Haloalkanes and halo arenes practice · Mixed difficulty

Option B is correct because the reaction of acetylene with HCl proceeds in two steps, eventually forming 1,1-dichloroethane (CH3CHCl2).

Option D is correct. S-I is wrong because Styrene on reaction with HBr gives 1-bromo-2-phenylethane, and S-II is correct because the benzyl radical is more stable than the alkyl radical due to resonance.

Option A (CHCl3) is correct because chloroform (CHCl3) reacts with a primary amine and KOH to produce an isocyanide, which has a foul smell, indicating a positive carbylamine test.

Option C (Triiodomethane) is correct because triiodomethane (iodoform) decomposes to release iodine, which has antiseptic properties.

Option A (Iodoform test) is correct because methyl ketones give a positive iodoform test, producing a yellow precipitate.

Option C is correct because the assertion is true, but the reason is false. 2-butene is the major product due to Zaitsev's rule, but the acidity of hydrogens is not the reason.

Option A ($HBr$) is correct because the reaction of iso-butylene with $HBr$ follows Markovnikov's rule, leading to the formation of tert-butyl bromide.

Option C is correct because the C-halogen bond length increases as the size of the halogen increases, following the order F < Cl < Br < I.

The correct answer is D (4). The compound has one chiral center and exhibits cis/trans isomerism around the double bond, leading to 4 stereoisomers.

The correct answer is C (A is correct; R is incorrect) because haloarenes are less reactive than benzene due to the -I effect of the halogen.

Option A (Chloroethane) is correct because chloroethane reacts with KCN to directly produce propanenitrile.

Option B (Iodine and aqueous potassium hydroxide) is correct because ethanol reacts with iodine and aqueous potassium hydroxide to form iodoform in the iodoform test.

Option C (1-phenylpropan-2-ol) is correct because the reaction of 1-phenyl, 2-chloropropane with aqueous KOH leads to nucleophilic substitution, replacing the chlorine with a hydroxyl group.

Option A (in the stratosphere, CFCs release chlorine free radicals $Cl,$ which reacts with ${O_3}$ to give chlorine dioxide radicals) is correct because the reaction of chlorine free radicals with ozone produces chlorine monoxide radicals, not chlorine dioxide radicals.

Option B (${\mathrm{I}_2}$ and aq. KOH ) is correct because the reaction of ethanol with iodine and aqueous KOH produces iodoform due to the formation of hypoiodite ion, which acts as an oxidizing agent.

Option B is correct because chlorine is an ortho-para directing group due to resonance (+R effect), and it's also an electron-withdrawing group due to its electronegativity (-I effect), but the electron-withdrawing nature doesn't explain its directing effect.

Option A is correct. Benzyl chloride undergoes SN1 mechanism due to the stability of the benzyl carbocation through resonance.

Option D (0.05) is correct because 0.1 Faraday of electricity will deposit 0.05 moles of Ni, given that Ni is in the +2 oxidation state.

Option D (stephen's reaction) is correct because Stephen's reaction involves the reduction of nitriles to aldehydes using $SnCl_2/HCl$ followed by hydrolysis.

Option A ($\mathrm{MeX}$) is correct because it experiences the least steric hindrance, making it the most reactive in $S_N2$ reactions.

Option A ($1 < 2 < 3$) is correct because the reactivity towards HCl follows the stability of the carbocation formed, with triphenyl carbocation being the most stable due to resonance.

Option D is correct. In ethanol, the nucleophilicity order is PhS- > I- > EtO- > PhO- due to the effects of solvation and polarizability.

Option D (${CCl}_4$) is correct because it is a common feedstock used in the synthesis of chlorofluorocarbons (CFCs) through reactions like Swarts reaction.

Option B (ii,iii) is correct because Alkyl fluorides are synthesized from alkyl chlorides/bromides in presence of $CoF_2$ or $Hg_2F_2$ via Swarts reaction.

Option A (Carbon tetrachloride) is correct because it was widely used as a cleaning fluid, degreasing agent, spot remover, and fire extinguisher in the mid-1960s.

Option A (Hydroxyl (OH) group) is correct because halogens can replace hydroxyl groups in reactions with reagents like PCl5 or SOCl2.

Option D ($\mathrm{CC}{\mathrm{l}_2}{\mathrm{F}_2}$) is correct because Dichlorodifluoromethane (Freon-12) is commonly used as a refrigerant in cooling systems.

Option B (${S}{{N}^1}$ and ${S}{{N}^2}$) is correct because reaction (i) with $C_2H_5OH$ favors $S_N1$ due to the formation of a carbocation, while reaction (ii) with $C_2H_5O^-$ favors $S_N2$ due to the strong nucleophile.

Option A ($CC{l_4}$) is correct because chlorination of $C{S_2}$ yields carbon tetrachloride and sulfur.

Option A is correct because both the assertion and reason are true, and the reason correctly explains why chloroform is more acidic than fluoroform due to the greater stability of its conjugate base.