Match the List-I with List-II

List-I (Chloro derivative)	List-II (Example)
A. Vinyl Chloride	I. CH_2=CH-CH_2Cl$CH_2=CH-CH_2Cl$
B. Benzyl chloride	II. CH_3-CH(Cl)CH_3$CH_3-CH(Cl)CH_3$
C. Alkyl chloride	III. CH_2=CHCl$CH_2=CHCl$
D. Allyl chloride	IV.

Choose the correct answer from the options given below :

### Core Logic Since P (C_4H_8$C_4H_8$) reacts with HCl to give an optically active compound Q (C_4H_9Cl$C_4H_9Cl$), P must be But-2-ene. Addition of HCl yields 2-chlorobutane, which possesses a chiral center. mathrmCH_3-mathrmCH=mathrmCH-mathrmCH_3 xrightarrowmathrmHCl mathrmCH_3-mathrmCH_2-mathrmC^*mathrmH(mathrmCl)-mathrmCH_3 quad text(P is But-2-ene, Q is 2-chlorobutane)$$\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3 \xrightarrow{\mathrm{HCl}} \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{C}^*\mathrm{H}(\mathrm{Cl})-\mathrm{CH}_3 \quad \text{(P is But-2-ene, Q is 2-chlorobutane)}$$

Compound Q reacts with mathrmNH_3$\mathrm{NH}_3$ to undergo nucleophilic substitution forming a primary amine R: mathrmCH_3-mathrmCH_2-mathrmCH(mathrmCl)-mathrmCH_3 xrightarrowmathrmNH_3 mathrmCH_3-mathrmCH_2-mathrmCH(mathrmNH_2)-mathrmCH_3 quad text(R is Butan-2-amine)$$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}(\mathrm{Cl})-\mathrm{CH}_3 \xrightarrow{\mathrm{NH}_3} \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}(\mathrm{NH}_2)-\mathrm{CH}_3 \quad \text{(R is Butan-2-amine)}$$

Diazotization of primary aliphatic amines followed by hydrolysis yields an alcohol via a carbocation intermediate (S): mathrmCH_3-mathrmCH_2-mathrmCH(mathrmNH_2)-mathrmCH_3 xrightarrowmathrmNaNO_2 / mathrmHCl / mathrmH_2mathrmO mathrmCH_3-mathrmCH_2-mathrmCH(mathrmOH)-mathrmCH_3 quad text(S is Butan-2-ol)$$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}(\mathrm{NH}_2)-\mathrm{CH}_3 \xrightarrow{\mathrm{NaNO}_2 / \mathrm{HCl} / \mathrm{H}_2\mathrm{O}} \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}_3 \quad \text{(S is Butan-2-ol)}$$

### Pattern Recognition An optically active alkyl halide formed from a C_4H_8$C_4H_8$ alkene with HX is a classic pointer to 2-halobutane originating from But-2-ene or But-1-ene. Diazotization of primary aliphatic amines (R-NH_2$R-NH_2$) with NaNO_2/HCl$NaNO_2/HCl$ yields alcohols (R-OH$R-OH$) with possible rearrangements, though here a secondary carbocation is already stable. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes Class 12 Chemistry: Amines

### Core Logic The reaction proceeds via an S_N1$S_N1$ mechanism for activated benzyl halides or nucleophilic substitution rate depends on carbocation stability / electronic effects of substituents (-OH, -textNH_2$-\text{NH}_2$, -textNO_2$-\text{NO}_2$, etc.). - Amino and hydroxy substituents strongly activate through +M effect. - Nitro groups strongly deactivate through -M effect. ### Step 1: Final Conclusion The correct order of reactivity is b > a > d > c$b > a > d > c$, corresponding to option (2). ### Pattern Recognition Sees: benzyl halide reactivity with cyanide. Trap: Confusing polar/inductive effects with resonance effects of substituents. ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes

### Core Logic Among the given compounds, sec-butyl chloride (2-chlorobutane) is optically active and contains a chiral center. Molecular formula of 2-chlorobutane rightarrow textC_4textH_9textCl$\rightarrow \text{C}_4\text{H}_9\text{Cl}$. Molar mass = 4(12) + 9(1) + 35.5 = 48 + 9 + 35.5 = 92.5 text g/mol$= 4(12) + 9(1) + 35.5 = 48 + 9 + 35.5 = 92.5 \text{ g/mol}$. ### Step 1: Calculating Percentage of Carbon text\% of C = frac4892.5 times 100 = 51.89\% approx 52\%$$\text{\% of C} = \frac{48}{92.5} \times 100 = 51.89\% \approx 52\%$$ ### Pattern Recognition Sees: optical isomerism identification combined with elemental percentage composition calculation. Trap: Selecting the wrong halogen derivative for optical activity. ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes

### Related Formula Rate propto textNucleophilicity$$Rate \propto \text{Nucleophilicity}$$ ### Core Logic The reaction of CH_3Br$CH_3Br$ (a primary halide) occurs via the S_N2$S_N2$ mechanism. The rate depends directly on the nucleophilicity of the attacking species. Methanol is a polar protic solvent. In polar protic solvents, larger halide ions are better nucleophiles because they are less solvated. Therefore, I^-$I^-$ is a stronger nucleophile than F^-$F^-$. For the alkoxide and phenoxide, C_2H_5O^-$C_2H_5O^-$ is a stronger nucleophile than C_6H_5O^-$C_6H_5O^-$ because the negative charge on phenoxide is delocalized over the benzene ring, reducing its electron-donating ability. ### Step 1: Final Order Combining these factors, the overall order of nucleophilicity in methanol is: I^- > C_2H_5O^- > C_6H_5O^- > F^-$$I^{-} > C_{2}H_{5}O^{-} > C_{6}H_{5}O^{-} > F^{-}$$ ### Pattern Recognition In polar protic solvents: Size dominates (down a group, nucleophilicity increases). For species of similar size, less resonance stabilization means stronger nucleophile. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes

### Related Formula textRate of S_N1 propto textStability of intermediate carbocation$$\text{Rate of } S_N1 \propto \text{Stability of intermediate carbocation}$$ ### Core Logic Evaluate the stability of the carbocations formed upon dissociation of the bromide ion: Structure (I) and (II) form carbocations at bridgehead positions of rigid bicyclic systems. According to Bredt's rule, these are highly unstable because they cannot adopt the necessary planar sp^2$sp^2$ geometry. Among them, (I) has an additional methyl group which provides slight +I stabilization compared to (II). So, (II) is even less stable than (I). Structure (III) forms a tertiary carbocation, but it's not a rigid bridgehead preventing planarity, so it's significantly more stable than I and II. Structure (IV) is a highly stable trityl-like carbocation where the positive charge is stabilized by extensive resonance with the adjacent phenyl ring(s). ### Step 1: Final Ordering Stability order: (II) < (I) < (III) < (IV)$(II) < (I) < (III) < (IV)$. Thus, the S_N1$S_{N}1$ rate follows the same order. ### Pattern Recognition Bridgehead halides effectively do not undergo S_N1$S_N1$ (or S_N2$S_N2$) reactions due to Bredt's rule. Resonance stabilized carbocations always dominate. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes