Structure and Synthesis of Alkenes (Chapter 7 Study Notes)

Structure and Synthesis of Alkenes

Introduction to Alkenes

Alkenes are a class of hydrocarbons characterized by the presence of at least one carbon–carbon double bond. This double bond imparts unique chemical and physical properties to alkenes, distinguishing them from alkanes and alkynes.

• Definition: Alkenes are hydrocarbons with the general formula $C_nH_{2n}$, containing at least one C=C double bond.

• Alternative Name: Alkenes are also called olefins ("oil-forming gas").

• Functional Group: The C=C double bond is the functional group responsible for alkene reactivity.

• Example: Ethene ($C_2H_4$) is the simplest alkene.

Sigma and Pi Bonds in Ethylene

The bonding in ethene (ethylene) illustrates the hybridization and geometry of alkenes.

• Each carbon in ethene is sp2 hybridized, forming three sp2 orbitals and one unhybridized p orbital.

• Two sp2 orbitals form sigma ($\sigma$) bonds with hydrogen atoms, and one forms a sigma bond with the other carbon.

• The unhybridized p orbitals on each carbon overlap side-by-side to form a pi ($\pi$) bond, completing the double bond.

• The molecular geometry is trigonal planar with bond angles of approximately 120°.

Bond Lengths and Angles

The presence of a double bond affects bond lengths and angles in alkenes compared to alkanes.

• sp2 hybrid orbitals have more s character than sp3 orbitals, resulting in shorter bonds.

• The C–C bond length in ethylene is 1.33 Å, shorter than the 1.54 Å in ethane (an alkane).

• Bond angles in ethylene are close to 120°, consistent with trigonal planar geometry.

Cis-Trans (Geometric) Isomerism

Alkenes can exhibit geometric isomerism due to restricted rotation around the double bond.

• Cis isomer: Similar groups are on the same side of the double bond.

• Trans isomer: Similar groups are on opposite sides of the double bond.

• Rotation around the C=C bond is not possible without breaking the $\pi$ bond (requires 264 kJ/mol).

• Cis and trans isomers cannot interconvert under normal conditions.

Elements of Unsaturation

Elements of unsaturation (also called the index of hydrogen deficiency) indicate the presence of rings and/or multiple bonds in a molecule.

• Each element of unsaturation reduces the number of hydrogens by two compared to a saturated hydrocarbon.

• Double bonds and rings each count as one element of unsaturation.

• Triple bonds count as two elements of unsaturation.

Calculating Elements of Unsaturation

1. Calculate the number of hydrogens for a saturated hydrocarbon ($C_nH_{2n+2}$).

2. Subtract the actual number of hydrogens in the formula.

3. Divide the difference by 2.

Formula: $\text{Elements of Unsaturation} = \frac{(2n + 2) - \text{actual H}}{2}$

• This calculation does not distinguish between rings and multiple bonds.

Adjustments for Heteroatoms

• Add one H for each group V element (N, P).

• Subtract one H for each group VII element (F, Cl, Br, I).

• No correction is needed for O or S.

Examples

• For $C_5H_8$: Saturated = 12 H; Actual = 8 H; $\frac{12-8}{2} = 2$ elements of unsaturation.

• For $C_4H_9Br$: Saturated = 10 H; Corrected = 9 H; Actual = 9 H; $\frac{10-9}{2} = 0.5$ (should be 1, as Br replaces one H).

• For $C_8H_{10}O_2$: Saturated = 18 H; Actual = 10 H; $\frac{18-10}{2} = 4$ elements of unsaturation.

IUPAC Nomenclature of Alkenes

Systematic naming of alkenes follows specific rules to ensure clarity and consistency.

• Find the longest continuous carbon chain containing the double bond.

• Change the -ane ending to -ene.

• Number the chain so the double bond has the lowest possible number.

• In rings, the double bond is assumed to be between carbons 1 and 2.

Multiple Double Bonds

• Use prefixes di-, tri-, tetra-, etc., before -ene to indicate the number of double bonds (e.g., diene, triene).

• Assign the lowest possible numbers to the double bonds.

Alkenes as Substituents

• Common alkene substituents include the methylene group ($=CH_2$) and the vinyl group ($-CH=CH_2$).

Geometric Isomerism and E/Z Nomenclature

When cis/trans nomenclature is insufficient (e.g., when all groups are different), the E/Z system is used.

• Assign priorities to groups attached to each carbon of the double bond based on atomic number (Cahn-Ingold-Prelog rules).

• Z (zusammen): High-priority groups are on the same side.

• E (entgegen): High-priority groups are on opposite sides.

• If more than one double bond is present, specify the stereochemistry for each.

Physical Properties of Alkenes

• Cis alkenes have a greater dipole moment and are slightly more polar than trans alkenes.

• The boiling point of cis alkenes is higher than that of trans alkenes due to increased polarity.

Double Bond Stability

The stability of alkenes can be assessed by their heat of hydrogenation and degree of substitution.

• The more substituted the double bond, the lower its heat of hydrogenation and the more stable it is.

