Foundations Section 4 of 123

Bonding

the why on every wrong step

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Ethane, CH₃–CH₃, and ethene, CH₂=CH₂, both hold two carbons together by sharing electrons. A shared pair of electrons held between two nuclei is a covalent bond. The other way to fill an octet, handing electrons over to make ions, is the route Atomic structure ruled out for carbon.

The two molecules still behave differently. The two ends of ethane spin freely around the bond between them. The two ends of ethene cannot turn at all.

The difference lies in how the orbitals overlap. Overlap decides whether the two atoms can rotate, how strong the bond is, and whether its electrons stay locked in place or are open to attack from another molecule. It sorts every bond into two kinds, sigma and pi, and the rest of the course leans on that sorting.

Sigma bonds

Start with one C–H bond of methane. Carbon points one of its sp³ orbitals (from Hybridization) straight at the hydrogen. Hydrogen's 1s orbital meets it end to end. The two orbitals overlap on the line that joins the two nuclei. That line is the bond axis, also called the internuclear axis (the axis between the nuclei).

A bond made this way is a sigma (σ) bond: two orbitals meet head-on, and their overlap sits on the bond axis.

Every connection between two atoms contains exactly one sigma bond, no more and no fewer. A single bond is that sigma bond and nothing else. A double or triple bond is the same sigma bond with one or two pi bonds added, as the next section shows.

The orbitals that overlap depend on hybridization, which is where bonding and hybridization meet. A C–H bond in methane is an sp³–s overlap: a hybrid on carbon meets hydrogen's 1s. The C–C bond in ethane is sp³–sp³. The sigma bond of a C=C is sp²–sp², and the sigma bond of a C≡C is sp–sp.

C–H of methane: sp³ meets 1sCHbond axis25% sC–C of ethane: sp³ meets sp³CC25% sσ part of ethene’s C=C: sp² meets sp²CC33% sσ part of ethyne’s C≡C: sp meets spCC50% s
Gray lobes are hybrid orbitals; the green disc is hydrogen’s 1s. In every row the overlap sits on the dashed line through both nuclei. The carbons in rows 3 and 4 sit closer mostly because of their π bonds, which are not drawn here; s-character adds a smaller part.

Down that list, the carbon orbitals gain s-character. More s-character holds electrons closer to the nucleus, as Hybridization showed, so an sp orbital reaches out less far than an sp³ one. Two sp orbitals therefore meet with the nuclei closer together, and they hold more firmly. You can measure this in single bonds: a C–H bond is about 106 pm on an sp carbon and about 109 pm on an sp³ carbon.

In the drawings from here on, a plain line lies in the plane of the page. A solid wedge is a bond coming toward you, and a hashed wedge is a bond going away from you.

Now turn one end of ethane around the C–C bond axis. The head-on overlap looks the same from every angle around the axis, so turning one end changes nothing about it. That is why rotation around a single bond is free.

beforeHHHHHHCCturnafter half a turn of the right carbonHHHHHHCCσ overlap unchanged: rotation is free
Follow the highlighted hydrogen: it has moved from top to bottom. The gray overlap between the two carbons has not changed at all.

Free rotation is the starting point of Alkanes & Conformations. There the question is not whether a molecule can rotate, but which of the arrangements it can freely reach it prefers.

Head-on overlap is also efficient overlap: the orbitals meet where both hold the most electron density. Sigma bonds are therefore strong. Their electrons sit between the nuclei, buried in the space the two atoms share, where a molecule approaching from outside cannot easily reach them.

So a C–C or C–H sigma bond is the frame a molecule is built on, and under ordinary conditions it is rarely the part that reacts. Sigma bonds to atoms that pull electrons strongly, such as C–Cl or O–H, are the exception: later chapters break them all the time, and Electronegativity explains why.

Pi bonds

Take ethene. Each carbon has three groups (two H and the other carbon), so it is sp², with one p orbital left over. A p orbital that took no part in hybridization is called an unhybridized p orbital. Its three sp² orbitals lie flat in one plane. The leftover p orbital stands at right angles to that plane, one lobe above it and one below.

