Alkene structure

Take ethene, CH₂=CH₂, the smallest alkene. An alkene is a hydrocarbon with a carbon–carbon double bond, C=C. The double bond holds two pairs of electrons. One pair sits on the line between the carbons. The other is spread out above and below that line, where it is loosely held and easy for other molecules to reach.

That exposed pair reverses the role carbon played in Substitution & Elimination. There, the carbon under attack was short of electrons, and a nucleophile brought them in. An alkene is itself the nucleophile, the electron-rich partner that reacts with an electron-poor one. This page covers the shape and stability of the C=C. Every alkene reaction in this chapter relies on both.

sp² carbons and the pi bond

Each carbon of ethene has three groups around it: two hydrogens and the other carbon. So each carbon is sp² hybridized, as you saw in Foundations. It uses its three sp² orbitals to make three sigma bonds. A sigma (σ) bond is one whose electron pair sits on the line between the two nuclei. The three sigma bonds lie flat in one plane, 120° apart, an arrangement called trigonal planar.

Each carbon still has one unhybridized p orbital, a p orbital that took no part in the hybrid set. It stands at right angles to the plane, with one lobe above and one below. The two p orbitals are parallel, so they overlap side by side, above and below the C–C axis. That side-by-side overlap is the pi (π) bond. A C=C is therefore one sigma bond plus one pi bond.

The two bonds are not equally strong. A C=C takes about 147 kcal/mol to break, against 83 for a C–C single bond. (It is also shorter: 134 pm against 154 pm.) If the sigma part is worth about as much as a single bond, the pi bond is worth the difference, roughly 65 kcal/mol. That split is only approximate, because a sigma bond between two sp² carbons is a little stronger than one between two sp³ carbons.

So the pi bond is the weaker of the two, and its electrons sit farther from the nuclei. That is why the pi bond is the part that breaks first in every addition reaction in this chapter.

Why the double bond is flat and rigid

Try to twist ethene. Hold the left CH₂ still and turn the right one a quarter turn about the C–C axis. The right p orbital turns with it and ends up at right angles to the left one. Two p orbitals at right angles do not overlap at all, so the pi bond is gone. The sigma bond, which lies on the axis, is untouched.

Twisting a C=C therefore means breaking its pi bond, which costs about 65 kcal/mol. Molecules colliding at room temperature have nowhere near that much energy, so the twist does not happen. A single bond has no such cost. Its sigma overlap looks the same from every angle, so its two ends turn freely.

The same overlap also fixes the shape. The pi bond needs the two p orbitals parallel, and that holds only while both carbons' sigma bonds lie in the same plane. So the two alkene carbons and the four atoms bonded to them all sit in one plane, as the dashed parallelogram shows.

Where the double bond is, and which way it faces

Take CH₃CH=CHCH₂CH₃. Number its five-carbon chain from the left and the double bond joins C2 and C3. Number from the right and it joins C3 and C4. The rule from IUPAC Nomenclature is to number from the end nearer the double bond and cite the lower of its two carbons. So the compound is pent-2-ene, not pent-3-ene.

In a ring, the two doubly bonded carbons are always C1 and C2. The numbering then carries on around the ring in whichever direction gives a substituent the lower number. In 1-methylcyclohexene the methyl sits on C1, an alkene carbon. In 3-methylcyclohexene it sits on C3, the first carbon past C2.

The locant alone does not describe the alkene fully. The Cis/trans and E/Z section of Stereochemistry named the two possible arrangements. Here is a short recap. In but-2-ene, CH₃CH=CHCH₃, each alkene carbon carries one CH₃ and one H. The two methyls can sit on the same side of the double bond or on opposite sides. As the box above says, these are two different compounds. cis-But-2-ene boils at 4 °C and has a small dipole moment, while trans-but-2-ene boils at 1 °C and has none, because its two C–CH₃ dipoles cancel. The locant cannot tell them apart, so a second label is needed.

Cis means the two groups that are not hydrogen sit on the same side of the double bond, and trans means they sit on opposite sides. These words work only when each alkene carbon carries one hydrogen and one other group. In 3-chloropent-2-ene, CH₃CH=C(Cl)CH₂CH₃, the right-hand alkene carbon carries Cl and CH₂CH₃ and no hydrogen, so there is no matching pair of groups to compare.

