Which of the following statements are TRUE about Haloform reaction?
A. Sodium hypochlorite reacts with KI to give KOI.
B. KOI is a reducing agent.
C. α, β$\alpha, \beta$-unsaturated methylketone (CH₃-CH=CH-CO-CH₃)$(CH_{3}-CH=CH-CO-CH_{3})$ will give iodoform reaction.
D. Isopropyl alcohol will not give iodoform test.
E. Methanoic acid will give positive iodoform test.
Choose the correct answer from the options given below:
A.A, C & E only$\text{A, C & E only}$
B.A, B & C only$\text{A, B & C only}$
C.A & C only$\text{A & C only}$
D.B, D & E only$\text{B, D & E only}$
Solution & Explanation
Core Logic
Evaluate each statement:
A. NaOCl$NaOCl$ (Sodium hypochlorite) reacts with KI$KI$ to produce KOI$KOI$ and NaCl$NaCl$. This statement is True.
NaOCl + KI arrow NaCl + KOI$$NaOCl + KI \rightarrow NaCl + KOI$$
B. KOI$KOI$ acts as a mild oxidizing agent, not a reducing agent. It is used to oxidize secondary alcohols to ketones in the iodoform reaction. This statement is False.
C. The molecule CH₃-CH=CH-CO-CH₃$CH_3-CH=CH-CO-CH_3$ has a terminal methyl ketone group (-CO-CH₃$-CO-CH_3$). The adjacent double bond does not interfere with the cleavage of the methyl group during the haloform test. It will yield yellow iodoform. This statement is True.
D. Isopropyl alcohol is CH₃-CH(OH)-CH₃$CH_3-CH(OH)-CH_3$. Since it is a secondary alcohol with an adjacent methyl group, it gets oxidized to acetone in situ and therefore gives a positive iodoform test. The statement says it will not give the test, which is False.
E. Methanoic acid (HCOOH$HCOOH$) does not have a terminal methyl group attached to a carbonyl carbon, so it cannot undergo the haloform reaction. This statement is False.
Step 1: Final Selection
Only statements A and C are True.
Pattern Recognition
Alcohols with the CH₃-CH(OH)-$CH_3-CH(OH)-$ moiety always test positive for iodoform. Reagents containing hypohalites (OCl^-$OCl^-$, OI^-$OI^-$) are strong oxidizing agents, frequently used in organic synthesis precisely for their oxidative power.
Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
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.
