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

Let's analyze each statement regarding chlorobenzene vs. chlorocyclohexane:

A. Magnitude of negative charge: In chlorobenzene, the lone pair of Cl undergoes +M resonance with the ring. This delocalizes electron density into the ring, reducing the net negative charge on chlorine compared to the purely -I withdrawing Cl in chlorocyclohexane. (False) B. C-Cl double bond character: The +M effect (resonance) imparts partial double bond character to the C-Cl bond in chlorobenzene. (True) C. Less polar bond: Because resonance acts opposite to the inductive (-I) effect, the net dipole moment of chlorobenzene (1.5 - 1.6 D) is lower than that of chlorocyclohexane (approx 2.1 D). So the bond is less polar. (True) D. C-Cl bond is longer: Because of the partial double bond character, the C-Cl bond in chlorobenzene is actually shorter (169 pm) than in chlorocyclohexane (177 pm). (False) E. Hybridization: The carbon bonded to Cl in chlorobenzene is sp²$sp^2$ hybridized, while in chlorocyclohexane it is sp³$sp^3$ hybridized. (True)

Step 1: Conclusion

The correct statements are B, C, and E only.

Pattern Recognition

Resonance decreases bond length and decreases polarity in aryl halides compared to alkyl halides.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes

Related Formula

Core Logic

Step 1: Compound [R] gives positive Iodoform test and negative Tollen's test $\implies$ [R] is a methyl ketone (acetophenone derivative, Ph-CO-CH₃$\text{Ph-CO-CH}_3$).

Step 2: Hydration of alkyne [Q] gives methyl ketone [R] $\implies$ [Q] is phenylacetylene (Ph-C$\text{Ph-C}\equiv\text{CH}$).

Step 3: Double elimination of [P] using excess NaNH₂$\text{NaNH}_2$ gives phenylacetylene [Q] $\implies$ [P] is geminal dibromide 1,1-dibromo-1-phenylethane (Ph-C(Br)₂CH₃$\text{Ph-C(Br)}_2\text{CH}_3$).

Pattern Recognition

Sees: Positive iodoform + negative Tollen's $\implies$ methyl ketone. Shortcut: Oxymercuration of terminal alkyne yields methyl ketone, identifying gem-dibromide at benzylic position.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

Core Logic

First, identify all structural isomers of C₅H₁₁Br$C_5H_{11}Br$. Aqueous KOH will simply replace the -Br$-Br$ with an -OH$-OH$ group via nucleophilic substitution without rearrangement. Then, check which of the resulting C₅H₁₁OH$C_5H_{11}OH$ products possess a chiral center.

Step 1: Structural Isomers and Products

Pentyl skeleton isomers:

• n$n$-pentyl: 1-bromopentane arrow$\rightarrow$ 1-pentanol (Achiral), 2-bromopentane arrow$\rightarrow$ 2-pentanol (Chiral), 3-bromopentane arrow$\rightarrow$ 3-pentanol (Achiral).
• Isopentyl skeleton: 1-bromo-3-methylbutane arrow$\rightarrow$ 3-methylbutan-1-ol (Achiral), 2-bromo-3-methylbutane arrow$\rightarrow$ 3-methylbutan-2-ol (Chiral), 2-bromo-2-methylbutane arrow$\rightarrow$ 2-methylbutan-2-ol (Achiral), 1-bromo-2-methylbutane arrow$\rightarrow$ 2-methylbutan-1-ol (Chiral).
• Neopentyl skeleton: 1-bromo-2,2-dimethylpropane arrow$\rightarrow$ 2,2-dimethylpropan-1-ol (Achiral).

Step 2: Counting Optically Active Products

The products exhibiting optical isomerism are:

• 2-pentanol
• 3-methylbutan-2-ol
• 2-methylbutan-1-ol

Total structurally distinct products exhibiting optical isomerism is 3.

Pattern Recognition

To quickly spot a chiral center in a standard alkane chain, look for a carbon bonded to -OH$-OH$, -H$-H$, a methyl group, and a longer alkyl chain. Any symmetry kills chirality (like 3-pentanol).

Chapter Mix

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

Core Logic

(A) Vinyl Chloride is a haloalkene where chlorine is directly attached to the sp²$sp^2$ carbon of a double bond: CH₂=CHCl$CH_2=CHCl$. So, A matches with III. (B) Benzyl chloride is a compound where chlorine is attached to an sp³$sp^3$ hybridized carbon directly linked to a benzene ring (C₆H₅CH₂Cl$C_6H_5CH_2Cl$). Structure IV represents this. So, B matches with IV. (C) Alkyl chloride represents a standard saturated aliphatic chain attached to a chlorine atom, such as isopropyl chloride: CH₃-CH(Cl)CH₃$CH_3-CH(Cl)CH_3$. So, C matches with II. (D) Allyl chloride features a chlorine attached to an sp³$sp^3$ carbon that is adjacent to a carbon-carbon double bond: CH₂=CH-CH₂Cl$CH_2=CH-CH_2Cl$. So, D matches with I.

Step 1: Final Conclusion

The correctly matched pairs are A-III, B-IV, C-II, D-I.

Pattern Recognition

Vinyl = directly on double bond. Allyl = one carbon away from double bond. Benzyl = one carbon away from phenyl ring.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes

Core Logic

Statement I: CH₂=CH-Cl$CH_2=CH-Cl$ is vinyl chloride. The lone pair on chlorine participates in resonance with the adjacent double bond. This imparts a partial double bond character to the C-Cl$\text{C-Cl}$ bond, making it shorter and stronger than the pure single C-Cl$\text{C-Cl}$ bond in CH₃-CH₂-Cl$CH_3-CH_2-Cl$ (ethyl chloride).

Statement II: The given molecule is a chiral, tertiary alkyl chloride. When it undergoes hydrolysis, it follows an SN1$S_N1$ mechanism because the resulting tertiary carbocation (stabilized by resonance from the phenyl ring and hyperconjugation) is very stable. An SN1$S_N1$ reaction proceeds through a planar carbocation intermediate. Attack by water occurs from both faces with near equal probability, resulting in a racemic mixture.

Step 1: Final Conclusion

Statement I is true but Statement II is false.

Pattern Recognition

Vinyl and aryl halides have partial double bond character due to resonance. SN1$S_N1$ at a chiral center leads to racemization (optically inactive product mix).

Chapter Mix

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