3,3-Dimethyl-2-butanol cannot be prepared by: A.

B.

C.

D.

E.

Choose the correct answer from the options given below:

Core Logic

Anisole (Ph-O-CH₃$Ph-O-CH_3$) contains an aryl-oxygen bond which possesses partial double bond character due to resonance stabilization with the aromatic ring. When treated with HBr$HBr$, protonation yields an oxonium ion. The nucleophile Br⁻$Br^{-}$ attacks via an SN2$S_N2$ pathway at the smaller, less hindered methyl group, cleaving the O-CH₃$O-CH_3$ bond to form Phenol (PhOH$PhOH$) and CH₃Br$CH_3Br$.

Pattern Recognition

Aromatic sp²$sp^2$ C-O$C-O$ bonds are exceptionally strong and cannot be cleaved by nucleophilic attack from Hal⁻$Hal^{-}$. Therefore, the oxygen always stays attached to the benzene ring, yielding phenol.

Chapter Mix

Class 12 Chemistry: Alcohols, Phenols and Ethers

Core Logic

Let's track the conversions row by row based on the description:

• Toxic compound A is methanol (CH₃OH$CH_3OH$). Reacting it via acidic NaCN sequences effectively converts it eventually into acetic acid (CH₃COOH$CH_3COOH$), which is an edible cooking component and food preservative (vinegar) B.
• Reduction of acetic acid (B) with diborane (B₂H₆$B_2H_6$) yields ethanol (CH₃CH₂OH$CH_3CH_2OH$) C. Ethanol is used as a biofuel additive in petrol.
• Ethanol (C) upon heating with oleum/concentrated sulfuric acid at 140°C$140^{\circ}\mathrm{C}$ undergoes intermolecular dehydration to yield diethyl ether (C₂H₅-O-C₂H₅$C_2H_5-O-C_2H_5$) D, which functions as an inhalable anesthetic.

Step 1: Chemical Reaction Sequence Flow

The full sequential progression corresponds exactly to:

Pattern Recognition

Key temperature indicator: Dehydration of ethanol at 140°C$140^{\circ}\mathrm{C}$ yields diethyl ether (ether synthesis), whereas 170°C$170^{\circ}\mathrm{C}$ yields ethene gas. Combined with the food preservative clue (acetic acid), the sequence locks instantly to option (3).

Chapter Mix

Class 12 Chemistry: Alcohols, Phenols and Ethers

Core Logic

An electrophile seeks electrons. Higher electron density in the benzene ring makes it more susceptible (reactive) to electrophilic attack. The ring's electron density is governed by the inductive (I) and mesomeric/resonance (M) effects of the attached groups.

Step 1: Evaluate Substituent Effects

• Cl$Cl$ (in chlorobenzene): shows weak +M effect but strong -I effect, causing net deactivation.
• CH₃$CH_3$ (in toluene): shows +I effect and hyperconjugation, slightly activating the ring.
• COOH$COOH$ (in benzoic acid): shows strong -M and -I effect, highly deactivating.
• OH$OH$ (in phenol): shows strong +M effect which dominates its weak -I effect, strongly activating the ring.

Step 2: Conclusion

Phenol has the highest electron density in the ring among the given options due to the strong +M effect of the -OH group. Thus, it is most easily attacked by an electrophile.

Pattern Recognition

Reactivity towards Electrophilic Aromatic Substitution (EAS): Strong +M (-OH, -NH2) > Weak +I/Hyperconjugation (-CH3) > Halogens (-Cl, net deactivating) > Strong -M (-COOH, -NO2).

Chapter Mix

Class 11 Chemistry: Hydrocarbons Class 12 Chemistry: Alcohols, Phenols and Ethers

Related Formula

Core Logic

This chemical sequences details the well-known Reimer-Tiemann reaction mechanism. The treatment of phenol with alkaline chloroform generates a dichlorocarbene intermediate (:CCl₂$:\text{CCl}_2$), which acts as an electrophile and specifically attacks the ortho position of the phenoxide ring system.

Subsequent basic hydrolysis converts the functional intermediate to an aldehyde block, providing 2-hydroxybenzaldehyde as the major final product.

Step 1: Visual Pathway Validation

The step-by-step schematic transforms structural blocks through standard intermediates:

Pattern Recognition

Chloroform + Base + Phenol yields formylation at the ortho site, producing salicylaldehyde (2-hydroxybenzaldehyde).

Chapter Mix

Class 12 Chemistry: Organic Compounds Containing Oxygen

Related Formula

Core Logic

Breaking down the multistep pathway:

• Step A: The ester group is selectively reduced to an aldehyde using DIBAL-H$\text{DIBAL-H}$ at low temperature without affecting other domains.
• Step B: An intramolecular Aldol condensation occurs in the presence of base (NaOH$\text{NaOH}$) to generate the bicyclic α,β$\alpha,\beta$-unsaturated carbonyl framework.
• Step C: Clemmensen reduction (using amalgamated zinc and hydrochloric acid, Zn(Hg)/HCl$\text{Zn(Hg)/HCl}$) reduces the ketone group to a hydrocarbon block.

Step 1: Verification

The step-by-step mechanism proceeds precisely as illustrated below:

Pattern Recognition

DIBAL-H stops cleanly at the aldehyde phase from an ester precursor, preparing the molecule perfectly for subsequent aldol ring closures.

Chapter Mix

Class 12 Chemistry: Organic Compounds Containing Oxygen