• Trans alkenes are generally more stable than cis alkenes.

• Stability order: tetrasubstituted > trisubstituted > disubstituted > monosubstituted.

Cycloalkenes

• Double bonds in five-membered or larger rings are relatively stable and behave like straight-chain alkenes.

• Small rings (e.g., cyclopropene) have significant angle strain, making their double bonds less stable.

• Trans cycloalkenes are not stable unless the ring has at least eight carbons.

• For cyclodecene and larger rings, the trans double bond is almost as stable as the cis.

Bredt's Rule

Bredt's Rule restricts the placement of double bonds in bridged bicyclic systems.

• A bridged bicyclic compound cannot have a double bond at a bridgehead position unless one of the rings contains at least eight carbon atoms.

Possible	Not Possible
Bicyclo[3.3.1]non-1-ene	Bicyclo[2.2.1]hept-1-ene
Bicyclo[4.4.1]undec-1-ene	Bicyclo[3.2.1]oct-1-ene

Alkene Synthesis Overview

Alkenes are commonly synthesized via elimination reactions.

• E1 Dehydrohalogenation: Removal of HX from alkyl halides via a two-step mechanism involving a carbocation intermediate.

• E2 Dehydrohalogenation: Removal of HX from alkyl halides in a single concerted step.

• Dehydration of Alcohols: Removal of H2O from alcohols, typically under acidic conditions.

• Dehalogenation of Vicinal Dibromides: Removal of X2 from adjacent carbons.

The E1 Reaction

• Unimolecular elimination: Involves a carbocation intermediate.

• Rate law: $\text{rate} = k[\text{alkyl halide}]$

• Two groups are lost: a hydrogen and a halide.

• Nucleophile acts as a base.

• E1 and SN1 reactions have the same conditions and first step (carbocation formation).

E1 Mechanism

1. Step 1: Ionization to form a carbocation intermediate.

2. Step 2: Solvent acts as a base and abstracts a proton to form the alkene.

E1 Energy Diagram

• Carbocation formation is the rate-determining step for both E1 and SN1 mechanisms.

Zaitsev's Rule

• If more than one elimination product is possible, the most substituted alkene (the most stable) is the major product.

• "The rich get richer": More alkyl groups on the double bond increase stability.

The E2 Reaction

• Bimolecular elimination: The proton is abstracted, the double bond forms, and the leaving group leaves in a single concerted step.

• Rate law: $\text{rate} = k[\text{alkyl halide}][\text{base}]$

• Requires a strong base (e.g., NaOCH3).

• Order of reactivity: 3° > 2° > 1° alkyl halides.

• E2 is not possible for 3° alkyl halides with poor bases.

• Zaitsev product usually predominates, but bulky bases can favor the less substituted (Hofmann) product.

Hofmann Orientation

• Bulky bases (e.g., tert-butoxide) favor elimination of less hindered hydrogens, leading to the less substituted alkene (Hofmann product).

E2 Stereochemical Requirement

• The hydrogen and leaving group must be anti-coplanar (180° apart) for elimination to occur.

• In cyclohexanes, both must be axial (trans-diaxial) for E2 elimination.

Comparison of E1 and E2 Mechanisms

Feature	E1	E2
Order	Unimolecular	Bimolecular
Intermediate	Carbocation	None (concerted)
Base Strength	Weak base	Strong base
Substrate	3° > 2°	3° > 2° > 1°
Stereochemistry	None	Anti-coplanar required

Substitution vs. Elimination

• Strong nucleophiles favor substitution (SN2), strong bases favor elimination (E2).

• High temperature favors elimination.

• Secondary alkyl halides can undergo both, depending on the reagent.

Dehydration of Alcohols

• Alcohols can be dehydrated to alkenes using an acid catalyst (e.g., concentrated H2SO4 or H3PO4) and heat.

• The mechanism is E1 and may involve carbocation rearrangement.

• The reaction follows Zaitsev's rule (most substituted alkene is major product).

1. Protonation of the hydroxyl group (fast, reversible).

2. Loss of water to form a carbocation (slow, rate-determining).

3. Deprotonation to give the alkene (fast).

Practice Problems (Selected)

1. How many elements of unsaturation do molecules with a molecular formula of $C_6H_6Cl_6$ have?

2. A newly isolated natural product was found to have the molecular formula $C_{15}H_{28}O_2$. By hydrogenating a sample, it was determined to possess one $\pi$ bond. How many rings are present?

3. Provide the IUPAC name for $(CH_3)_2CHCH=C(CH_3)CH_2CH_3$.

4. Name the compound shown below (cyclopentene derivative).

5. Provide the proper IUPAC name for the alkene shown below (substituted cyclohexene).

6. For which of the following alkenes will cis- and trans- isomers not exist? (Structures provided.)

7. Assign E/Z configuration to given alkenes.

8. Predict the major product of elimination reactions using Zaitsev's rule.

Additional info: For full mechanisms, stereochemical drawings, and more practice, refer to your textbook or lecture notes.