Alkenes and alkynes

Hydrocarbons with only CC single bonds are known as saturated hydrocarbons. If a hydrocarbon contains double and/or triple bonds, it is known as unsaturated. The molecule ethylene (C$_2$H$_4$) contains a carbon double bond and is officially referred to as ethene, although the name ``ethylene'' is more commonly used. Similarly, the molecule acetylene (C$_2$H$_2$) contains a carbon triple bond and has the official name ethyne. In general, alkenes contain double bonds and alkynes contain triple bonds. If the main carbon chain contains more than three carbons, it is necessary to specify where on the chain the double or triple bond occurs. Here, we follow the same IUPAC rules as specified above. The molecule H$_2$C=C-CH$_2$-CH$_3$ is known as 1-butene while CH$_3$-CH=CH-CH$_3$ is known as 2-butene.

In order to describe bonding in alkenes, we need a combination of the valence bond theory and LCAO. Let us consider ethylene (ethene). The two carbons are $sp^2$ hybridized, and there is one $2p_z$ orbital that is not hybridized. The book uses the convention that the unhybridized orbital is a $p_z$ orbital and combines it with another $p_z$ orbital to form a $\pi$ orbital, which contradicts the definition of a $\pi$ orbital as having $m=1$ as the angular momentum quantum number (recall the exact treatment of H$_2^+$!). However, this is all just a matter of how the coordinate system is defined, so as along as one is consistent, is does not matter which $p$ orbitals are hybridized and which are combined to give $\pi$ orbitals. In this case, if the bond axis between the carbon atoms is the $x$-axis, then $L_x$ would be conserved, and the $p_z$ orbital would be an $m=1$ orbital! In any case, adopting the book's convention, the $sp^2$ orbitals all lie in the $xy$-plane and the $p_z$ orbital points out of the plane as shown in the figure below:

Figure: Illustration of sp$^2$ orbitals.

The CH bonds form in the same way as they do in alkanes. An $sp^2$ hybrid orbital from a carbon combines with the $1s$ orbital of H, and the two-electron wave function is:

$$\Psi({\bf x}_1,{\bf x}_2) = C\left[\psi_{1s}^{\rm H}({\bf r}... ...rrow}(s_2) - \psi_{\uparrow}(s_2)\psi_{\downarrow}(s_1)\right]$$

For the double bond, however, something rather interesting happens. Two of the sp$^2$ hybrids on the carbons combine to give a $\sigma$ molecular orbital via the LCAO procedure. However, this $\sigma$ bond only describes one of the electron pairs in the bond. The other electron pair ends up in a $\pi$ bond that forms by applying the LCAO procedure to the two unhybridized $p_z$ orbitals on the carbons. We will not write down the full four-electron wave function (it contains too many terms), however, it is worth noting that a double bond contains both $\sigma$ and $\pi$ character. This is something that is completely hidden when writing down the Lewis structures. The orbitals for ethylene can be visualized as below:

Figure: Bonding orbitals in ethene (ethylene).

In alkynes, the hybridization is simply $sp$. Thus, in acetylene (ethyne), there are two unhybridized orbitals, which, by convention are $p_x$ and $p_y$ (see figure below):

Figure: Illustrating the sp orbitals.

The CH bonds in acetylene are formed by combining the $1s$ orbital in H with one of the $sp$ orbitals in carbon to form a $\sigma$ bond. The other $sp$ combines with an $sp$ from the other carbon to form another $\sigma$ bond that contains two of the 6 electrons in the triple bond. The remaining 4 electrons are placed in the MOs formed by mixing the $p_x$ and $p_y$ unhybridized orbitals. These orbitals combine via the LCAO procedure to form two $\pi$ bonds, each of which contains a pair of electrons. Thus, a triple bond contains both $\sigma$ and $\pi$ character. The bonding in acetylene is pictured below:

Figure: Bonding in ethyne (acetylene).

Rotation around double and triple bonds is energetically unfavorable. Thus, ethylene prefers to be planar while acetylene prefers to be linear. Moreover, alkynes tend to be less stable than alkenes because of the larger number of $\pi$ electrons. The energetic spacing between the degenerate pair of $\pi$ orbitals and the corresponding $\pi^*$ orbitals is smaller than the spacing between the nondegenerate $\pi$ and $\pi^*$ orbitals in alkenes.

