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Alkenes. C n H 2n. Alkenes. contain carbon - carbon double bonds. 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. Degree of unsaturation.
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Alkenes CnH2n
Alkenes • contain carbon - carbon double bonds • called unsaturated hydrocarbons • also known as olefins (oleum, latin, oil; facere, latin, make) • CnH2n • CnH2n + H2 CnH2n+2 - one degree of unsaturation
Degree of unsaturation Degree of unsaturation = (2NC - NX + NN – NH + 2)/2 NC = number of carbons NX = number of halogens NN = number of nitrogens NH = number of hydrogens
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
Nomenclature – the E/Z system 2. Number this chain so that the first carbon of the double bond has the lowest number possible.
Nomenclature – the E/Z system trans cis
cis/trans problems This molecule is a 1-bromo-1-chloropropene but is it cis or trans!
Nomenclature – the E/Z system (Z)-1-bromo-1-chloropropene • use the Cahn-Ingold-Prelog system to assign priorities to the two groups on each carbon of the double bond. • 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).
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.
Synthesis of alkenes by elimination reactions dehydrohalogenation: dehydration:
Dehydrohalogenation of alkyl halides Reactivity: RX 3o > 2o > 1o a 1,2 elimination reaction
Dehydrohalogenation of alkyl halides - no rearrangement
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.
E1 mechanism slow fast
Evidence for the E1 mechanism • Follows first order kinetics • Same structural effects on reactivity as for SN1 reactions - 3 > 2 > 1 • Rearrangements can occur indicative of the formation of carbocations
Evidence for the E2 mechanism • The reaction follows second order kinetics • There are no rearrangements • There is a large deuterium isotope effect • There is an anti periplanar geometry requirement
Isotope effects A difference in rate due to a difference in the isotope present in the reaction system is called an isotope effect.
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.
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.
Further evidence for the E2 mechanism RI > RBr > RCl > RF
Orientation and reactivity KOH CH CH CHCH CH CH=CHCH 3 2 3 3 3 C H OH Cl 2 5 80% The ease of alkene formation follows the sequence:- R2C=CR2 > R2C=CHR > R2C=CH2, RHC=CHR > RHC=CH2 This is also the order of alkene stability. Therefore the more stable the alkene formed, the faster it is formed. Why?
Orientation and reactivity 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.
anti elimination KOH ?
Formation of the less substituted alkene Dehydrohalogenation using a bulky base favours the formation of the less substituted alkene:
Substitution vs elimination SN2 v E2 substitution elimination
Substitution vs elimination SN1 v E1
The Zaitsev product predominates The transition state explains the orientation: