Assertion (A): D(+)-Glucose is dextrorotatory in nature. Reason (R): 'D' represents its dextrorotatory nature.

NCERT Exemplar · Q65

Q. Assertion (A): D(+)-Glucose is dextrorotatory in nature. Reason (R): 'D' represents its dextrorotatory nature.

(i) Both A and R are true and R is the correct explanation of A.

(ii) Both A and R are true but R is not the correct explanation of A.

(iii) A is true but R is false.

(iv) Both A and R are false.

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Concept understanding — De Broglie Wavelength

De Broglie Wavelength: When Particles Start Acting Like Waves

Imagine you're holding a cricket ball. You know exactly where it is, and if you throw it, you can predict its path. That's a particle — localised, definite, following Newton's laws. Now think of light. You can't "hold" a beam of light; it spreads out, bends around corners, creates interference patterns. That's a wave — spread out, not localised.

For centuries, physics kept these two worlds separate. Particles were particles. Waves were waves. Never the twain shall meet.

Then came a young French physicist, Louis de Broglie, in 1924. He asked a question that seemed almost absurd: If light — which we thought was a wave — can behave like a particle (the photoelectric effect), then why can't a particle — say, an electron — behave like a wave?

That question turned physics upside down.

The Core Idea

De Broglie proposed that every moving particle has a wave associated with it. The wavelength of that wave depends on the particle's momentum. The faster or heavier the particle, the shorter the wavelength.

\lambda = \frac{h}{p} = \frac{h}{mv}

This is not a mathematical trick. It's a physical reality. An electron moving through a crystal actually behaves like a wave of this wavelength — it can diffract, interfere, and form patterns just like light does.

Why You Don't See It in Daily Life

Here's the crucial point: the de Broglie wavelength is incredibly tiny for everyday objects.

Take a cricket ball of mass 0.16 kg moving at 30 m/s. Its de Broglie wavelength is:

\lambda = \frac{6.626 \times 10^{-34}}{0.16 \times 30} \approx 1.38 \times 10^{-34} \ \text{m}

That's about a hundred trillion trillion times smaller than the nucleus of an atom. No experiment can detect such a wave — it's effectively zero for all practical purposes.

Now take an electron (mass $9.1 \times 10^{-31}$ kg) accelerated through 100 volts. Its speed is about $5.9 \times 10^6$ m/s. Its de Broglie wavelength:

\lambda = \frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 5.9 \times 10^6} \approx 1.23 \times 10^{-10} \ \text{m}

That's about 0.12 nanometres — comparable to the spacing between atoms in a crystal. This is measurable. And indeed, in 1927, Davisson and Germer fired electrons at a nickel crystal and observed diffraction — the unmistakable signature of a wave.

Note

The de Broglie wavelength is only observable when it is comparable to the size of objects the particle interacts with. For macroscopic objects, it's far too small to matter. For subatomic particles, it's the key to understanding their behaviour.

What This Means Physically

The wave is not a physical wave in space like a water wave. It's a probability wave — its amplitude at any point tells you the probability of finding the particle there. Where the wave amplitude is large, you're likely to find the particle; where it's zero, you won't.

This wave-particle duality is not a compromise. It's the actual nature of reality. An electron is neither a pure particle nor a pure wave — it's something that shows particle-like behaviour in some experiments (like hitting a screen at a point) and wave-like behaviour in others (like passing through two slits and interfering with itself). …

Why this formula?

De Broglie Wavelength: Why the Formula Holds

Let's build this from the ground up — understanding why matter has a wavelength, not just memorizing $\lambda = \frac{h}{p}$ .

The Core Insight: Nature's Symmetry

Before de Broglie, physics had two separate worlds:

Light — showed wave behaviour (diffraction, interference) but also particle behaviour (photoelectric effect)
Matter — showed particle behaviour (momentum, collisions) but no wave behaviour yet

De Broglie asked a daring question in his 1924 PhD thesis:

If light (a wave) can behave like a particle, why can't a particle (like an electron) behave like a wave?

Nature should be symmetric — what applies to one should apply to the other.

Step 1: Start with Light (What We Already Knew)

For a photon, Einstein had given us two key relations:

Energy: $E = hf$ (Planck's relation)
Momentum: $p = \frac{h}{\lambda}$ (from $E = pc$ for light, combined with $c = f\lambda$ )

So for light:

\lambda = \frac{h}{p}

This was experimentally verified for photons.

Step 2: De Broglie's Bold Hypothesis

De Broglie said: This relation is not special to light. It is universal.

For any particle with momentum $p$ :

\boxed{\lambda = \frac{h}{p}}

Where:

$\lambda$ = de Broglie wavelength
$h$ = Planck's constant ( $6.626 \times 10^{-34} \text{ J·s}$ )
$p$ = momentum of the particle

Step 3: Why Momentum and Not Velocity?

This is crucial. The formula uses momentum ( $p = mv$ ), not just velocity.

For a non-relativistic particle (slow compared to light):

\lambda = \frac{h}{mv}

For a relativistic particle (like an electron at high speed):

p = \gamma mv \quad \text{where} \quad \gamma = \frac{1}{\sqrt{1 - v^2/c^2}}

\lambda = \frac{h}{\gamma mv}

Why momentum? Because momentum is the more fundamental quantity — it's conserved, it's frame-independent in a deeper sense, and it connects directly to the wave's phase.

