Consider the following reactions :
(i) Glucose + ROH - Acetal - (x eq. of (CH3CO)2O) Acetyl deriva

Q: Why doesn't the aldehyde group in open-chain glucose react with acetic anhydride?

Acetic anhydride specifically targets nucleophilic groups like alcohols (-OH) and amines (-NH2) to form esters and amides. The carbonyl carbon of an aldehyde is electrophilic, not nucleophilic, so it does not react with acetic anhydride under these conditions.

Q: Why do we consider the cyclic form of glucose for reaction (i) but the open-chain form for reaction (iii)?

Glucose is in a dynamic equilibrium between its open-chain and cyclic forms. Acetic anhydride reacts with all free -OH groups, shifting the equilibrium until the open-chain form is fully acetylated (or the cyclic form is fully acetylated, yielding the same number of acetate groups: 5). However, reaction (i) specifically targets the highly reactive anomeric -OH of the cyclic hemiacetal to form a stable acetal (glycoside), locking the ring closed.

Q: What exactly is the difference between a hemiacetal and an acetal?

A hemiacetal has one -OH group and one -OR group attached to the same carbon atom (the anomeric carbon in sugars). It is relatively unstable and can open up. An acetal has two -OR groups attached to the same carbon. It is stable in neutral and basic conditions and locks the sugar in its cyclic form.

Q: Why does reduction with Ni/H2 increase the number of -OH groups?

The aldehyde group (-CHO) at the top of the glucose molecule contains a carbon-oxygen double bond. Catalytic hydrogenation (Ni/H2) adds hydrogen across this double bond, converting the aldehyde into a primary alcohol (-CH2OH). This adds one more -OH group to the molecule.

Q: Can the acetal formed in reaction (i) revert back to glucose during acetylation?

No, acetals are stable under the neutral or slightly basic conditions typically used for acetylation (like using pyridine as a solvent). Therefore, the -OR group remains intact, and only the remaining four free -OH groups are acetylated.

Question

Consider the following reactions :
(i) $	ext{Glucose} + 	ext{ROH} \xrightarrow{	ext{Dry HCl}} 	ext{Acetal} \xrightarrow[	ext{(CH}_3	ext{CO)}_2	ext{O}]{x 	ext{ eq. of}} 	ext{Acetyl derivative}$
(ii) $	ext{Glucose} \xrightarrow{	ext{Ni/H}_2} 	ext{A} \xrightarrow[	ext{(CH}_3	ext{CO)}_2	ext{O}]{y 	ext{ eq.of}} 	ext{Acetyl derivative}$
(iii) $	ext{Glucose} \xrightarrow[	ext{(CH}_3	ext{CO)}_2	ext{O}]{z 	ext{ eq.of}} 	ext{Acetyl derivative}$
$x$, '$y$' and '$z$' in these reactions are respectively

Options:

(A) 5, 4 and 5

(B) 4, 5 and 5

(C) 5, 6 and 5

(D) 4, 6 and 5

Sigma Percentile · Accepted Answer

The chemistry of carbohydrates is a fascinating puzzle of structures and functional groups. In this problem, we are tasked with determining the number of equivalents of acetic anhydride consumed by glucose and its derivatives in three distinct reactions.

To solve this, we must understand the dual nature of glucose—its open-chain and cyclic forms—and how different reagents interact with its specific functional groups.

### The Detective Reagent: Acetic Anhydride

Before we dive into the specific reactions, let's establish our primary tool. Acetic anhydride, $(	ext{CH}_3	ext{CO})_2	ext{O}$, acts as a chemical detective that specifically hunts down and reacts with free hydroxyl ($-	ext{OH}$) groups.

Through an acetylation reaction, each free $-	ext{OH}$ group is converted into an acetate ester ($-	ext{OCOCH}_3$). Crucially, the stoichiometry is perfectly one-to-one: **one equivalent of acetic anhydride is consumed for every free $-	ext{OH}$ group present in the molecule.**

### Reaction (iii): The Baseline Acetylation

Let's begin with the most straightforward case, reaction (iii), where glucose is directly treated with acetic anhydride.

