Explain why (i) the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride? (ii) alkyl halid

Exercises · 6.12

Q. Explain why

(i) the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride?

(ii) alkyl halides, though polar, are immiscible with water?

(iii) Grignard reagents should be prepared under anhydrous conditions?

Nagaland NbseTextbookSubjective· 3mImportance★★★★★

20% · 30/147 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Dipole Moment Trends

Dipole Moment Trends

A dipole moment measures how unevenly the electron density is shared across a bond or molecule — it is the product of the magnitude of separated charge and the distance between the charge centres, and it has a direction (from the positive end to the negative end, by convention pointing toward the more electronegative atom).

What Decides a Bond's Dipole Moment

Two things, working together, not just one:

Electronegativity difference — a bigger gap between the two bonded atoms means a more polarised bond, in principle a bigger dipole.
Bond length — dipole moment is charge × distance, so even a modestly polarised bond can have a sizeable moment if the atoms are far apart, and a strongly polarised but very short bond can have a smaller moment than you'd expect.

These two factors often pull in OPPOSITE directions down a group, which is why dipole-moment trends are frequently non-monotonic even when electronegativity itself decreases smoothly.

The Carbon–Halogen Trend

Going down Group 17 (F → Cl → Br → I), electronegativity decreases steadily, which would suggest the C–X dipole moment should decrease steadily too. But bond length increases at the same time (the halogen atom gets bigger), which pushes the dipole moment back up. The two effects partially cancel, so the C–X dipole moment does NOT fall in a simple, uniform way down the group — it is often close to level from C–Cl to C–I, or even shows a local maximum, rather than tracking electronegativity alone. Always check real values rather than assuming "more electronegative always means bigger dipole."

Molecular Dipole Moment: Vector Addition, Not Just Bond Polarity

A molecule's OVERALL dipole moment is the vector sum of all its individual bond dipoles — geometry matters as much as bond polarity.

$CCl_4$ (tetrahedral, symmetric): four polar C–Cl bonds, but their vectors cancel exactly by symmetry, so the molecule has ZERO net dipole moment despite having polar bonds.
$CHCl_3$ (one bond replaced by C–H): the symmetry is broken, the vectors no longer cancel, and the molecule has a real net dipole moment. …

Why this formula?

Great! Let's build a deep, concept-first understanding of Dipole Moment Trends — not just the formula, but why it works the way it does.

1. What is a Dipole Moment?

A dipole moment arises when there is a separation of charge within a molecule — one end is slightly positive ( $\delta^+$ ), the other slightly negative ( $\delta^-$ ). This is called a polar bond.

The dipole moment is a vector quantity: it has both magnitude and direction.

2. The Key Formula

The magnitude of the dipole moment ( $\mu$ ) for a pair of charges is given by:

\mu = q \times d

Where:

$q$ = magnitude of charge separation (in coulombs, but often in Debye units)
$d$ = distance between the centers of positive and negative charge (in meters or Ångströms)

Why this formula?

It comes directly from the definition: dipole moment measures how much the charges are "pulled apart" and how big those charges are. If you double the charge or double the separation, the dipole moment doubles. It's a product, not a sum.

3. Why the Formula Makes Physical Sense

Imagine a simple diatomic molecule like HCl:

The H atom has $\delta^+$
The Cl atom has $\delta^-$

The dipole moment points from positive to negative (by convention in chemistry, from $\delta^+$ to $\delta^-$ ).

If the bond length $d$ increases, the charges are farther apart → larger $\mu$ .
If the electronegativity difference increases → larger $q$ → larger $\mu$ .

So the formula $\mu = q \times d$ captures both the "how much charge" and "how far apart" factors.

4. The Vector Nature — Why Direction Matters

In molecules with more than two atoms, dipole moments add as vectors, not scalars.

For example, in water ( $H_2O$ ):

Each O–H bond has its own dipole moment (pointing from H to O).
The molecule is bent (~104.5°), so the two bond dipoles partially cancel but not completely.

The net dipole moment is:

\mu_{\text{net}} = \sqrt{\mu_1^2 + \mu_2^2 + 2\mu_1\mu_2\cos\theta}

Where $\theta$ is the angle between the two bond dipoles.

Why this formula?

It's just the law of vector addition (parallelogram law). If the dipoles point in exactly opposite directions ( $\theta = 180^\circ$ ), they cancel completely. If they point in the same direction ( $\theta = 0^\circ$ ), they add fully.

5. Trends You Can Now Explain

Trend	Why? (Based on formula)
Larger bond length → larger $\mu$	$d$ increases in $\mu = q \times d$

Start your 14-day free trial to unlock the full solution →

The key idea is that molecular structure and intermolecular forces govern these three distinct phenomena: (i) resonance delocalisation in chlorobenzene reduces its dipole moment compared to the purely inductive effect in cyclohexyl chloride;

(ii) alkyl halides cannot form strong hydrogen bonds with water despite being polar, so they remain immiscible;

(iii) Grignard reagents are violently reactive with water because the carbon–magnesium bond is highly polarised and carbanionic, so anhydrous conditions are essential to prevent decomposition.

(i) Why is the dipole moment of chlorobenzene lower than that of cyclohexyl chloride?

Concept and intuition

Dipole moment depends on both the magnitude of charge separation and the distance between the partial charges. In cyclohexyl chloride, the C–Cl bond is essentially a pure sigma bond with chlorine pulling electron density away from the sp³ carbon — a straightforward inductive effect. In chlorobenzene, the chlorine is attached to an sp² carbon of an aromatic ring. Here, the lone pairs on chlorine can participate in resonance with the π-system of the benzene ring, which changes the electron distribution significantly.

Step-by-step reasoning

Resonance in chlorobenzene

The chlorine atom has three lone pairs. One of these lone pairs can be delocalised into the aromatic ring via resonance, creating structures where the C–Cl bond acquires partial double-bond character. This resonance hybrid has a significant contribution from a structure where the chlorine bears a positive charge and the ortho and para positions of the ring bear negative charges.
Effect on bond polarity

In the resonance hybrid, the partial positive charge on chlorine partially cancels the partial negative charge that would normally reside on chlorine due to its electronegativity. The net result is that the C–Cl bond in chlorobenzene is less polar than a typical C–Cl single bond.
Comparison with cyclohexyl chloride

Cyclohexyl chloride has no such resonance possibility — the chlorine is on a saturated carbon. The C–Cl bond here is a pure sigma bond with full inductive electron withdrawal. The charge separation is larger, and the dipole moment is therefore higher.
Quantitative result

The dipole moment of chlorobenzene is about 1.69 D, while that of cyclohexyl chloride is about 2.20 D. The difference is directly attributable to resonance delocalisation reducing the effective polarity of the C–Cl bond in the aromatic compound.

Watch out

A common mistake is to think that because chlorine is more electronegative, chlorobenzene should have a higher dipole moment. But resonance can reduce the dipole moment by spreading charge over the ring and even placing a partial positive charge on chlorine.

(ii) Why are alkyl halides, though polar, immiscible with water?

Concept and intuition

Miscibility is not simply about polarity — it is about the ability to form strong intermolecular attractions with the solvent. Water is a highly hydrogen-bonded solvent. For a solute to dissolve, it must be able to break water–water hydrogen bonds and replace them with comparable solute–water interactions. Alkyl halides fail this test.

Step-by-step reasoning

Nature of polarity in alkyl halides

The C–X bond (X = F, Cl, Br, I) is indeed polar, giving alkyl halides a net dipole moment. However, the molecule as a whole is largely nonpolar — the alkyl chain is hydrophobic, and the polar C–X bond is only a small part of the molecule.
Hydrogen bonding inability

For a molecule to be water-miscible, it must be able to form hydrogen bonds with water. Alkyl halides have no hydrogen atom attached to a highly electronegative atom (like O, N, or F), and the halogen atoms themselves, though electronegative, have lone pairs that are poor hydrogen bond acceptors compared to oxygen. The C–H bonds are not sufficiently polarised to participate in hydrogen bonding.
Energy penalty

When an alkyl halide is added to water, the water molecules must reorganise around the nonpolar alkyl chain, forming a cage-like structure (the hydrophobic effect). This costs energy. The weak dipole–dipole interactions between the alkyl halide and water are insufficient to compensate for the loss of water–water hydrogen bonds.
Result

The alkyl halide remains as a separate layer — it is immiscible. The polar C–X bond is simply not enough to overcome the hydrophobic nature of the rest of the molecule and the inability to form hydrogen bonds.

