The correct match between item 'I' and item 'II' is
Item 'I' (Compound)
(A) Lysine
(B) Furfural
(C)

Q: Why does Furfural react with 1-naphthol?

This reaction is the chemical basis of the Molisch test. Carbohydrates are dehydrated by strong acids to form furfural (or its derivatives). The furfural then undergoes a condensation reaction with two molecules of 1-naphthol to produce a highly conjugated, violet-colored dye.

Q: Can the Ceric Ammonium Nitrate (CAN) test be used to identify phenols?

No, the CAN test is generally specific for aliphatic alcohols, producing a red or pink complex. Phenols, on the other hand, typically give a brown or black precipitate with CAN, or are better tested using neutral Ferric Chloride ($FeCl_3$), which gives a characteristic violet color.

Q: What exactly is Baeyer's reagent?

Baeyer's reagent is a cold, dilute, and alkaline solution of Potassium Permanganate ($KMnO_4$). It is used as a qualitative test for the presence of carbon-carbon double or triple bonds (unsaturation). The reaction results in the disappearance of the purple color and the formation of a brown precipitate of $MnO_2$.

Q: Does the Ninhydrin test work for all types of amines?

The Ninhydrin test is highly specific for primary amines and $\alpha$-amino acids, producing the characteristic deep purple color known as Ruhemann's purple. Secondary amines (like proline) react differently, typically producing a yellow color instead.

Q: What happens to the Styrene molecule during Baeyer's test?

During Baeyer's test, the carbon-carbon double bond in the vinyl group of Styrene is oxidized. The $KMnO_4$ adds two hydroxyl ($-OH$) groups across the double bond in a 'syn' fashion, converting the alkene into a vicinal diol (1-phenyl-1,2-ethanediol).

Question

The correct match between item 'I' and item 'II' is
\begin{array}{clcl} \hline & 	extbf{Item 'I'} & & 	extbf{Item 'II'} \ & 	ext{(Compound)} & & 	ext{(Reagent)} \ \hline 	ext{(A)} & 	ext{Lysine} & 	ext{(P)} & 	ext{1-naphthol} \ 	ext{(B)} & 	ext{Furfural} & 	ext{(Q)} & 	ext{Ninhydrin} \ 	ext{(C)} & 	ext{Benzyl alcohol} & 	ext{(R)} & 	ext{KMnO}_4 \ 	ext{(D)} & 	ext{Styrene} & 	ext{(S)} & 	ext{Ceric ammonium nitrate} \ \hline \end{array}

Options:

(A) (A) $ightarrow$ Q, (B) $ightarrow$ R, (C) $ightarrow$ S, (D) $ightarrow$ P

(B) (A) $ightarrow$ R, (B) $ightarrow$ P, (C) $ightarrow$ Q, (D) $ightarrow$ S

(C) (A) $ightarrow$ Q, (B) $ightarrow$ P, (C) $ightarrow$ S, (D) $ightarrow$ R

(D) (A) $ightarrow$ Q, (B) $ightarrow$ P, (C) $ightarrow$ R, (D) $ightarrow$ S

Sigma Percentile · Accepted Answer

### The Colorful World of Qualitative Organic Analysis

Welcome to the fascinating realm of qualitative organic chemistry! When you walk into a chemistry laboratory, you aren't just looking at clear liquids and white powders; you are looking at a world of hidden identities waiting to be revealed through the magic of chemical tests. This problem is a beautiful exercise in matching organic compounds with their signature chemical reagents. Let's break down the chemistry behind each of these colorful reactions.

### The Ninhydrin Test: Unmasking Amino Acids

Our first compound is **Lysine**, a fundamental building block of proteins. If you look closely at its structure, you will notice it is an $\alpha$-amino acid, possessing both an amine group ($-NH_2$) and a carboxylic acid group ($-COOH$).

