Cis-trans isomerism is not limited to disubstituted alkenes. It can occur whenever both double-bond carbons are attached to two different groups. If one of the double-bond carbons is attached to two identical groups, however, cis-trans isomerism is not possible (FIGURE 7.5).
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0206.jpg?height=145&width=895&top_left_y=768&top_left_x=585)
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STEP 1 - 7.4 Cis-Trans Isomerism in Alkenes

These two compounds are not identical; they are cis-trans isomers.


FIGURE 7.5 The requirement for cis-trans isomerism in alkenes. Compounds that have one of their carbons bonded to two identical groups can't exist as cis-trans isomers. Only when both carbons are bonded to two different groups is cis-trans isomerism possible.


PROBLEM The sex attractant of the common housefly is an alkene named cis-9-tricosene. Draw its structure.
7-8 (Tricosane is the straight-chain alkane $\mathrm{C}_{23} \mathrm{H}_{48}$.)
PROBLEM Which of the following compounds can exist as pairs of cis-trans isomers? Draw each pair, and
7-9 indicate the geometry of each isomer.
(a) $\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}_{2}$
(b) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CHCH}_{3}$
(c) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CHCH}_{3}$
(d) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{C}\left(\mathrm{CH}_{3}\right) \mathrm{CH}_{2} \mathrm{CH}_{3}$
(e) $\mathrm{ClCH}=\mathrm{CHCl}$ (f) $\mathrm{BrCH}=\mathrm{CHCl}$


PROBLEM Name the following alkenes, including a cis or trans designation:


7-10 (a) -


7-10 (a) - 7.5 Alkene Stereochemistry and the E,Z Designation

The cis-trans naming system used in the previous section works only with disubstituted alkenes-compounds that have two substituents other than hydrogen on the double bond. With trisubstituted and tetrasubstituted double bonds, a more general method is needed for describing double-bond geometry. (Trisubstituted means three substituents other than hydrogen on the double bond; tetrasubstituted means four substituents other than hydrogen.)


The method used for describing alkene stereochemistry is called the $\boldsymbol{E}, \boldsymbol{Z}$ system and employs the same Cahn-Ingold-Prelog sequence rules given in Section 5.5 for specifying the configuration of a chirality center. Let's briefly review the sequence rules and then see how they're used to specify double-bond geometry. For a more thorough review, reread Section 5.5.


RULE 1 -

Considering each of the double-bond carbons separately, look at the two substituents attached and rank them according to the atomic number of the first atom in each ( 8 for oxygen, 6 for carbon, 1 for hydrogen, and so forth). An atom with higher atomic number ranks higher than an atom with lower atomic number.


RULE 1 -

RULE 2
If a decision can't be reached by ranking the first atoms in the two substituents, look at the second, third, or fourth atoms away from the double-bond until the first difference is found.


RULE 3 -

Multiple-bonded atoms are equivalent to the same number of single-bonded atoms.
Once the two groups attached to each double-bonded carbon have been ranked as either higher or lower, look at the entire molecule. If the higher-ranked groups on each carbon are on the same side of the double bond, the alkene is said to have a $\boldsymbol{Z}$ configuration, for the German zusammen, meaning "together." If the higher-ranked groups are on opposite sides, the alkene has an $\boldsymbol{E}$ configuration, for the German entgegen, meaning "opposite." (For a simple way to remember which is which, note that the groups are on "ze zame zide" in the $Z$ isomer.)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0207.jpg?height=169&width=224&top_left_y=889&top_left_x=758)
$E$ double bond
(Higher-ranked groups
are on opposite sides.)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0207.jpg?height=169&width=210&top_left_y=1149&top_left_x=762)


RULE 3 -

Z double bond
(Higher-ranked groups are on the same side.)


RULE 3 -

As an example, look at the following two isomers of 2-chloro-2-butene. Because chlorine has a higher atomic number than carbon, a -Cl substituent is ranked higher than a $-\mathrm{CH}_{3}$ group. Methyl is ranked higher than hydrogen, however, so isomer (a) is designated $E$ because the higher-ranked groups are on opposite sides of the double bond. Isomer (b) has a $Z$ configuration because its higher-ranked groups are on ze zame zide of the double bond.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0207.jpg?height=231&width=979&top_left_y=1554&top_left_x=573)
(a) (E)-2-Chloro-2-butene
(b) (Z)-2-Chloro-2-butene


RULE 3 -


WORKED EXAMPLE 7.1 -


Assigning E and Z Configurations to Alkenes -


Strategy -

Look at the two substituents connected to each double-bonded carbon, and determine their ranking using the Cahn-Ingold-Prelog rules. Then, check whether the two higher-ranked groups are on the same or opposite sides of the double bond.


Solution -

The left-hand carbon has -H and $-\mathrm{CH}_{3}$ substituents, of which $-\mathrm{CH}_{3}$ ranks higher by sequence rule 1 . The righthand carbon has $-\mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}$ and $-\mathrm{CH}_{2} \mathrm{OH}$ substituents, which are equivalent by rule 1 . By rule 2 , however, $-\mathrm{CH}_{2} \mathrm{OH}$ ranks higher than $-\mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}$ because the substituent $-\mathrm{CH}_{2} \mathrm{OH}$ has an oxygen as its highest second atom, but $-\mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}$ has a carbon as its highest second atom. The two higher-ranked groups are on the same side of the double bond, so we assign a $Z$ configuration.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0208.jpg?height=304&width=792&top_left_y=840&top_left_x=664)


Solution -

Z configuration


Solution -

PROBLEM Which member in each of the following sets ranks higher?
7-11 (a) -H or $-\mathrm{CH}_{3}$
(b) -Cl or $-\mathrm{CH}_{2} \mathrm{Cl}$
(c) $-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}$ or $-\mathrm{CH}=\mathrm{CH}_{2}$
(d) $-\mathrm{NHCH}_{3}$ or $-\mathrm{OCH}_{3}$
(e) $-\mathrm{CH}_{2} \mathrm{OH}$ or $-\mathrm{CH}=\mathrm{O}$ (f) $-\mathrm{CH}_{2} \mathrm{OCH}_{3}$ or $-\mathrm{CH}=\mathrm{O}$

PROBLEM Rank the substituents in each of the following sets according to the sequence rules:
7-12 (a) $-\mathrm{CH}_{3},-\mathrm{OH},-\mathrm{H},-\mathrm{Cl}$ (b) $-\mathrm{CH}_{3},-\mathrm{CH}_{2} \mathrm{CH}_{3},-\mathrm{CH}=\mathrm{CH}_{2},-\mathrm{CH}_{2} \mathrm{OH}$
(c) $-\mathrm{CO}_{2} \mathrm{H},-\mathrm{CH}_{2} \mathrm{OH},-\mathrm{C} \equiv \mathrm{N},-\mathrm{CH}_{2} \mathrm{NH}_{2}$ (d) $-\mathrm{CH}_{2} \mathrm{CH}_{3},-\mathrm{C} \equiv \mathrm{CH},-\mathrm{C} \equiv \mathrm{N},-\mathrm{CH}_{2} \mathrm{OCH}_{3}$


