Foundations Section 9 of 123

Bond polarity

the why on every wrong step

15 min readfree

Carbon dioxide and water each have two strongly polar bonds. Yet a carbon dioxide molecule has no positive end and no negative end, and a water molecule has both. The difference is shape: carbon dioxide is straight, and water is bent.

This section puts two earlier ideas together. The Electronegativity section tells you which bonds are polar. The Molecular Geometry section tells you which way each bond points. Together they tell you whether a whole molecule is polar. That decides what it dissolves in, what it boils at, and how strongly it holds on to the molecules around it.

Bond dipoles

Start with one bond, C–O. Oxygen's electronegativity is 3.44 and carbon's is 2.55, so the shared electrons spend more of their time near the oxygen. The oxygen takes a partial negative charge, δ−, and the carbon a partial positive charge, δ+. The Electronegativity section explains where these partial charges come from.

Two opposite charges held a distance apart make a dipole. The dipole of a single bond is its bond dipole. A dipole has a size and a direction, so it is a vector: a quantity drawn as an arrow, whose length shows the size and whose head shows the direction. The size of a dipole is its dipole moment, written μ (the Greek letter mu) and measured in units called debyes (D).

Chemists draw a bond dipole as an arrow beside the bond. The arrowhead points at the δ− atom. A short cross bar marks the tail, at the δ+ atom, so the tail looks like a plus sign. For C–O, the arrow points at oxygen.

A C–Li bond reverses the arrow. Lithium's electronegativity is 0.98, far below carbon's 2.55, so here carbon is the δ− atom and the arrow points at carbon. Compounds with a C–Li bond are organolithium reagents. The closely related Grignard reagents, such as CH3MgBr, have a C–Mg bond, and magnesium (1.31) is also far below carbon. They are written R–MgBr in general, where R stands for any carbon chain.

That reversal is what makes both kinds of reagent useful. Their carbon carries spare electron density, and it reacts by giving that density to an electron-poor atom in another molecule. Organometallic Bonding & Grignard Reagents puts them to work.

Two common bonds get no arrow. A C–C bond joins two atoms of identical electronegativity, so neither end pulls harder. A C–H bond has a difference of only 0.35 (2.55 against 2.20). The course treats both as nonpolar, so no arrow is drawn for either.

C–O: oxygen pulls harderδ+δ−COEN 2.55EN 3.44cross bar at the δ+ end,arrowhead at the δ− endC–Li: the arrow reversesδ+δ−CLiEN 2.55EN 0.98lithium pulls far less,so carbon is the δ− endC–H: treated as nonpolarCHEN 2.55EN 2.20difference 0.35:no arrow drawn
Three carbon bonds. In each panel, compare the two electronegativity values with the direction of the arrow.
Two conventions for the arrow. Organic chemistry points the dipole arrow at the δ− end, and courses and exams grade by that convention, so this course uses it. Physics, and the formal IUPAC definition of the dipole moment, point the same vector the other way, from the negative end to the positive end. If an arrow in a physics text looks backward, this is why.

Adding bond dipoles: the molecular dipole

Take carbon dioxide, O=C=O. Each C=O bond has a dipole pointing from the carbon out to an oxygen. The molecule is linear, so the two arrows point in exactly opposite directions. They are also the same size, because the two bonds are identical. The two pulls cancel, and the molecule as a whole has no dipole: μ = 0 D.

That is one case of a general rule. A molecule's overall dipole, its molecular dipole, is the vector sum of all its bond dipoles. To add vectors, slide the arrows, without turning them, so that each one starts where the one before it ended: tip to tail. The arrow from the first tail to the last tip is the sum.

Water shows the other outcome. Each O–H bond is polar, with its arrow pointing from hydrogen to oxygen. But water is bent, with an H–O–H angle of 104.5°, so the two arrows do not point in opposite directions. Their sideways parts cancel and their upward parts add. The sum points from between the two hydrogens toward the oxygen, and water is polar: μ = 1.85 D.

CO₂: linearCOOδ+δ−δ−tip to tailends where it begansum = 0, so μ = 0H₂O: bent12OHHδ−δ+δ+arrow 1arrow 2sumsum points to the O side: μ = 1.85 D
Each panel draws the molecule with its bond dipoles, then slides the same arrows tip to tail. Compare where the last arrow ends in each case.

