Given below are two statements : Statement (I): Vanillin

will react with NaOH and also with Tollen's reagent. Statement (II) : Vanillin

will undergo self aldol condensation very easily. In the light of the above statements, choose the most appropriate answer from the options given below:

### Core Logic Reactivity in nucleophilic addition reactions is governed by a combination of **steric hindrance** and **electronic effects** around the electrophilic carbonyl carbon: 1. Ketones are significantly less reactive than aldehydes due to the bulkiness and electron-donating inductive effect (+I$+I$) of their two alkyl/aryl groups. Thus, **acetophenone** has the lowest reactivity. 2. For substituted benzaldehydes, electron-withdrawing groups heighten the partial positive charge on the carbonyl carbon, accelerating nucleophilic attack. Conversely, electron-donating groups suppress reactivity. Symmetry breakdown structures are shown below: - Acetophenone:

- p-tolualdehyde:

- Benzaldehyde:

- p-nitrobenzaldehyde:

- Methoxy/methyl donors decrease reactivity: textp-tolualdehyde < textbenzaldehyde$\text{p-tolualdehyde} < \text{benzaldehyde}$ - Nitro group (-mathrmNO_2$-\mathrm{NO}_2$) acts as a strong electron-withdrawing agent via both -M$-M$ and -I$-I$ pathways, maximizing the electrophilic nature of the carbonyl site. Therefore, **p-nitrobenzaldehyde** is the most reactive. Thus, the correct order of reactivity is: textacetophenone < textp-tolualdehyde < textbenzaldehyde < textp-nitrobenzaldehyde$$\text{acetophenone} < \text{p-tolualdehyde} < \text{benzaldehyde} < \text{p-nitrobenzaldehyde}$$ ### Pattern Recognition Aldehydes naturally exhibit higher reactivity than ketones. Electron-withdrawing groups (-mathrmNO_2$-\mathrm{NO}_2$) accelerate addition pathways, whereas electron-donating groups (-mathrmCH_3$-\mathrm{CH}_3$) impede them. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

### Core Logic Analyzing the retro-synthesis path step-by-step: 1. The objective compound is 1-acetylcyclopentene. 2. Performing reductive ozonolysis (O_3, Zn/H_2O$O_3, Zn/H_2O$) on 1-methylcyclohexene (Option A) symmetrically breaks the internal endocyclic double bond to form heptane-2,6-dione, a dicarbonyl system.

3. Adding a base intermediate trigger (OH^-, Delta$OH^{-}, \Delta$) drives an intramolecular aldol condensation: the methyl group carbanion at position 1 attacks the carbon 6 ketone site. This ring-closing event effectively drops water to synthesize the 5-membered cyclopentene core molecule attached to the acetyl unit.

### Pattern Recognition Counting carbon coordinates is essential. Reductive cleavage transforms a 6-membered ring into an open heptane system, which easily self-condenses into a stable 5-membered ring attached to a methyl ketone side chain. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

### Core Logic Breaking down the multi-step reaction path sequence: 1. Hydration of propyne using Kucherov's trigger condition (Hg^2+, H_2SO_4$Hg^{2+}, H_2SO_4$) adds water across the triple bond via Markovnikov's rule. The intermediate enol undergoes tautomerization to yield **acetone** (CH_3-CO-CH_3$CH_3-CO-CH_3$). 2. Reacting acetone with HCN$HCN$ drives nucleophilic addition at the carbonyl carbon, forming a **cyanohydrin** intermediate: CH_3-C(OH)(CH_3)-CN$CH_3-C(OH)(CH_3)-CN$. 3. Introducing a reducing agent (H_2/Ni$H_2/Ni$) selectively converts the nitrile group (-CN$-CN$) into a primary amine side chain (-CH_2-NH_2$-CH_2-NH_2$). The final synthesized structure is: mathrmCH_3-mathrmC(mathrmOH)(mathrmCH_3)-mathrmCH_2-mathrmNH_2$\mathrm{CH}_3-\mathrm{C}(\mathrm{OH})(\mathrm{CH}_3)-\mathrm{CH}_2-\mathrm{NH}_2$.

