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. alpha, beta$\alpha, \beta$-unsaturated methylketone (CH_3-CH=CH-CO-CH_3)$(CH_{3}-CH=CH-CO-CH_{3})$ 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.textA, C & E only$\text{A, C & E only}$
B.textA, B & C only$\text{A, B & C only}$
C.textA & C only$\text{A & C only}$
D.textB, 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 rightarrow 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 CH_3-CH=CH-CO-CH_3$CH_3-CH=CH-CO-CH_3$ has a terminal methyl ketone group (-CO-CH_3$-CO-CH_3$). 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 CH_3-CH(OH)-CH_3$CH_3-CH(OH)-CH_3$. 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 CH_3-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.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Keywords:#TRUE about Haloform reaction#JEE Main 2026 Evening Q62#Aldehydes, Ketones and Carboxylic Acids JEE Main 2026#Haloform Reaction JEE Main 2026
More Aldehydes, Ketones and Carboxylic Acids Previous-Year Questions
Q61jee_main_2026_21_jan_morningChemical Reactions of Aldehydes and Ketones
An organic compound “P” of molecular formula C_6H_12O_3$C_{6}H_{12}O_{3}$ gives positive Iodoform test but negative Tollen’s test. When “P” is treated with dilute acid, it produces “Q”. “Q” gives positive Tollen’s test and also iodoform test. The structure of “P” is :
D.mathrmCH_3-CO-CO-CH_3 text (with acetal structure)$\mathrm{CH_3-CO-CO-CH_3} \text{ (with acetal structure)}$
Solution
### Core Logic
Compound P (C_6H_12O_3$C_6H_{12}O_3$) gives a positive iodoform test, indicating it has a methyl ketone group (CH_3CO-$CH_3CO-$). It gives a negative Tollen's test, indicating no free aldehyde group.
On acidic hydrolysis, P yields Q. Compound Q gives both positive iodoform test and Tollen's test, meaning Q contains both a methyl ketone and an aldehyde group.
Looking at the options, if P is an acetal of an aldehyde, acidic hydrolysis will regenerate the aldehyde.
Option 2 is mathrmCH_3-CO-CH_2-CH(OCH_3)_2$\mathrm{CH_3-CO-CH_2-CH(OCH_3)_2}$ (an acetal of aldehyde).
This compound 'P' has a CH_3CO-$CH_3CO-$ group (positive iodoform) and an acetal (protected aldehyde, negative Tollen's).
On hydrolysis:
mathrmCH_3-CO-CH_2-CH(OCH_3)_2 xrightarrowmathrmH_2O/H^+ mathrmCH_3-CO-CH_2-CHO + 2mathrmCH_3OH$$\mathrm{CH_3-CO-CH_2-CH(OCH_3)_2} \xrightarrow{\mathrm{H_2O/H^+}} \mathrm{CH_3-CO-CH_2-CHO} + 2\mathrm{CH_3OH}$$
The product Q (mathrmCH_3-CO-CH_2-CHO$\mathrm{CH_3-CO-CH_2-CHO}$) has a methyl ketone (positive iodoform test) and an aldehyde (positive Tollen's test).
Chemical Reactions of Aldehydes and Ketones diagram for Q61 - JEE Main 2026 Morning
### Pattern Recognition
Whenever an aldehyde test becomes positive *after* hydrolysis, it points to a protected aldehyde, usually an acetal or hemiacetal. A compound with molecular formula C_n H_2n O_3$C_n H_{2n} O_3$ often represents a keto-acetal.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Class 12 Chemistry: Alcohols, Phenols and Ethers
Q57jee_main_2026_21_jan_eveningName Reactions
Match List-I with List-II.
List-I (Reagents)
List-II (Reaction Name Involving Aldehydes)
A. textH_2, textPd-textBaSO_4$\text{H}_2, \text{Pd}-\text{BaSO}_4$
I. Etard Reaction
B. textSnCl_2, textHCl$\text{SnCl}_2, \text{HCl}$
II. Rosenmund Reduction
C. textCrO_2textCl_2, textCS_2$\text{CrO}_2\text{Cl}_2, \text{CS}_2$
III. Gatterman–Koch Reaction
D. textCO, textHCl, textAnhyd. textAlCl_3$\text{CO}, \text{HCl}, \text{Anhyd. } \text{AlCl}_3$
IV. Stephen Reaction
Choose the correct answer from the options given below:
### Core Logic
- A. textH_2, textPd-textBaSO_4 rightarrow$\text{H}_2, \text{Pd}-\text{BaSO}_4 \rightarrow$ II. Rosenmund Reduction
- B. textSnCl_2, textHCl rightarrow$\text{SnCl}_2, \text{HCl} \rightarrow$ IV. Stephen Reaction
- C. textCrO_2textCl_2, textCS_2 rightarrow$\text{CrO}_2\text{Cl}_2, \text{CS}_2 \rightarrow$ I. Etard Reaction
- D. textCO, textHCl, textAnhyd. textAlCl_3 rightarrow$\text{CO}, \text{HCl}, \text{Anhyd. } \text{AlCl}_3 \rightarrow$ III. Gatterman–Koch Reaction
### Step 1: Final Conclusion
Combining the correct matching gives A-II, B-IV, C-I, D-III, which is option (4).
### Pattern Recognition
Sees: standard name reactions for aldehyde preparation.
Trap: Confusing Etard reagent with Gatterman-Koch or Stephen reduction.
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
### Core Logic
(A) NH_2-NH_2 / KOH$NH_2-NH_2 / KOH$: Hydrazine with base is used in Wolff-Kishner Reduction to reduce carbonyls to alkanes. (A rightarrow$\rightarrow$ III)
(B) [Ag(NH_3)_2]OH$[Ag(NH_3)_2]OH$: Ammoniacal silver nitrate is Tollen's reagent, used in Tollen's Test to differentiate aldehydes from ketones. (B rightarrow$\rightarrow$ I)
(C) Aqueous CuSO_4$CuSO_4$ + Sodium Potassium Tartrate + KOH: This is Fehling's solution (mixture of Fehling A and Fehling B). Used in Fehling's Test. (C rightarrow$\rightarrow$ IV)
(D) Zn-Hg / HCl$Zn-Hg / HCl$: Zinc amalgam with concentrated hydrochloric acid is the reagent for Clemmensen Reduction. (D rightarrow$\rightarrow$ II)
### Step 1: Final Matching
The correct matches are A-III, B-I, C-IV, D-II.
### Pattern Recognition
Pure factual matching strictly based on NCERT name reactions and distinguishing tests for carbonyl compounds.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Aldehydes Ketones and Carboxylic Acids
Q68jee_main_2026_22_january_eveningCross-Aldol and Cannizzaro Reactions
The compound A, textC_8textH_8textO_2$\text{C}_8\text{H}_8\text{O}_2$ reacts with acetophenone to form a single product via cross-Aldol condensation. The compound A on reaction with conc. NaOH forms a substituted benzyl alcohol as
A. 2-hydroxy acetophenone
B. 4-methoxy benzaldehyde
C. 4-hydroxy benzaldehyde
D. 4-methyl benzoic acid
Solution
### Related Formula
textAr-CHO + textconc. NaOH xrightarrowtextCannizzaro textAr-CH_2textOH + textAr-COO^-$$\text{Ar-CHO} + \text{conc. NaOH} \xrightarrow{\text{Cannizzaro}} \text{Ar-CH}_2\text{OH} + \text{Ar-COO}^-$$
### Core Logic
Step 1: Compound A has molecular formula textC_8textH_8textO_2$\text{C}_8\text{H}_8\text{O}_2$ and undergoes Cannizzaro reaction with conc. NaOH implies$\implies$ A lacks alpha$\alpha$-hydrogen and is an aromatic aldehyde.
