Choose the correct set of reagents for the following conversion: Ethyl benzene 4-bromostyrene$$\text{Ethyl benzene} \longrightarrow \text{4-bromostyrene}$$ {{Q_IMG1}}

Core Logic

Since P (C₄H₈$C_4H_8$) reacts with HCl to give an optically active compound Q (C₄H₉Cl$C_4H_9Cl$), P must be But-2-ene. Addition of HCl yields 2-chlorobutane, which possesses a chiral center.

Compound Q reacts with NH₃$\mathrm{NH}_3$ to undergo nucleophilic substitution forming a primary amine R:

Diazotization of primary aliphatic amines followed by hydrolysis yields an alcohol via a carbocation intermediate (S):

Pattern Recognition

An optically active alkyl halide formed from a C₄H₈$C_4H_8$ 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-NH₂$R-NH_2$) with NaNO₂/HCl$NaNO_2/HCl$ yields alcohols (R-OH$R-OH$) with possible rearrangements, though here a secondary carbocation is already stable.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes Class 12 Chemistry: Amines

Core Logic

The reaction proceeds via an SN1$S_N1$ mechanism for activated benzyl halides or nucleophilic substitution rate depends on carbocation stability / electronic effects of substituents (-OH, -NH₂$-\text{NH}_2$, -NO₂$-\text{NO}_2$, 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 arrow C₄H₉Cl$\rightarrow \text{C}_4\text{H}_9\text{Cl}$. Molar mass = 4(12) + 9(1) + 35.5 = 48 + 9 + 35.5 = 92.5 g/mol$= 4(12) + 9(1) + 35.5 = 48 + 9 + 35.5 = 92.5 \text{ g/mol}$.

Step 1: Calculating Percentage of Carbon

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

Core Logic

The reaction of CH₃Br$CH_3Br$ (a primary halide) occurs via the SN2$S_N2$ 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, C₂H₅O^-$C_2H_5O^-$ is a stronger nucleophile than C₆H₅O^-$C_6H_5O^-$ 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:

Pattern Recognition

In polar protic solvents: Size dominates (down a group, nucleophilicity increases). For species of similar size, less resonance stabilization means stronger nucleophile.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes

Related Formula

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 sp²$sp^2$ 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 SN1$S_{N}1$ rate follows the same order.

Pattern Recognition

Bridgehead halides effectively do not undergo SN1$S_N1$ (or SN2$S_N2$) reactions due to Bredt's rule. Resonance stabilized carbocations always dominate.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes