### Core Logic The balanced chemical equation for the synthesis of dibenzalacetone is: 2 Ph-CHO + CH3COCH3 xrightarrowtextStrong Base Ph-CH=CH-CO-CH=CH-Ph + 2H2O$2 Ph-CHO + CH{3}COCH{3} xrightarrow{\text{Strong Base}} Ph-CH=CH-CO-CH=CH-Ph + 2H{2}O$ Let's calculate the moles of reactants: textMolar mass of benzaldehyde (PhCHO) = 106text g/mol$\text{Molar mass of benzaldehyde } (PhCHO) = 106\text{ g/mol}$ textMoles of benzaldehyde used = frac5.3106 = 0.05text mol = frac120text mol$\text{Moles of benzaldehyde used} = \frac{5.3}{106} = 0.05\text{ mol} = \frac{1}{20}\text{ mol}$

According to the reaction stoichiometry, 2text moles of PhCHO$2\text{ moles of } PhCHO$ yield 1text mole of dibenzalacetone$1\text{ mole of dibenzalacetone}$. textTheoretical moles of product = frac0.052 = 0.025text mol$\text{Theoretical moles of product} = \frac{0.05}{2} = 0.025\text{ mol}$ ### Step 1: Yield Evaluation $textMolar mass of dibenzalacetone (C17H14O) = 234text g/mol$$\text{Molar mass of dibenzalacetone } (C{17}H{14}O) = 234\text{ g/mol}$ textTheoretical mass = 0.025 times 234 = 5.85text g$\text{Theoretical mass} = 0.025 \times 234 = 5.85\text{ g}$ textPercentage yield = fractextActual masstextTheoretical mass times 100 = frac3.515.85 times 100 = 60%$\text{Percentage yield} = \frac{\text{Actual mass}}{\text{Theoretical mass}} \times 100 = \frac{3.51}{5.85} \times 100 = 60%$ ### Pattern Recognition Always remember the stoichiometric ratio: It takes 2 moles of benzaldehyde to condense with 1 mole of acetone to form the symmetrical dibenzalacetone product. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

### Core Logic The reaction sequence indicates that molecule "P" undergoes an acid-catalyzed rearrangement to produce alkene/alcohol intermediate "Q". Subsequent ozonolysis breaks down the double bond system, and alkaline reflux sets up an intramolecular aldol condensation sequence to form the cyclic ketone system "R". Following the detailed ring contraction/expansion step templates outlined below:

### Pattern Recognition Sees: Acidic rearrangement rightarrow$\rightarrow$ ozonolysis rightarrow$\rightarrow$ intramolecular aldol condensation. Shortcut: Work backwards from the dicarbonyl fragments formed after opening the final product ring system. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

### Core Logic Let us check each option pathway: - **A. Iodoform Reaction:** Positive for both because both contain the mathrmCH_3-mathrmC=mathrmO$\mathrm{CH}_3-\mathrm{C}=\mathrm{O}$ methyl ketone fragment. - **B. Cannizaro Reaction:** Negative for both because both contain alpha$\alpha$-hydrogens. - **C. Aldol Condensation:** Positive for both because they have alpha$\alpha$-hydrogens available for enolization. - **D. Pollen's Test (Tollen's Test):** Positive only for acetaldehyde (aldehyde); negative for acetone (ketone). - **E. Clemmensen Reduction:** Positive for both as they contain reducible carbonyl groups. Thus, both react via A, C, and E. ### Pattern Recognition Sees: Functional comparison of Acetaldehyde and Acetone. Shortcut: Ketones do not respond to Tollen's test, which instantly eliminates choices featuring statement D. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

