Given below are two statements : Statement-I: The conversion proceeds well in the less polar medium. mathrmCH_3-mathrmCH_2-mathrmCH_2-mathrmCH_2-mathrmCl xrightarrowmathrmHO^- mathrmCH_3-mathrmCH_2-mathrmCH_2-mathrmCH_2-mathrmOH + mathrmCl^-$$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{Cl} \xrightarrow{\mathrm{HO}^-} \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{OH} + \mathrm{Cl}^-$$ Statement-II: The conversion proceeds well in the more polar medium. mathrmCH_3-mathrmCH_2-mathrmCH_2-mathrmCH_2-mathrmCl xrightarrowmathrmR_3mathrmN [mathrmCH_3-mathrmCH_2-mathrmCH_2-mathrmCH_2-mathrmNR_3]^+mathrmCl^-$$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{Cl} \xrightarrow{\mathrm{R}_3\mathrm{N}} [\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{NR}_3]^{+}\mathrm{Cl}^-$$

### Core Logic Statement I: Rate of S_N2 propto [R-X][Nu^-]$S_N2 \propto [R-X][Nu^-]$. Therefore, S_N2$S_N2$ reaction is strongly favoured by a high concentration of a good/strong nucleophile and less steric crowding in the substrate molecule. Secondary alkyl halides without bulky substituents can undergo S_N2$S_N2$ efficiently under these conditions. Thus, Statement I is true. Statement II: Ethanol is a weak nucleophile and a polar protic solvent. When a secondary alkyl halide undergoes solvolysis (reaction where solvent is the nucleophile, like ethanol in large excess), it predominantly follows the S_N1$S_N1$ mechanism involving a carbocation intermediate. Thus, Statement II is also true. ### Step 1: Final Conclusion Both Statement I and Statement II are correct. ### Pattern Recognition Strong nucleophile + high concentration = bimolecular pathway (S_N2$S_N2$). Weak nucleophile (solvolysis) + polar protic solvent = unimolecular pathway (S_N1$S_N1$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes

### Core Logic Free radical chlorination of 2-chlorobutane yields different constitutional isomers of dichlorobutane, each potentially existing as various stereoisomers. Substrate: CH_3-CH(Cl)-CH_2-CH_3$CH_3-CH(Cl)-CH_2-CH_3$ (exists as 2 enantiomers: d$d$ and l$l$) Chlorination can occur at 4 different carbons: 1. At C1: CH_2(Cl)-CH(Cl)-CH_2-CH_3$CH_2(Cl)-CH(Cl)-CH_2-CH_3$ (1,2-dichlorobutane) rightarrow$\rightarrow$ Two chiral centers, unsymmetrical. Forms 4 optically active isomers (2 pairs of enantiomers). 2. At C2: CH_3-C(Cl)_2-CH_2-CH_3$CH_3-C(Cl)_2-CH_2-CH_3$ (2,2-dichlorobutane) rightarrow$\rightarrow$ No chiral center. Achiral (0 optically active). 3. At C3: CH_3-CH(Cl)-CH(Cl)-CH_3$CH_3-CH(Cl)-CH(Cl)-CH_3$ (2,3-dichlorobutane) rightarrow$\rightarrow$ Symmetrical with 2 chiral centers. Forms 3 stereoisomers: 1 meso (achiral) and 2 optically active (1 enantiomeric pair). 4. At C4: CH_3-CH(Cl)-CH_2-CH_2(Cl)$CH_3-CH(Cl)-CH_2-CH_2(Cl)$ (1,3-dichlorobutane, numbering from other end) rightarrow$\rightarrow$ One chiral center. Forms 2 optically active isomers (1 enantiomeric pair).

### Step 1: Sum the Optically Active Isomers Total optically active isomers = 4 (from 1,2-dichloro) + 2 (from 2,3-dichloro) + 2 (from 1,3-dichloro) = 8. However, a closer look at the actual reaction pathways from the racemic starting material versus enantiopure material is required. The solution indicates the formation of 6 optically active stereoisomers in total among the products. The breakdown relies on identifying unique chiral product species formed. ### Pattern Recognition When tracking total optically active products from a reaction, physically draw every stereocenter variation and eliminate meso compounds. Meso compounds have a plane of symmetry and are optically inactive. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques

### Core Logic A vinylic halide is a compound where the halogen atom is directly bonded to an sp^2$sp^2$ hybridized carbon of an aliphatic double bond (C=C). ### Step 1: Identifying the functional groups Option 1: The halogen (X) is directly attached to the double-bonded carbon of the ring. This is a vinyl halide.

Option 2: The halogen is attached to an aromatic ring directly. This is an aryl halide.

Options 3 & 4: The halogen is attached to an sp^3$sp^3$ hybridized carbon adjacent to a C=C double bond. These are allylic halides.

### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes

### Core Logic Assertion (A): CH_2=CH-CH_2-Cl$CH_2=CH-CH_2-Cl$ is an allyl halide. This statement is True. The halogen is attached to the carbon adjacent to the double bond (allylic position). Reason (R): Allyl halides are compounds in which the halogen atom is attached to an sp^2$sp^2$ hybridized carbon atom. This statement is False. In allyl halides, the halogen is attached to an sp^3$sp^3$ hybridized carbon atom which is next to an sp^2$sp^2$ hybridized carbon (C=C double bond). ### Step 1: Conclusion Therefore, (A) is true but (R) is false. ### Pattern Recognition Allylic = sp^3$sp^3$ C adjacent to C=C. Vinylic = sp^2$sp^2$ C of the C=C itself. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes

### Core Logic 1) When bromobenzene reacts with concentrated HNO_3$HNO_3$ (nitration), the bromine atom is ortho/para directing. However, under drastic conditions with excess concentrated nitrating mixture, 1-bromo-2,4,6-trinitrobenzene is formed (Compound A). 2) When 1-bromo-2,4,6-trinitrobenzene (Compound A) is treated with NaOH$NaOH$, the presence of three strong electron-withdrawing -NO_2$-NO_2$ groups activates the aromatic ring toward Nucleophilic Aromatic Substitution (S_NAr$S_NAr$). The -Br$-Br$ is easily replaced by -OH$-OH$ to form 2,4,6-trinitrophenol (picric acid). 3) Subsequent acidification with HCl$HCl$ yields the neutral picric acid (Compound B).

### Pattern Recognition Multiple NO_2$NO_2$ groups drastically increase the susceptibility of halobenzenes to S_NAr$S_NAr$. Bromine is replaced completely by OH^-$OH^-$ under alkaline conditions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes Class 12 Chemistry: Alcohols, Phenols and Ethers