Which among the following compounds give yellow solid when reacted with NaOI/NaOH?
(A) CH_3 - CH(OH) - C_2H_5$CH_3 - CH(OH) - C_2H_5$
(B) CH_3 - CH_2 - CH_2 - OH$CH_3 - CH_2 - CH_2 - OH$
(C) CH_3 - CO - C_2H_5$CH_3 - CO - C_2H_5$
(D) CH_3 - OH$CH_3 - OH$
(E) CH_3 - CH_2 - H$CH_3 - CH_2 - H$
Choose the correct answer from the options given below:
A.(B), (C) and (E) Only
B.(A) and (C) Only
C.(C) and (D) Only
D.(A), (C) and (D) Only
Solution & Explanation
### Related Formula
textCompounds with CH_3-CH(OH)- text or CH_3-CO- text groups undergo the iodoform reaction to form CHI_3 downarrow text (Yellow Solid)$\text{Compounds with } CH_3-CH(OH)- \text{ or } CH_3-CO- \text{ groups undergo the iodoform reaction to form } CHI_3 \downarrow \text{ (Yellow Solid)}$
### Core Logic
Let's check the structural groups of each given option:
- **(A)** CH_3 - CH(OH) - C_2H_5$CH_3 - CH(OH) - C_2H_5$: Contains the methylcarbinol group (CH_3-CH(OH)-$CH_3-CH(OH)-$). Gives a positive iodoform test.
- **(B)** CH_3 - CH_2 - CH_2 - OH$CH_3 - CH_2 - CH_2 - OH$: Linear primary alcohol, does not contain the required group.
- **(C)** CH_3 - CO - C_2H_5$CH_3 - CO - C_2H_5$: Contains the methyl ketone group (CH_3-CO-$CH_3-CO-$). Gives a positive iodoform test.
- **(D)** CH_3 - OH$CH_3 - OH$: Methanol does not give the test.
- **(E)** CH_3 - CH_2 - H$CH_3 - CH_2 - H$: Ethane does not give the test.
Thus, only **(A)** and **(C)** yield the yellow precipitate of iodoform (CHI_3$CHI_3$).
### Step 1: Chemical Equations
The balanced haloform pathways occur as follows:
CH_3-CH(OH)-CH_2-CH_3 xrightarrowtextNaOI/NaOH CHI_3downarrow + textCH_3text-CH_2text-COO^-textNa^+$CH_3-CH(OH)-CH_2-CH_3 \xrightarrow{\text{NaOI/NaOH}} CHI_3\downarrow + \text{CH}_3\text{-CH}_2\text{-COO}^-\text{Na}^+$CH_3-CO-CH_2-CH_3 xrightarrowtextNaOI/NaOH CHI_3downarrow + textCH_3text-CH_2text-COO^-textNa^+$CH_3-CO-CH_2-CH_3 \xrightarrow{\text{NaOI/NaOH}} CHI_3\downarrow + \text{CH}_3\text{-CH}_2\text{-COO}^-\text{Na}^+$
### Pattern Recognition
The iodoform test specifically isolates methyl ketones or secondary methyl alcohols. Scan dynamically for a terminal -CH_3$-CH_3$ affixed directly to a carbonyl oxygen index (C=O$C=O$) or a hydroxyl carbon (CH-OH$CH-OH$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Class 12 Chemistry: Alcohols, Phenols and Ethers
Keywords:#yellow solid iodoform test#JEE Main 2025 Evening Q28#Aldehydes Ketones and Carboxylic Acids iodoform#methyl ketone functional group
More Aldehydes, Ketones and Carboxylic Acids Previous-Year Questions — Page 5
Q502025Chemical Reactions of Carbonyl Compounds
Based on the reaction sequence:
A structural flow charts the transformation of a cyclic dialcohol into final product S.
0.1 mole of compound 'S' will weigh ________ g.
(Given molar mass in mathrmg \, mol^-1$\mathrm{g \, mol^{-1}}$ C:12, H:1, O:16)
Numerical Answer.Answer: 13 to 13
Solution
### Related Formula
textMass = textMoles cdot textMolar Mass$\text{Mass} = \text{Moles} \cdot \text{Molar Mass}$
### Core Logic
Let us trace the reaction scheme row-by-row:
1. Reactant: 2-(hydroxymethyl)cyclopentan-1-ol.
2. Step 1: Excess mathrmCrO_3$\mathrm{CrO}_3$ (Jones oxidation) oxidizes the secondary alcohol to a ketone and the primary alcohol to a carboxylic acid
ightarrow$
ightarrow$ Product P.
3. Step 2: Reaction with 1 mole of glycol protects the ketone selectively as a cyclic ketal
ightarrow$
ightarrow$ Product Q.
4. Step 3: Treatment with mathrmCH_3mathrmMgI$\mathrm{CH}_3\mathrm{MgI}$ targets the free carboxylic acid (or ester equivalent) to form a methyl ketone after workup
ightarrow$
ightarrow$ Product R.
5. Step 4: mathrmNaBH_4$\mathrm{NaBH}_4$ reduces the methyl ketone to a secondary alcohol, and acid workup deprotects the ketal back to the original ketone
ightarrow$
ightarrow$ Compound S .
Chemical structural formula of S: 2-(1-hydroxyethyl)cyclopentan-1-one (mathrmC_7mathrmH_12mathrmO_2$\mathrm{C}_7\mathrm{H}_{12}\mathrm{O}_2$) .
Molar mass of S (mathrmC_7mathrmH_12mathrmO_2$\mathrm{C}_7\mathrm{H}_{12}\mathrm{O}_2$) :
M = (7 cdot 12) + (12 cdot 1) + (2 cdot 16) = 84 + 12 + 32 = 130 \, mathrmg/mol$M = (7 \cdot 12) + (12 \cdot 1) + (2 \cdot 16) = 84 + 12 + 32 = 130 \, \mathrm{g/mol}$textMass of 0.1 mole = 0.1 cdot 130 = 13 \, mathrmg$\text{Mass of 0.1 mole} = 0.1 \cdot 130 = 13 \, \mathrm{g}$
### Pattern Recognition
Ketal groups serve as stable protective masks that shield ketones, allowing selective organometallic reactions to take place elsewhere on the molecule.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
More Aldehydes, Ketones and Carboxylic Acids Questions — jee_main_2025_04_april_evening
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