Number of molecules from below which cannot give iodoform reaction is:
Ethanol, Isopropyl alcohol, Bromoacetone, 2-Butanol, 2-Butanone, Butanal, 2-Pentanone, 3-Pentanone, Pentanal and 3-Pentanol
A.5
B.4
C.3
D.2
Solution & Explanation
### Core Logic
The iodoform test requires compounds containing either a methyl ketone group (textCH_3textCO-$\text{CH}_3\text{CO}-$) or a methyl carbinol structural unit (textCH_3textCH(OH)-$\text{CH}_3\text{CH(OH)}-$).
Let us audit the provided compounds:
1. **Ethanol** (textCH_3textCH_2textOH$\text{CH}_3\text{CH}_2\text{OH}$): Contains textCH_3textCH(OH)-$\text{CH}_3\text{CH(OH)}-$
ightarrow$
ightarrow$ **Positive**
2. **Isopropyl alcohol** (textCH_3textCH(OH)CH_3$\text{CH}_3\text{CH(OH)CH}_3$): Contains textCH_3textCH(OH)-$\text{CH}_3\text{CH(OH)}-$
ightarrow$
ightarrow$ **Positive**
3. **Bromoacetone** (textCH_3textCOCH_2textBr$\text{CH}_3\text{COCH}_2\text{Br}$): Contains textCH_3textCO-$\text{CH}_3\text{CO}-$
ightarrow$
ightarrow$ **Positive**
4. **2-Butanol** (textCH_3textCH(OH)CH_2textCH_3$\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3$): Contains textCH_3textCH(OH)-$\text{CH}_3\text{CH(OH)}-$
ightarrow$
ightarrow$ **Positive**
5. **2-Butanone** (textCH_3textCOCH_2textCH_3$\text{CH}_3\text{COCH}_2\text{CH}_3$): Contains textCH_3textCO-$\text{CH}_3\text{CO}-$
ightarrow$
ightarrow$ **Positive**
6. **Butanal** (textCH_3textCH_2textCH_2textCHO$\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO}$): **Negative**
7. **2-Pentanone** (textCH_3textCOCH_2textCH_2textCH_3$\text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_3$): Contains textCH_3textCO-$\text{CH}_3\text{CO}-$
ightarrow$
ightarrow$ **Positive**
8. **3-Pentanone** (textCH_3textCH_2textCOCH_2textCH_3$\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3$): **Negative**
9. **Pentanal** (textCH_3textCH_2textCH_2textCH_2textCHO$\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CHO}$): **Negative**
10. **3-Pentanol** (textCH_3textCH_2textCH(OH)CH_2textCH_3$\text{CH}_3\text{CH}_2\text{CH(OH)CH}_2\text{CH}_3$): **Negative**
### Step 1: Summation
The molecules that **cannot** give the iodoform reaction are: Butanal, 3-Pentanone, Pentanal, and 3-Pentanol. This gives a total count of exactly 4 molecules.
### Pattern Recognition
Shortcut: Filter for names ending with '-anal' or having ketones/alcohols at positions higher than 2 (e.g., 3-pentanone, 3-pentanol). These lack the vital terminal methyl group adjacent to the functional group.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Keywords:#Iodoform reaction positive criteria#Methyl ketone structure functional test#Aldehydes and Ketones JEE Main 2025#JEE Main 2025 Morning Q36
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_03_april_morning
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