### Core Logic The iodoform test requires compounds containing either a methyl ketone group (textCH_3textCO-$\text{CH}_3\text{CO}-$) or a methyl carbinol structural unit (textCH_3textCH(OH)-$\text{CH}_3\text{CH(OH)}-$). Let us audit the provided compounds: 1. **Ethanol** (textCH_3textCH_2textOH$\text{CH}_3\text{CH}_2\text{OH}$): Contains textCH_3textCH(OH)-$\text{CH}_3\text{CH(OH)}-$ ightarrow$ ightarrow$ **Positive** 2. **Isopropyl alcohol** (textCH_3textCH(OH)CH_3$\text{CH}_3\text{CH(OH)CH}_3$): Contains textCH_3textCH(OH)-$\text{CH}_3\text{CH(OH)}-$ ightarrow$ ightarrow$ **Positive** 3. **Bromoacetone** (textCH_3textCOCH_2textBr$\text{CH}_3\text{COCH}_2\text{Br}$): Contains textCH_3textCO-$\text{CH}_3\text{CO}-$ ightarrow$ ightarrow$ **Positive** 4. **2-Butanol** (textCH_3textCH(OH)CH_2textCH_3$\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3$): Contains textCH_3textCH(OH)-$\text{CH}_3\text{CH(OH)}-$ ightarrow$ ightarrow$ **Positive** 5. **2-Butanone** (textCH_3textCOCH_2textCH_3$\text{CH}_3\text{COCH}_2\text{CH}_3$): Contains textCH_3textCO-$\text{CH}_3\text{CO}-$ ightarrow$ ightarrow$ **Positive** 6. **Butanal** (textCH_3textCH_2textCH_2textCHO$\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO}$): **Negative** 7. **2-Pentanone** (textCH_3textCOCH_2textCH_2textCH_3$\text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_3$): Contains textCH_3textCO-$\text{CH}_3\text{CO}-$ ightarrow$ ightarrow$ **Positive** 8. **3-Pentanone** (textCH_3textCH_2textCOCH_2textCH_3$\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3$): **Negative** 9. **Pentanal** (textCH_3textCH_2textCH_2textCH_2textCHO$\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CHO}$): **Negative** 10. **3-Pentanol** (textCH_3textCH_2textCH(OH)CH_2textCH_3$\text{CH}_3\text{CH}_2\text{CH(OH)CH}_2\text{CH}_3$): **Negative** ### Step 1: Summation The molecules that **cannot** give the iodoform reaction are: Butanal, 3-Pentanone, Pentanal, and 3-Pentanol. This gives a total count of exactly 4 molecules. ### Pattern Recognition Shortcut: Filter for names ending with '-anal' or having ketones/alcohols at positions higher than 2 (e.g., 3-pentanone, 3-pentanol). These lack the vital terminal methyl group adjacent to the functional group. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

### Related Formula textCompounds with CH_3-CH(OH)- text or CH_3-CO- text groups undergo the iodoform reaction to form CHI_3 downarrow text (Yellow Solid)$\text{Compounds with } CH_3-CH(OH)- \text{ or } CH_3-CO- \text{ groups undergo the iodoform reaction to form } CHI_3 \downarrow \text{ (Yellow Solid)}$ ### Core Logic Let's check the structural groups of each given option: - **(A)** CH_3 - CH(OH) - C_2H_5$CH_3 - CH(OH) - C_2H_5$: Contains the methylcarbinol group (CH_3-CH(OH)-$CH_3-CH(OH)-$). Gives a positive iodoform test. - **(B)** CH_3 - CH_2 - CH_2 - OH$CH_3 - CH_2 - CH_2 - OH$: Linear primary alcohol, does not contain the required group. - **(C)** CH_3 - CO - C_2H_5$CH_3 - CO - C_2H_5$: Contains the methyl ketone group (CH_3-CO-$CH_3-CO-$). Gives a positive iodoform test. - **(D)** CH_3 - OH$CH_3 - OH$: Methanol does not give the test. - **(E)** CH_3 - CH_2 - H$CH_3 - CH_2 - H$: Ethane does not give the test. Thus, only **(A)** and **(C)** yield the yellow precipitate of iodoform (CHI_3$CHI_3$). ### Step 1: Chemical Equations The balanced haloform pathways occur as follows: CH_3-CH(OH)-CH_2-CH_3 xrightarrowtextNaOI/NaOH CHI_3downarrow + textCH_3text-CH_2text-COO^-textNa^+$CH_3-CH(OH)-CH_2-CH_3 \xrightarrow{\text{NaOI/NaOH}} CHI_3\downarrow + \text{CH}_3\text{-CH}_2\text{-COO}^-\text{Na}^+$ CH_3-CO-CH_2-CH_3 xrightarrowtextNaOI/NaOH CHI_3downarrow + textCH_3text-CH_2text-COO^-textNa^+$CH_3-CO-CH_2-CH_3 \xrightarrow{\text{NaOI/NaOH}} CHI_3\downarrow + \text{CH}_3\text{-CH}_2\text{-COO}^-\text{Na}^+$ ### Pattern Recognition The iodoform test specifically isolates methyl ketones or secondary methyl alcohols. Scan dynamically for a terminal -CH_3$-CH_3$ affixed directly to a carbonyl oxygen index (C=O$C=O$) or a hydroxyl carbon (CH-OH$CH-OH$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids Class 12 Chemistry: Alcohols, Phenols and Ethers