Solution -


Solution -

PROBLEM Assign stereochemistry ( $E$ or $Z$ ) to the double bond in the following compound, and convert the 7-14 drawing into a skeletal structure (red = O):
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0209.jpg?height=444&width=498&top_left_y=190&top_left_x=423)


Solution - 7.6 Stability of Alkenes

Although the cis-trans interconversion of alkene isomers does not occur spontaneously, it can often be made to happen by treating the alkene with a strong acid catalyst. If we interconvert cis-2-butene with trans-2-butene and allow them to reach equilibrium, we find that they aren't of equal stability. The trans isomer is more stable than the cis isomer by $2.8 \mathrm{~kJ} / \mathrm{mol}(0.66 \mathrm{kcal} / \mathrm{mol})$ at room temperature, corresponding to a $76: 24$ ratio.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0209.jpg?height=221&width=647&top_left_y=906&top_left_x=736)

Cis alkenes are less stable than their trans isomers because of steric strain between the two larger substituents on the same side of the double bond. This is the same kind of steric interference that we saw previously in the axial conformation of methylcyclohexane (Section 4.7).
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0209.jpg?height=421&width=432&top_left_y=1269&top_left_x=613)
cis-2-Butene
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0209.jpg?height=411&width=411&top_left_y=1345&top_left_x=1101)
trans-2-Butene


Solution - 7.6 Stability of Alkenes

Although it's sometimes possible to find relative stabilities of alkene isomers by establishing a cis-trans equilibrium through treatment with strong acid, a more general method is to take advantage of the fact that alkenes undergo a hydrogenation reaction to give the corresponding alkane when treated with $\mathrm{H}_{2}$ gas in the presence of a catalyst such as palladium or platinum.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0209.jpg?height=251&width=1009&top_left_y=1994&top_left_x=555)


Solution - 7.6 Stability of Alkenes

Energy diagrams for the hydrogenation reactions of cis- and trans-2-butene are shown in FIGURE 7.6. Because cis-2-butene is less stable than trans-2-butene by $2.8 \mathrm{~kJ} / \mathrm{mol}$, the energy diagram shows the cis alkene at a higher energy level. After reaction, however, both curves are at the same energy level (butane). It therefore follows that $\Delta G^{\circ}$ for reaction of the cis isomer must be larger than $\Delta G^{\circ}$ for reaction of the trans isomer by 2.8 $\mathrm{kJ} / \mathrm{mol}$. In other words, more energy is released in the hydrogenation of the cis isomer than the trans isomer because the cis isomer has more energy to begin with.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0210.jpg?height=644&width=784&top_left_y=193&top_left_x=665)

FIGURE 7.6 Energy diagrams for hydrogenation of cis- and trans-2-butene. The cis isomer is higher in energy than the trans isomer by about $2.8 \mathrm{~kJ} / \mathrm{mol}$ and therefore releases more energy when hydrogenated.


Solution - 7.6 Stability of Alkenes

If we were to measure the so-called heats of hydrogenation ( $\Delta H^{\circ}{ }_{\text {hydrog }}$ ) for two double-bond isomers and find their difference, we could determine the relative stabilities of cis and trans isomers without having to measure an equilibrium position. cis-2-Butene, for instance, has $\Delta H^{\circ}{ }_{\text {hydrog }}=-119 \mathrm{~kJ} / \mathrm{mol}(-28.3 \mathrm{kcal} / \mathrm{mol})$, while trans-2-butene has $\Delta H^{\circ}$ hydrog $=-115 \mathrm{~kJ} / \mathrm{mol}(-27.4 \mathrm{kcal} / \mathrm{mol})-$ a difference of $4 \mathrm{~kJ} / \mathrm{mol}$.
<img class="imgSvg" id = "m6sn3n2cx0mic070qk8" src="data:image/svg+xml;base64,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"/>

Cis isomer
$\Delta H^{\circ}{ }_{\text {hydrog }}=-120 \mathrm{~kJ} / \mathrm{mol}$
<img class="imgSvg" id = "m6sn3n2cj2yodgrlxo" src="data:image/svg+xml;base64,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"/>


Solution - 7.6 Stability of Alkenes

Trans isomer
$\Delta H^{\circ}$ hydrog $=-116 \mathrm{~kJ} / \mathrm{mol}$


Solution - 7.6 Stability of Alkenes

The $4 \mathrm{~kJ} / \mathrm{mol}$ energy difference between the 2-butene isomers calculated from heats of hydrogenation agrees reasonably well with the $2.8 \mathrm{~kJ} / \mathrm{mol}$ energy difference calculated from equilibrium data, but the values aren't exactly the same for two reasons. First, there is probably some experimental error, because heats of hydrogenation are difficult to measure accurately. Second, heats of reaction and equilibrium constants don't measure exactly the same thing. Heats of reaction measure enthalpy changes, $\Delta H^{\circ}$, whereas equilibrium constants measure free-energy changes, $\Delta G^{\circ}$, so we might expect a slight difference between the two.


Solution - 7.6 Stability of Alkenes

TABLE 7.2 Heats of Hydrogenation of Some Alkenes


Solution - 7.6 Stability of Alkenes

| Substitution | Alkene | $\Delta H^{\circ}$ hydrog |  |
| :--- | :--- | :--- | :--- | :--- |
|  |  | $(\mathrm{kJ} / \mathrm{mol})$ | $(\mathrm{kcal} / \mathrm{mol})$ |
| Ethylene | $\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2}$ | -136 | -32.6 |
| Monosubstituted | $\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}_{2}$ | -125 | -29.9 |
| Disubstituted | $\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCH}_{3}$ (cis)
 $\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCH}_{3}$ (trans)
 $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CH}_{2}$ | -119
 -115
 -118 | -28.3
 -28.4 |
| Trisubstituted | $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CHCH}_{3}$ | -112 | -26.7 |
| Tetrasubstituted | $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{C}\left(\mathrm{CH}_{3}\right)_{2}$ | -110 | -26.4 |


Solution - 7.6 Stability of Alkenes

TABLE 7.2 lists some representative data for the hydrogenation of different alkenes and shows that alkenes become more stable with increasing substitution. That is, alkenes follow the stability order:
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0211.jpg?height=220&width=1204&top_left_y=188&top_left_x=455)


Solution - 7.6 Stability of Alkenes

The stability order of substituted alkenes is due to a combination of two factors. One is a stabilizing interaction between the $\mathrm{C}=\mathrm{C} \pi$ orbital and adjacent $\mathrm{C}-\mathrm{H} \sigma$ bonds on substituents. In valence-bond language, the interaction is called hyperconjugation. In a molecular orbital description, there is a bonding MO that extends over the four-atom $\mathrm{C}=\mathrm{C}-\mathrm{C}-\mathrm{H}$ grouping, as shown in FIGURE 7.7. The more substituents present on the double bond, the more hyperconjugation occurs and the more stable the alkene.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0211.jpg?height=260&width=795&top_left_y=666&top_left_x=656)

FIGURE 7.7 Hyperconjugation is a stabilizing interaction between the $\mathbf{C}=\mathbf{C} \boldsymbol{\pi}$ orbital and adjacent $\mathbf{C}-\mathbf{H} \boldsymbol{\sigma}$ bonds on substituents. The more substituents there are, the greater the stabilization of the alkene.
A second factor that contributes to alkene stability involves bond strengths. A bond between a $s p^{2}$ carbon and a $s p^{3}$ carbon is somewhat stronger than a bond between two $s p^{3}$ carbons. Thus, in comparing 1-butene and 2-butene, the monosubstituted isomer has one $s p^{3}-s p^{3}$ bond and one $s p^{3}-s p^{2}$ bond, while the disubstituted isomer has two $s p^{3}-s p^{2}$ bonds. More highly substituted alkenes always have a higher ratio of $s p^{3}-s p^{2}$ bonds to $s p^{3}-s p^{3}$ bonds than less highly substituted alkenes and are therefore more stable.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0211.jpg?height=145&width=293&top_left_y=1280&top_left_x=691)


Solution - 7.6 Stability of Alkenes

2-Butene
(more stable)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0211.jpg?height=128&width=299&top_left_y=1286&top_left_x=1127)

1-Butene
(less stable)


Solution - 7.6 Stability of Alkenes


Solution - 7.7 Electrophilic Addition Reactions of Alkenes

Before beginning a detailed discussion of alkene reactions, let's review briefly some conclusions from the previous chapter. We said in Section 6.5 that alkenes behave as nucleophiles (Lewis bases) in polar reactions, donating a pair of electrons from their electron-rich $\mathrm{C}=\mathrm{C}$ bond to an electrophile (Lewis acid). For example, reaction of 2-methylpropene with HBr yields 2-bromo-2-methylpropane. A careful study of this and similar reactions by the British chemist Christopher Ingold and others in the 1930s led to the generally accepted mechanism shown in FIGURE 7.8 for an electrophilic addition reaction.


FIGURE 7.8 MECHANISM -

Mechanism of the electrophilic addition of HBr to 2-methylpropene. The reaction occurs in two steps, protonation and bromide addition, and involves a carbocation intermediate.
(1) A hydrogen atom on the electrophile HBr is attacked by $\pi$ electrons from the nucleophilic double bond, forming a new $\mathrm{C}-\mathrm{H}$ bond. This leaves the other carbon atom with a + charge and a vacant $p$ orbital. Simultaneously, two electrons from the $\mathrm{H}-\mathrm{Br}$ bond move onto bromine, giving bromide anion.
(2) The bromide ion donates an electron pair to the positively charged carbon atom, forming a $\mathrm{C}-\mathrm{Br}$ bond and yielding the neutral addition product.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0212.jpg?height=694&width=302&top_left_y=455&top_left_x=1291)


FIGURE 7.8 MECHANISM -


FIGURE 7.8 MECHANISM -

The reaction begins with an attack on the hydrogen of the electrophile HBr by the electrons of the nucleophilic $\pi$ bond. Two electrons from the $\pi$ bond form a new $\sigma$ bond between the entering hydrogen and an alkene carbon, as shown by the curved arrow at the top of FIGURE 7.8. The resulting carbocation intermediate is itself an electrophile, which can accept an electron pair from nucleophilic $\mathrm{Br}^{-}$ion to form a $\mathrm{C}-\mathrm{Br}$ bond and yield a neutral addition product.

An energy diagram for the overall electrophilic addition reaction (FIGURE 7.9) has two peaks (transition states) separated by a valley (carbocation intermediate). The energy level of the intermediate is higher than that of the starting alkene, but the reaction as a whole is exergonic (negative $\Delta G^{\circ}$ ). The first step, protonation of the alkene to yield the intermediate cation, is relatively slow. But once the cation intermediate is formed, it rapidly reacts to yield the final alkyl bromide product. The relative rates of the two steps are indicated in FIGURE 7.9 by the fact that $\Delta G_{1}{ }^{\ddagger}$ is larger than $\Delta G_{2}{ }^{\dagger}$.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0213.jpg?height=868&width=933&top_left_y=192&top_left_x=596)

FIGURE 7.9 Energy diagram for the two-step electrophilic addition of $\mathbf{H B r}$ to 2-methylpropene. The first step is slower than the second step.


FIGURE 7.8 MECHANISM -

Electrophilic addition to alkenes is successful not only with HBr but with $\mathrm{HCl}, \mathrm{HI}$, and $\mathrm{H}_{2} \mathrm{O}$ as well. Note that HI is usually generated in the reaction mixture by treating potassium iodide with phosphoric acid and that a strong acid catalyst is needed for the addition of water.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0213.jpg?height=150&width=643&top_left_y=1302&top_left_x=730)


FIGURE 7.8 MECHANISM -

2-Methylpropene
2-Chloro-2-methylpropane
(94\%)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0213.jpg?height=132&width=722&top_left_y=1582&top_left_x=685)


FIGURE 7.8 MECHANISM -

1-Pentene
(HI)
2-Iodopentane
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0213.jpg?height=167&width=668&top_left_y=1803&top_left_x=731)


FIGURE 7.8 MECHANISM -

1-Methylcyclohexene
1-Methylcyclohexanol


WRITING ORGANIC REACTIONS -

This is a good place to mention that the equations for organic reactions are sometimes written in different ways to emphasize different points. In describing a laboratory process, for instance, the reaction of 2-methylpropene with HCl might be written in the format $\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}$ to emphasize that both reactants are equally important for the purposes of the discussion. The solvent and notes about other reaction conditions such as temperature are written either above or below the reaction arrow.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0214.jpg?height=312&width=744&top_left_y=191&top_left_x=688)

Alternatively, we might write the same reaction in a format to emphasize that 2 -methylpropene is the reactant whose chemistry is of greater interest. The second reactant, HCl , is placed above the reaction arrow together with notes about solvent and reaction conditions.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0214.jpg?height=348&width=760&top_left_y=636&top_left_x=680)


WRITING ORGANIC REACTIONS -

In describing a biological process, the reaction is almost always written to show only the structures of the primary reactant and product, while abbreviating the structures of various biological "reagents" and byproducts with a curved arrow that intersects the straight reaction arrow. As discussed in Section 6.11, the reaction of glucose with ATP to give glucose 6-phosphate plus ADP would then be written as
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0214.jpg?height=346&width=879&top_left_y=1166&top_left_x=623)


WRITING ORGANIC REACTIONS - 7.8 Orientation of Electrophilic Additions: Markovnikov's Rule

Look carefully at the electrophilic addition reactions shown in the previous section. In each case, an unsymmetrically substituted alkene has given a single addition product rather than the mixture that might be expected. For example, 2-methylpropene might react with HCl to give both 2-chloro-2-methylpropane and 1-chloro-2-methylpropane, but it doesn't. It gives only 2-chloro-2-methylpropane as the sole product. Similarly, it's invariably the case in biological alkene addition reactions that only a single product is formed. We say that such reactions are regiospecific (ree-jee-oh-specific) when only one of two possible orientations of an addition occurs.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0214.jpg?height=337&width=1069&top_left_y=1927&top_left_x=523)


After looking at the results of many such reactions, the Russian chemist Vladimir Markovnikov proposed in 1869 what has become known as:


Markovnikov's rule
In the addition of HX to an alkene, the H attaches to the carbon with fewer alkyl substituents and the X
attaches to the carbon with more alkyl substituents.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0215.jpg?height=312&width=993&top_left_y=191&top_left_x=539)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0215.jpg?height=358&width=700&top_left_y=542&top_left_x=753)


1-Methylcyclohexene
1-Bromo-1-methylcyclohexane
When both double-bonded carbon atoms have the same degree of substitution, a mixture of addition products results.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0215.jpg?height=261&width=1189&top_left_y=1062&top_left_x=468)


Because carbocations are involved as intermediates in these electrophilic addition reactions, Markovnikov's rule can be restated in the following way:


Markovnikov's rule restated
In the addition of HX to an alkene, the more highly substituted carbocation is formed as the intermediate rather than the less highly substituted one.


For example, addition of $\mathrm{H}^{+}$to 2-methylpropene yields the intermediate tertiary carbocation rather than the alternative primary carbocation, and addition to 1-methylcyclohexene yields a tertiary cation rather than a secondary one. Why should this be?
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0216.jpg?height=1262&width=1321&top_left_y=182&top_left_x=402)


WORKED EXAMPLE 7.2 -


Predicting the Product of an Electrophilic Addition Reaction -

What product would you expect from reaction of HCl with 1-ethylcyclopentene?
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0216.jpg?height=107&width=559&top_left_y=1670&top_left_x=780)


Strategy -

When solving a problem that asks you to predict a reaction product, begin by looking at the functional group(s) in the reactants and deciding what kind of reaction is likely to occur. In the present instance, the reactant is an alkene that will probably undergo an electrophilic addition reaction with HCl . Next, recall what you know about electrophilic addition reactions to predict the product. You know that electrophilic addition reactions follow Markovnikov's rule, so $\mathrm{H}^{+}$will add to the double-bond carbon that has one alkyl group ( C 2 on the ring) and the Cl will add to the double-bond carbon that has two alkyl groups ( C 1 on the ring).


Solution -

The expected product is 1-chloro-1-ethylcyclopentane.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0217.jpg?height=351&width=861&top_left_y=188&top_left_x=632)


WORKED EXAMPLE 7.3 -


Synthesizing a Specific Compound -

What alkene would you start with to prepare the following alkyl halide? There may be more than one possibility.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0217.jpg?height=156&width=464&top_left_y=819&top_left_x=828)


Strategy -

When solving a problem that asks how to prepare a given product, always work backward. Look at the product, identify the functional group(s) it contains, and ask yourself, "How can I prepare that functional group?" In the present instance, the product is a tertiary alkyl chloride, which can be prepared by reaction of an alkene with HCl . The carbon atom bearing the -Cl atom in the product must be one of the double-bond carbons in the reactant. Draw and evaluate all possibilities.


Solution -


Solution -

PROBLEM What alkenes would you start with to prepare the following products?


Solution - 7.9 Carbocation Structure and Stability

To understand why Markovnikov's rule works, we need to learn more about the structure and stability of carbocations and about the general nature of reactions and transition states. The first point to explore involves structure.


A great deal of experimental evidence has shown that carbocations are planar. The trivalent carbon is $s p^{2}$-hybridized, and the three substituents are oriented toward the corners of an equilateral triangle, as indicated in FIGURE 7.10. Because there are only six valence electrons on carbon and all six are used in the three $\sigma$ bonds, the $p$ orbital extending above and below the plane is unoccupied.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0218.jpg?height=537&width=391&top_left_y=661&top_left_x=883)

FIGURE 7.10 The structure of a carbocation. The trivalent carbon is $s p^{2}$-hybridized and has a vacant $p$ orbital perpendicular to the plane of the carbon and three attached groups.


The second point to explore involves carbocation stability. 2-Methylpropene might react with $\mathrm{H}^{+}$to form a carbocation having three alkyl substituents (a tertiary ion, $3^{\circ}$ ), or it might react to form a carbocation having one alkyl substituent (a primary ion, $1^{\circ}$ ). Since the tertiary alkyl chloride, 2 -chloro-2-methylpropane, is the only product observed, formation of the tertiary cation is evidently favored over formation of the primary cation. Thermodynamic measurements show that, indeed, the stability of carbocations increases with increasing substitution so that the stability order is tertiary > secondary > primary > methyl.
<img class="imgSvg" id = "m6sn3n34xn1piunditj" src="data:image/svg+xml;base64,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"/>


Tertiary ( $\mathbf{3}^{\circ}$ )


Stability -

One way of determining carbocation stabilities is to measure the amount of energy required to form a carbocation by dissociation of the corresponding alkyl halide, $R-X \rightarrow R^{+}+: X^{-}$. As shown in FIGURE 7.11, tertiary alkyl halides dissociate to give carbocations more easily than secondary or primary ones. Thus, trisubstituted carbocations are more stable than disubstituted ones, which are more stable than monosubstituted ones. The data in FIGURE 7.11 are taken from measurements made in the gas phase, but a similar stability order is found for carbocations in solution. The dissociation enthalpies are much lower in solution because polar solvents can stabilize the ions, but the order of carbocation stability remains the same.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0219.jpg?height=456&width=998&top_left_y=192&top_left_x=561)


Stability -

FIGURE 7.11 A plot of dissociation enthalpy versus substitution pattern for the gas-phase dissociation of alkyl chlorides to yield carbocations. More highly substituted alkyl halides dissociate more easily than less highly substituted ones.


Stability -

Why are more highly substituted carbocations more stable than less highly substituted ones? There are at least two reasons. Part of the answer has to do with inductive effects, and part has to do with hyperconjugation. Inductive effects, discussed in Section 2.1 in connection with polar covalent bonds, result from the shifting of electrons in a $\sigma$ bond in response to the electronegativity of nearby atoms. In the present instance, electrons from a relatively larger and more polarizable alkyl group can shift toward a neighboring positive charge more easily than the electron from a hydrogen. Thus, the more alkyl groups attached to the positively charged carbon, the more electron density shifts toward the charge and the more inductive stabilization of the cation occurs (FIGURE 7.12).
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0219.jpg?height=318&width=302&top_left_y=1218&top_left_x=245)
<img class="imgSvg" id = "m6sn3n366y1ger1izdt" src="data:image/svg+xml;base64,PHN2ZyBpZD0ic21pbGVzLW02c24zbjM2NnkxZ2VyMWl6ZHQiIHhtbG5zPSJodHRwOi8vd3d3LnczLm9yZy8yMDAwL3N2ZyIgdmlld0JveD0iMCAwIDExNSA0Mi4wMDAwMTQxMzczMDcxNzYiIHN0eWxlPSJ3aWR0aDogMTE1LjQ5OTk5OTk5OTk5NjgycHg7IGhlaWdodDogNDIuMDAwMDE0MTM3MzA3MTc2cHg7IG92ZXJmbG93OiB2aXNpYmxlOyI+PGRlZnM+PGxpbmVhckdyYWRpZW50IGlkPSJsaW5lLW02c24zbjM2NnkxZ2VyMWl6ZHQtMSIgZ3JhZGllbnRVbml0cz0idXNlclNwYWNlT25Vc2UiIHgxPSI0MiIgeTE9IjIxIiB4Mj0iNzMuNDk5OTk5OTk5OTk2ODIiIHkyPSIyMS4wMDAwMTQxMzczMDcxOCI+PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjIwJSI+PC9zdG9wPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIxMDAlIj48L3N0b3A+PC9saW5lYXJHcmFkaWVudD48L2RlZnM+PG1hc2sgaWQ9InRleHQtbWFzay1tNnNuM24zNjZ5MWdlcjFpemR0Ij48cmVjdCB4PSIwIiB5PSIwIiB3aWR0aD0iMTAwJSIgaGVpZ2h0PSIxMDAlIiBmaWxsPSJ3aGl0ZSI+PC9yZWN0PjwvbWFzaz48c3R5bGU+CiAgICAgICAgICAgICAgICAuZWxlbWVudC1tNnNuM24zNjZ5MWdlcjFpemR0IHsKICAgICAgICAgICAgICAgICAgICBmb250OiAxNHB4IEhlbHZldGljYSwgQXJpYWwsIHNhbnMtc2VyaWY7CiAgICAgICAgICAgICAgICAgICAgYWxpZ25tZW50LWJhc2VsaW5lOiAnbWlkZGxlJzsKICAgICAgICAgICAgICAgIH0KICAgICAgICAgICAgICAgIC5zdWItbTZzbjNuMzY2eTFnZXIxaXpkdCB7CiAgICAgICAgICAgICAgICAgICAgZm9udDogOC40cHggSGVsdmV0aWNhLCBBcmlhbCwgc2Fucy1zZXJpZjsKICAgICAgICAgICAgICAgIH0KICAgICAgICAgICAgPC9zdHlsZT48ZyBtYXNrPSJ1cmwoI3RleHQtbWFzay1tNnNuM24zNjZ5MWdlcjFpemR0KSI+PGxpbmUgeDE9IjQyIiB5MT0iMjEiIHgyPSI3My40OTk5OTk5OTk5OTY4MiIgeTI9IjIxLjAwMDAxNDEzNzMwNzE4IiBzdHlsZT0ic3Ryb2tlLWxpbmVjYXA6cm91bmQ7c3Ryb2tlLWRhc2hhcnJheTpub25lO3N0cm9rZS13aWR0aDoxLjI2IiBzdHJva2U9InVybCgnI2xpbmUtbTZzbjNuMzY2eTFnZXIxaXpkdC0xJykiPjwvbGluZT48L2c+PGc+PHRleHQgeD0iNzMuNDk5OTk5OTk5OTk2ODIiIHk9IjIxLjAwMDAxNDEzNzMwNzE4IiBjbGFzcz0iZGVidWciIGZpbGw9IiNmZjAwMDAiIHN0eWxlPSIKICAgICAgICAgICAgICAgIGZvbnQ6IDVweCBEcm9pZCBTYW5zLCBzYW5zLXNlcmlmOwogICAgICAgICAgICAiPjwvdGV4dD48dGV4dCB4PSI0MiIgeT0iMjEiIGNsYXNzPSJkZWJ1ZyIgZmlsbD0iI2ZmMDAwMCIgc3R5bGU9IgogICAgICAgICAgICAgICAgZm9udDogNXB4IERyb2lkIFNhbnMsIHNhbnMtc2VyaWY7CiAgICAgICAgICAgICI+PC90ZXh0PjwvZz48L3N2Zz4="/>


Stability -


Stability -

Tertiary:
Three alkyl groups donating electrons


Stability -

FIGURE 7.12 A comparison of inductive stabilization for methyl, primary, secondary, and tertiary carbocations. The more alkyl groups that are bonded to the positively charged carbon, the more electron density shifts toward the charge, making the charged carbon less electron-poor (blue in electrostatic potential maps).


Stability -

Hyperconjugation, discussed in Section 7.6 in connection with the stabilities of substituted alkenes, is the stabilizing interaction between a $p$ orbital and properly oriented $\mathrm{C}-\mathrm{H} \sigma$ bonds on neighboring carbons that are roughly parallel to the $p$ orbital. The more alkyl groups there are on the carbocation, the more possibilities there are for hyperconjugation and the more stable the carbocation. FIGURE 7.13 shows the molecular orbital for the ethyl carbocation, $\mathrm{CH}_{3} \mathrm{CH}_{2}{ }^{+}$, and indicates the difference between the $\mathrm{C}-\mathrm{H}$ bond perpendicular to the cation $p$ orbital and the two $\mathrm{C}-\mathrm{H}$ bonds more parallel to the cation $p$ orbital. Only these roughly parallel $\mathrm{C}-\mathrm{H}$ bonds are oriented properly to take part in hyperconjugation.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0220.jpg?height=262&width=555&top_left_y=205&top_left_x=775)

FIGURE 7.13 Stabilization of the ethyl carbocation, $\mathbf{C H}_{\mathbf{3}} \mathbf{C H}_{\mathbf{2}}{ }^{\boldsymbol{+}}$, through hyperconjugation. Interaction of neighboring $\mathrm{C}-\mathrm{H} \sigma$ bonds with the vacant $p$ orbital stabilizes the cation and lowers its energy. The molecular orbital shows that only the two $\mathrm{C}-\mathrm{H}$ bonds more parallel to the cation $p$ orbital are oriented properly. The $\mathrm{C}-\mathrm{H}$ bond perpendicular to the cation $p$ orbital can't take part properly.


Stability -

PROBLEM Show the structures of the carbocation intermediates you would expect in the following reactions:


Stability -

PROBLEM Draw a skeletal structure of the following carbocation. Identify it as primary, secondary, or tertiary, 7-19 and identify the hydrogen atoms that have the proper orientation for hyperconjugation in the conformation shown.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0220.jpg?height=313&width=324&top_left_y=944&top_left_x=429)


Stability - 7.10 The Hammond Postulate

Let's summarize what we've learned of electrophilic addition reactions to this point:


Stability - 7.10 The Hammond Postulate

- Electrophilic addition to an unsymmetrically substituted alkene gives the more substituted carbocation intermediate. A more substituted carbocation forms faster than a less substituted one and, once formed, rapidly goes on to give the final product.
- A more substituted carbocation is more stable than a less substituted one. That is, the stability order of carbocations is tertiary > secondary > primary > methyl.


Stability - 7.10 The Hammond Postulate

What we have not yet seen is how these two points are related. Why does the stability of the carbocation intermediate affect the rate at which it's formed and thereby determine the structure of the final product? After all, carbocation stability is determined by the free-energy change $\Delta G^{\circ}$, but reaction rate is determined by the activation energy $\Delta G^{\ddagger}$. The two quantities aren’t directly related.


Stability - 7.10 The Hammond Postulate

Although there is no simple quantitative relationship between the stability of a carbocation intermediate and the rate of its formation, there is an intuitive relationship. It's generally true when comparing two similar reactions that the more stable intermediate forms faster than the less stable one. The situation is shown graphically in FIGURE 7.14, where the energy profile in part (a) represents the typical situation, as opposed to the profile in part (b). That is, the curves for two similar reactions don't cross one another.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0221.jpg?height=511&width=1502&top_left_y=178&top_left_x=306)


Stability - 7.10 The Hammond Postulate

FIGURE 7.14 Energy diagrams for two similar competing reactions. In (a), the faster reaction yields the more stable intermediate. In (b), the slower reaction yields the more stable intermediate. The curves shown in (a) represent the typical situation.


Stability - 7.10 The Hammond Postulate

Called the Hammond postulate, the explanation of the relationship between reaction rate and intermediate stability goes like this: Transition states represent energy maxima. They are high-energy activated complexes that occur transiently during the course of a reaction and immediately go on to a more stable species. Although we can't actually observe transition states because they have no finite lifetime, the Hammond postulate says that we can get an idea of a particular transition state's structure by looking at the structure of the nearest stable species. Imagine the two cases shown in FIGURE 7.15, for example. The reaction profile in part (a) shows the energy curve for an endergonic reaction step, and the profile in part (b) shows the curve for an exergonic step.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0221.jpg?height=486&width=1240&top_left_y=1080&top_left_x=437)


Stability - 7.10 The Hammond Postulate

FIGURE 7.15 Energy diagrams for endergonic and exergonic steps. (a) In an endergonic step, the energy levels of transition state and product are closer. (b) In an exergonic step, the energy levels of transition state and reactant are closer.

In an endergonic reaction (FIGURE 7.15a), the energy level of the transition state is closer to that of the product than that of the reactant. Since the transition state is closer energetically to the product, we make the natural assumption that it's also closer structurally. In other words, the transition state for an endergonic reaction step structurally resembles the product of that step. Conversely, the transition state for an exergonic reaction (FIGURE 7.15b), is closer energetically, and thus structurally, to the reactant than to the product. We therefore say that the transition state for an exergonic reaction step structurally resembles the reactant for that step.


Hammond postulate: -

The structure of a transition state resembles the structure of the nearest stable species. Transition states for endergonic steps structurally resemble products, and transition states for exergonic steps structurally resemble reactants.


Hammond postulate: -

How does the Hammond postulate apply to electrophilic addition reactions? The formation of a carbocation by protonation of an alkene is an endergonic step. Thus, the transition state for alkene protonation structurally resembles the carbocation intermediate, and any factor that stabilizes the carbocation will also stabilize the nearby transition state. Since increasing alkyl substitution stabilizes carbocations, it also stabilizes the transition states leading to those ions, thus resulting in a faster reaction. In other words, more stable carbocations form faster because their greater stability is reflected in the lower-energy transition state leading to them (FIGURE 7.16).
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0222.jpg?height=630&width=1023&top_left_y=192&top_left_x=546)


Hammond postulate: -

FIGURE 7.16 Energy diagrams for carbocation formation. The more stable tertiary carbocation is formed faster (green curve) because its increased stability lowers the energy of the transition state leading to it.


Hammond postulate: -

We can imagine the transition state for alkene protonation to be a structure in which one of the alkene carbon atoms has almost completely rehybridized from $s p^{2}$ to $s p^{3}$ and the remaining alkene carbon bears much of the positive charge (FIGURE 7.17). This transition state is stabilized by hyperconjugation and inductive effects in the same way as the product carbocation. The more alkyl groups that are present, the greater the extent of stabilization and the faster the transition state forms.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0222.jpg?height=435&width=1500&top_left_y=1135&top_left_x=307)

FIGURE 7.17 The hypothetical structure of a transition state for alkene protonation. The transition state is closer in both energy and structure to the carbocation than to the alkene. Thus, an increase in carbocation stability (lower $\Delta G^{\circ}$ ) also causes an increase in transitionstate stability (lower $\Delta G^{\ddagger}$ ), thereby increasing the rate of its formation.


Hammond postulate: -

PROBLEM What about the second step in the electrophilic addition of HCl to an alkene-the reaction of
$\mathbf{7 - 2 0}$ chloride ion with the carbocation intermediate? Is this step exergonic or endergonic? Does the transition state for this second step resemble the reactant (carbocation) or product (alkyl chloride)? Make a rough drawing of what the transition-state structure might look like.


Hammond postulate: - 7.11 Evidence for the Mechanism of Electrophilic Additions: Carbocation Rearrangements

How do we know that the carbocation mechanism for electrophilic addition reactions of alkenes is correct? The answer is that we don't know it's correct; at least we don't know with complete certainty. Although an incorrect reaction mechanism can be disproved by demonstrating that it doesn't account for observed data, a correct reaction mechanism can never be entirely proven. The best we can do is to show that a proposed mechanism is consistent with all known facts. If enough facts are accounted for, the mechanism is probably correct.


