Question

Which from following reagents is used to identify straight chain of glucose?

• A. $\mathrm{HI}$
• B. $\mathrm{dil.\ HNO_3}$
• C. $\mathrm{HCN}$
• D. Acetic anhydride

Hint:

To easily remember the chemical tests for glucose structure: 1. $\mathrm{HI}$ confirms the six-carbon straight chain ($n$-hexane). 2. $\mathrm{NH_2OH}$ or $\mathrm{HCN}$ confirms the presence of a carbonyl group. 3. Bromine water ($\mathrm{Br_2/H_2O}$) confirms that the carbonyl is an aldehyde group. 4. Acetic anhydride confirms the presence of five hydroxyl groups.

Answer 1

The Correct Option is A

Step 1: Understand what we must prove.
We need a reagent that shows the six carbons of glucose lie in one straight unbranched chain.
Step 2: Pick the right kind of reaction.
To reveal the carbon skeleton we must strip away every oxygen function and look at the bare hydrocarbon left behind.
Step 3: Recall the reagent that strips oxygen.
Hot hydroiodic acid $HI$ with red phosphorus is a powerful reducing agent that removes all $-OH$ and carbonyl oxygens.
Step 4: Apply it to glucose.
On prolonged heating with $HI$, glucose $C_6H_{12}O_6$ is fully reduced to a six carbon alkane, $n$-hexane. \[ C_6H_{12}O_6 + HI \xrightarrow{\Delta} CH_3(CH_2)_4CH_3 \]
Step 5: Read the skeleton.
Getting straight chain $n$-hexane, rather than a branched isomer, proves the original six carbons were in one continuous open chain.
Step 6: Reject the others briefly.
Dilute $HNO_3$, $HCN$ and acetic anhydride probe specific groups, not the whole carbon backbone, so $HI$ is the answer.
\[ \boxed{HI \text{ (option A)}} \]