Alkenes
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Alkenes CnH2n © E.V. Blackburn, 2011 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 © E.V. Blackburn, 2011 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 NH 2 H Cl O © 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. 4 2 3 1 4-butyl-2-octene © E.V. Blackburn, 2011 Nomenclature – the E/Z system H3C H C C H CH3 trans H3C CH3 C C H H cis diastereoisomers geometric isomers © E.V. Blackburn, 2011 cis/trans problems H 3C Br C C H Cl This molecule is a 1-bromo-1-chloropropene but is it cis or trans! © E.V. Blackburn, 2011 Nomenclature – the E/Z system H 3C Br C C H Cl (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). © E.V. Blackburn, 2011 E-Z designations Br H3CH2C OCH 3 CH3 CH2Br CH2CH2CH3 H3C H3CH2C CH(CH 3)2 C(CH3)3 © 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. H3C CH3 H H H CH3 H3C H H2 catalyst H2 catalyst CH3CH2CH2CH3 H = -119.7 kJ/mol CH3CH2CH2CH3 H = -115.5 kJ/mol © E.V. Blackburn, 2011 Overall relative stabilities of alkenes H3C CH3 > H3C CH3 H3C H3C C2H5 H > H C2H5 C2H5 H H CH2CH2CH2CH3 H H > > C 2H 5 H C 2H 5 H © E.V. Blackburn, 2011 Synthesis of alkenes by elimination reactions X base - HX OH H+/ dehydrohalogenation: H dehydration: H - H2O © E.V. Blackburn, 2011 Dehydrohalogenation of alkyl halides H C C X + KOH Reactivity: RX C2H5OH + KX + H2O 3o > 2o > 1o a 1,2 elimination reaction © E.V. Blackburn, 2011 Dehydrohalogenation H C C X Br + KOH C2H5OH C2H5ONa o C2H5OH, 55 C Br + H2O + NaBr + C 2H5OH 79% C2H5ONa C2H5OH, 55 oC + KX + NaBr + C 2H5OH 91% © E.V. Blackburn, 2011 Alkoxide ions – bases used in dehydrohalogenation CH3CH2OH Na CH3CH2O- Na + sodium ethoxide (CH3)2CHOH Al ((CH3)2CHO -)3 Al 3+ aluminum isopropoxide (CH3)3COH K (CH3)3CO- K+ potassium t-butoxide © E.V. Blackburn, 2011 Dehydrohalogenation of alkyl halides CH3CH2CH2Cl CH3CH2CH2CH2Cl KOH C2H5OH KOH C2H5OH - no rearrangement KOH CH3CH2CHCH 3 C2H5OH Cl CH3CH=CH 2 CH3CH2CH=CH 2 CH3CH=CHCH 3 + 80% CH3CH2CH=CH 2 20% © 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 X C C H :B X C C H B - + HB + X - © E.V. Blackburn, 2011 E1 mechanism X C C H + C C H :B + C C H + X - slow + HB fast © E.V. Blackburn, 2011 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 © E.V. Blackburn, 2011 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 © 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. H C H + Z k C H Z C + HZ 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. C H + Z C D + Z kH kD C H Z C + HZ C D Z C + DZ kH D = 5-8 k 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 kH CH3CHCH 3 NaOEt Br kD CD3CHCD 3 NaOEt Br CH3CH=CH 2 CD3CH=CD 2 kH/kD = 7 © E.V. Blackburn, 2011 Further evidence for the E2 mechanism X C C H :B X C C H B - + HB + X - RI > RBr > RCl > RF © E.V. Blackburn, 2011 Orientation and reactivity KOH CH3CH2CHCH3 C2H5OH Cl CH3CH=CHCH3 + CH3CH2CH=CH 2 80% 20% 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? © E.V. Blackburn, 2011 Orientation and reactivity Let’s look at the transition state for the reaction: X C C H :B X C C H + HB + X B - 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. © E.V. Blackburn, 2011 anti elimination :B X H B: H X © E.V. Blackburn, 2011 anti elimination anti H3C H Cl H H anti CH(CH 3)2 KOH ? H H neomenthyl chloride + © E.V. Blackburn, 2011 anti elimination H H 3C Cl H H H KOH H ? CH(CH 3)2 menthyl chloride © E.V. Blackburn, 2011 Formation of the less substituted alkene Dehydrohalogenation using a bulky base favours the formation of the less substituted alkene: CH3 o 75 C (CH3)3CO + CH3CH2-C-Br (CH3)3COH CH3 CH3 H3CHC C CH3 + H3CH2CC CH3 27.5% 72.5% CH2 © E.V. Blackburn, 2011 Substitution vs elimination SN2 v E2 X H Nu: substitution X X H Nu: SN2 elimination H :Nu E2 © E.V. Blackburn, 2011 Substitution vs elimination SN1 v E1 X C C H + C C H - + X Nu SN1 Nu: + H H :Nu E1 © E.V. Blackburn, 2011 Substitution vs elimination RX + -OR' 1o RX + 'R C C 1o RX + CN 1o R-O-R' + X'R C C R + X- SN2 RCN + X- elimination RX = 1o 2o 3o substitution © E.V. Blackburn, 2011 Dehydration of alcohols H C C OH acid + H 2O H2SO4, Al 2O3 or H3PO4 CH3 H3C C OH CH3 H3C H H3C H + H2O © E.V. Blackburn, 2011 Dehydration of alcohols - the mechanism 1. (CH3)3C-OH + H + 2. (CH3)3C-O-H H H3C 3. H3C + H C H H + (CH3)3C-O-H H + (CH3)3C+ + H2O H3C H H3C H + + ROH2 ROH © E.V. Blackburn, 2011 Dehydration of alcohols orientation CH3CH2CH2CH2OH CH3 CH3CH2CHCH 2OH H+ + H CH3CH=CHCH 3 CH3 CH3CH=CCH 3 © 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 C C X X Zn + ZnX 2 © E.V. Blackburn, 2011 Hydrogenation of alkynes R H2 H H Lindlar's catalyst R R Na or Li H R NH 3 R H R Lindlar catalyst: Pd/CaCO 3/Pb(OAc) 2/quinoline © E.V. Blackburn, 2011 Synthesis of alkynes by elimination reactions H H C C X X H NaNH 2 KOH alcohol CH3CH=CH 2 X Br2 CH3CHBrCH 2Br C C CH3CHBrCH 2Br NaNH 2 CH3-C C H © E.V. Blackburn, 2011 1. H2, Pd/CaCO3 quinoline Na, NH3, -78oC H H H H © E.V. Blackburn, 2011