For 3-chloropent-2-ene, look at each alkene carbon on its own. On the left carbon, CH₃ outranks H. On the right carbon, Cl outranks CH₂CH₃, because chlorine has a higher atomic number than carbon. In the isomer drawn below, both higher-ranked groups sit above the double bond, so it is the Z isomer, from the German zusammen, "together". With them on opposite sides it would be E, from entgegen, "opposite".

That is the whole E/Z method, and it works for any alkene that has two different groups on each alkene carbon. (But-1-ene, with two hydrogens on C1, has no E or Z isomers.) Rank the two groups on each alkene carbon by the CIP priority rules you used for R and S, then compare only the higher-ranked group on each carbon.

Z is not another word for cis. In the Z isomer above, the carbon chain runs from the upper left to the lower right, so the main chain is trans across the double bond, even though the isomer is Z. The rest of the course uses E and Z for alkene geometry.

Degrees of unsaturation

Butane, C₄H₁₀, has only single bonds and no ring. Four carbons cannot hold any more hydrogens than that, so butane is saturated. But-1-ene and cyclobutane are both C₄H₈, two hydrogens short of butane. But-1-ene gave up those two hydrogens to make a pi bond, and cyclobutane gave them up to close a ring.

The pattern is general. Every pi bond (in C=C, C=O or any other multiple bond) and every ring removes two hydrogens from the saturated formula. For a hydrocarbon that formula is CₙH₂ₙ₊₂. Each missing pair of hydrogens is one degree of unsaturation (DoU). C₄H₈ has one: a pi bond or a ring, and the formula alone cannot say which.

Counting the missing hydrogens gives this equation, where C, H, N and X are the numbers of carbon, hydrogen, nitrogen and halogen atoms:

A halogen counts like a hydrogen, because it also makes one bond and ends a chain. Each nitrogen adds one hydrogen to the saturated count, because its third bond carries one extra hydrogen. Oxygen is left out: slipping an O into a chain, as in CH₃–O–CH₃, changes no hydrogen count.

Try benzene, C₆H₆: (12 + 2 − 6)/2 = 4. Four degrees fit three pi bonds and one ring, which is what benzene has. This count is the first step in almost every structure problem in IR & Mass Spectrometry and NMR Spectroscopy, because it tells you at once whether to look for rings, C=O groups or a benzene ring.

Alkene stability: more substituted is more stable

Take three of the butenes: but-1-ene, cis-but-2-ene and trans-but-2-ene. Add H₂ across the double bond of any one of them and it becomes butane, the same product every time. This reaction is hydrogenation. It needs a metal catalyst, such as Pd, Pt or Ni, which speeds the reaction up without being used up. The catalyst changes neither the energy of the alkene nor that of the butane. The full reaction, which happens on the metal's surface, comes later in this chapter under Catalytic hydrogenation.

Each hydrogenation releases heat, the heat of hydrogenation. All three reactions end at the same butane, so any difference in the heat released comes from where each alkene started. Less heat released means the alkene started lower in energy, and lower in energy means more stable.

The table adds two more alkenes, 2-methylbut-2-ene and 2,3-dimethylbut-2-ene. They give other alkanes, and those differ slightly in energy from butane, so their heats can be compared with the butenes' only roughly.

The table gives ΔH°hyd. The minus sign means heat is released.

The first trend in the table is substitution. Count the carbon groups bonded directly to the two alkene carbons, the same count you used for Zaitsev's rule in E1 (Substitution & Elimination). But-1-ene has one, so it is monosubstituted. The but-2-enes have two (disubstituted), 2-methylbut-2-ene has three (trisubstituted) and 2,3-dimethylbut-2-ene has four (tetrasubstituted). Going from one group to four lowers the heat of hydrogenation by 3.7 kcal/mol in all, a little over 1 kcal/mol per group on average. Isomers that share a product show the same trend. The three methylbutenes all give 2-methylbutane, and the most substituted of them, 2-methylbut-2-ene, releases the least heat.

The usual explanation is hyperconjugation, the same effect that stabilizes carbocations in How Reactions Happen. Take propene, CH₂=CH–CH₃. One C–H bond of its methyl group can line up parallel to the p orbitals of the double bond. Its electron pair then spreads slightly over the double bond. Every alkyl group on an alkene carbon brings more C–H bonds that can do this, so more groups mean more stabilization.

Some courses give a second reason alongside it. A bond from an sp² carbon to an sp³ carbon is a little stronger than a bond between two sp³ carbons, and a more substituted alkene has more of these stronger bonds. Both effects favor the more substituted alkene, so they predict the same order.

The second trend is geometry. trans-But-2-ene releases 1.0 kcal/mol less heat than cis-but-2-ene, so the trans isomer is more stable. In the cis isomer the two methyls sit on the same side of the double bond and crowd each other. In the trans isomer they are held apart.

This order is the reason behind Zaitsev's rule from E1 and E2 in Substitution & Elimination. When an elimination can form either of two alkenes, the more substituted one is more stable, and it is usually the major product.

Rings: why small ones are always cis

Cyclohexene has its double bond inside a six-membered ring. On each alkene carbon, one of the two groups is the rest of the ring, and the ring has to reach from one alkene carbon to the other. When both ring bonds leave on the same side of the double bond, the chain only has to span that side, which even a short chain can do. So cyclohexene, cyclopentene and every other cycloalkene with a small ring is cis, whether or not the name says so.

Now put the two ring bonds on opposite sides, trans. The chain leaves one alkene carbon on the upper side and has to return to the other on the lower side, so it must pass across one face of the double bond. Doing that twists the p orbitals out of line. With fewer than about eight atoms in the ring, the chain is too short to make the trip without twisting the pi bond so far that the compound cannot be kept. trans-Cyclohexene has never been isolated, and trans-cyclooctene is the smallest trans cycloalkene that can be kept in a bottle.

trans-Cyclooctene is also chiral: it cannot be superimposed on its mirror image. Its chain crosses one face of the double bond, and in the mirror image the chain crosses the other face. Turning one into the other would mean swinging the double bond through the loop of the ring, and the loop is too tight for that.

Bridgehead alkenes and Bredt's rule

Norbornane is a bridged bicyclic compound: two rings that share more than one atom. Its two shared carbons are the bridgeheads. Three chains of carbons, called bridges, join one bridgehead to the other: two bridges of two carbons and one bridge of one carbon.

The systematic name records the bridges. Bicyclo says there are two rings. The numbers in square brackets give the carbons in each bridge, largest first, not counting the bridgeheads. The parent name gives the total number of carbons. So norbornane is bicyclo[2.2.1]heptane: bridges of 2, 2 and 1 carbons plus the two bridgeheads make seven.

Numbering starts at one bridgehead, which is C1. It runs along the largest bridge to the other bridgehead, back along the next largest bridge, and finishes on the smallest. In bicyclo[2.2.1]hept-1-ene, the 1-ene puts the double bond between bridgehead C1 and C2.

A double bond at a bridgehead meets the same problem as trans-cyclooctene, only worse. An alkene carbon needs its three sigma bonds flat in one plane, with its p orbital at right angles to them. A bridgehead carbon is held by three bridges that pull its bonds into a pyramid, so the orbital it has left over points out sideways and cannot line up with the p orbital next door. Bredt's rule says that a small bicyclic compound cannot have a double bond at a bridgehead.

The rule has a size limit. Find the largest ring that contains the double bond. The double bond is trans in that ring: the ring leaves C1 and C2 on opposite sides, like the strap in trans-cyclooctene above. So, like a trans cycloalkene, the ring needs about eight atoms or more. In bicyclo[2.2.1]hept-1-ene that ring has six atoms, and the compound cannot be isolated. In bicyclo[3.3.1]non-1-ene it has eight, and the compound can be isolated, though it is reactive.

Bredt's rule matters most in elimination. If the Zaitsev product of an E1 or E2 would put a double bond at the bridgehead of a small bicyclic compound, that product does not form. The reaction gives a less substituted alkene instead, or no alkene at all. So check whether an alkene can exist before asking how stable it is.

What carries forward

The exposed pi electrons make the alkene a nucleophile, and the addition reactions that follow in this chapter start there. Alkene stability decided the elimination products in Substitution & Elimination, and it will decide them again for the alcohol dehydrations in Alcohols, Ethers & Related Chemistry. E/Z labels report the geometry of every alkene made in this course. Degrees of unsaturation is the first number to work out in every structure problem, starting with IR & Mass Spectrometry, the chapter after this one.