The two carbons sit side by side, so their p orbitals are parallel. They overlap sideways, above the bond axis and below it. That sideways overlap is a pi (π) bond. It is one bond whose electron density comes in two clouds, one above the plane and one below. Nothing of it lies on the bond axis itself.

HHHHCCp orbitalupper π cloudlower π cloudthe planeone π bond = both clouds together
The dashed parallelogram is the flat plane of the six atoms. The two dashed outlines mark the two clouds of the one π bond. The two shades mark the two halves (phases) of each p orbital.

A pi bond needs a leftover p orbital on each atom. An sp² atom has one, so it can make one pi bond. An sp atom has two, so it can make two. The triple-bond section below takes that case apart.

Pi bonds block rotation. Try to twist one CH₂ of ethene a quarter turn around the C–C axis. Its p orbital turns with it, until it stands at right angles to the p orbital on the other carbon. Two p orbitals at right angles do not overlap, so the pi bond is gone. Breaking a C=C pi bond this way costs roughly 65 kcal/mol, far more than molecules get from collisions at room temperature. So the two ends of a double bond stay locked.

before: both p orbitals point out of the pageHHHHCCπ overlap (dashed)after a quarter turn of the right carbonHHHHCCp orbitals at 90°: no overlap, no π bond
Top: the discs are p orbitals seen end-on, pointing at you. Bottom: the right carbon’s p orbital now runs up and down the page, at right angles to the left one.

Pi electrons are exposed. Sideways overlap is less efficient than head-on overlap. The pi electrons sit farther from the nuclei, in the clouds above and below the plane, instead of between the atoms. They are held more loosely, and they are the first electrons a molecule approaching from outside meets.

A molecule that is short of electrons and looking for some finds them here. That is why a double or triple bond is usually the part of a molecule that reacts. Nearly every alkene mechanism in Alkenes & Alkynes begins with these pi electrons.

A triple bond: one sigma and two pi

Take ethyne, HC≡CH. Each carbon has two groups, the H and the other carbon, so it is sp. Its two sp orbitals point in opposite directions along one straight line, and it has two p orbitals left over. Those two p orbitals stand at right angles to each other and to the bond axis.

The sp orbitals of the two carbons meet head-on: that is the sigma bond. One pair of p orbitals overlaps above and below the axis: the first pi bond. The other pair overlaps in front of the axis and behind it: the second pi bond, at right angles to the first.

the σ bond: sp meets sp on the axisHHCCthe first π: p orbitals above and belowHHCClooking straight down the H–C≡C–H axisCfirst πsecond πthe second C and both H sit straight behind
Rows 1 and 2 are side views. Row 3 turns the molecule to look along the axis, which shows the second π bond: its lobes point left and right, at 90° to the first.

The right angle between the two pi bonds is forced. Each sp carbon has exactly two leftover p orbitals, already perpendicular to each other, so the second pi bond has nowhere else to go. A triple bond is therefore not three copies of one bond. It is one strong sigma bond plus two weaker, more exposed pi bonds.

Together, the two pi bonds wrap the axis in a sleeve of electron density. A triple-bonded carbon's only other group lies on that same axis, so there is nothing to twist into a different arrangement. Alkynes have no cis/trans forms.

Counting sigma and pi bonds

Worked example — counting sigma and pi in acetonitrile, CH₃C≡N

Three C–H bonds: 3 σ. One C–C bond: 1 σ. The C≡N triple bond: 1 σ + 2 π.

Total: 5 σ and 2 π.

Hybridization, as a check. The methyl carbon has 4 groups, so it is sp³. The nitrile carbon has 2 groups (the methyl and the nitrogen), so it is sp. The nitrogen also has 2 groups (its bond to carbon and one lone pair), so it is sp too. The C–C≡N unit is therefore one straight line. Nitrogen's lone pair sits in an sp orbital, pointing straight out along that axis.

HHHCCN3 C–H bonds: 3 σ1 σ1 σ + 2 πsp lone pairsp³spspC–C≡N: one straight line
Every connection and every extra line counted.

That count used the rule from the Sigma section. In short:

single bond = 1σ  ·  double bond = 1σ + 1π  ·  triple bond = 1σ + 2π
Worked example — a C=C: propene, CH₃CH=CH₂

Sigma. Six C–H bonds (three on the methyl, one on the middle carbon, two on the end CH₂): 6 σ. Two carbon–carbon connections, each with one sigma: 2 σ. Total 8 σ.

Pi. Only the C=C carries a second line: 1 π.

Hybridization, as a check. The methyl carbon has 4 groups, so it is sp³, and its bonds are sp³–s (to H) and sp³–sp² (to the middle carbon). The two alkene carbons have 3 groups each, so they are sp². Each has one leftover p orbital, and those two p orbitals make the pi bond.

What lies in one plane. Two sp² carbons joined by a double bond hold everything attached to them in one plane. Here that means the CH₃ carbon, both alkene carbons and the three hydrogens on the alkene carbons. The methyl hydrogens are different. At most one of them lies in that plane at any moment, and the methyl group spins about its single bond.

the shaded panel is the pageHHHHHHCCCleave the planesp³sp²sp²always in plane: 3 C + 3 H on the C=C carbons
The two coral hydrogens leave the plane of the page. The CH₃ group spins; in the position drawn, its third hydrogen happens to lie in the plane.
Worked example — a C=O: propenal, CH₂=CH–CHO

Sigma. Four C–H bonds (two on the CH₂, one on the CH, one on the CHO carbon): 4 σ. Two C–C connections: 2 σ. One C–O connection: 1 σ. Total 7 σ.

Pi. The C=C gives 1 and the C=O gives 1: 2 π.

What the count tells you. Every carbon here has three groups and is sp², and so is the oxygen, so each of the four has a leftover p orbital. In its preferred shapes the molecule is flat, and all four p orbitals stand parallel in one continuous row.

propenal, flat in the pageHHHHCCCOthe same molecule edge-on (H atoms left out)CCCOfour parallel p orbitals in one row
Top: every atom, in the plane of the page. Bottom: the four p orbitals, one on each carbon and one on oxygen, stand side by side and parallel.

Nothing in this chapter uses that row yet. It is the structural fact behind conjugation, where pi electrons spread along a row of parallel p orbitals, in Drawing Molecules & Moving Electrons. You can already spot it from a sigma/pi count.

Common slip: counting every line of a multiple bond as a sigma bond. A double bond holds one sigma bond, and a triple bond still holds only one; every extra line is a pi bond.

Bond length and strength

Compare a carbon–carbon single, double and triple bond. More bonds between the same two atoms pull them closer together and hold them more firmly.

BondLengthStrengthMade of
C–C~154 pm~83 kcal/mol1σ
C=C~134 pm~147 kcal/mol1σ + 1π
C≡C~120 pm~200 kcal/mol1σ + 2π

Read the increments rather than the totals. Going from one bond to two adds about 64 kcal/mol, which is roughly the strength of the first pi bond. Going from two bonds to three adds only about 53, so the second pi bond is weaker still. Each pi bond is worth clearly less than the sigma bond it sits on, as sideways overlap is less efficient than head-on overlap.

The increments are rough, not exact. The sigma bond underneath also changes a little as the carbons go from sp³ to sp² to sp, and the subtraction cannot separate that change out.

This course quotes energies in kcal/mol. Other sources use kJ/mol, where 1 kcal = 4.184 kJ, so the three rows above are about 347, 615 and 837 kJ/mol.

average C–C bond: 83 kcal/mol to breakCC154 pmσ 83average C=C bond: 147 kcal/mol to breakCC134 pmσ 83π ≈64average C≡C bond: 200 kcal/mol to breakCC120 pmσ 83π ≈64π ≈53
The carbon–carbon distances are drawn to scale. Each bar is the average energy to break that kind of bond, split into the σ bond and the rough share of each π bond.

What "strength" measures: breaking a bond

Split ethane's C–C bond so that each CH₃ keeps one of the two shared electrons. That is homolytic cleavage, or homolysis: the pair splits down the middle, one electron to each fragment, leaving two neutral radicals. A radical is an atom or group with an unpaired electron.

