Haloalkanes And Haloarenes

Step 1: Identify Alcohol X:
Formula . Condition: Reacts with conc. HCl at room temperature to form Chloride Y. This indicates the Lucas Test. Tertiary alcohols react immediately at room temperature. Secondary take minutes, Primary require heat/ZnCl2. Structure of X (Tertiary Butyl Alcohol): (2-Methylpropan-2-ol).
Step 2: Identify Chloride Y:
Reaction: . Y is t-Butyl Chloride (2-Chloro-2-methylpropane).
Step 3: Identify Z:
Reaction of Y with Mg/Ether Grignard Reagent ( ). Reaction with Water Hydrolysis. . Z is Isobutane (2-Methylpropane).
Step 4: Check Options:
Y = 2-Chloro-2-methylpropane. Z = 2-Methylpropane. Matches Option (B) (Diagram corresponds to t-butyl chloride and isobutane).

Alkene reacts with KMnO under mild conditions → syn dihydroxylation → vicinal diol (X). Alkyne reacts with Na/NH → anti addition → trans-alkene (Y).

1. Chain isomers have the same molecular formula but different carbon skeletons. All options are C H .
2. (1) Pent-1-ene (CH =CH-CH -CH -CH ) and Pent-2-ene (CH -CH=CH-CH -CH ): Same chain, position isomers.
3. (2) 2-Methylbut-1-ene ((CH ) C=CH ) and 2-Methylbut-2-ene (CH -C(CH )=CH-CH ): Same branched chain, position isomers.
4. (3) Pent-2-ene and 2-Methylbut-2-ene: Different skeletons (straight vs branched), but not chain isomers in strict sense.
5. (4) 2-Methylbut-1-ene and 3-Methylbut-1-ene (CH =C(CH )-CH -CH vs CH =CH-CH(CH ) ): Different branching, chain isomers.
6. Thus, the answer is (4).

1. 2-Methylpropene ((CH ) C=CH ) reacts with HBr via electrophilic addition, not nucleophilic substitution, but the question likely refers to the mechanism of Br addition.
2. Without peroxide: HBr adds via Markovnikov rule, forming (CH ) CBr (X) via carbocation (tertiary, stable), so -like addition.
3. With peroxide ((CH COO) ): Free radical mechanism gives anti-Markovnikov product (CH ) CHCH Br (Y) via primary radical, -like addition.
4. Thus, X (Markovnikov) is , Y (anti-Markovnikov) is , so the answer is (4).

Isopentane (\ce{C5H12}) undergoes oxidation with \ce{KMnO4} to form a diol, X, which is \ce{HO-CH2-CH(CH3)-CH2OH} (a 1,3-diol). This diol reacts to form an intermediate \ce{C5H12O}, and Y is formed with Conc. HCl, leading to \ce{C5H11Cl} through chlorination.