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Substitution and Elimination Competing Reactions S N 1 & S N 2 vs. E1 and E2 Formation of Alkenes. –. +. :. H. Y. X. C. C. –. :. Y. C. C. –. +. :. X. C. C. Two Reaction Types.
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Substitution and EliminationCompeting ReactionsSN1 & SN2 vs. E1 and E2Formation of Alkenes
– + : H Y X C C – : Y C C – + : X C C Two Reaction Types Alkyl halides can react with Lewis bases in two different ways; nucleophilic substitution or elimination. b-elimination + H + H X Y nucleophilic substitution
– + : H Y X C C – : Y C C – + : X C C Two Reaction Types How can we tell which reaction pathway is followedfor a particular alkyl halide? b-elimination + H + H X Y nucleophilic substitution
Elimination versus Substitution A systematic approach is to choose as a referencepoint the reaction followed by a typical alkyl halide(secondary) with a typical Lewis base (an alkoxideion). • The major reaction of a secondary alkyl halidewith an alkoxide ion is elimination by the E2mechanism.
CH3CHCH3 Br CH3CHCH3 OCH2CH3 Example NaOCH2CH3 ethanol, 55°C + CH3CH=CH2 (87%) (13%)
– •• CH3CH2 O •• •• Figure 1 E2 Br
Regioselectivity of the E2 Reaction The major product of an E2 reaction is the most stable alkene The greater the number of substituents, the more stable is the alkene
Reaction coordinatediagram for the E2 reaction of 2-bromobutane and methoxide ion
– •• CH3CH2 O •• •• Figure 2 SN2 Br
When is substitution favored? Given that the major reaction of a secondaryalkyl halide with an alkoxide ion is elimination by the E2 mechanism, we can expect the proportion of substitution to increase with: • 1) decreased crowding at the carbon that bears the leaving group
Uncrowded Alkyl Halides Decreased crowding at carbon that bears the leaving group increases substitution relative to elimination. • primary alkyl halide CH3CH2CH2Br NaOCH2CH3 ethanol, 55°C + CH3CH2CH2OCH2CH3 CH3CH=CH2 (9%) (91%)
But a crowded alkoxide base can favor elimination even with a primary alkyl halide. • primary alkyl halide + bulky base CH3(CH2)15CH2CH2Br KOC(CH3)3 tert-butyl alcohol, 40°C CH3(CH2)15CH2CH2OC(CH3)3 + CH3(CH2)15CH=CH2 (13%) (87%)
When is substitution favored? Given that the major reaction of a secondaryalkyl halide with an alkoxide ion is elimination by the E2 mechanism, we can expect the proportion of substitution to increase with: • 1) decreased crowding at the carbon that bears the leaving group • 2) decreased basicity of the nucleophile
CH3CH(CH2)5CH3 Cl CH3CH(CH2)5CH3 CN Weakly Basic Nucleophile Weakly basic nucleophile increases substitution relative to elimination secondary alkyl halide + weakly basic nucleophile KCN pKa (HCN) = 9.1 DMSO (70%)
N3 Weakly Basic Nucleophile Weakly basic nucleophile increases substitution relative to elimination secondary alkyl halide + weakly basic nucleophile I NaN3 pKa (HN3) = 4.6 (75%)
Tertiary Alkyl Halides Tertiary alkyl halides are so sterically hinderedthat elimination is the major reaction with allanionic nucleophiles. Only in solvolysis reactionsdoes substitution predominate over eliminationwith tertiary alkyl halides.
(CH3)2CCH2CH3 Br CH3 CH3 CH3 CH3CCH2CH3 CH3C=CHCH3 CH2=CCH2CH3 OCH2CH3 ethanol, 25°C 36% 64% Example + + 2M sodium ethoxide in ethanol, 25°C 99% 1%