The product (A) formed in the following reaction sequence is : mathrm C H _ 3 - mathrm C equiv mathrm C H xrightarrow [ (mathrm i i) mathrm H _ 2 / mathrm N i ](mathrm i) mathrm H g ^ 2 +, mathrm H _ 2 mathrm S O _ 4 (A)$$\mathrm {C H} _ {3} - \mathrm {C} \equiv \mathrm {C H} \xrightarrow [ (\mathrm {i i}) \mathrm {H} _ {2} / \mathrm {N i} ]{(\mathrm {i}) \mathrm {H} g ^ {2 +}, \mathrm {H} _ {2} \mathrm {S O} _ {4}} (A)$$

### Core Logic Fehling's test is a mild oxidizing test used primarily to distinguish aliphatic aldehydes from ketones and aromatic aldehydes. - **Aliphatic aldehydes** (like methanal, acetaldehyde, cyclohexane carbaldehyde) give a positive Fehling's test (formation of red-brown precipitate of Cu_2O$Cu_2O$). - **Aromatic aldehydes** (like benzaldehyde, 4-nitrobenzaldehyde) lack alpha-hydrogens in a purely aliphatic environment and are not sufficiently easily oxidized to give a positive Fehling's test. - **Ketones** (like acetone, acetophenone) generally do not give a positive Fehling's test (except alpha$\alpha$-hydroxy ketones). ### Step 1: Evaluation of Given Compounds 1. Benzaldehyde: Aromatic aldehyde rightarrow$\rightarrow$ Negative 2. Acetaldehyde (CH_3CHO$CH_3CHO$): Aliphatic aldehyde rightarrow$\rightarrow$ Positive 3. Acetone: Ketone rightarrow$\rightarrow$ Negative 4. Acetophenone: Ketone rightarrow$\rightarrow$ Negative 5. Methanal (HCHO$HCHO$): Aliphatic aldehyde rightarrow$\rightarrow$ Positive 6. 4-nitrobenzaldehyde: Aromatic aldehyde rightarrow$\rightarrow$ Negative 7. Cyclohexane carbaldehyde: Aliphatic aldehyde rightarrow$\rightarrow$ Positive The compounds giving a positive test are Acetaldehyde, Methanal, and Cyclohexane carbaldehyde. ### Pattern Recognition Tollens' reagent oxidizes ALL aldehydes (aliphatic + aromatic). Fehling's reagent is weaker and only oxidizes ALIPHATIC aldehydes. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes Ketones and Carboxylic Acids

### Core Logic m-chlorobenzaldehyde lacks alpha-hydrogen atoms. Therefore, when treated with concentrated base like 50\%$50\%$ KOH, it undergoes a disproportionation redox reaction known as the Cannizzaro reaction. Two molecules of the aldehyde react: one gets oxidized to the corresponding carboxylate ion (m-chlorobenzoate ion), and the other gets reduced to the corresponding alcohol (m-chlorobenzyl alcohol). ### Step 1: Reaction The reaction proceeds as:

### Pattern Recognition No alpha$\alpha$-hydrogen + Conc. Alkali (50\%$50\%$ KOH/NaOH) = Cannizzaro (Oxidation to salt of carboxylic acid + Reduction to primary alcohol). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes Ketones and Carboxylic Acids

### Related Formula R-COCl + H_2 xrightarrowPd/BaSO_4 R-CHO + HCl$$R-COCl + H_2 \xrightarrow{Pd/BaSO_4} R-CHO + HCl$$ ### Core Logic The reaction depicts the partial reduction of an acid chloride (benzoyl chloride) to an aldehyde (benzaldehyde) using hydrogen gas in the presence of a poisoned palladium catalyst (Pd supported on BaSO_4$BaSO_4$).

This specific reaction is known as the Rosenmund reduction. ### Pattern Recognition Acid Chloride + H_2, Pd/BaSO_4$H_2, Pd/BaSO_4$ rightarrow$\rightarrow$ Aldehyde is strictly the Rosenmund reduction. The BaSO_4$BaSO_4$ poisons the catalyst to prevent over-reduction to an alcohol. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

### Core Logic Decode the IUPAC name: 4-Methylpent-2-enal 1. Word root: 'pent' rightarrow$\rightarrow$ 5 carbon principal chain. 2. Primary suffix: '2-en' rightarrow$\rightarrow$ Double bond starting at carbon 2. 3. Secondary suffix: 'al' rightarrow$\rightarrow$ Aldehyde group (-CHO) at carbon 1. 4. Substituent: '4-Methyl' rightarrow$\rightarrow$ A methyl group (-CH3) at carbon 4. ### Step 1: Drafting the structure Numbering starts from the aldehyde carbon. C^5 - C^4 - C^3 = C^2 - C^1(=O)H$$C^5 - C^4 - C^3 = C^2 - C^1(=O)H$$ Attach the methyl at C^4$C^4$: CH_3 - CH(CH_3) - CH = CH - CHO$$CH_3 - CH(CH_3) - CH = CH - CHO$$ This matches Option 4. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Core Logic Pathway 1 (Upper): Toluene is treated with Reagent 'A' and acetic anhydride (CH_3CO)_2O$(CH_3CO)_2O$ to form an intermediate (benzylidene diacetate), which on hydrolysis gives benzaldehyde. The reagent used here is Chromic oxide (CrO_3$CrO_3$). Thus, A is CrO_3$CrO_3$. Pathway 2 (Lower): Toluene is treated with Reagent 'B' in CS_2$CS_2$ to form a chromium complex intermediate, which on hydrolysis yields benzaldehyde. This is the Etard reaction, and the reagent used is Chromyl chloride (CrO_2Cl_2$CrO_2Cl_2$). Thus, B is CrO_2Cl_2$CrO_2Cl_2$.

### Step 1: Selection Therefore, A = CrO_3$CrO_3$ and B = CrO_2Cl_2$CrO_2Cl_2$. ### Pattern Recognition Etard reaction always uses Chromyl chloride (CrO_2Cl_2$CrO_2Cl_2$). Oxidation of toluene with acetic anhydride uses Chromic acid (CrO_3$CrO_3$). Both stop the oxidation at the aldehyde stage via intermediate formation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids