Reference: McMurry Ch 3 & 4, George et al Ch 1.2

• "Unsaturated" hydrocarbons
• Contain at least one C=C bond

• Each carbon atom of the C=C bond is sp² hybridized
• The sp² hybridized carbon centres are planar with bond angles of 120°
• The double bond is composed of overlap of two sp² hybridized orbitals to form a sigma (s ) bond and overlap of two p orbitals to form a pi (p ) bond.

• p bonds are weaker and more reactive than s bonds as a result of less efficient orbital overlap.

• p bond electron density concentrated above and below plane of molecule
• Rotation about a carbon-carbon double bond is NOT possible, as this requires the p bond to break (280 kJ mol^-1 of energy).

• Stem derived from the number of carbon atoms in longest chain which includes the double bond
• Ending is ene
• Give double bond lowest possible number (always between carbons 1, 2 in cyclic compounds)
• Name the substituents as usual

Examples:

IUPAC name of CH₂=CH₂ is ethylene, and of CH₂=CHCH₃ is propylene

Constitutional isomers are possible, e.g. C₄H₈, (write names)

 Stereoisomers are also possible as there is no rotation about the double bond. Stereoisomers of this sort are called diastereoisomers or diastereomers; they have different physical properties.

• Each end of the double bond must have groups that can be distinguished from one another.

In general,

 Example:

• Z/E determined by assigning a priority to each of the pairs of groups on each carbon of the double bond
• The higher the atomic number of the atom attached, the higher the priority
• If identical atoms are attached to each carbon of C=C, work outwards along the chain until the first point of difference is reached
• If groups of high priority are on the same side of the double bond the alkene is denoted (Z) from the German zusammen "together", if groups of high priority are on the opposite side of the double bond the alkene is denoted (E) from the German entgegen "opposite".

 Example:

• Ethylene and propylene from refining
• Lab methods for other alkenes

•  Non-polar, hydrophobic compounds

General type of reaction is an electrophilic addition reaction. The electrons of the p bond are available for reaction, which has the net effect of replacing a p bond with a s bond and, therefore, is energetically favorable.

• Require a catalyst (Pt/C or Pd/C) to help break strong H-H bond
• Form a saturated hydrocarbon from an unsaturated one

 Example:

2. Halogenation

• No catalyst required for addition of Cl₂ or Br₂
• Addition of I₂ is reversible and F₂ also reacts with C-H bonds so not so useful
• Solvent is usually CCl₄

Example:

 3. Hydrohalogenation

• HCl, HBr, HI
• Solvent is usually anhydrous CCl₄

 Example:

4. Hydration

• Effectively addition of H-OH but require H⁺ catalyst
• Use a dilute aqueous solution of H₂SO₄

 It is convenient to classify reactions according to the overall type – addition, substitution, elimination and, where appropriate, the nature of the reagent which first interacts with the organic substrate. We need to define two new terms, which describe this nature: electrophile and nucleophile. These terms are complimentary, just like acid and base. Note the terms acid/base may also apply but generally have more limited use in organic chemistry.

Electrophile: A species that seeks and an electron pair

 Nucleophile: A species that supplies and an electron pair

For example: H⁺ + H₂O ® H₃O⁺

Curly arrow notation is used to indicate the movement of electrons in a chemical reaction and thus shows how bonds form and break. A "curly arrow" indicates the movement of an electron pair. It starts at the electron pair that moves and ends at the atom to which the electron pair has moved.

For example:

 Here the lone pair on the oxygen is forming a new O-H covalent bond with the proton.

 and similarly

Here the O-H bond is breaking and the electron pair from the bond ends up as the lone pair on the oxygen atom of water.

The addition of HX to a symmetric alkene gives only one possible product.

 If an unsymmetrical alkene is used, there are two possible products, one of which dominates the product mixture.

 This was observed before it was understood and formulated as Markovnikov’s Rule: The hydrogen of an unsymmetrical reagent adds to the carbon of the double bond that has the greatest number of hydrogens already directly attached to it.

The explanation of this selectivity (regioselectivity) is as follows:

• The reaction is stepwise

Reactants ® Intermediate ® Product

• The hydrogen of HX, as H⁺, attacks the p bond of the alkene to form a new s bond and a charged intermediate called a carbocation

• The carbocation reacts with the negative part of the reagent (Br^–) to form a neutral compound.
• Any other nucleophile may also react.

• The selectivity results from the most stable carbocation intermediate being the one which forms: tertiary > secondary > primary

  Hence:

The evidence for the mechanism of the reaction is as follows:

• The reaction is selective
• The rate of reaction increases with acidity of HX (HI > HBr > HCl > HF)
• The rate of reaction increases with a greater number of alkyl groups on the alkene

• The reaction in the presence of HBr, H₂O and NaCl gives three products

• There is no reaction in the absence of H⁺

Questions on Alkenes

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