Which is the most suitable reagent for the following transformation ?
CH3-CH=CH-CH2-CH(OH)-CH3 - CH

Q: Why doesn't Tollen's reagent work here?

Tollen's reagent is a very mild oxidizing agent specifically designed to oxidize aldehydes to carboxylic acids. It is not strong enough to oxidize secondary alcohols.

Q: What would happen if we used alkaline $KMnO_4$?

Alkaline $KMnO_4$ is a powerful oxidizing agent. While it would oxidize the alcohol, it would also oxidatively cleave the $C=C$ double bond, breaking the molecule into smaller fragments.

Q: Does the haloform reaction affect double bonds?

No, the haloform reaction ($I_2/NaOH$) is highly selective. It specifically targets methyl ketones and methyl carbinols, leaving isolated carbon-carbon double bonds completely unaffected.

Q: Why is the product a carboxylic acid and not a carboxylate salt?

The immediate product of the haloform reaction in a basic medium is indeed a carboxylate salt. However, in such transformation questions, a subsequent acidic workup is implicitly assumed to yield the final neutral carboxylic acid.

Q: What is the yellow precipitate formed in this reaction?

The yellow precipitate is iodoform ($CHI_3$), which is a characteristic byproduct of the haloform reaction when iodine is used as the halogen.

Question

Which is the most suitable reagent for the following transformation ?
$\mathrm{CH_3-CH=CH-CH_2-CH(OH)-CH_3 \longrightarrow CH_3-CH=CH-CH_2CO_2H}$

Options:

(A) Tollen's reagent

(B) $\mathrm{I_2/NaOH}$

(C) Alkaline $\mathrm{KMnO_4}$

(D) $\mathrm{CrO_2Cl_2 / CS_2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Look closely at the given transformation. In the reactant, we have a secondary alcohol, specifically a methyl carbinol group ($-CH(OH)-CH_3$). In the product, this group has been transformed into a carboxylic acid ($-COOH$).

But the most important detail to notice is the structural change: the carbon chain has become shorter by exactly one carbon atom, and the $C=C$ double bond in the middle of the chain is completely safe and intact! This specific pattern of losing a terminal methyl group while oxidizing the adjacent carbon is a massive hint.

### Evaluating the Reagents

Let's systematically check the options provided to see which one fits our requirements.

**1. Tollen's Reagent ($AgNO_3 + NH_4OH$):**
This is a mild oxidizing agent. It is famous for oxidizing aldehydes to carboxylic acids (the silver mirror test), but it doesn't even touch secondary alcohols. So, this cannot be our answer.

**2. Alkaline $KMnO_4$:**
This is a very strong oxidizing agent. While it would certainly oxidize the secondary alcohol, it is too aggressive. It would also attack our precious $C=C$ double bond, leading to oxidative cleavage or hydroxylation. Since we need the double bond intact, this reagent is rejected.

**3. $CrO_2Cl_2 / CS_2$ (Etard Reagent):**
This reagent is highly specific. It is used in the Etard reaction to convert a methyl group attached to a benzene ring ($Ar-CH_3$) into an aldehyde ($Ar-CHO$). It has absolutely no use in our current aliphatic system.

### The Master Equation: The Haloform Reaction

Now we are left with Iodine and $NaOH$ ($I_2/NaOH$). Remember, this is the classic reagent combination for the **Haloform Reaction**!

Our reactant possesses a methyl carbinol group ($-CH(OH)-CH_3$). This is a perfect candidate for the haloform reaction.

Here is how the magic happens:
1. **Initial Oxidation:** The iodine and base first oxidize the secondary alcohol into a methyl ketone ($-CO-CH_3$).
2. **Haloform Cleavage:** The methyl ketone then undergoes the haloform reaction. The terminal methyl group is fully halogenated and leaves as a yellow precipitate of iodoform ($CHI_3$). 
3. **Carboxylate Formation:** The rest of the carbon chain is converted into a carboxylate ion ($-COO^-$).

$$R-CH(OH)-CH_3 \xrightarrow{I_2/NaOH} R-CO-CH_3 \xrightarrow{I_2/NaOH} R-COO^- + CHI_3 \downarrow$$

And the best part? The haloform reaction is highly selective and does absolutely nothing to isolated carbon-carbon double bonds!

### Final Calculation

Finally, a standard acidic workup (which is implicitly assumed in such transformation sequences) protonates the carboxylate ion, giving us our desired carboxylic acid.

$$R-COO^- \xrightarrow{H^+} R-COOH$$

Therefore, $I_2/NaOH$ is the perfect reagent to achieve this highly specific transformation.

Animated Solution for Chemistry - Organic Chemistry: Which is the most suitable reagent for the following transformation ? $C H_{3} - CH = CH - C H_{2} - CH (OH) - C H_{3} ⟶ C H_{3} - CH = CH - C H_{2} C O_{2} H$

Select Answer:

Visualized Solution

$Analyzing the Transformation$

$Evaluating Tollen’s Reagent$

$Evaluating Alkaline KMn O_{4}$

$Evaluating Etard Reagent$

$The Haloform Reagent$

$Haloform Reaction Mechanism$

$Final Conclusion$

The Sigma Insight: Carbonyl Compounds

Analyzing the Setup

Evaluating the Reagents

The Master Equation

Final Calculation

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