To determine the equivalents consumed ($z$), we simply need to count the free $-	ext{OH}$ groups in the standard open-chain structure of D-glucose.

Looking at the Fischer projection, glucose is an aldohexose. It contains an aldehyde group at the top, four secondary alcohol groups along the chain, and one primary alcohol group at the bottom.

Counting the hydroxyls, we find exactly **five free $-	ext{OH}$ groups**. Therefore, direct acetylation yields glucose pentaacetate, consuming 5 equivalents of the reagent.

This gives us our first value: **$z = 5$**.

### Reaction (i): The Hemiacetal Trap

Reaction (i) introduces a classic trap. We are first reacting glucose with an alcohol ($	ext{ROH}$) in the presence of dry $	ext{HCl}$.

To understand this, we must remember that in aqueous solution, glucose does not predominantly exist as an open chain. Instead, it undergoes intramolecular cyclization to form a six-membered pyranose ring, specifically a **cyclic hemiacetal**.

In this cyclic form, the $-	ext{OH}$ group on the anomeric carbon (C1) is part of the hemiacetal linkage. This specific $-	ext{OH}$ is highly reactive. When treated with $	ext{ROH}/	ext{HCl}$, this hemiacetal $-	ext{OH}$ undergoes a substitution reaction, being replaced by an $-	ext{OR}$ group to form an **acetal**, commonly known as a glycoside.

Now, we must acetylate this newly formed acetal. Let's count the remaining free $-	ext{OH}$ groups. The anomeric position is now an ether linkage ($-	ext{OR}$), which does not react with acetic anhydride.

This leaves only the other **four $-	ext{OH}$ groups** on the ring available for acetylation. Consequently, the acetal will consume only 4 equivalents of acetic anhydride.

This gives us our second value: **$x = 4$**.

### Reaction (ii): The Reduction to Sorbitol

Finally, let's decode reaction (ii). Glucose is treated with hydrogen gas and a nickel catalyst ($	ext{Ni/H}_2$).

This is a standard catalytic reduction. The reagent targets the carbonyl double bond of the aldehyde group ($-	ext{CHO}$) at the top of the open-chain glucose molecule.

The reduction converts the aldehyde into a primary alcohol ($-	ext{CH}_2	ext{OH}$). The resulting molecule is a sugar alcohol known as **sorbitol** (or glucitol).

Now, we must acetylate sorbitol. Let's count its free $-	ext{OH}$ groups. We have the original five $-	ext{OH}$ groups from the glucose chain, plus the brand new $-	ext{OH}$ group generated from the reduced aldehyde.

This gives sorbitol a total of **six free $-	ext{OH}$ groups**. Therefore, complete acetylation of sorbitol will consume 6 equivalents of acetic anhydride.

This gives us our final value: **$y = 6$**.

### Bringing It All Together

We have systematically analyzed each transformation by tracking the functional groups:
- The acetal has 4 free $-	ext{OH}$ groups, so **$x = 4$**.
- The reduced sorbitol has 6 free $-	ext{OH}$ groups, so **$y = 6$**.
- The original glucose has 5 free $-	ext{OH}$ groups, so **$z = 5$**.

The correct sequence for $x$, $y$, and $z$ is **4, 6, and 5**, which perfectly matches option (d). This problem beautifully illustrates the importance of understanding both the open-chain and cyclic structures of carbohydrates!

Select Answer:

Visualized Solution

\text{Acetylation of Carbohydrates}

\text{Reaction (iii): Direct Acetylation}

\text{Reaction (i): Acetal Formation}

\text{Reaction (i): Acetylation of Acetal}

\text{Reaction (ii): Reduction}

\text{Reaction (ii): Acetylation of Sorbitol}

\text{Final Answer}

The Sigma Insight: Carbohydrates

The Detective Reagent

Reaction (iii)

Reaction (i)

Reaction (ii)

Bringing It All Together

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