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

B.

C.

D.

E.

Choose the correct answer from the options given below:

### Related Formula Reimer-Tiemann Reaction: Phenol xrightarrowCHCl_3, KOH$\xrightarrow{CHCl_3, KOH}$ o-hydroxybenzaldehyde (Salicylaldehyde) as major product. ### Core Logic Statement I claims that p-hydroxybenzaldehyde is the major product. This is incorrect. In the Reimer-Tiemann reaction, ortho-hydroxybenzaldehyde (salicylaldehyde) is the major product due to the stabilization of the intermediate and product via intramolecular hydrogen bonding.

Statement II states that the mixture of p-hydroxybenzaldehyde and o-hydroxybenzaldehyde can be separated by steam distillation. This is true. Ortho-hydroxybenzaldehyde forms intramolecular hydrogen bonds, making it more volatile (steam volatile), whereas the para-isomer forms intermolecular hydrogen bonds, increasing its boiling point and making it non-steam volatile. ### Step 1: Final Conclusion Therefore, Statement I is false but Statement II is true. ### Pattern Recognition Always remember: Ortho-isomers capable of intramolecular H-bonding are steam volatile. Para-isomers exhibit intermolecular H-bonding and are not steam volatile. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Alcohols Phenols and Ethers

### Core Logic For a compound to release mathrmCO_2$\mathrm{CO}_2$ upon reaction with aqueous mathrmNaHCO_3$\mathrm{NaHCO}_3$, its acidity must be strictly greater than that of carbonic acid (mathrmH_2CO_3$\mathrm{H_2CO_3}$). Let's evaluate the acidity of the options: 1. **Picric Acid** (2,4,6-trinitrophenol - Option A): Extremely acidic (pK_a approx 0.38$pK_a \approx 0.38$) due to three strong electron-withdrawing nitro groups. Reacts with mathrmNaHCO_3$\mathrm{NaHCO}_3$ easily. 2. **4-Nitrobenzoic Acid** (Option B): Carboxylic acids generally have pK_a approx 4text--5$pK_a \approx 4\text{--}5$. More acidic than carbonic acid (pK_a approx 6.3$pK_a \approx 6.3$). Reacts with mathrmNaHCO_3$\mathrm{NaHCO}_3$. 3. **Anilinium Chloride** (Option 3): A salt of a strong acid and weak base. The anilinium ion is quite acidic (pK_a approx 4.6$pK_a \approx 4.6$) and easily decomposes mathrmNaHCO_3$\mathrm{NaHCO}_3$. 4. **4-Nitrophenol** (Option D): Only one nitro group is present. Its acidity (pK_a approx 7.15$pK_a \approx 7.15$) is weaker than carbonic acid. Hence, it does not react with sodium bicarbonate to yield effervescence of mathrmCO_2$\mathrm{CO}_2$. ### Pattern Recognition Acid-bicarbonate test shortcut: - All carboxylic acids and picric acid give a positive bicarbonate test. - Normal phenols and mono/di-nitrophenols are too weak to decompose bicarbonate. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Alcohols, Phenols and Ethers Class 12 Chemistry: Carboxylic Acids

### Core Logic Statement I: Dimethyl ether (CH_3OCH_3$CH_3OCH_3$) is highly soluble in water because its smaller alkyl chain allows substantial hydrogen bonding with water molecules. In contrast, diethyl ether has a larger hydrophobic ethyl group which drastically reduces its water solubility (to about 7.5text g$7.5\text{ g}$ per 100text mL$100\text{ mL}$). Thus, Statement I is true. Statement II: Sodium metal (Na$Na$) reacts violently with alcohols like ethyl alcohol to release hydrogen gas: 2textC_2textH_5textOH + 2textNa rightarrow 2textC_2textH_5textONa + textH_2uparrow$$2\text{C}_2\text{H}_5\text{OH} + 2\text{Na} \rightarrow 2\text{C}_2\text{H}_5\text{ONa} + \text{H}_2\uparrow$$ Since diethyl ether has no active acidic hydrogen, it does not react with sodium metal. Hence, sodium can dry diethyl ether but cannot be used for ethyl alcohol. Statement II is true. ### Pattern Recognition Ethers are miscible with water primarily when the non-polar alkyl parts are very small. Active hydrogen presence (-textOH$-\text{OH}$ group in alcohols) prevents the use of alkali metals like sodium for moisture removal. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Alcohols, Phenols and Ethers

### Core Logic Williamson Ether Synthesis operates strictly via an **S_N2$S_N2$ mechanism**. To optimize ether formation yields, the alkyl halide component MUST be primary (1^circ$1^\circ$) or methyl to evade spatial shielding restrictions. Let us review the alkyl halide across the options: * Option 1: Ethyl bromide (textEtBr$\text{EtBr}$) is 1^circ$1^\circ$ rightarrow$\rightarrow$ Excellent ether yield via substitution. * Option 2: Methyl bromide (textCH_3textBr$\text{CH}_3\text{Br}$) is a unhindered methyl halide rightarrow$\rightarrow$ High substitution efficiency. * Option 3: n-Propyl bromide (textn-PrBr$\text{n-PrBr}$) is 1^circ$1^\circ$ rightarrow$\rightarrow$ Clean substitution pathway. * Option 4: sec-Butyl bromide (textsec-BuBr$\text{sec-BuBr}$) is a secondary (2^circ$2^\circ$) alkyl halide. When a sterically hindered 2^circ$2^\circ$ halide is treated with a strongly basic alkoxide reagent like isobutoxide, **alkene elimination (E2$E2$) competes aggressively and dominates** as the major pathway over nucleophilic substitution (S_N2$S_N2$).

### Pattern Recognition Williamson Synthesis Rule: Alkoxide can be as massive and complex as desired (3^circ$3^\circ$ or branched), but the Alkyl Halide MUST be unhindered (1^circ$1^\circ$ or methyl). If the halide is 2^circ$2^\circ$ or 3^circ$3^\circ$, elimination wins, forming an alkene instead of an ether. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Alcohols, Phenols and Ethers