Tip

A useful rule of thumb: "like dissolves like" is too simplistic. The real criterion is whether the solute can participate in the same type of intermolecular forces as the solvent. For water, that means hydrogen bonding — polarity alone is not enough.

(iii) Why must Grignard reagents be prepared under anhydrous conditions?

Concept and intuition …

Start your 14-day free trial to unlock the full solution →

Here is the clear solution method for the given questions, following a Concept-First approach.

Method: Hybridization & Solvation Analysis

This method uses two core ideas:

Hybridization effect on bond polarity: The $s$ -character of the carbon atom bonded to the halogen changes the electronegativity of carbon, altering the dipole moment.
Intermolecular forces & solvation: The ability of a molecule to dissolve in water depends on its ability to overcome water's hydrogen bonding (energy required) versus the energy released by forming new interactions.

(i) Why is the dipole moment of chlorobenzene lower than that of cyclohexyl chloride?

Step 1: Identify the carbon hybridization.

In chlorobenzene, the C–Cl bond is attached to an $sp^2$ hybridized carbon (part of the aromatic ring).
In cyclohexyl chloride, the C–Cl bond is attached to an $sp^3$ hybridized carbon.

Step 2: Apply the hybridization effect.

$sp^2$ carbon has more s-character (33%) than $sp^3$ carbon (25%).
More s-character means the carbon atom is more electronegative (it holds its electrons tighter).
In chlorobenzene, the more electronegative $sp^2$ carbon pulls electron density away from chlorine, reducing the bond polarity ( $\delta^+$ on C is smaller).
In cyclohexyl chloride, the less electronegative $sp^3$ carbon allows chlorine to hold the electron density more strongly, giving a larger bond polarity.

Step 3: Consider resonance (the secondary effect).

In chlorobenzene, lone pairs on chlorine can delocalize into the aromatic ring via resonance, giving partial double-bond character. This further reduces the net dipole moment.

Result: The dipole moment of chlorobenzene is lower because the $sp^2$ carbon is more electronegative and resonance delocalization reduces polarity.

(ii) Why are alkyl halides, though polar, immiscible with water?

Step 1: Identify the dominant force in water.

Water molecules are held together by strong hydrogen bonds (energy ≈ 20–40 kJ/mol).

Step 2: Analyze the alkyl halide's ability to interact.

Alkyl halides (R–X) are polar due to the C–X bond, but they cannot form hydrogen bonds with water. They lack a hydrogen atom bonded to N, O, or F, and the halogen (though electronegative) is not sufficiently electronegative or small enough to form strong H-bonds with water.

Step 3: Apply the "like dissolves like" principle with energy.

For mixing to occur, water must break its own H-bonds to make room for alkyl halide molecules. The energy required to break these H-bonds is not compensated by the weak dipole-dipole or van der Waals interactions between water and the alkyl halide.
The large non-polar alkyl chain (R group) further disrupts water structure, making the process energetically unfavorable. …

Start your 14-day free trial to unlock the full solution →

Here are the common mistakes students make on this question, broken down by each sub-part, along with the correct conceptual approach to avoid them.

General Mistake: Memorizing without understanding the "why"

Students often memorize the final answer (e.g., "chlorobenzene has lower dipole moment") but fail to explain the mechanism (resonance vs. inductive effect). In exams, the explanation carries more marks than the result.

(i) Dipole Moment: Chlorobenzene vs. Cyclohexyl Chloride

Common Mistake 1: Ignoring the role of hybridization.

The Error: Students say "chlorobenzene has lower dipole moment because it is aromatic" without specifying why aromaticity changes the dipole.
The Fix: Recognize that the carbon attached to chlorine in chlorobenzene is $sp^2$ hybridized (more s-character, 33%), while in cyclohexyl chloride it is $sp^3$ hybridized (25% s-character).
- Why it matters: Greater s-character in $sp^2$ carbon pulls the bonding electrons closer to itself, reducing the electron density on chlorine. This shortens the C–Cl bond and reduces the bond dipole moment.
- Result: The $\mu$ of chlorobenzene is lower than that of cyclohexyl chloride.

Common Mistake 2: Confusing resonance with inductive effect.