Water's two lone pairs push the same way. They sit on the far side of the oxygen from the hydrogens, and they add electron density on that side. That is part of why water's dipole is as large as it is.

The working method has four steps. Draw the Lewis structure. Find the shape at the central atom, as in the Molecular Geometry section. Draw an arrow on each polar bond. Add the arrows.

Cancelling in three dimensions

Carbon tetrachloride, CCl4, has four polar C–Cl bonds (2.55 against 3.16). The carbon has four groups, so the bonds point to the corners of a tetrahedron. Draw the molecule with one chlorine straight up. The other three then point down and outward, evenly spaced around the vertical.

Add the three lower arrows first. Their sideways parts point three ways, evenly spaced, and cancel. Their downward parts add. On a regular tetrahedron the result is one arrow pointing straight down, exactly as long as a single C–Cl arrow. The top C–Cl arrow points straight up and has the same length. The two cancel, and CCl4 is nonpolar: μ = 0 D.

Now swap the top chlorine for a hydrogen. That gives chloroform, CHCl3. The three lower C–Cl arrows still add to one arrow pointing straight down. But the top corner now holds a C–H bond, which carries no arrow. Nothing opposes the downward arrow, so the sum no longer reaches zero.

CCl₄: four identical cornersCClClClClδ+δ−δ−δ−δ−topC–Cllower 3,addedequal length, opposite wayssum = 0, so μ = 0CHCl₃: one corner swappedCHClClClδ+δ−δ−δ−C–H: no arrowlower three,nothing opposingnet arrow toward the chlorinesμ = 1.04 D
The same tetrahedral frame twice. Compare the two sums on the right of each panel: in CCl₄ the top arrow cancels the lower three; in CHCl₃ nothing is left at the top to do it.

Chloroform is polar, with its negative end toward the chlorines and μ = 1.04 D. One substitution, and chloroform behaves as a completely different kind of solvent from carbon tetrachloride.

Worked example — two more tetrahedral carbons

Methane, CH4. Four C–H bonds and no arrows at all. There is nothing to add, so methane is nonpolar.

Dichloromethane, CH2Cl2. Two corners hold chlorine and two hold hydrogen. Only the two C–Cl bonds carry arrows, and with the corners no longer all alike, nothing can balance them. Dichloromethane is polar, μ = 1.60 D.

For the molecules in this course, polar bonds around a central atom cancel only when that atom has no lone pairs and every atom around it is the same. So linear CO2, trigonal planar BF3 and tetrahedral CCl4 cancel. Bent water has its lone pairs on one side and its hydrogens on the other. Trigonal pyramidal ammonia has its lone pair above and its hydrogens below. Both keep a net dipole. A few molecules outside this course, such as square planar XeF4, cancel even with lone pairs on the central atom.
"Has polar bonds" and "is a polar molecule" are different claims. A question about a molecule's polarity cannot be answered from electronegativity alone. Find the shape first, then add the arrows.

Dipole moments: the measured values

A dipole moment is charge times distance: the size of the partial charges multiplied by how far apart they sit. One debye is 3.34 × 10−30 coulomb-meters. For scale, a full proton charge and a full electron charge 100 pm apart would give 4.8 D. Real partial charges are much smaller, so most organic molecules land between 0 and about 4 D.

CompoundFormulaμ (D)
Carbon dioxideO=C=O0
Carbon tetrachlorideCCl40
Toluene, a hydrocarbonC7H80.36
ChloroformCHCl31.04
EthanolCH3CH2OH1.69
WaterH2O1.85
ChloromethaneCH3Cl1.87
AcetoneCH3–C(=O)–CH32.88
AcetonitrileCH3–C≡N3.92
Dimethyl sulfoxide (DMSO)(CH3)2S=O3.96

Acetone sits high on the list because of its C=O group, called a carbonyl. A C=O bond is strongly polar, and in most molecules no bond points the opposite way to cancel it. A molecule with a carbonyl is polar almost regardless of what else it carries.

Water is not at the top. Acetone, acetonitrile and DMSO all beat it. A dipole moment measures one thing: the charge separation inside a single isolated molecule. What makes liquid water unusual is something else, hydrogen bonding, and the water paragraph under the next heading comes back to it.

