Question

During the structural analysis of an unknown aldohexose, a chemist treats a sample with periodic acid ($\text{HIO}_4$). If the carbohydrate is completely cleaved to yield five molecules of formic acid ($\text{HCOOH}$) and one molecule of formaldehyde ($\text{HCHO}$), this diagnostic breakdown directly proves the presence of:

• A. A ketohexose structure with a carbonyl at C-2
• B. A cyclic pyranose ring configuration
• C. A continuous straight-chain structure containing five $-\text{CHOH}$ groups and one $-\text{CH}_2\text{OH}$ group
• D. Three isolated, non-adjacent primary alcohol branches

Hint:

Count your carbons to eliminate wrong choices instantly! A hexose sugar has 6 carbons. Formic acid (\(\text{HCOOH}\)) and formaldehyde (\(\text{HCHO}\)) both contain exactly 1 carbon atom. Since \(5 \times 1 \text{ (from HCOOH)} + 1 \times 1 \text{ (from HCHO)} = 6\) carbons total, it confirms that the entire open-chain backbone cracked completely into single-carbon pieces.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks us to identify the structural characteristics of an unknown aldohexose based on its complete oxidative cleavage by periodic acid (\(\text{HIO}_4\)).
The reaction produces five molecules of formic acid (\(\text{HCOOH}\)) and one molecule of formaldehyde (\(\text{HCHO}\)).
Step 2: Key Formula or Approach:
Periodic acid (\(\text{HIO}_4\)) selectively cleaves carbon-carbon single bonds in vicinal glycols (1,2-diols) and \(\alpha\)-hydroxy carbonyl compounds.
The stoichiometry of the oxidation products is:
- A terminal primary alcohol group (\(-\text{CH}_2\text{OH}\)) yields formaldehyde (\(\text{HCHO}\)).
- A secondary alcohol group (\(-\text{CHOH}-\)) within a chain of adjacent cleavable groups yields formic acid (\(\text{HCOOH}\)).
- A terminal aldehyde group (\(-\text{CHO}\)) adjacent to a hydroxyl group yields formic acid (\(\text{HCOOH}\)).
Step 3: Detailed Explanation:
1. An aldohexose contains 6 carbon atoms.
2. Let us analyze the cleavage products:
- One molecule of \(\text{HCHO}\): This confirms the presence of exactly one terminal primary alcohol group (\(-\text{CH}_2\text{OH}\)) at one end of the carbon chain (C-6).
- Five molecules of \(\text{HCOOH}\): These must come from the remaining 5 carbon atoms.
3. In an open-chain aldohexose (such as D-glucose):
- The aldehyde group (\(-\text{CHO}\)) at C-1 oxidizes to 1 \(\text{HCOOH}\).
- The four internal secondary alcohol groups (\(-\text{CHOH}-\)) at C-2, C-3, C-4, and C-5 each oxidize to 1 \(\text{HCOOH}\), yielding 4 \(\text{HCOOH}\).
- Together, these yield exactly 5 molecules of \(\text{HCOOH}\).
4. The complete breakdown into single-carbon fragments indicates that all carbon-carbon bonds in the 6-carbon backbone were cleaved.
5. This is only possible if all 6 carbon atoms are in a continuous, unbranched straight chain containing contiguous hydroxyl or carbonyl groups.
6. This matches the description in Option (C).
Step 4: Final Answer:
The cleavage products confirm a continuous straight-chain structure containing five \(-\text{CHOH}\) groups and one \(-\text{CH}_2\text{OH}\) group, corresponding to Option (C).