Haloalkanes and halo arenes practice · Mixed difficulty

Option D ($\mathrm{C}_2 \mathrm{Cl}_2 \mathrm{~F}_2$) is correct because Freons are chlorofluorocarbons, and this compound contains both chlorine and fluorine.

Option A ($\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Cl} < \mathrm{CH}_3 \mathrm{CH}(\mathrm{Cl}) \mathrm{CH}_3 < (\mathrm{CH}_3)_3 \mathrm{CCl}$) is correct because the ease of dehydrohalogenation follows the stability of the carbocation formed, which is tertiary > secondary > primary.

Option B (a > d > c > b) is correct because $S{N^2}$ reactivity depends on steric hindrance, with less hindered primary alkyl halides reacting faster.

Option D (Methane) is correct because Grignard reagents react with water to form alkanes, and in this case, methylmagnesium bromide reacts to form methane.

Option B (Carbonyl chloride) is correct because chloroform reacts with oxygen in the presence of light to form carbonyl chloride (phosgene).

Option C (Both (1) and (2)) is correct because red phosphorus reacts with bromine to form phosphorus tribromide and with iodine to form phosphorus triiodide.

Option D ($CC{l_2} = CHCl$) is correct because Westrosol is the common name for trichloroethylene, which has the formula $CC{l_2} = CHCl$.

Option D ($Neo$-pentane) is correct because it's the only alkane that yields only one monochloroalkane due to all its hydrogen atoms being equivalent.

Option C (Darzen's procedure) is correct because the reaction of an alcohol with thionyl chloride (SOCl₂) in the presence of pyridine to form an alkyl chloride is known as the Darzen's procedure.

The correct answer is D (Position isomers) because propyl iodide and isopropyl iodide differ only in the position of the iodine atom on the carbon chain.

Option A is correct because both the assertion and reason are true, and the reason correctly explains why alkyl halides are insoluble in water.

Option D (${ClCH}_2-{CH}={CH}_2$) is correct because the carbocation formed after the departure of Cl is allylic and resonance stabilized, making it more reactive towards nucleophilic substitution.

Option C is correct because Alkyl fluorides can be prepared by treating alkyl chlorides with AgF (Swarts reaction), making the assertion true, but the reason is false because Swarts reaction *is* used to prepare alkyl fluorides.

The correct answer is A ($\mathrm{C}{\mathrm{H}_3} - \mathrm{\;C}{\mathrm{H}_2}\mathrm{\;}\left( {\mathrm{OH}} \right)$) because compounds with the $CH_3CH(OH)-$ or $CH_3C(O)-$ group give a positive iodoform test.

Option A (Diphenyl) is correct because Ulmann's reaction is a coupling reaction specifically used to synthesize biaryls, such as diphenyl, from aryl halides using copper.

Option B (${CH}_3 {COCH}_2 {CH}_2 {Br}$) is correct because the presence of the electron-withdrawing carbonyl group makes the beta-hydrogens more acidic, facilitating elimination by alcoholic KOH.

Option C (Phosgene) is correct because chloroform, in the presence of light and air, undergoes oxidation to form phosgene (carbonyl chloride), a poisonous gas.

Option B (2,1) is correct because chlorination is less selective, leading to two possible products, while bromination is highly selective, resulting in only one major product.