Step 2: Compound A is 4-methoxy benzaldehyde (textCH_3textO-C_6textH_4text-CHO$\text{CH}_3\text{O-C}_6\text{H}_4\text{-CHO}$).
Step 3: Cross-Aldol condensation with acetophenone yields a single aldol condensation product (B)$(B)$ structure.
Step 4: Cannizzaro reaction with conc. NaOH produces 4-methoxybenzyl alcohol and 4-methoxybenzoate anion.
Cross-aldol and Cannizzaro reaction diagram for Q68 - JEE Main 2026 EveningCross-aldol and Cannizzaro reaction diagram for Q68 - JEE Main 2026 Evening
### Pattern Recognition
Sees: textC_8textH_8textO_2$\text{C}_8\text{H}_8\text{O}_2$ undergoing Cannizzaro to form substituted benzyl alcohol.
Shortcut: Presence of -textCHO$-\text{CHO}$ without alpha$\alpha$-H and -textOCH_3$-\text{OCH}_3$ ring substituent points directly to 4-methoxy benzaldehyde.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Q54jee_main_2026_23_january_morningOxidation of Alcohols and Alkene Reactions
The image shows an organic conversion from a bromo-substituted cyclopentene to a carboxylic acid derivative.
The correct sequence of reagents for the above conversion of X to Y is :
A.text(i) NaOH (aq) (ii) Jones reagent (iii) H_3O^+$\text{(i) NaOH (aq) (ii) Jones reagent (iii) }H_{3}O^{+}$
B.text(i) B_2H_6 / H_2O_2text (ii) NaOEt (iii) Jones reagent$\text{(i) }B_{2}H_{6} / H_{2}O_{2}\text{ (ii) NaOEt (iii) Jones reagent}$
C.text(i) Jones reagent (ii) NaOEt (iii) Hot KMnO_4text /KOH$\text{(i) Jones reagent (ii) NaOEt (iii) Hot }KMnO_{4}\text{ /KOH}$
D.text(i) NaOEt (ii) B_2H_6/H_2O_2text (iii) Jones reagent$\text{(i) NaOEt (ii) }B_{2}H_{6}/H_{2}O_{2}\text{ (iii) Jones reagent}$
Solution
### Core Logic
The transformation requires converting a cyclic bromoalkane into a conjugated cyclic carboxylic acid. The steps required are elimination to form an alkene, anti-Markovnikov hydration to an alcohol, and finally oxidation.
### Step 1: Elimination
Treating the starting material with NaOEt (a strong base) promotes an E2 elimination, forming a conjugated diene in the cyclopentane ring.
The image shows an organic conversion from a bromo-substituted cyclopentene to a carboxylic acid derivative.
### Step 2: Hydroboration-Oxidation
Using B_2H_6 / H_2O_2$B_2H_6 / H_2O_2$ (Hydroboration-oxidation) adds water across the less hindered double bond in an anti-Markovnikov fashion, placing an -OH group on the terminal carbon attached to the ring.
### Step 3: Strong Oxidation
Finally, utilizing Jones reagent (CrO_3 / H^+$CrO_3 / H^+$) strongly oxidizes the primary alcohol directly into a carboxylic acid (-COOH$-COOH$), yielding the desired product Y.
The image shows an organic conversion from a bromo-substituted cyclopentene to a carboxylic acid derivative.
### Pattern Recognition
Formation of a new double bond alongside chain functionalization generally implies Base-catalyzed E2 followed by functional group addition (like hydroboration) and terminal oxidation.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Aldehydes Ketones and Carboxylic Acids
Class 11 Chemistry: Hydrocarbons
Class 12 Chemistry: Alcohols Phenols and Ethers
More Aldehydes, Ketones and Carboxylic Acids Questions — jee_main_2026_23_january_evening
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