The Error: Students claim "chlorobenzene has lower dipole moment because chlorine donates electrons via resonance." This is partially wrong for the net dipole direction.
The Fix: Understand the two opposing effects in chlorobenzene:
1. -I effect (Inductive): Chlorine withdraws electrons through sigma bonds.
2. +R effect (Resonance): Chlorine donates its lone pairs into the ring (creating partial double bond character).
- Key insight: The +R effect increases electron density on the ring (especially ortho/para positions), which opposes the direction of the -I effect. This reduces the net dipole moment compared to cyclohexyl chloride, where only the -I effect operates (no resonance).

How to avoid: Always draw the resonance structures of chlorobenzene to show the +R effect. Then state: "The +R effect of chlorine in chlorobenzene opposes the -I effect, reducing the net dipole moment. In cyclohexyl chloride, only the -I effect is present, so the dipole is larger."

(ii) Alkyl Halides: Polar but Immiscible with Water

Common Mistake 1: Thinking "like dissolves like" is the only rule.

The Error: Students say "alkyl halides are non-polar" (which is false) or "they are polar but water is polar, so they should mix."
The Fix: Understand that miscibility requires more than just polarity — it requires hydrogen bonding.
- Alkyl halides (R–X) are polar due to the C–X bond dipole, but they cannot form hydrogen bonds with water (no –OH or –NH group, and the halogen is not sufficiently electronegative to act as a strong H-bond acceptor in this context).
- Water molecules are held together by strong intermolecular H-bonds. To dissolve an alkyl halide, water must break its own H-bonds and form new ones with the solute. Since alkyl halides cannot offer strong H-bonding, the energy cost of breaking water's structure is not compensated, so they remain immiscible.

Common Mistake 2: Confusing "polar" with "ionic".

The Error: Students think that because alkyl halides are polar covalent, they should behave like ionic salts (which dissolve in water).
The Fix: Ionic compounds dissolve because of ion-dipole interactions (strong electrostatic attraction). Alkyl halides are covalent molecules with only dipole-dipole and London forces. These are too weak to overcome water's cohesive H-bonding.

How to avoid: Write: "Alkyl halides are polar but cannot form hydrogen bonds with water. The energy required to break water's H-bond network is not compensated, so they are immiscible."

(iii) Grignard Reagents: Anhydrous Conditions

Common Mistake 1: Saying "water destroys the reagent" without mechanism.

The Error: Students just state "Grignard reagents react with water" but don't explain what they form or why it's a problem.
The Fix: Know the exact reaction: …

Start your 14-day free trial to unlock the full solution →

←Previous question6.11 — How will you bring about the following conversions? (i) Ethanol to but…→Next question6.13 — Give the uses of freon 12, DDT, carbon tetrachloride and iodoform.

Q.Explain why

Concept understanding — Dipole Moment Trends

Dipole Moment Trends

What Decides a Bond's Dipole Moment

The Carbon–Halogen Trend

Molecular Dipole Moment: Vector Addition, Not Just Bond Polarity

Why this formula?

1. What is a Dipole Moment?

2. The Key Formula

3. Why the Formula Makes Physical Sense

4. The Vector Nature — Why Direction Matters

5. Trends You Can Now Explain

(i) Dipole moment of chlorobenzene vs. cyclohexyl chloride

(i) Why is the dipole moment of chlorobenzene lower than that of cyclohexyl chloride?

(ii) Why are alkyl halides, though polar, immiscible with water?

(iii) Why must Grignard reagents be prepared under anhydrous conditions?

Method: Hybridization & Solvation Analysis

(i) Why is the dipole moment of chlorobenzene lower than that of cyclohexyl chloride?

(ii) Why are alkyl halides, though polar, immiscible with water?

General Mistake: Memorizing without understanding the "why"

(i) Dipole Moment: Chlorobenzene vs. Cyclohexyl Chloride

(ii) Alkyl Halides: Polar but Immiscible with Water

(iii) Grignard Reagents: Anhydrous Conditions

Q.Explain why

(i) Why is the dipole moment of chlorobenzene lower than that of cyclohexyl chloride?

(ii) Why are alkyl halides, though polar, immiscible with water?

(iii) Why must Grignard reagents be prepared under anhydrous conditions?

🎓Unlock everything free for 14 days

Unlock everything free for 14 days