Step 4: The Deeper Reasoning — Wave-Particle Duality

De Broglie didn't just guess. He reasoned:

Every moving particle has an associated wave — called the "matter wave" or "pilot wave"
The frequency of this wave comes from energy: $f = \frac{E}{h}$
The wavelength comes from momentum: $\lambda = \frac{h}{p}$

These two relations are linked by the phase velocity of the wave:

v_{\text{phase}} = f\lambda = \frac{E}{h} \cdot \frac{h}{p} = \frac{E}{p}

For a free particle with kinetic energy $E = \frac{p^2}{2m}$ :

v_{\text{phase}} = \frac{p}{2m} = \frac{v}{2}

This is half the particle's speed — a strange but mathematically consistent result.

Step 5: Experimental Confirmation (Why We Believe It)

De Broglie's idea was confirmed when electrons showed wave behaviour: …

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The assertion is true (D(+)-glucose is dextrorotatory), but the reason is false — the 'D' in D-glucose refers to the configuration at the chiral carbon farthest from the aldehyde group, not to its optical rotation. The correct option is (C).

Understanding the Concept: Dextrorotatory vs. D-Configuration

This question tests a classic confusion in carbohydrate chemistry. You need to separate two ideas that sound similar but are completely independent:

Optical rotation — whether a compound rotates plane-polarised light to the right (dextrorotatory, +) or left (laevorotatory, −). This is a physical property measured experimentally.
D/L configuration — a naming convention based on the orientation of the −OH group on the chiral carbon farthest from the carbonyl group (the penultimate carbon). This is a structural assignment, not a measurement of rotation.

The 'D' in D-glucose has nothing to do with dextrorotation. It comes from the Fischer projection: if the −OH on the bottom-most chiral centre points to the right, the sugar is D; if it points left, it is L. The actual optical rotation is indicated separately by a (+) or (−) sign.

Watch out

Many students assume 'D' stands for dextrorotatory. This is the single most common mistake in carbohydrate nomenclature. D-glucose happens to be dextrorotatory, but D-fructose is laevorotatory — proving the D/L label and rotation are unrelated.

Step-by-Step Analysis

1. Check the Assertion (A): D(+)-Glucose is dextrorotatory in nature.

D(+)-Glucose is the naturally occurring form of glucose. The (+) sign explicitly tells us it rotates plane-polarised light to the right. This is a well-established experimental fact. So Assertion (A) is true.

2. Check the Reason (R): 'D' represents its dextrorotatory nature. …

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Here are the common mistakes students make with this Assertion-Reason question, along with how to avoid each.

Mistake 1: Confusing ‘D’ with Dextrorotatory

The Mistake:

Students assume that the ‘D’ in D(+)-Glucose stands for dextrorotatory (i.e., it rotates plane-polarized light to the right). They then think the Reason (R) is correct, leading them to choose option (A) or (B).

Why it’s wrong:

The ‘D’ in carbohydrate nomenclature refers to the configuration at the chiral carbon farthest from the aldehyde group (C-5 in glucose). It indicates that the -OH group on that carbon is on the right in the Fischer projection.
The ‘(+)’ sign (or ‘+’) is what actually denotes dextrorotatory (clockwise rotation of light).
So, ‘D’ does not mean dextrorotatory — it is a configurational label.

How to Avoid:

Memorise the distinction:
- D/L = configuration (relative to glyceraldehyde).
- (+)/(-) = observed optical rotation (experimental).
Always check the symbols in the question: D(+)-Glucose has both a configurational label (D) and a rotation sign (+). They are independent.

Mistake 2: Thinking the Reason Explains the Assertion

The Mistake:

Even if a student knows that D(+)-Glucose is dextrorotatory (Assertion is true), they might still think the Reason correctly explains why it is dextrorotatory — because they confuse ‘D’ with rotation.

Why it’s wrong:

The Assertion (A) is true: D(+)-Glucose is indeed dextrorotatory (it rotates light to the right).
The Reason (R) is false because ‘D’ does not represent dextrorotatory nature.
Since R is false, it cannot be the correct explanation of A.

How to Avoid:

In Assertion-Reason questions, first independently check the truth of each statement.
- A: True (D(+)-Glucose is dextrorotatory).
- R: False (D does not mean dextrorotatory).
Then check if R explains A — if R is false, it cannot explain anything.
The correct answer here is (C) A is true but R is false.

Mistake 3: Forgetting that ‘D’ and ‘L’ are Configurational, Not Rotational

The Mistake: …

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Q.Assertion (A): D(+)-Glucose is dextrorotatory in nature.
Reason (R): 'D' represents its dextrorotatory nature.

Concept understanding — De Broglie Wavelength

De Broglie Wavelength: When Particles Start Acting Like Waves

The Core Idea

Why You Don't See It in Daily Life

What This Means Physically

Why this formula?

De Broglie Wavelength: Why the Formula Holds

The Core Insight: Nature's Symmetry

Step 1: Start with Light (What We Already Knew)

Step 2: De Broglie's Bold Hypothesis

Step 3: Why Momentum and Not Velocity?

Step 4: The Deeper Reasoning — Wave-Particle Duality

Step 5: Experimental Confirmation (Why We Believe It)

Understanding the Concept: Dextrorotatory vs. D-Configuration

Step-by-Step Analysis

Method: Conceptual Analysis of D/L and (+)/(−) Nomenclature in Carbohydrates

Relevant Concept

Steps to Solve

Mistake 1: Confusing ‘D’ with Dextrorotatory

Mistake 2: Thinking the Reason Explains the Assertion

Mistake 3: Forgetting that ‘D’ and ‘L’ are Configurational, Not Rotational

Q.Assertion (A): D(+)-Glucose is dextrorotatory in nature. Reason (R): 'D' represents its dextrorotatory nature.

Understanding the Concept: Dextrorotatory vs. D-Configuration

Step-by-Step Analysis

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Q.Assertion (A): D(+)-Glucose is dextrorotatory in nature.
Reason (R): 'D' represents its dextrorotatory nature.

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