The molecules having dipole moment are _____. (Note: more than one of the given options may be correct.) (i) 2

NCERT Exemplar · Q19

Q. The molecules having dipole moment are __________. (Note: more than one of the given options may be correct.)

(i) 2,2-Dimethylpropane

(ii) trans-Pent-2-ene

(iii) cis-Hex-3-ene

(iv) 2, 2, 3, 3 - Tetramethylbutane.

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Concept understanding — Boiling Point Trends

Boiling Point Trends (Organic Compounds)

A substance's boiling point is set by how much energy is needed to overcome the attractive forces HOLDING its molecules together in the liquid — the stronger those intermolecular forces, the higher the boiling point.

The Forces, Weakest to Strongest

Van der Waals (London dispersion) forces — present in every molecule, and they grow stronger as the molecule gets bigger (more electrons, larger surface area of contact between neighbouring molecules) and more polarisable.
Dipole–dipole forces — present in polar molecules, add an extra attraction on top of dispersion forces.
Hydrogen bonding — present when H is bonded directly to N, O, or F; much stronger than ordinary dipole–dipole attraction, and it raises the boiling point sharply compared to a similarly-sized molecule without it.

Trend 1: Down a Series of Halogens (Same Alkyl Group)

For a fixed R group, boiling point rises as the halogen gets heavier: $R{-}I > R{-}Br > R{-}Cl > R{-}F$ . This looks surprising at first, since electronegativity (and so bond polarity/dipole moment) actually DECREASES down the group — but boiling point here is dominated by the growing size and polarisability of the halogen atom (stronger dispersion forces), which outweighs the shrinking dipole contribution.

The measured values for the methyl, ethyl and propyl halides show this rise clearly:

Measured boiling points of methyl, ethyl and propyl chlorides, bromides and iodides (NCERT Fig 6.1), rising steadily from R-Cl to R-Br to R-I for every alkyl group

Trend 2: Chain Length and Branching

Longer chains (more carbons) have more surface area for van der Waals contact between neighbouring molecules, so boiling point rises with chain length within a homologous series.
Branching LOWERS boiling point compared to a straight-chain isomer of the same molecular formula — a more compact, spherical shape has less surface-to-surface contact with neighbouring molecules, weakening the dispersion forces. (E.g. neopentane boils well below n-pentane.)

Trend 3: Hydrogen Bonding Beats Molecular Mass …

Why this formula?

Boiling Point Trends: Why They Happen

Boiling point is the temperature at which a liquid's vapor pressure equals the external atmospheric pressure. To understand why boiling points follow certain trends, we must first understand what determines vapor pressure.

The Core Idea: Intermolecular Forces

A liquid boils when its molecules have enough kinetic energy to overcome the intermolecular forces (IMFs) holding them together in the liquid phase. Stronger IMFs → harder to escape → lower vapor pressure at a given temperature → higher boiling point.

There is no single "formula" for boiling point, but the relationship is captured by the Clausius–Clapeyron equation, which links vapor pressure ( $P$ ) to temperature ( $T$ ) and the enthalpy of vaporization ( $\Delta H_{\text{vap}}$ ):

\ln P = -\frac{\Delta H_{\text{vap}}}{R} \cdot \frac{1}{T} + C

Where:

$P$ = vapor pressure
$\Delta H_{\text{vap}}$ = enthalpy of vaporization (energy needed to vaporize 1 mole)
$R$ = gas constant
$T$ = absolute temperature (Kelvin)
$C$ = constant (depends on substance)

Why this formula makes sense

$\Delta H_{\text{vap}}$ is large when IMFs are strong — more energy is needed to separate molecules.
At boiling point, $P = P_{\text{atm}}$ (usually 1 atm). So a substance with larger $\Delta H_{\text{vap}}$ needs a higher $T$ to reach that pressure.

Thus, boiling point ∝ strength of intermolecular forces.