The energy for a homolytic break of one particular bond is its bond dissociation energy. The Strength column above averages these energies over many compounds. A bond in one particular molecule can differ from the average by several kcal/mol, and a multiple bond by 20 or more: ethane's own C–C bond takes about 90 kcal/mol, not 83.

A bond can also break the other way. In heterolytic cleavage, or heterolysis, the whole pair leaves with one fragment, giving a cation and an anion. That is what happens in almost every reaction mechanism in this course that moves electrons in pairs, and it is how the carbocation of Atomic structure forms.

homolysis of ethane’s C–C bondH₃CCH₃H₃C+CH₃each CH₃ keeps one electron: two radicalsheterolysis of a C–Cl bond(CH₃)₃CCl(CH₃)₃C⁺+Cl⁻Cl takes both electrons: a cation and an anioneach CH₃ stands for one C with three H
The two dots between the atoms are the shared pair. Follow them: in the top row they split one each way, in the bottom row both go to chlorine.

Tables list the homolytic number because it belongs to the bond alone. The cost of a heterolytic break depends heavily on the solvent around the ions it makes. So read the table as "how hard is this bond to pull apart", not "how easily does this bond ionize". The homolytic numbers come into their own in radical halogenation, in How Reactions Happen.

Locked rotation makes cis and trans

Take CH₃CH=CHCH₃, 2-butene. Each alkene carbon carries one CH₃ and one H. The two CH₃ groups can sit on the same side of the double bond or on opposite sides. The pi bond stops either carbon turning, so neither arrangement can change into the other.

cis-2-buteneCH₃CH₃HHCCboth CH₃ on the same sidetrans-2-buteneCH₃CH₃HHCCthe CH₃ groups on opposite sideseach CH₃ circle is one C with three H
Look at which side of the C=C each highlighted CH₃ sits on.

The two arrangements are therefore two different compounds that can be bottled apart. The one with both CH₃ groups on the same side is cis-2-butene. The one with them on opposite sides is trans-2-butene.

Restricted rotation around a pi bond is why alkenes have cis/trans isomers. Rings, which also stop rotation, are the other source; Stereochemistry covers both. The two compounds have the same atoms joined in the same order, yet they have different melting and boiling points, different dipole moments and different biology. It is why the double bond in retinal can act as a light-triggered switch in your eye.

Put the same four carbons on single bonds, as in butane, and the ends spin freely billions of times a second. There is nothing to separate. This is the first place in the course where a drawing carries information a formula does not: both 2-butenes are C₄H₈, and both are written CH₃CH=CHCH₃. The Cis/trans and E/Z section of Stereochemistry gives the pair their systematic names and the rule for when cis/trans applies at all.

Bond strength is not the same as reactivity

The triple bond is the strongest carbon–carbon bond in the table. Yet alkynes react readily with many reagents that leave C–C single bonds alone. Alkynes are one of the functional groups: the recurring arrangements of atoms that give a molecule its chemistry, collected in Functional groups at the end of this chapter.

That is not a contradiction. Total bond strength measures what it costs to pull the two atoms entirely apart. Reactivity is about what it costs to break the weakest part you can reach, and what you get back.

An addition reaction adds a new pair of atoms across a multiple bond. In an addition, the bond that breaks is a pi bond, which is both weaker and easier to reach than the sigma bond. The sigma bond stays. The molecule gives up one pi bond and gains two new sigma bonds, which is a good trade. Keep strength and reactivity apart, and many apparent paradoxes in Alkenes & Alkynes and Aromatic Chemistry go away.

What carries forward

Three habits start here. Count sigma and pi correctly, because hybridization depends on the count of sigma bonds plus lone pairs, not on the total number of lines. Remember that pi electrons are the reachable ones: most mechanisms in Alkenes & Alkynes, Carbonyl Chemistry and Aromatic Chemistry begin with a pi bond attacking something or being attacked. And remember that pi bonds lock geometry, which is the door into stereochemistry.