How do we detect such a molecule? Enter the **Ninhydrin test**. Ninhydrin is a powerful oxidizing agent. When it reacts with an $\alpha$-amino acid, it causes oxidative deamination and decarboxylation. The amino acid is broken down into an aldehyde, ammonia, and carbon dioxide. The liberated ammonia then reacts with another molecule of ninhydrin and its reduced form (hydrindantin) to form a deeply colored blue-purple complex known as **Ruhemann's purple**.

Because Lysine produces this striking violet colouration, we confidently match **(A) Lysine** with **(Q) Ninhydrin**.

### The Molisch Test: The Signature of Carbohydrates

Next, we encounter **Furfural**. Furfural is an organic compound derived from the dehydration of sugars, specifically pentoses. It is the central player in the **Molisch test**, which is the universal test for carbohydrates.

In the Molisch test, a carbohydrate is treated with the Molisch reagent, which is simply a solution of **1-naphthol** in ethanol. When concentrated sulfuric acid is carefully added down the side of the test tube, it dehydrates the carbohydrate to form furfural (or hydroxymethylfurfural for hexoses). This newly formed furfural immediately condenses with two molecules of 1-naphthol to form a stunning violet or purple ring at the junction of the two liquids.

Thus, the presence of furfural directly links to the reagent 1-naphthol, giving us the match **(B) Furfural** with **(P) 1-naphthol**.

### The CAN Test: Identifying Alcohols

Our third molecule is **Benzyl alcohol**, a classic primary alcohol. To identify the presence of an alcoholic hydroxyl group ($-OH$), chemists frequently rely on the **Ceric Ammonium Nitrate (CAN)** test.

Ceric Ammonium Nitrate, $(NH_4)_2Ce(NO_3)_6$, is a yellow-orange coordination complex. When an alcohol is added to a solution of CAN, the alcohol molecules displace the nitrate ligands around the central Cerium(IV) ion. This ligand exchange alters the electronic transitions within the complex, resulting in a dramatic color shift from yellow to a deep **red or pink complex**.

This color change is a definitive positive test for alcohols. Therefore, we match **(C) Benzyl alcohol** with **(S) Ceric ammonium nitrate**.

### Baeyer's Test: Exposing Unsaturation

Finally, we look at **Styrene**. Styrene features a benzene ring attached to a vinyl group ($-CH=CH_2$). The presence of this carbon-carbon double bond means the molecule is **unsaturated**.

To test for unsaturation, we use **Baeyer's reagent**, which is a cold, dilute, alkaline solution of potassium permanganate ($KMnO_4$). Potassium permanganate is a brilliant purple-pink color. When it encounters a double bond, it undergoes a syn-dihydroxylation reaction, adding two hydroxyl groups across the double bond to form a vicinal diol. In the process, the $Mn(VII)$ in permanganate is reduced to $Mn(IV)$, precipitating as brown manganese dioxide ($MnO_2$).

The visual result is the rapid **discharge of the pink color**, confirming the presence of the double bond. Hence, we match **(D) Styrene** with **(R) $KMnO_4$**.

### Bringing It All Together

By understanding the specific functional groups and their corresponding chemical reactions, we have successfully decoded the matrix:

*   **A $\rightarrow$ Q**
*   **B $\rightarrow$ P**
*   **C $\rightarrow$ S**
*   **D $\rightarrow$ R**

This perfectly aligns with option **(c)**. Mastering these qualitative tests not only helps in solving complex matrix matches but also builds a strong intuitive foundation for practical organic chemistry.

Select Answer:

Visualized Solution

$Qualitative Organic Analysis$

$Test for Amino Acids$

$Test for Carbohydrates$

$Test for Alcohols$

$Test for Unsaturation$

$Final Conclusion$

The Sigma Insight: Nomenclature and Characterisation

The Colorful World of Qualitative Organic Analysis

The Ninhydrin Test

The Molisch Test

The CAN Test

Baeyer's Test

Bringing It All Together

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