The Four Key Trends (with Reasoning)

1. Trend across a period (e.g., Period 2: CH₄ → NH₃ → H₂O → HF)

Molecule	IMFs present	Boiling point (°C)
CH₄	London dispersion only	-161
NH₃	Dispersion + H-bonding	-33
H₂O	Dispersion + H-bonding (2 per molecule)	100
HF	Dispersion + H-bonding	19

Why?

CH₄ is nonpolar — only weak London dispersion forces.
NH₃, H₂O, HF have hydrogen bonding (strongest IMF).
H₂O forms two H-bonds per molecule (donor + acceptor), while NH₃ forms one and HF forms one — hence H₂O has the highest boiling point.

Key insight: Hydrogen bonding dominates over molecular mass in small molecules.

2. Trend down a group (e.g., Halogens: F₂ → Cl₂ → Br₂ → I₂)

Molecule	Molar mass (g/mol)	Boiling point (°C)
F₂	38	-188
Cl₂	71	-34
Br₂	160	59
I₂	254	184

Why?

All are nonpolar — only London dispersion forces.
Dispersion force strength increases with number of electrons (larger molar mass → more polarizable electron cloud → stronger temporary dipoles).
So boiling point increases down the group.

Key insight: For nonpolar molecules, molar mass (electron count) is the primary factor.

3. Branching in alkanes (e.g., C₅H₁₂ isomers)

Isomer	Boiling point (°C)
n-pentane (straight chain)	36
2-methylbutane (branched)	28
2,2-dimethylpropane (highly branched)	10

Why?

All have same molecular formula — same molar mass. …

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Dipole moment requires a net molecular polarity; symmetric molecules cancel their bond dipoles. Only cis-Hex-3-ene has an asymmetric geometry that produces a permanent dipole moment.

Why molecular symmetry determines dipole moment

A molecule has a dipole moment when the vector sum of all its bond dipoles does not cancel to zero. Even if individual bonds are polar, a highly symmetric molecular geometry can arrange them so that their dipole vectors point in opposite directions and cancel completely. The key is to examine the three-dimensional structure and look for planes or centres of symmetry.

For hydrocarbons, C–H and C–C bonds have very small individual dipole moments (carbon and hydrogen have similar electronegativities), so we focus on the overall molecular shape and any substituent arrangement that breaks symmetry.

Step-by-step analysis of each molecule

1. 2,2-Dimethylpropane (neopentane)

The structure is:

\ce{C(CH3)4}

A central carbon bonded to four methyl groups in a perfect tetrahedral arrangement. This molecule has $T_d$ symmetry—every direction is balanced by an equivalent group in the opposite direction. The vector sum of all bond dipoles is zero.

Dipole moment: zero.

2. trans-Pent-2-ene

The structure places the double bond between C2 and C3, with substituents on opposite sides:

\ce{CH3-CH=CH-CH2-CH3}

In the trans configuration, the methyl group (on C1) and the ethyl group (on C4–C5) lie on opposite sides of the double bond. The molecule has a plane of symmetry perpendicular to the double bond, and the two halves mirror each other. Any small dipole from one side is canceled by the corresponding dipole on the other.

Dipole moment: zero.

3. cis-Hex-3-ene

The structure places the double bond between C3 and C4, with substituents on the same side:

\ce{CH3-CH2-CH=CH-CH2-CH3}

In the cis configuration, both ethyl groups (C1–C2 and C5–C6) are on the same side of the double bond. This breaks the symmetry: there is no plane or centre that would cancel the dipoles. The electron density is slightly skewed toward one side of the molecule, creating a small but nonzero net dipole moment.

Tip

cis isomers of alkenes generally have dipole moments because the substituents cluster on one side, while trans isomers are symmetric and have zero dipole moment.

Dipole moment: nonzero.

4. 2,2,3,3-Tetramethylbutane …

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Q.The molecules having dipole moment are __________. (Note: more than one of the given options may be correct.)

Why molecular symmetry determines dipole moment

Step-by-step analysis of each molecule

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