### Pattern Recognition Alkyne hydration produces a ketone carbonyl system. Cyanohydrin synthesis introduces a carbon coordinate, which subsequently reduces to a primary aliphatic amine functional group. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

### Related Formula textMoles of gas at STP = fracVtext in Liters22.4 text L/mol$$\text{Moles of gas at STP} = \frac{V\text{ in Liters}}{22.4 \text{ L/mol}}$$ ### Core Logic The balanced acid-base reaction equation is: C_6H_5COOH + NaHCO_3 rightarrow C_6H_5COO^-Na^+ + H_2O + CO_2$$C_{6}H_{5}COOH + NaHCO_{3} \rightarrow C_{6}H_{5}COO^{-}Na^{+} + H_{2}O + CO_{2}$$ This stoichiometry shows a 1:1$1:1$ molar ratio between benzoic acid and the released carbon dioxide gas. Calculate the total moles of evolved CO_2$CO_2$ gas at standard conditions: textmoles of CO_2 = frac11.2text L22.4text L mol^-1 = 0.5text moles$$\text{moles of } CO_2 = \frac{11.2\text{ L}}{22.4\text{ L mol}^{-1}} = 0.5\text{ moles}$$ Thus, the reaction consumed exactly 0.5text moles$0.5\text{ moles}$ of benzoic acid (C_6H_5COOH$C_6H_5COOH$, molecular weight = 122text g/mol$122\text{ g/mol}$): textmass consumed (X) = 0.5 times 122 = 61text grams$$\text{mass consumed (X)} = 0.5 \times 122 = 61\text{ grams}$$ ### Pattern Recognition Carboxylic acids react with sodium bicarbonate in a straightforward 1:1$1:1$ molar ratio, releasing exactly 1 mole of carbon dioxide gas per mole of acid group. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

### Related Formula Side-chain oxidation criteria: textAr-CH(R)_2 xrightarrowtexthot KMnO_4 textAr-COOH$$\text{Ar-CH(R)}_2 \xrightarrow{\text{hot } KMnO_4} \text{Ar-COOH}$$ Requires the presence of at least one benzylic hydrogen atom on the aromatic side chain structure. ### Core Logic Alkyl side chains on a benzene ring are oxidized entirely down to a carboxylic acid group (benzoic acid) by strong oxidizing agents like hot alkaline KMnO_4$KMnO_4$, provided the benzylic carbon contains at least one hydrogen atom. Evaluating the structures from the diagram: 1. Toluene (contains 3$3$ benzylic H) rightarrow$\rightarrow$ **Yields benzoic acid** 2. Ethylbenzene (contains 2$2$ benzylic H) rightarrow$\rightarrow$ **Yields benzoic acid** 3. Isopropylbenzene / Cumene (contains 1$1$ benzylic H) rightarrow$\rightarrow$ **Yields benzoic acid** 4. tert-Butylbenzene (contains 0$0$ benzylic H) rightarrow$\rightarrow$ **Resists oxidation** 5. Isobutylbenzene (contains 2$2$ benzylic H) rightarrow$\rightarrow$ **Yields benzoic acid** 6. 2-Phenylpropan-2-ol (contains 0$0$ benzylic H, tertiary alcohol center) rightarrow$\rightarrow$ **Resists oxidation** 7. n-Propylbenzene (contains 2$2$ benzylic H) rightarrow$\rightarrow$ **Yields benzoic acid** 8. The last biphenyl derivative undergoes complex disruption or ring cleavages and does not cleanly yield simple benzoic acid under standard monocyclic oxidation definitions. ### Step 1: Counting Valid Targets The compounds that undergo oxidation to form benzoic acid are toluene, ethylbenzene, isopropylbenzene, isobutylbenzene, and n-propylbenzene. Total count = 5 compounds.

### Pattern Recognition Look immediately at the benzylic carbon (the carbon bonded directly to the ring). If it is a quaternary center (like in tert-butylbenzene) or lacks a hydrogen atom entirely, mark it as unreactive to hot KMnO_4$KMnO_4$ side-chain oxidation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids