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

Concept: Periodic acid (\(\text{HIO}_4\)) is a highly specific oxidizing agent utilized in carbohydrate chemistry to perform vicinal glycol cleavage. It selectively cuts carbon-carbon single bonds if and only if both adjacent carbon atoms carry hydroxyl groups (\(-\text{CHOH}-\text{CHOH}-\)) or a carbonyl adjacent to a hydroxyl group (\(-\text{CHO}-\text{CHOH}-\)). The structural fragments oxidize according to these strict stoichiometric diagnostic rules: -CH_2OH (Terminal primary alcohol) &\xrightarrow{HIO_4} HCHO (Formaldehyde)
-CHOH- (Secondary internal alcohol) &\xrightarrow{HIO_4} HCOOH (Formic Acid)
-CHO (Terminal aldehyde carbonyl) &\xrightarrow{HIO_4} HCOOH (Formic Acid)

Step 1: Interpreting the fragment ratios back into functional groups.
The experimental cleavage generated:
• 1 molecule of \(\text{HCHO}\): This establishes that the molecule contains exactly one terminal primary alcohol group (\(-\text{CH}_2\text{OH}\)) at one end of the chain.
• 5 molecules of \(\text{HCOOH}\): For an aldohexose (like Glucose), the top terminal aldehyde (\(-\text{CHO}\)) yields 1 \(\text{HCOOH}\), and the remaining 4 secondary alcohol links (\(-\text{CHOH}-\)) yield 4 \(\text{HCOOH}\), summing up precisely to 5 molecules.

Step 2: Deducing the structural assembly.
Because all 5 carbon-carbon bonds were broken to release independent single-carbon molecules, every single carbon atom in the hexose sugar must have been directly bonded to its neighbor in a continuous, unbranched open chain containing contiguous vicinal glycol functional groups. This confirms the classic open-chain structure of D-glucose.