© E.V. Blackburn, 2011 Alkenes C n H 2n

© E.V. Blackburn, 2011 Alkenes called unsaturated hydrocarbons also known as olefins (oleum, latin, oil; facere, latin, make) C n H 2n C n H 2n + H 2  C n H 2n+2 - one degree of unsaturation contain carbon - carbon double bonds

© E.V. Blackburn, 2011 Degree of unsaturation Degree of unsaturation = (2N C - N X + N N – N H + 2)/2 N C = number of carbons N X = number of halogens N N = number of nitrogens N H = number of hydrogens

© E.V. Blackburn, 2011 Nomenclature – the E/Z system 1. To name alkenes, select the longest carbon chain which includes the carbons of the double bond. Remove the -ane suffix from the name of the alkane which corresponds to this chain. Add the suffix -ene. a derivative of octene not nonane

© E.V. Blackburn, 2011 Nomenclature – the E/Z system 2. Number this chain so that the first carbon of the double bond has the lowest number possible.

© E.V. Blackburn, 2011 Nomenclature – the E/Z system transcis

© E.V. Blackburn, 2011 This molecule is a 1-bromo-1-chloropropene but is it cis or trans! cis/trans problems

© E.V. Blackburn, 2011 then compare the relative positions of the groups of higher priority on these two carbons. if the two groups are on the same side, the compound has the Z configuration (zusammen, German, together). if the two groups are on opposite sides, the compound has the E configuration (entgegen, German, across). (Z)-1-bromo-1-chloropropene Nomenclature – the E/Z system use the Cahn-Ingold-Prelog system to assign priorities to the two groups on each carbon of the double bond.

© E.V. Blackburn, 2011 E-Z designations

© E.V. Blackburn, 2011 Relative stabilities of alkenes Cis isomers are generally less stable than trans isomers due to strain caused by crowding of the two alkyl groups on the same side of the double bond Stabilities can be compared by measuring heats of hydrogenation of alkenes.

© E.V. Blackburn, 2011 Overall relative stabilities of alkenes

© E.V. Blackburn, 2011 Synthesis of alkenes by elimination reactions dehydrohalogenation: dehydration:

© E.V. Blackburn, 2011 Dehydrohalogenation of alkyl halides a 1,2 elimination reaction Reactivity: RX 3 o > 2 o > 1 o

© E.V. Blackburn, 2011 Dehydrohalogenation

© E.V. Blackburn, 2011 Alkoxide ions – bases used in dehydrohalogenation

© E.V. Blackburn, 2011 Dehydrohalogenation of alkyl halides - no rearrangement

© E.V. Blackburn, 2011 The mechanism In the presence of a strong base, the reaction follows second order kinetics: rate = k[RX][B - ] However, with weak bases at low concentrations and as we move from a primary halide to a secondary and a tertiary, the reaction becomes first order. There are two mechanisms for this elimination: E1 and E2.

© E.V. Blackburn, 2011 E2 mechanism

© E.V. Blackburn, 2011 E1 mechanism slow fast

© E.V. Blackburn, 2011 Evidence for the E1 mechanism Same structural effects on reactivity as for S N 1 reactions - 3  > 2  > 1  Rearrangements can occur indicative of the formation of carbocations Follows first order kinetics

© E.V. Blackburn, 2011 Evidence for the E2 mechanism There are no rearrangements There is a large deuterium isotope effect There is an anti periplanar geometry requirement The reaction follows second order kinetics

© E.V. Blackburn, 2011 Isotope effects A difference in rate due to a difference in the isotope present in the reaction system is called an isotope effect.

© E.V. Blackburn, 2011 Isotope effects If an atom is less strongly bonded in the transition state than in the starting material, the reaction involving the heavier isotope will proceed more slowly. The isotopes of hydrogen have the greatest mass differences. Deuterium has twice and tritium three times the mass of protium. Therefore deuterium and tritium isotope effects are the largest and easiest to determine.

© E.V. Blackburn, 2011 Primary isotope effects These effects are due to breaking the bond to the isotope. Thus the reaction with protium is 5 to 8 times faster than the reaction with deuterium.

© E.V. Blackburn, 2011 Evidence for the E2 mechanism - a large isotope effect

© E.V. Blackburn, 2011 RI > RBr > RCl > RF Further evidence for the E2 mechanism

© E.V. Blackburn, 2011 Orientation and reactivity CH 3 2 CHCH 3 Cl KOH C 2 H 5 OH CH 3 CH=CHCH 3 80% The ease of alkene formation follows the sequence:- R 2 C=CR 2 > R 2 C=CHR > R 2 C=CH 2, RHC=CHR > RHC=CH 2 This is also the order of alkene stability. Therefore the more stable the alkene formed, the faster it is formed. Why?

© E.V. Blackburn, 2011 Let’s look at the transition state for the reaction: The double bond is partially formed in the transition state and therefore the transition state resembles an alkene. Thus the factors which stabilize alkenes will stabilize this nascent alkene. A Zaitsev elimination. Orientation and reactivity

© E.V. Blackburn, 2011 anti elimination

© E.V. Blackburn, 2011 ? anti elimination KOH

© E.V. Blackburn, 2011 anti elimination

© E.V. Blackburn, 2011 Formation of the less substituted alkene Dehydrohalogenation using a bulky base favours the formation of the less substituted alkene:

© E.V. Blackburn, 2011 Substitution vs elimination substitutionelimination S N 2 v E2

© E.V. Blackburn, 2011 Substitution vs elimination S N 1 v E1

© E.V. Blackburn, 2011 Substitution vs elimination

© E.V. Blackburn, 2011 Dehydration of alcohols

© E.V. Blackburn, 2011 Dehydration of alcohols - the mechanism

© E.V. Blackburn, 2011 Dehydration of alcohols - orientation

© E.V. Blackburn, 2011 The Zaitsev product predominates

© E.V. Blackburn, 2011 The Zaitsev product predominates The transition state explains the orientation:

© E.V. Blackburn, 2011 Dehalogenation of vicinal dihalides

© E.V. Blackburn, 2011 Hydrogenation of alkynes

© E.V. Blackburn, 2011 Synthesis of alkynes by elimination reactions

© E.V. Blackburn, 2011