Bond length counts as well as the electronegativity difference, and the two can pull against each other. C–F has the largest difference of the carbon–halogen bonds, but it is also by far the shortest, about 139 pm against 178 pm for C–Cl. The two effects nearly cancel: fluoromethane, CH3F, has μ = 1.85 D and chloromethane 1.87 D. C–Cl edges ahead, but by far too little to build an argument on.

Further down the group the difference keeps shrinking (0.41 for C–Br, 0.11 for C–I) while the bond keeps getting longer. The dipole falls, but slowly: CH3Br is 1.81 D and CH3I 1.62 D. The longer bond makes up for much of the lost difference. That is why CH3Br, with a difference under the 0.5 line, is nearly as polar as CH3Cl.

So an electronegativity difference is a rough guide, not a measurement. Even C–I, at 0.11, gives CH3I a real dipole, because its bond is long and iodine's large electron cloud is easily pushed toward one end. C–H is the opposite case: a short bond with a small difference, so its dipole is small. Treating C–H as nonpolar is a close approximation, not an exact truth. Toluene, built only from C–C and C–H bonds, still shows a small dipole of 0.36 D, because its bonds are not all exactly even: its ring carbons and its methyl carbon are not identical partners either.

How polarity decides boiling points

Boiling a liquid breaks no covalent bonds. It pulls whole molecules away from their neighbors. The attractions between separate molecules are intermolecular forces, and the stronger they are, the higher the boiling point. Three kinds matter for the neutral molecules in this course, and polarity decides which ones a compound can use.

Pentane, C5H12, boils at 36 °C and decane, C10H22, at 174 °C, and neither has a dipole. Both hold together by London dispersion forces. Electrons move constantly, so at any instant one side of a molecule can carry a little more electron density than the other. That brief lopsidedness induces the opposite one in a neighbor, and the two attract. Every molecule has these forces. Each contact is weak, but the total grows with the molecule's surface area, so the longer chain boils higher.

Acetone and butane (C4H10) have the same molecular weight, 58. Butane boils at −1 °C and acetone at 56 °C. Acetone has a permanent dipole and butane has none. In liquid acetone the molecules line up so that the δ− oxygen of one sits near the δ+ carbon of the next. These dipole–dipole forces add to dispersion. Nearly sixty degrees of difference at the same molecular weight comes from polarity alone.

acetone, μ = 2.88 DCOCH₃CH₃δ+δ−COCH₃CH₃δ+δ−δ− oxygen of one next to the δ+ carbon of the nextdipole–dipole: boils at 56 °Cbutane, μ ≈ 0CH₃CH₂CH₂CH₃C–C and C–H only: no dipoledispersion only: boils at −1 °C
Both molecules weigh 58. Look at what lines up between the two acetones, and at what butane has to offer in its place.

Ethanol, CH3CH2OH, and dimethyl ether, CH3OCH3, share one formula, C2H6O, and a molecular weight of 46. Ethanol boils at 78 °C and dimethyl ether at −24 °C.

The difference is the hydrogen bond: an unusually strong dipole–dipole attraction between a hydrogen bonded to N, O or F and a lone pair on an N, O or F of another molecule. Oxygen pulls so hard on the electrons of an O–H bond that the hydrogen is strongly δ+. The hydrogen is also tiny, so it can sit very close to a lone pair on the next molecule's oxygen.

A hydrogen bond needs two partners. The donor supplies the hydrogen on N, O or F. The acceptor supplies the lone pair. Ethanol has both. Dimethyl ether has lone pairs to accept with, but every hydrogen sits on a carbon, and a C–H hydrogen is not δ+ enough to donate. That one hydrogen, on oxygen rather than on carbon, is worth a hundred degrees.

So for molecules of similar size, dispersion forces are the weakest of the three, dipole–dipole forces are stronger, and hydrogen bonds are the strongest.

ethanol, CH₃CH₂OHCH₃CH₂OHCH₃CH₂OHδ+δ−donor: H on Ohydrogen bondacceptor: lonepair on OMW 46 · boils at 78 °Cdimethyl ether, CH₃OCH₃OCH₃CH₃lone pairs: can acceptevery H is on a carbon:nothing to donateMW 46 · boils at −24 °C
Same formula, C₂H₆O. Follow the dashed line in ethanol from the hydrogen to the lone pair it points at, then look for a hydrogen in dimethyl ether that could do the same.

