Question

Match Reaction with Reagent Required:

Choose the correct answer from the options given below:

• A. (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
• B. (A)-(I), (B)-(II), (C)-(IV), (D)-(III)
• C. (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
• D. (A)-(III), (B)-(I), (C)-(II), (D)-(IV)

Hint:

Remember: \(Br_2/H_2O\) gives gluconic acid, \(HNO_3\) gives saccharic acid, \(NH_2OH\) gives oxime and excess HI converts glucose into \(n\)-hexane.

Answer 1

The Correct Option is A

Step 1: Match Glucose to \(n\)-Hexane.
When glucose is heated with excess hydroiodic acid in the presence of red phosphorus, complete reduction takes place and all oxygen atoms are removed. \[ \text{Glucose} \xrightarrow[\Delta]{HI} n\text{-Hexane} \] Thus, \[ (A)\rightarrow(III) \]

Step 2: Match Glucose to Oxime.
The aldehyde group of glucose reacts with hydroxylamine to form glucose oxime. \[ \text{Glucose}+NH_2OH \rightarrow \text{Glucose Oxime} \] Therefore, \[ (B)\rightarrow(I) \]

Step 3: Match Glucose to Gluconic Acid.
Bromine water is a mild oxidizing agent and oxidizes only the aldehyde group to carboxylic acid. \[ \text{Glucose} \xrightarrow{Br_2/H_2O} \text{Gluconic Acid} \] Hence, \[ (C)\rightarrow(IV) \]

Step 4: Match Glucose to Saccharic Acid.
Nitric acid is a strong oxidizing agent. It oxidizes both the aldehyde group and terminal alcohol group into carboxylic acids. \[ \text{Glucose} \xrightarrow{HNO_3} \text{Saccharic Acid} \] Thus, \[ (D)\rightarrow(II) \]

Step 5: Write the final matching.
\[ (A)-(III),\ (B)-(I),\ (C)-(IV),\ (D)-(II) \] Hence the correct option is \[ {\text{Option (A)}} \]