One of the best pieces of evidence supporting the carbocation mechanism for the electrophilic addition reaction was discovered during the 1930s by F. C. Whitmore of Pennsylvania State University, who found that structural rearrangements often occur during the reaction of HX with an alkene. For example, reaction of HCl with 3 -methyl-1-butene yields a substantial amount of 2 -chloro-2-methylbutane in addition to the "expected" product, 2-chloro-3-methylbutane.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0223.jpg?height=294&width=1232&top_left_y=184&top_left_x=441)


If the reaction takes place in a single step, it would be difficult to account for rearrangement, but if the reaction takes place in several steps, rearrangement is more easily explained. Whitmore suggested that it is a carbocation intermediate that undergoes rearrangement. The secondary carbocation intermediate formed by protonation of 3-methyl-1-butene rearranges to a more stable tertiary carbocation by a hydride shift-the shift of a hydrogen atom and its electron pair (a hydride ion, $: \mathrm{H}^{-}$) between neighboring carbons.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0223.jpg?height=565&width=1326&top_left_y=707&top_left_x=383)

2-Chloro-3-methylbutane
2-Chloro-2-methylbutane
Carbocation rearrangements can also occur by the shift of an alkyl group with its electron pair. For example, reaction of 3,3-dimethyl-1-butene with HCl leads to an equal mixture of unrearranged 3 -chloro-2,2-dimethylbutane and rearranged 2 -chloro-2,3-dimethylbutane. In this instance, a secondary carbocation rearranges to a more stable tertiary carbocation by the shift of a methyl group.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0223.jpg?height=575&width=1345&top_left_y=1515&top_left_x=363)

3-Chloro-2,2-dimethylbutane 2-Chloro-2,3-dimethylbutane
Note the similarities between the two carbocation rearrangements: in both cases, a group (: $\mathrm{H}^{-}$or : $\mathrm{CH}_{3}{ }^{-}$) moves to an adjacent positively charged carbon, taking its bonding electron pair with it. Also in both cases, a less stable carbocation rearranges to a more stable ion. Rearrangements of this kind are a common feature of carbocation chemistry and are particularly important in the biological pathways by which steroids and related substances are synthesized. An example is the following hydride shift that occurs during the biosynthesis of cholesterol.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0224.jpg?height=486&width=1402&top_left_y=185&top_left_x=359)


As always, when looking at any complex chemical transformation, whether biochemical or not, focus on the part of the molecule where the change is occurring and don't worry about the rest. The tertiary carbocation just pictured looks complicated, but all the chemistry is taking place in the small part of the molecule inside the red circle.


PROBLEM On treatment with HBr , vinylcyclohexane undergoes addition and rearrangement to yield
7-21 1-bromo-1-ethylcyclohexane. Using curved arrows, propose a mechanism to account for this result.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0224.jpg?height=180&width=593&top_left_y=1013&top_left_x=427)


Vinylcyclohexane 1-Bromo-1-ethylcyclohexane


(80) CHEMISTRY MATTERS -


Bioprospecting: Hunting for Natural Products -

Most people know the names of the common classes of biomolecules-proteins, carbohydrates, lipids, and nucleic acids-but there are many more kinds of compounds in living organisms than just those four. All living organisms also contain a vast diversity of substances usually grouped under the heading natural products. The term natural product really refers to any naturally occurring substance but is generally taken to mean a so-called secondary metabolite-a small molecule that is not essential to the growth and development of the producing organism and is not classified by structure.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0225.jpg?height=693&width=1391&top_left_y=716&top_left_x=367)


FIGURE 7.18 Rapamycin, an immunosuppressant natural product used during organ transplants, was originally isolated from a soil sample found on Rapa Nui (Easter Island), an island 2200 miles off the coast of Chile known for its giant Moai statues. (credit: modification of work "Moai facing inland at Ahu Tongariki" by Ian Sewell/Wikimedia Commons, CC BY 2.5)


It has been estimated that well over 300,000 secondary metabolites exist, and it's thought that their primary function is to increase the likelihood of an organism's survival by repelling or attracting other organisms. Alkaloids, such as morphine; antibiotics, such as erythromycin and the penicillins; and immunosuppressive agents, such as rapamycin (sirolimus) prescribed for liver transplant recipients, are examples.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0225.jpg?height=643&width=912&top_left_y=1714&top_left_x=604)


Where do these natural products come from, and how are they found? Although most chemists and biologists spend their working time in the laboratory, a few spend their days scuba diving on South Pacific islands or trekking through the rainforests of South America and Southeast Asia at work as bioprospectors. Their job is to
hunt for new and unusual natural products that might be useful as drugs.
As noted in the Chapter 6 Chemistry Matters, more than half of all new drug candidates come either directly or indirectly from natural products. Morphine from the opium poppy, prostaglandin $\mathrm{E}_{1}$ from sheep prostate glands, erythromycin A from a Streptomyces erythreus bacterium cultured from a Philippine soil sample, and benzylpenicillin from the mold Penicillium notatum are examples. The immunosuppressive agent rapamycin, whose structure is shown previously, was first isolated from a Streptomyces hygroscopicus bacterium found in a soil sample from Rapa Nui (Easter Island), located 2200 miles off the coast of Chile.


With less than $1 \%$ of living organisms yet investigated, bioprospectors have a lot of work to do. But there is a race going on. Rainforests throughout the world are being destroyed at an alarming rate, causing many species of both plants and animals to become extinct before they can even be examined. The governments in many countries seem aware of the problem, but there is as yet no international treaty on biodiversity that could help preserve vanishing species.


Key Terms -

- alkene $\left(\mathrm{R}_{2} \mathrm{C}=\mathrm{CR}_{2}\right)$
- allyl group
- degree of unsaturation
- Econfiguration
- E,Z system
- electrophilic addition reaction
- Hammond postulate
- hydride shift
- hyperconjugation
- Markovnikov's rule
- methylene group
- regiospecific reaction
- unsaturated
- vinyl group
- Z configuration


Key Terms -


Summary -

Carbon-carbon double bonds are present in most organic and biological molecules, so a good understanding of their behavior is needed. In this chapter, we've looked at some consequences of alkene stereoisomerism and at the details of the broadest class of alkene reactions-the electrophilic addition reaction.


Summary -

An alkene is a hydrocarbon that contains a carbon-carbon double bond. Because they contain fewer hydrogens than alkanes with the same number of carbons, alkenes are said to be unsaturated.


Summary -

Because rotation around the double bond can't occur, substituted alkenes can exist as cis-trans stereoisomers. The configuration of a double bond can be specified by applying the Cahn-Ingold-Prelog sequence rules, which rank the substituents on each double-bond carbon. If the higher-ranking groups on each carbon are on the same side of the double bond, the configuration is $\boldsymbol{Z}$ (zusammen, "together"); if the higher-ranking groups on each carbon are on opposite sides of the double bond, the configuration is $\boldsymbol{E}$ (entgegen, "apart").


