A toxic compound “A” when reacted with NaCN in aqueous acidic medium yields an edible cooking component and food preservative “B”. “B” is converted to “C” by diborane and can be used as an additive to petrol to reduce emission. “C” upon reaction with oleum at 140°C$140^{\circ}\mathrm{C}$ yields an inhalable anesthetic “D”. Identify “A”, “B”, “C” and “D”, respectively.

Core Logic

A secondary alcohol features a hydroxyl group (-OH) attached to a carbon atom that is bonded to two other carbon atoms (2°$2^{\circ}$ carbon). Evaluating the structures based on visual analysis: Compounds II, IV, and V contain at least one hydroxyl group on a 2°$2^{\circ}$ carbon. Therefore, 3 compounds feature a secondary alcohol.

Pattern Recognition

Spot the carbon carrying the -OH group. Count the C-C bonds attached to it. 1 = Primary, 2 = Secondary, 3 = Tertiary. Phenols are aromatic and distinct from secondary alcohols.

Chapter Mix

Class 12 Chemistry: Alcohols, Phenols and Ethers

Core Logic

Based on empirical data for (x): Molar Mass = 2 × Vapour Density = 2 × 47 = 94~g/mol$\text{Molar Mass} = 2 \times \text{Vapour Density} = 2 \times 47 = 94\mathrm{~g/mol}$. C = 76.6% (76.6/12) ≈ 6.38$\text{C} = 76.6\% \implies (76.6/12) \approx 6.38$ H = 6.38% (6.38/1) ≈ 6.38$\text{H} = 6.38\% \implies (6.38/1) \approx 6.38$ O = 17.02% (17.02/16) ≈ 1.06$\text{O} = 17.02\% \implies (17.02/16) \approx 1.06$ Ratio C:H:O ≈ 6:6:1 C₆H₆O$\text{C:H:O} \approx 6:6:1 \implies \mathrm{C_6H_6O}$ (Phenol).

Step 1: Identifying Compounds

Compound (x) is Phenol. It reacts with CO₂/NaOH$\mathrm{CO}_2/\mathrm{NaOH}$ via Kolbe-Schmitt reaction to form Salicylic Acid as the major product (Compound y).

Step 2: Checking Options

(1) Both Phenol and Salicylic acid dissolve in NaOH$\mathrm{NaOH}$. (True) (2) Salicylic acid (y) has a -COOH$-COOH$ group and dissolves in NaHCO₃$\mathrm{NaHCO}_3$, evolving CO₂$\mathrm{CO}_2$ gas. (True) (3) Salicylic acid (y) is much more acidic than Phenol (x) due to the presence of the carboxylic acid group and intramolecular hydrogen bonding in its conjugate base. Therefore, the statement "x is more acidic than y" is INCORRECT. (4) Being highly aromatic compounds, both burn with a sooty flame. (True)

Pattern Recognition

The reaction sequence Phenol + CO₂/NaOH arrow Salicylic acid$\mathrm{Phenol} + \mathrm{CO_2/NaOH} \rightarrow \mathrm{Salicylic \; acid}$ is a staple (Kolbe's Reaction). Always test functional group properties: Phenols dissolve in NaOH$\mathrm{NaOH}$ but not NaHCO₃$\mathrm{NaHCO}_3$, while Carboxylic acids dissolve in both.

Chapter Mix

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

Core Logic

For a compound to release CO₂$\mathrm{CO}_2$ upon reaction with aqueous NaHCO₃$\mathrm{NaHCO}_3$, its acidity must be strictly greater than that of carbonic acid (H₂CO₃$\mathrm{H_2CO_3}$).

Let's evaluate the acidity of the options:

• Picric Acid (2,4,6-trinitrophenol - Option A): Extremely acidic (pKₐ ≈ 0.38$pK_a \approx 0.38$) due to three strong electron-withdrawing nitro groups. Reacts with NaHCO₃$\mathrm{NaHCO}_3$ easily.
• 4-Nitrobenzoic Acid (Option B): Carboxylic acids generally have pKₐ ≈ 4--5$pK_a \approx 4\text{--}5$. More acidic than carbonic acid (pKₐ ≈ 6.3$pK_a \approx 6.3$). Reacts with NaHCO₃$\mathrm{NaHCO}_3$.
• Anilinium Chloride (Option 3): A salt of a strong acid and weak base. The anilinium ion is quite acidic (pKₐ ≈ 4.6$pK_a \approx 4.6$) and easily decomposes NaHCO₃$\mathrm{NaHCO}_3$.
• 4-Nitrophenol (Option D): Only one nitro group is present. Its acidity (pKₐ ≈ 7.15$pK_a \approx 7.15$) is weaker than carbonic acid. Hence, it does not react with sodium bicarbonate to yield effervescence of CO₂$\mathrm{CO}_2$.

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.

Chapter Mix

Class 12 Chemistry: Alcohols, Phenols and Ethers Class 12 Chemistry: Carboxylic Acids

Core Logic

Statement I: Dimethyl ether (CH₃OCH₃$CH_3OCH_3$) 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 7.5 g$7.5\text{ g}$ per 100 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:

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 (-OH$-\text{OH}$ group in alcohols) prevents the use of alkali metals like sodium for moisture removal.

Chapter Mix

Class 12 Chemistry: Alcohols, Phenols and Ethers

Core Logic

Williamson Ether Synthesis operates strictly via an SN2$S_N2$ mechanism. To optimize ether formation yields, the alkyl halide component MUST be primary (1^°$1^\circ$) or methyl to evade spatial shielding restrictions.

Let us review the alkyl halide across the options:

• Option 1: Ethyl bromide (EtBr$\text{EtBr}$) is 1^°$1^\circ$ arrow$\rightarrow$ Excellent ether yield via substitution.
• Option 2: Methyl bromide (CH₃Br$\text{CH}_3\text{Br}$) is a unhindered methyl halide arrow$\rightarrow$ High substitution efficiency.
• Option 3: n-Propyl bromide (n-PrBr$\text{n-PrBr}$) is 1^°$1^\circ$ arrow$\rightarrow$ Clean substitution pathway.
• Option 4: sec-Butyl bromide (sec-BuBr$\text{sec-BuBr}$) is a secondary (2^°$2^\circ$) alkyl halide. When a sterically hindered 2^°$2^\circ$ 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 (SN2$S_N2$).

  

Pattern Recognition

Williamson Synthesis Rule: Alkoxide can be as massive and complex as desired (3^°$3^\circ$ or branched), but the Alkyl Halide MUST be unhindered (1^°$1^\circ$ or methyl). If the halide is 2^°$2^\circ$ or 3^°$3^\circ$, elimination wins, forming an alkene instead of an ether.

Chapter Mix

Class 12 Chemistry: Alcohols, Phenols and Ethers