What happens when D-glucose is treated with the following reagents- (a) HI (b) Bromine Water (c) HNO3 OR Explain the following- (a) Zwitter ion (b) Nucleotide (c) Glycosidic linkage

> > > [!TLDR] > Strong reduction (HI) removes all oxygen; mild oxidation (Br2 water) oxidises only the -CHO; strong oxidation (HNO3) oxidises both terminal groups. > > D-glucose is $CH_2OH-(CHOH)_4-CHO$. > > **(a) With HI (strong reducing conditions, prolonged heating):** all the oxygen-containing functional groups are removed by reduction, converting glucose fully into the straight-chain hydrocarbon: > > $CH_2OH-(CHOH)_4-CHO \xrightarrow{HI,\ \Delta} CH_3-(CH_2)_4-CH_3$ (n-hexane) > > **(b) With Bromine water (mild oxidising agent):** selectively oxidises only the aldehyde ($-CHO$) group to a carboxylic acid ($-COOH$), leaving the $-OH$ groups untouched: > > $CH_2OH-(CHOH)_4-CHO \xrightarrow{Br_2/H_2O} CH_2OH-(CHOH)_4-COOH$ (D-gluconic acid) > > **(c) With HNO3 (strong oxidising agent):** oxidises *both* the terminal groups — the aldehyde ($-CHO \to -COOH$) *and* the terminal primary alcohol ($-CH_2OH \to -COOH$): > > $CH_2OH-(CHOH)_4-CHO \xrightarrow{HNO_3} HOOC-(CHOH)_4-COOH$ (D-glucaric acid / saccharic acid) > > --- > > **OR:** > > **(a) Zwitter ion:** In amino acids, the carboxyl group ($-COOH$) loses a proton to become $-COO^-$ while the amino group ($-NH_2$) simultaneously accepts a proton to become $-NH_3^+$. The resulting dipolar structure, carrying both a positive and a negative charge (net charge zero), is called a **zwitterion**. > …

What happens when D-glucose is treated with the following reagents- (a) HI (b) Bromine Water (c) HNO3 OR Expla

Question of 110

Q. What happens when D-glucose is treated with the following reagents-

(a) HI

(b) Bromine Water

(OR)

Explain the following-

(a) Zwitter ion

(b) Nucleotide

Uttarakhand UbseUttarakhand Board Intermediate (Class 12) 2025Subjective· 3mImportance★★★★★

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Concept understanding — Glucose Cyclization

Glucose Cyclization: From a Straight Chain to a Ring

Imagine you have a long, flexible chain with a hook at one end and an eyelet at the other. If you swing that chain around, the hook can snap into the eyelet, forming a loop. That's the core idea behind glucose cyclization.

Glucose in its simplest written form is a straight chain of six carbon atoms with an aldehyde group ( $-CHO$ ) at one end. But in water (like in your blood), this chain doesn't stay straight. The aldehyde group reacts with the alcohol group on the fifth carbon, forming a stable six-membered ring.

Why does this happen?

The aldehyde carbon is electrophilic (electron-poor), and the oxygen on carbon-5 has lone pairs (nucleophilic). They attack each other, forming a new bond. This creates a hemiacetal — a carbon bonded to both an $-OH$ and an $-OR$ group. The ring is more stable than the open chain in solution.

Note

The open-chain form of glucose exists in equilibrium with the cyclic form, but at any given time, over 99% of glucose molecules are in the cyclic form.

The precise statement

Glucose cyclization is an intramolecular nucleophilic addition where the aldehyde group at C1 reacts with the hydroxyl group at C5, forming a six-membered pyranose ring (named after pyran, a six-membered oxygen heterocycle). This reaction creates a new chiral center at C1, giving two possible stereoisomers called anomers: $\alpha$ and $\beta$ .

