Why do ketoses undergo dehydration more rapidly than aldoses?

Question

I was reading about Seliwanoff's test and Wikipedia mentions the principle behind this test

This test relies on the principle that, when heated, ketoses are more rapidly dehydrated than aldoses.

Well, as to what might be the reason behind the rapid reaction of ketoses, my conjecture is that as 5HMF is the final product of the dehydration of both aldoses and ketoses is a 5 membered compound, so there will be a extra step of ring contraction in dehydration of aldoses which increases the activation energy and thus decreases the rate of the reaction.

Answer 1

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Seliwanoff's Test - The Favouritism for Ketoses

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TL;DR

• The Seliwanoff's Test is a method to distinguish ketoses and aldoses.
• The test uses resorcinol + dil. $\boldsymbol{\ce{HCl}}$ as the reagent.
• The test involves saccharide undergoes triple dehydration, forming furfural (5-hydroxymethylfurfural), and the furan derivative reacts with resorcinol to form the red colouration.
• Francisco Sánchez-Viesca and Reina Gómez's paper "Reactivities Involved in the Seliwanoff Reaction" explains the difference in reaction times between glucose and fructose.
• They found evidence for a C-2 to C-1 intramolecular hydrogen-transfer during the acid-catalized (sic) isomerization of D-glucose to D-fructose.
• The dehydration of fructose to 5-hydroxymethylfurfural occurs through the fructofuranose form, which can be originated either by glucose isomerization to fructose or more directly from the fructopyranose form.
• Fructose presents common reactivities in the synthesis of hydroxymethylfurfural, not being the case with glucose or glucose isomerization to fructose.
• Slower reactions resulted in accord with less common reaction steps, i.e., with especial reactivities.

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Quoting the NCERT Lab Manual:

Ketohexoses give red colour and ketopentoses give blue-green colour. Aldoses take longer time to produce colour because under the same conditions, aldoses form furfural slowly, probably because β -elimination is required before dehydration to furfural. Therefore prolonged heating should be avoided.

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Introduction

This is a test to distinguish ketoses and aldoses. The reagent used here is resorcinol + dil. $\boldsymbol{\ce{HCl}}$. The ketoses show a positive test (red colouration of furfural derivative).

Mechanism is as given below:

The saccharide undergoes triple dehydration and forms furfural (5-hydroxymethylfurfural as OP has said) and the furan derivative finally reacts with the resorcinol to form the red colouration.

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The answer

Francisco Sánchez-Viesca and Reina Gómez had written a paper titled "Reactivities Involved in the Seliwanoff Reaction" (source: Google Scholar) that had explained everything in concern.

Quoting:

Since glucose reacts much slower than fructose as observed in the Seliwanoff’s Test, we thought that this delay could be attributed to a previous isomerization of glucose to fructose. Thus, we looked for the current trends on this isomerization. We found that there is evidence for a C-2 to C-1 intramolecular hydrogen-transfer during the acid-catalized (sic) isomerization of D-glucose to D-fructose. The reaction from β-D-glucopyranose to α-D-fructofuranose is depicted in the below figure (sic), from a novel method of direct degradation of cellulose in to 5-hydroxymethylfurfural with sodium or potassium salts of acidic salts (sulphates and phosphates), a green HMF production.

The dehydration of fructose to 5-hydroxymethylfurfural occurs through the fructofuranose form. This can be originated either by glucose isomerization to fructose or more directly from the fructopyranose form. Dehydration starts at the hemiketal, and the intermediate enol rearranges to the aldehyde. The second dehydration gives the α, β-unsaturated derivative. A third water loss affords hydroxymethylfurfural.

These mechanism hold high significance with respect to ketoses and aldoses, and the hydride shift here makes the huge difference.

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What are the consequences?

The article says a lot of things, but here's a table summarising everything, comparing and contrasting glucose and fructose.

Reactant	Key Steps	Distinctive Features
Fructose	Protonation of anomeric OH	Initiated by protonation of more reactive OH
-do-	Tautomerism	Resonance Stabilisation
-do-	Dehydration of tetrahydrofuran	Allylic OH dehydration prior to tautomerism
Glucose	Protonation at O-2	Not most basic OH
-do-	Cationic rearrangement contraction
-do-	Two dehydrations to HMF	Distinctive ring contraction step

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In a nutshell: Quoting the article once again:

Fructose presents common reactivities in the synthesis of hydroxymethylfurfural, this not being the case with glucose or glucose isomerization to fructose. This explains the different reaction times observed in the test and it’s utility for ketose identification. Slower reactions resulted in accord with less common reaction steps, i.e., with especial reactivities.

In common man language: compare it to climbing a ramp and a staircase. The ramp is easy to climb, needs less physical stress, while the staircase needs more physical exertion. Strike an analogy for this as: Ketoses just need to climb a ramp, while glucose has to climb a steep staircase of a mechanism.

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Sources:

1. Sánchez-Viesca, Francisco, and Reina Gómez. "Reactivities involved in the Seliwanoff reaction." Modern Chemistry 6.1 (2018): 1-5.

Answer 2

Ketoses are more rapidly dehydrated than aldoses due to their molecular structure. In ketoses, the carbonyl group is located at the second carbon atom, making it more reactive and prone to dehydration reactions. When heated with a strong acid, ketoses can quickly form furfural derivatives through dehydration.

Aldoses, on the other hand, have the carbonyl group at the end of the molecule, which makes them less reactive in comparison. They require a β-elimination step before dehydration can occur, which increases the activation energy and slows down the reaction.