Chemical structures related to the options provided for the iodoform differentiation test.Iodoform test can differentiate between
A. Methanol and Ethanol
B. CH₃COOH$CH_{3}COOH$ and CH₃CH₂COOH$CH_{3}CH_{2}COOH$
C. Cyclohexene and cyclohexanone
D. Diethyl ether and Pentan-3-one
E. Anisole and acetone
Choose the correct answer from the options given below:
The iodoform test gives a positive result for compounds containing a methyl ketone group (-CO-CH₃$-CO-CH_3$) or a secondary alcohol with an adjacent methyl group (-CH(OH)-CH₃$-CH(OH)-CH_3$), which gets oxidized to a methyl ketone under the reaction conditions.
A. Methanol vs Ethanol: Ethanol (CH₃CH₂OH$CH_3CH_2OH$) contains the -CH(OH)-CH₃$-CH(OH)-CH_3$ terminal structure and gives a positive test. Methanol (CH₃OH$CH_3OH$) does not. (Can differentiate)
B. Neither CH₃COOH$CH_3COOH$ nor CH₃CH₂COOH$CH_3CH_2COOH$ gives the iodoform test since carboxylic acids do not undergo this reaction despite having a terminal methyl in acetic acid. (Cannot differentiate)
C. Neither cyclohexene nor cyclohexanone has a methyl group adjacent to the carbonyl/alcohol carbon. (Cannot differentiate)
D. Neither diethyl ether (CH₃CH₂-O-CH₂CH₃$CH_3CH_2-O-CH_2CH_3$) nor pentan-3-one (CH₃CH₂-CO-CH₂CH₃$CH_3CH_2-CO-CH_2CH_3$) has a methyl ketone group. (Cannot differentiate)
E. Acetone (CH₃-CO-CH₃$CH_3-CO-CH_3$) gives a positive test. Anisole (Ph-O-CH₃$Ph-O-CH_3$) does not. (Can differentiate)
Step 1: Final Selection
Only pairs A and E contain exactly one compound that gives a positive iodoform test, allowing for chemical differentiation.
Pattern Recognition
Look strictly for CH₃-C=O$CH_3-C=O$ or CH₃-CH-OH$CH_3-CH-OH$. If a molecule lacks these exact terminal fragments, it will fail the iodoform test.
Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Class 12 Chemistry: Alcohols, Phenols and Ethers
Chemical structures related to the options provided for the iodoform differentiation test.Chemical structures related to the options provided for the iodoform differentiation test.
Keywords:#Iodoform test can differentiate between#JEE Main 2026 Evening Q58#Aldehydes, Ketones and Carboxylic Acids JEE Main 2026#Iodoform Reaction JEE Main 2026#Iodoform test#Methanol#Ethanol#Cyclohexene
More Aldehydes, Ketones and Carboxylic Acids Previous-Year Questions — Page 9
Qjee_main_2024_30_jan_morningPreparation of Aldehydes
This reduction reaction is known as:
The image shows the catalytic hydrogenation of benzoyl chloride to benzaldehyde in the presence of Pd-BaSO4.
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 BaSO₄$BaSO_4$).
The image shows the catalytic hydrogenation of benzoyl chloride to benzaldehyde in the presence of Pd-BaSO4.
This specific reaction is known as the Rosenmund reduction.
Pattern Recognition
Acid Chloride + H₂, Pd/BaSO₄$H_2, Pd/BaSO_4$arrow$\rightarrow$ Aldehyde is strictly the Rosenmund reduction. The BaSO₄$BaSO_4$ poisons the catalyst to prevent over-reduction to an alcohol.
Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Pathway 1 (Upper): Toluene is treated with Reagent 'A' and acetic anhydride (CH₃CO)₂O$(CH_3CO)_2O$ to form an intermediate (benzylidene diacetate), which on hydrolysis gives benzaldehyde. The reagent used here is Chromic oxide (CrO₃$CrO_3$). Thus, A is CrO₃$CrO_3$.
Pathway 2 (Lower): Toluene is treated with Reagent 'B' in CS₂$CS_2$ to form a chromium complex intermediate, which on hydrolysis yields benzaldehyde. This is the Etard reaction, and the reagent used is Chromyl chloride (CrO₂Cl₂$CrO_2Cl_2$). Thus, B is CrO₂Cl₂$CrO_2Cl_2$.
The image shows the oxidation of toluene to benzaldehyde via two different pathways requiring reagents A and B.
Step 1: Selection
Therefore, A = CrO₃$CrO_3$ and B = CrO₂Cl₂$CrO_2Cl_2$.
Pattern Recognition
Etard reaction always uses Chromyl chloride (CrO₂Cl₂$CrO_2Cl_2$). Oxidation of toluene with acetic anhydride uses Chromic acid (CrO₃$CrO_3$). Both stop the oxidation at the aldehyde stage via intermediate formation.
Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Ethanal (CH₃CHO$CH_3CHO$) reacts with semicarbazide (H₂N-NH-CO-NH₂$H_2N-NH-CO-NH_2$) via nucleophilic addition followed by elimination of water to form a semicarbazone.
Step 1: Product Analysis
The product is Ethanal semicarbazone: CH₃-CH=N-NH-CO-NH₂$CH_3-CH=N-NH-CO-NH_2$.
Counting the nitrogen atoms in this structure:
The imine nitrogen (=N-$=N-$)
The amine nitrogen (-NH-$-NH-$)
The amide nitrogen (-NH₂$-NH_2$)
Total = 3 Nitrogen atoms.
Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Qjee_main_2024_31_jan_eveningPreparation of Aldehydes and Ketones
Identify the name reaction.
The image shows the conversion of benzene to benzaldehyde using CO, HCl, and Anhydrous AlCl3/CuCl.
The reaction of benzene with carbon monoxide (CO$CO$) and hydrogen chloride (HCl$HCl$) in the presence of anhydrous aluminium chloride (AlCl₃$AlCl_3$) and cuprous chloride (CuCl$CuCl$) to give benzaldehyde is known as the Gatterman-Koch reaction.
The image shows the conversion of benzene to benzaldehyde using CO, HCl, and Anhydrous AlCl3/CuCl.
Pattern Recognition
CO + HCl arrow$CO + HCl \rightarrow$ Formyl chloride intermediate (in situ) with Lewis acid arrow$\rightarrow$ formylation of benzene. This is definitively the Gatterman-Koch formylation.
Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
More Aldehydes, Ketones and Carboxylic Acids Questions — jee_main_2026_23_january_evening
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