Now back to water. Each water molecule has two O–H hydrogens to donate and two lone pairs to accept, so one molecule can hold four neighbors by hydrogen bonds at once. Pulling that network apart takes a lot of energy. Water, with a molecular weight of only 18, boils at 100 °C, well above acetone's 56 °C, although acetone has the larger dipole moment.

OHHOHHOHHOHHOHHthe central water:gives 2 hydrogen bonds(its two H)takes 2 hydrogen bonds(its two lone pairs)boils at 100 °C
Count the dashed lines around the highlighted molecule, and check which end of each one sits on it.
Worked example — rank four compounds by boiling point

Butane (C4H10, MW 58), acetone (C3H6O, MW 58), propan-1-ol (CH3CH2CH2OH, MW 60) and propane (C3H8, MW 44). Three of the four weigh almost the same, on purpose. That takes molecular weight out of the comparison, so only the forces are left.

Step 1: ask which forces each one has. Rank by the strongest attraction available to each. Propan-1-ol has an O–H, so it can hydrogen bond: top tier. Acetone has a strong C=O dipole and no O–H, so dipole–dipole: middle tier. Butane and propane are hydrocarbons with only dispersion forces: bottom tier.

Step 2: break ties within a tier by size. Dispersion grows with surface area, so butane (4 carbons) beats propane (3).

Step 3: read off the order. Propane < butane < acetone < propan-1-ol. Measured: −42 °C, −1 °C, 56 °C, 97 °C. Every step goes the predicted way, and the two big jumps come exactly where a new tier of force switches on.

The trap is answering from molecular weight. Butane and acetone weigh the same to within a unit and boil 57° apart. Propan-1-ol is only two units heavier than acetone and boils 41° higher still. Weight decides things only within a tier.

Like dissolves like

Shake hexane, C6H14, with water, and the two separate into layers, as oil and water do. Shake ethanol with water, and they mix in any proportion.

Water molecules hold each other by hydrogen bonds. To dissolve, a solute has to push in between them, and that breaks some of those bonds. Ethanol repays the cost, because its O–H forms new hydrogen bonds with water. Hexane has no polar bond and nothing to offer, so the water molecules hold on to each other and push the hexane out.

"Like dissolves like." A polar solvent dissolves polar solutes, because they form new attractions that repay the cost of separating the solvent's own molecules. A nonpolar solute cannot repay it, so it stays out. A nonpolar solute dissolves instead in a nonpolar solvent, where only weak dispersion forces are broken and replaced.

The same rule explains why a long hydrocarbon chain makes an alcohol less soluble in water, as the example below shows. It also explains extraction: shaking a mixture with two liquids that do not mix, such as water and diethyl ether, so that each compound moves into the layer it resembles. Organic chemists use extraction to purify almost every product they make.

Worked example — which of these dissolves in water?

Ethanol (CH3CH2OH), diethyl ether (CH3CH2OCH2CH3), hexane (C6H14), acetic acid (CH3–C(=O)–OH).

Ethanol: an O–H that can both donate and accept a hydrogen bond, on a two-carbon chain. It mixes with water in any proportion.

Acetic acid: the same argument, and its C=O oxygen is a second acceptor. It also mixes in any proportion.

Diethyl ether: the oxygen has lone pairs, so it can accept a hydrogen bond from water. But it has no O–H to donate one, and it carries four carbons of hydrocarbon. It is only slightly soluble, about 6 g per 100 mL, which is why ether and water separate into two layers in an extraction.

Hexane: no polar bond anywhere, and nothing to offer water in exchange for breaking water's own hydrogen bonds. Insoluble.

The pattern to carry forward: solubility in water needs a polar group, and it is lost as the hydrocarbon part grows. Methanol, ethanol and propanol mix with water in any proportion. Hexan-1-ol, with the same OH on a six-carbon chain, is barely soluble at all.

What carries forward

The next section, Functional Groups, reads the δ+ and δ− ends of each group's bonds to predict where it reacts. How Reactions Happen gives the δ+ carbon of a polar bond its proper name, an electrophile, and in most of the mechanisms ahead it is the atom that gets attacked.

Polarity also decides solvent choice in every reaction you will run. In Substitution & Elimination, the solvent is what tips a reaction between its two substitution pathways, SN1 and SN2. And the boiling-point and solubility rankings above come back for alcohols, ethers, aldehydes and carboxylic acids.