Given below are two statements for the following reaction sequence.

Statement I: Compound 'Z' will give yellow precipitate with NaOI. Statement II: Compound 'Q' has two different types of 'H' atoms (aromatic : aliphatic) in the ratio 1 : 3. In the light of the above statements, choose the correct answer from the option given below:

Core Logic

This is the classic Peroxide Effect (Kharasch effect) showing Anti-Markovnikov addition of HBr$\mathrm{HBr}$ across an alkene. Ph-CH=CH₂ + HBr Peroxide Ph-CH₂-CH₂-Br$\mathrm{Ph-CH=CH_2 + HBr} \xrightarrow{\text{Peroxide}} \mathrm{Ph-CH_2-CH_2-Br}$ (1-Bromo-2-phenylethane is the MAJOR product).

Step 1: Analyze Radical Mechanism (A, B, C)

$\bullet$ The reaction proceeds via the most stable free radical intermediate (a secondary benzylic radical). Statement A is correct. $\bullet$ Peroxide (Benzoyl peroxide) undergoes homolytic cleavage to form benzoyloxy radicals, which lose CO₂$CO_2$ to form phenyl radicals (Ph$\dot{Ph}$). Phenyl radical abstracts H from HBr to form a Bromine radical (Br$\dot{Br}$), not a Hydrogen radical (H$\dot{H}$). Statement B is false. $\bullet$ The abstraction of H from HBr by the phenyl radical (Ph$\dot{Ph}$) yields Benzene (Ph-H$Ph-H$) as a byproduct. Statement C is correct.

Step 2: Product Check (D, E)

$\bullet$ 1-Bromo-2-phenylethane is the major product, not the minor product. Statement D is false. $\bullet$ In the absence of peroxide, standard electrophilic addition occurs via a stable carbocation intermediate (Markovnikov addition). Statement E is correct.

Final Conclusion

Statements A, C, and E are correct.

Pattern Recognition

Peroxides generate Br$\dot{Br}$, never H$\dot{H}$. The initiating radical pulls H from H-Br since H-Br is weaker than forming C-H bond, leaving Br$\dot{Br}$ to attack the alkene.

Chapter Mix

Class 11 Chemistry: Hydrocarbons

Core Logic

Presence of -NO₂$-NO_{2}$ group in Benzene ring strongly deactivates the ring towards electrophilic substitution reaction due to -M$-M$ effect. Conversely, it activates the ring towards nucleophilic substitution reactions by stabilizing the intermediate carbanion.

Step 1: Final Conclusion

Statements B and C are correct.

Pattern Recognition

Electron-withdrawing groups (EWG) like -NO₂$-NO_2$ withdraw electron density, raising the activation energy for electrophilic attacks while lowering it for nucleophilic attacks.

Chapter Mix

Class 11 Chemistry: Hydrocarbons Class 12 Chemistry: Haloalkanes and Haloarenes

Core Logic

Analyzing each oxidation reaction: (A) CH₄ + O₂ [Δ]Mo₂O₃ HCHO + H₂O$CH_{4} + O_{2} \xrightarrow[\Delta]{Mo_{2}O_{3}} HCHO + H_2O$ (Yields formaldehyde, an aldehyde)

(B) 2CH₃CH₃ + 3O₂ [Δ](CH₃COO)₂Mn 2CH₃COOH + 2H₂O$2CH_{3}CH_{3} + 3O_{2} \xrightarrow[\Delta]{(CH_{3}COO)_{2}Mn} 2CH_{3}COOH + 2H_2O$ (Yields acetic acid, a carboxylic acid)

(C) (CH₃)₃CH KMnO₄ (CH₃)₃C-OH$(CH_{3})_{3}CH \xrightarrow{KMnO_{4}} (CH_3)_3C-OH$ (Oxidation of 3°$3^{\circ}$ alkane yields tert-butyl alcohol)

(D) 2CH₄ + O₂ [523K, 100 atm]Cu 2CH₃OH$2CH_{4} + O_{2} \xrightarrow[523\text{K, } 100\text{ atm}]{Cu} 2CH_{3}OH$ (Controlled oxidation yields methanol)

(E) CH₃-CH=CH-CH₃ KMnO₄/H⁺ 2CH₃COOH$CH_{3}-CH=CH-CH_{3} \xrightarrow{KMnO_{4}/H^{+}} 2CH_{3}COOH$ (Strong oxidative cleavage yields acetic acid)

Step 1: Final Conclusion

Reactions (C) and (D) yield alcohols as products. Therefore, C and D only.

Pattern Recognition

Molybdenum catalyst yields aldehydes. Copper catalyst yields alcohols. KMnO₄$KMnO_4$ attacks tertiary hydrogens to give tertiary alcohols, while vigorously cleaving alkenes to carboxylic acids.

Chapter Mix

Class 11 Chemistry: Hydrocarbons

Related Formula

Core Logic

Statement (I) is correct: Neopentane (2,2-dimethylpropane) contains a quaternary carbon bonded to four methyl groups. There are 12 hydrogen atoms, all of which are primary and chemically equivalent. Halogenation yields exactly one monosubstituted derivative:

Statement (II) is correct: Neopentane has a compact, symmetrical, nearly spherical molecular structure. In the solid crystal lattice, these spherical molecules pack much more efficiently compared to the floppy, linear n-pentane. This robust crystalline packing dramatically increases its melting point (256.4~K$256.4~\mathrm{K}$) compared to that of n-pentane (143.4~K$143.4~\mathrm{K}$).

Step 1: Final Verification

Since both statements are theoretically and experimentally correct, the correct option is (2).

Pattern Recognition

While branching decreases the boiling point (due to decreased surface area and weaker van der Waals forces), branching that creates a highly symmetric structure increases the melting point because of close-packing efficiency in the solid phase.

Chapter Mix

Class 11 Chemistry: Hydrocarbons