial double bond character due to resonance, reducing its polarity.

Option B (Electrophilic substitution reaction) is correct because the reaction of toluene with a halogen in the presence of iron(III) chloride is a characteristic electrophilic aromatic substitution.

Option A is correct because chloroform reacts with primary amines in the presence of alcoholic KOH to form isocyanides (carbylamines) which have a foul smell, while carbon tetrachloride does not react.

Option A is correct because the reaction involves a tertiary alkyl halide, which favors SN1 mechanism due to the formation of a stable tertiary carbocation.

Option C (Nature of both alkyl group and halogen atom) is correct because the reactivity of alkyl halides depends on the stability of the carbocation formed (alkyl group) and the leaving group ability of the halogen.

Option B ($\mathrm{NaHS}$) is correct because NaHS (sodium hydrosulfide) will react with ethyl bromide to form ethanethiol through a nucleophilic substitution reaction.

Option D (Propyl chloride with KCN) is correct because the reaction of propyl chloride with KCN via an S\N2 mechanism will yield butane nitrile.

Option B (Petrol engines) is correct because 1,2-dibromoethane is added to petrol to convert lead deposits into volatile lead bromide, preventing lead fouling in engines.

Option C ((A), (C) and (D)) is correct because Carbon dioxide, Water vapour, and Methane are all well-known greenhouse gases that contribute to global warming.

Option C ($\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{COC}{\mathrm{H}_2}\mathrm{C}{\mathrm{H}_3}$) is correct because it does not contain a $CH_3CO-$ group or a $CH_3CH(OH)-$ group, both of which are necessary for a positive iodoform test.

Option D (R-I) is correct because the C-I bond is the weakest, making it easiest to break during dehydrohalogenation.

Option A (${\mathrm{S}_\mathrm{N}}1$ mechanism) is correct because isopropyl chloride, being a secondary alkyl halide, can undergo hydrolysis via both $S_N1$ and $S_N2$ mechanisms, with $S_N1$ being more favored due to the formation of a relatively stable secondary carbocation.

Option A (Alcoholic $\mathrm{AgN}{\mathrm{O}_3}$) is correct because the rate of formation of AgBr and AgCl will be different due to the different nature of the alkyl halides (primary vs. secondary).

Option A (Carboxylic radical represented by COOH) is correct because Grignard reagents have the general formula RMgX, where R is an alkyl or aryl group, Mg is magnesium, and X is a halogen. They do not contain a carboxylic acid group.

Option D is correct because SN2 mechanism results in inversion of configuration, not racemization. The nucleophile attacks from the opposite side of the leaving group.

Option A (Nitromethane) is correct because it reacts with chlorine and sodium hydroxide to produce chloropicrin ($CCl_3NO_2$).

Option D (Ethanol) is correct because protic solvents are those that contain an O-H or N-H bond, and ethanol ($C_2H_5OH$) has an O-H bond.

Option D ((ii), (iii)) is correct because S\N2 reactions are bimolecular, meaning the rate depends on both the substrate and nucleophile, and they proceed with inversion of configuration through a transition state.

Option A (A= aqueous KOH; B= AgOH) is correct because aqueous KOH leads to substitution and AgOH, being a source of OH-, also leads to substitution.

Option D (DDT – Fat insoluble) is the incorrect match because DDT is actually fat-soluble, which is one of the reasons it accumulates in the environment and in living organisms.

Option D (II $>$ I $>$ III $>$ IV) is correct because benzyl halides are more reactive than primary alkyl halides, and bromides are more reactive than chlorides. Electron-donating groups on the aromatic ring decrease the reactivity.

Option B ($\left({CH}_3\right)_2 {C}({Br})-{CH}({NO}) {CH}_3$) is correct because the addition of $NOBr$ to the alkene follows the anti-Markovnikov rule, with $Br$ adding to the more substituted carbon and $NO$ to the less substituted carbon.

Option A (Diphenyl) is correct because Ullmann's reaction involves the coupling of aryl halides in the presence of copper to form biaryls, such as diphenyl.

Option A is correct because the reactivity of nucleophiles in SN2 reactions depends on their basicity and steric hindrance. More basic and less hindered nucleophiles are more reactive.

Option B (DDT, Aldrin) is correct because both DDT (dichlorodiphenyltrichloroethane) and Aldrin are well-known organochlorine pesticides.