If we consider a molecule like 2-butene, which is CH$_3$-CH=CH-CH$_3$, the outer electrons are clearly $sp^3$ hybridized, while the inner carbons forming the double bond are $sp^2$ hybridized. Hence, while there can be rotation of the outer methyl groups, but not rotation about the C=C bond axis. This means that the four carbons tend to lie in a single plane. However, within this configuration two isomers are possible, which are the cis and trans conformers shown in the figure below (cis-top, trans-bottom):

Figure: Isomers of 2-butene.

Because rotation about the C=C double bond is energetically unfavorable, interconversion between these to isomers at room temperature is very slow.

Compounds that contain more than one double bond are called polyenes. As with alkenes, if the backbone contains four or more carbons, we need to specify where on the chain the double bonds occur. Thus, the molecule CH$_2$=CH-CH=CH$_2$ is known as 1,3-butadiene (see Figure below):

Figure: 1,3-butadiene.

The di is used here because the molecule contains two double bonds. 1,3-butadiene is particularly interesting because all of the carbons are $sp^2$ hybridized. Because of this, each of the four carbons has an unhybridized $p_z$ orbital, and these four $p_z$ orbitals can be combined to give four new MOs for carbons A, B, C, D:

$$\psi_1({\bf r}) \propto \psi_{2p_z}^A({\bf r}) + \psi_{2p_z}^B({\bf r}) + \psi_{2p_z}^C({\bf r}) + \psi_{2p_z}^D({\bf r})$$

$$\psi_2({\bf r}) \propto \psi_{2p_z}^A({\bf r}) + \psi_{2p_z}^B({\bf r}) - \psi_{2p_z}^C({\bf r}) - \psi_{2p_z}^D({\bf r})$$

$$\psi_3({\bf r}) \propto \psi_{2p_z}^A({\bf r}) - \psi_{2p_z}^B({\bf r}) - \psi_{2p_z}^C({\bf r}) + \psi_{2p_z}^D({\bf r})$$

$$\psi_4({\bf r}) \propto \psi_{2p_z}^A({\bf r}) - \psi_{2p_z}^B({\bf r}) + \psi_{2p_z}^C({\bf r}) - \psi_{2p_z}^D({\bf r})$$

For $\psi_1$, the four $p_z$ orbitals have the positive lobes above the plane of the carbons and the negative lobes below the plane, so the resulting MO has a single nodal plane, which is the plane of the carbons. This is the lowest energy MO. For $\psi_2$, The leftmost pair of $p_z$ orbitals have their positive lobes above the plane and negative lobes below the plane, but the rightmost pair of $p_z$ orbitals is oriented in the opposite sense, with the positive lobes below and plane and the negative lobes above the plane. This MO will have an additional nodal plane cutting the C-C single bond in the middle of the molecule. For $\psi_3$, the central pair of $p_z$ orbitals is oriented so that both positive lobes are below the plane, while for the outer $p_z$ orbitals, the positive lobes are above the plane. This MO will, therefore, have two additional nodal planes cutting through the two double bonds. Finally, for $\psi_4$, the highest energy orbital, the $p_z$ orbitals altenate in their orientation with respect to the carbon plane. The resulting MO has three additional nodal planes cutting through each of the carbon-carbon bonds. The MOs are shown in the figure below:

Figure: The four MOs formed from the unhybridized $2p_z$ orbitals in 1,3-butadiene.

Thus, 1,3-butadiene has two short CC bonds and one long bond between the two short bonds. When short and long bonds alternate in this way, the molecule is known as a conjugated $\pi$ system. Note that the lowest energy MO is an extended orbital that is delocalized over the entire molecule. This is typical of conjugated $\pi$ systems and indicates that the electrons in such an orbital are really delocalized over the entire molecule. This type of electron delocalization makes such systems ideal in molecular electronics, as it is possible to pass an electrical current through the molecule, with electrons passing through the delocalized $\pi$ system between two electrodes.