The two anomers

When the ring forms, the oxygen from C5 becomes part of the ring. The carbon that was the aldehyde (now C1) becomes a new chiral center. The $-OH$ group at this new center can point:

Down (relative to the ring plane) → $\alpha$ -D-glucose
Up → $\beta$ -D-glucose

Important

The $\alpha$ and $\beta$ anomers are diastereomers, not enantiomers. They differ only at the anomeric carbon (C1). In solution, they interconvert through the open-chain form — a process called mutarotation.

How to draw it (Haworth projection)

Draw a hexagon with an oxygen atom at the top-right corner.
Number the carbons clockwise from the oxygen: C1 is the carbon to the right of oxygen, C2 next, and so on.
For $\alpha$ -D-glucose, the $-OH$ at C1 points down (opposite to the CH $_2$ OH group at C5).
For $\beta$ -D-glucose, the $-OH$ at C1 points up (same side as the CH $_2$ OH group). …

Why this formula?

Glucose Cyclization: Why the Ring Forms

Glucose cyclization is a classic example of an intramolecular reaction — a molecule reacting with itself. Let's build the understanding step by step.

1. The Starting Point: Open-Chain Glucose

Glucose ( $\text{C}_6\text{H}_{12}\text{O}_6$ ) in its open-chain form has:

An aldehyde group ( $-\text{CHO}$ ) at carbon 1 (C1)
A hydroxyl group ( $-\text{OH}$ ) at carbon 5 (C5)

The aldehyde is electrophilic (electron-deficient at the carbonyl carbon), and the hydroxyl is nucleophilic (electron-rich oxygen with lone pairs).

2. Why Cyclization Happens: Thermodynamic & Kinetic Favorability

The Key Insight

The molecule is flexible — it can bend so that the C5 hydroxyl oxygen approaches the C1 aldehyde carbon. This brings two reactive groups into close proximity.

Entropy is favorable: One molecule becomes one ring — no loss of translational entropy (unlike two separate molecules reacting).
Ring strain is manageable: A 5- or 6-membered ring (furanose or pyranose) has minimal angle strain (close to tetrahedral angles).

Result: The reaction is reversible but strongly favors the cyclic form — in aqueous solution, >99% of glucose exists as the ring.

3. The Reaction: Hemiacetal Formation

The nucleophilic oxygen of the C5 hydroxyl attacks the electrophilic carbonyl carbon of C1:

\text{R-CHO} + \text{R'-OH} \rightleftharpoons \text{R-CH(OH)(OR')}

This is a hemiacetal — a carbon bonded to both an $-\text{OH}$ and an $-\text{OR}$ group.

Mechanism (simplified)

Protonation of the carbonyl oxygen (acid-catalyzed) makes C1 more electrophilic.
Nucleophilic attack by the C5 oxygen.
Deprotonation yields the cyclic hemiacetal.

4. The Key Formula(e): Ring Size & Anomeric Carbon

Ring Size Determination

The ring size depends on which hydroxyl attacks:

C5 hydroxyl → pyranose (6-membered ring: 5 carbons + 1 oxygen)
C4 hydroxyl → furanose (5-membered ring: 4 carbons + 1 oxygen)

For D-glucose, the C5 attack is overwhelmingly favored, giving the pyranose form.

The Anomeric Carbon

The new chiral center formed at C1 is called the anomeric carbon. Two stereoisomers arise:

$\alpha$ -anomer: $-\text{OH}$ at C1 is trans to the $-\text{CH}_2\text{OH}$ group (axial in the chair conformation)
$\beta$ -anomer: $-\text{OH}$ at C1 is cis to the $-\text{CH}_2\text{OH}$ group (equatorial in the chair conformation)

Why two forms? The attack can occur from either face of the planar carbonyl group — leading to two possible configurations at the new stereocenter.

5. The Equilibrium Constant & Mutarotation

The interconversion between $\alpha$ and $\beta$ anomers is called mutarotation:

\alpha\text{-D-glucopyranose} \rightleftharpoons \text{open chain} \rightleftharpoons \beta\text{-D-glucopyranose}

At equilibrium (in water at 20°C):

$\beta$ -D-glucopyranose: ~64%
$\alpha$ -D-glucopyranose: ~36%
Open chain: <0.1% …