Summary -

Alkene chemistry is dominated by electrophilic addition reactions. When HX reacts with an unsymmetrically substituted alkene, Markovnikov's rule predicts that the $H$ will add to the carbon having fewer alkyl substituents and the X group will add to the carbon having more alkyl substituents. Electrophilic additions to alkenes take place through carbocation intermediates formed by reaction of the nucleophilic alkene $\pi$ bond with electrophilic $\mathrm{H}^{+}$. Carbocation stability follows the order

Tertiary $\left(3^{\circ}\right)>$ Secondary $\left(2^{\circ}\right)>\operatorname{Primary}\left(1^{\circ}\right)>$ Methyl
$\mathrm{R}_{3} \mathrm{C}^{+}>\mathrm{R}_{2} \mathrm{CH}^{+}>\mathrm{RCH}_{2}^{+}>\mathrm{CH}_{3}{ }^{+}$
Markovnikov's rule can be restated by saying that, in the addition of HX to an alkene, a more stable carbocation intermediate is formed. This result is explained by the Hammond postulate, which says that the transition state of an exergonic reaction step structurally resembles the reactant, whereas the transition state of an endergonic reaction step structurally resembles the product. Since an alkene protonation step is endergonic, the stability of the more highly substituted carbocation is reflected in the stability of the transition state leading to its formation.


Summary -

Evidence in support of a carbocation mechanism for electrophilic additions comes from the observation that structural rearrangements often take place during reaction. Rearrangements occur by shift of either a hydride ion, : $\mathrm{H}^{-}$(a hydride shift), or an alkyl anion, $: \mathrm{R}^{-}$, from a carbon atom to the neighboring positively charged carbon. This results in isomerization of a less stable carbocation to a more stable one.


Additional Problems -


Visualizing Chemistry -

PROBLEM Name the following alkenes, and convert each drawing into a skeletal structure:


Visualizing Chemistry -


Visualizing Chemistry -

PROBLEM Assign $E$ or $Z$ stereochemistry to the double bonds in each of the following alkenes, and convert
7-23 each drawing into a skeletal structure (red $=0$, green $=\mathrm{Cl}$ ):
(a)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0227.jpg?height=326&width=445&top_left_y=1103&top_left_x=482)
(b)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0227.jpg?height=400&width=473&top_left_y=1101&top_left_x=997)


Visualizing Chemistry -

PROBLEM The following carbocation is an intermediate in the electrophilic addition reaction of HCl with
7-24 two different alkenes. Identify both, and tell which $\mathrm{C}-\mathrm{H}$ bonds in the carbocation are aligned for hyperconjugation with the vacant $p$ orbital on the positively charged carbon.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0227.jpg?height=367&width=362&top_left_y=1681&top_left_x=426)


Visualizing Chemistry -

PROBLEM The following alkyl bromide can be made by HBr addition to three different alkenes. Show their 7-25 structures.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0228.jpg?height=375&width=486&top_left_y=189&top_left_x=429)


Mechanism Problems -

PROBLEM Predict the major product and show the complete mechanism for each of the following electrophilic 7-26 addition reactions.
(a)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0228.jpg?height=134&width=374&top_left_y=768&top_left_x=480)
(b)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0228.jpg?height=123&width=367&top_left_y=768&top_left_x=928)
(c)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0228.jpg?height=112&width=318&top_left_y=773&top_left_x=1370)

PROBLEM Each of the following electrophilic addition reactions involves a carbocation rearrangement.
7-27 Predict the product and draw the complete mechanism of each using curved arrows.
(a)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0228.jpg?height=113&width=426&top_left_y=1025&top_left_x=478)
(b)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0228.jpg?height=133&width=434&top_left_y=1034&top_left_x=989)
(c)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0228.jpg?height=143&width=663&top_left_y=1175&top_left_x=476)


Mechanism Problems -

PROBLEM When 1,3-butadiene reacts with 1 mol of HBr , two isolable products result. Propose mechanisms 7-28 for both.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0228.jpg?height=142&width=798&top_left_y=1439&top_left_x=414)


Mechanism Problems -


Methyl vinyl ether -

PROBLEM Addition of HCl to 1-isopropylcyclohexene yields a rearranged product. Propose a mechanism,
7-30 showing the structures of the intermediates and using curved arrows to indicate electron flow in each step.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0228.jpg?height=190&width=722&top_left_y=2106&top_left_x=409)

PROBLEM Addition of HCl to 1-isopropenyl-1-methylcyclopentane yields
7-31 1-chloro-1,2,2-trimethylcyclohexane. Propose a mechanism, showing the structures of the intermediates and using curved arrows to indicate electron flow in each step.
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0229.jpg?height=220&width=679&top_left_y=188&top_left_x=411)


Methyl vinyl ether -

PROBLEM Limonene, a fragrant hydrocarbon found in lemons and oranges, is biosynthesized from geranyl
7-32 diphosphate by the following pathway. Add curved arrows to show the mechanism of each step. Which step involves an alkene electrophilic addition? (The ion $\mathrm{OP}_{2} \mathrm{O}_{6}{ }^{4-}$ is the diphosphate ion, and "Base" is an unspecified base in the enzyme that catalyzes the reaction.)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0229.jpg?height=389&width=1194&top_left_y=613&top_left_x=406)


Methyl vinyl ether -

PROBLEM epi-Aristolochene, a hydrocarbon found in both pepper and tobacco, is biosynthesized by the
7-33 following pathway. Add curved arrows to show the mechanism of each step. Which steps involve alkene electrophilic addition(s), and which involve carbocation rearrangement(s)? (The abbreviation $\mathrm{H}-\mathrm{A}$ stands for an unspecified acid, and "Base" is an unspecified base in the enzyme.)
![](https://cdn.mathpix.com/cropped/2025_02_06_5556f7656cbdcaa0f9a7g-0229.jpg?height=594&width=1332&top_left_y=1210&top_left_x=407)


Calculating a Degree of Unsaturation -

PROBLEM Calculate the degree of unsaturation in the following formulas, and draw five possible structures for
7-34 each:
(a) $\mathrm{C}_{10} \mathrm{H}_{16}$
(b) $\mathrm{C}_{8} \mathrm{H}_{8} \mathrm{O}$
(c) $\mathrm{C}_{7} \mathrm{H}_{10} \mathrm{Cl}_{2}$
(d) $\mathrm{C}_{10} \mathrm{H}_{16} \mathrm{O}_{2}$
(e) $\mathrm{C}_{5} \mathrm{H}_{9} \mathrm{NO}_{2}$
(f) $\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{ClNO}$


Calculating a Degree of Unsaturation -

PROBLEM How many hydrogens does each of the following compounds have?
7-35 (a) $\mathrm{C}_{8} \mathrm{H}_{?} \mathrm{O}_{2}$, has two rings and one double bond (b) $\mathrm{C}_{7} \mathrm{H}_{?} \mathrm{~N}$, has two double bonds
(c) $\mathrm{C}_{9} \mathrm{H}_{?} \mathrm{NO}$, has one ring and three double bonds