Question

For the detection of sucrose in milk, ______________ reacts with resorcinol forming cherry red colour.

• A. 5-hydroxymethylfurfural
• B. Glucose
• C. Fructose
• D. Galactose

Hint:

Remember the pathway:
\[ \text{Sucrose} \xrightarrow{\text{Acid}} \text{Fructose} \xrightarrow{\text{Dehydration}} \text{5-HMF} \xrightarrow{\text{Resorcinol}} \text{Cherry Red Complex} \]

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The presence of cane sugar (sucrose) in milk is an adulteration defect detected using the resorcinol test (often called Seliwanoff's test), which selectively identifies ketose sugars under acidic conditions.

Step 2: Detailed Explanation:
- When milk containing sucrose is heated with concentrated hydrochloric acid (\( \text{HCl} \)), the sucrose is hydrolyzed into its constituent monosaccharides: glucose and fructose.
- Under the influence of hot, concentrated acid, the keto-hexose sugar (fructose) undergoes rapid dehydration to form 5-hydroxymethylfurfural (5-HMF).
- This newly formed 5-hydroxymethylfurfural intermediate then directly condenses with resorcinol present in the reagent to form a characteristic cherry-red colored complex.
- While fructose is the precursor sugar, it is the dehydrated intermediate, 5-hydroxymethylfurfural, that directly reacts with resorcinol to generate the colored complex.

Step 3: Final Answer
The compound that directly reacts with resorcinol to form the cherry-red color is 5-hydroxymethylfurfural.