The product B formed in the following reaction sequence is :

Core Logic

In an SN1$\text{S}_\text{N}1$ pathway, a planar carbocation intermediate is produced. Attack by the nucleophile can take place with equal probability from either side, resulting in complete/partial racemisation. In an SN2$\text{S}_\text{N}2$ pathway, the nucleophile attacks exclusively from the backside opposite the leaving group, causing an absolute structural inversion (Walden inversion).

Pattern Recognition

SN1 arrow$S_N1 \rightarrow$ planar intermediate carbocation arrow$\rightarrow$ Racemisation. SN2 arrow$S_N2 \rightarrow$ direct backside launch arrow$\rightarrow$ Inversion.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes

Core Logic

Assertion (A): In phenols, the C-O$C-O$ bond possesses partial double bond character due to resonance (the lone pair of oxygen delocalizes into the benzene ring). Because of this strong C-O$C-O$ bond, nucleophilic substitution reactions where a halide ion would replace the hydroxyl group do not occur under normal conditions. Thus, aryl halides cannot be prepared directly from phenols by reaction with HX$HX$. The statement is True.

Reason (R): Phenols do NOT react violently with halogen acids. In fact, they practically do not react with halogen acids (HX$HX$) to form aryl halides because the C-O$C-O$ bond is difficult to break. The statement is False.

Step 1: Visualization

Given reason is false.

Step 2: Conclusion

Assertion (A) is correct but Reason (R) is false.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes Class 12 Chemistry: Alcohols Phenols and Ethers

Core Logic

Free radical chlorination of 2-chlorobutane yields different constitutional isomers of dichlorobutane, each potentially existing as various stereoisomers.

Substrate: CH₃-CH(Cl)-CH₂-CH₃$CH_3-CH(Cl)-CH_2-CH_3$ (exists as 2 enantiomers: d$d$ and l$l$) Chlorination can occur at 4 different carbons:

• At C1: CH₂(Cl)-CH(Cl)-CH₂-CH₃$CH_2(Cl)-CH(Cl)-CH_2-CH_3$ (1,2-dichlorobutane) arrow$\rightarrow$ Two chiral centers, unsymmetrical. Forms 4 optically active isomers (2 pairs of enantiomers).
• At C2: CH₃-C(Cl)₂-CH₂-CH₃$CH_3-C(Cl)_2-CH_2-CH_3$ (2,2-dichlorobutane) arrow$\rightarrow$ No chiral center. Achiral (0 optically active).
• At C3: CH₃-CH(Cl)-CH(Cl)-CH₃$CH_3-CH(Cl)-CH(Cl)-CH_3$ (2,3-dichlorobutane) arrow$\rightarrow$ Symmetrical with 2 chiral centers. Forms 3 stereoisomers: 1 meso (achiral) and 2 optically active (1 enantiomeric pair).
• At C4: CH₃-CH(Cl)-CH₂-CH₂(Cl)$CH_3-CH(Cl)-CH_2-CH_2(Cl)$ (1,3-dichlorobutane, numbering from other end) arrow$\rightarrow$ 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.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques

Core Logic

Statement I: Rate of SN2 ∝ [R-X][Nu^-]$S_N2 \propto [R-X][Nu^-]$. Therefore, SN2$S_N2$ 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 SN2$S_N2$ 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 SN1$S_N1$ 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 (SN2$S_N2$). Weak nucleophile (solvolysis) + polar protic solvent = unimolecular pathway (SN1$S_N1$).

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes

Core Logic

A vinylic halide is a compound where the halogen atom is directly bonded to an sp²$sp^2$ 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.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes