Both acetaldehyde and acetone (individually) undergo which of the following reactions? A. Iodoform Reaction B. Cannizaro Reaction C. Aldol condensation D. Pollen's Test E. Clemmensen Reduction Choose the correct answer from the options given below:

### 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