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

Step 1: Free radical side-chain halogenation of toluene. Toluene reacts with Cl₂$Cl_2$ in the presence of light (hν$h\nu$) to undergo substitution on the methyl group. Under typical conditions intended to yield an aldehyde later, di-chlorination occurs forming benzal chloride (A).

Step 2: Hydrolysis. Benzal chloride upon hydrolysis with H₂O$H_2O$ at 373 K yields a gem-diol intermediate which is unstable and loses water to form Benzaldehyde (B).

Step 1: Identify Structures

(A) = Benzal chloride (C₆H₅CHCl₂$C_6H_5CHCl_2$) (B) = Benzaldehyde (C₆H₅CHO$C_6H_5CHO$)

Pattern Recognition

Toluene Cl₂, hν$\xrightarrow{Cl_2, h\nu}$ targets the side chain. If the next step is hydrolysis to an aldehyde, you must have stopped at the gem-dihalide stage (CHCl₂$CHCl_2$).

Chapter Mix

Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids Class 11 Chemistry: Hydrocarbons

Core Logic

Let's analyze the transformation happening in each reaction:

(A) CH₃(CH₂)₅COOC₂H₅ arrow CH₃(CH₂)₅CHO$CH_3(CH_2)_5COOC_2H_5 \rightarrow CH_3(CH_2)_5CHO$ An ester is reduced to an aldehyde. This is a selective reduction achieved using DIBAL-H (Diisobutylaluminium hydride) followed by hydrolysis. Thus, (A) arrow$\rightarrow$ (IV).

(B) C₆H₅COC₆H₅ arrow C₆H₅CH₂C₆H₅$C_6H_5COC_6H_5 \rightarrow C_6H_5CH_2C_6H_5$ A ketone (carbonyl group >C=O$>C=O$) is fully reduced to an alkane (>CH₂$>CH_2$) methylene group. This is the Clemmensen reduction, which uses Zinc amalgam and concentrated HCl. Thus, (B) arrow$\rightarrow$ (II).

(C) C₆H₅CHO arrow C₆H₅CH(OH)CH₃$C_6H_5CHO \rightarrow C_6H_5CH(OH)CH_3$ Benzaldehyde (aldehyde) is converted into a secondary alcohol with an extra methyl group. This is a nucleophilic addition of a Grignard reagent (CH₃MgBr$CH_3MgBr$) followed by hydrolysis. Thus, (C) arrow$\rightarrow$ (I).

(D) CH₃COCH₂COOC₂H₅ arrow CH₃CH(OH)CH₂COOC₂H₅$CH_3COCH_2COOC_2H_5 \rightarrow CH_3CH(OH)CH_2COOC_2H_5$ A ketone group is reduced to a secondary alcohol while the ester group remains intact. NaBH₄$NaBH_4$ is a mild reducing agent that reduces aldehydes and ketones but generally does not touch esters. Thus, (D) arrow$\rightarrow$ (III).

Pattern Recognition

Ester arrow$\rightarrow$ Aldehyde = DIBAL-H Ketone arrow$\rightarrow$ Alkane = Clemmensen (Zn(Hg)/HCl$Zn(Hg)/HCl$) or Wolff-Kishner Carbonyl arrow$\rightarrow$ Alcohol with carbon chain extension = Grignard Reagent Ketone arrow$\rightarrow$ Alcohol (leaving ester intact) = NaBH₄$NaBH_4$

Chapter Mix

Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

Core Logic

The reaction sequence proceeds in three distinct steps from the starting material, styrene (Ph-CH=CH₂$Ph-CH=CH_2$).

Step 1: Acid-catalyzed Hydration Styrene reacts with H₂O, H^+$H_2O, H^+$ to undergo electrophilic addition. Protonation yields the more stable secondary benzylic carbocation. Attack by water followed by deprotonation gives 1-phenylethanol (Ph-CH(OH)-CH₃$Ph-CH(OH)-CH_3$).

Step 2: Oxidation 1-phenylethanol is a secondary alcohol. Treatment with chromium trioxide (CrO₃$CrO_3$, Jones reagent condition) oxidizes the secondary alcohol to a ketone. This yields acetophenone (Ph-CO-CH₃$Ph-CO-CH_3$).

Step 3: Wolff-Kishner Reduction Acetophenone is treated with hydrazine (NH₂-NH₂$NH_2-NH_2$) and a strong base (KOH$KOH$) under heating. This is the classic Wolff-Kishner reduction, which completely reduces the carbonyl group (C=O$C=O$) to a methylene group (-CH₂-$-CH_2-$). The final product is ethylbenzene (Ph-CH₂-CH₃$Ph-CH_2-CH_3$).

Step 1: Overall Reaction Pathway

The final product A is ethylbenzene.

Chapter Mix

Class 11 Chemistry: Hydrocarbons Class 12 Chemistry: Alcohols Phenols and Ethers Class 12 Chemistry: Aldehydes Ketones and Carboxylic Acids

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 Cu₂O$Cu_2O$).
• 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 α$\alpha$-hydroxy ketones).

Step 1: Evaluation of Given Compounds

• Benzaldehyde: Aromatic aldehyde arrow$\rightarrow$ Negative
• Acetaldehyde (CH₃CHO$CH_3CHO$): Aliphatic aldehyde arrow$\rightarrow$ Positive
• Acetone: Ketone arrow$\rightarrow$ Negative
• Acetophenone: Ketone arrow$\rightarrow$ Negative
• Methanal (HCHO$HCHO$): Aliphatic aldehyde arrow$\rightarrow$ Positive
• 4-nitrobenzaldehyde: Aromatic aldehyde arrow$\rightarrow$ Negative
• Cyclohexane carbaldehyde: Aliphatic aldehyde arrow$\rightarrow$ 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.

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 α$\alpha$-hydrogen + Conc. Alkali (50%$50\%$ KOH/NaOH) = Cannizzaro (Oxidation to salt of carboxylic acid + Reduction to primary alcohol).

Chapter Mix

Class 12 Chemistry: Aldehydes Ketones and Carboxylic Acids