Section 11.5 Stereochemistry of E2 and E1 Reactions 415Progress of the reactionFree energy(Z)-2-pentene
(E)-2-pentene2- bromopentaneCH 3 CH 2 O−+ CH
++ 3 CH 2 OHH HH 3 C CH 2 CH 3HHH 3 CCH 2 CH 3CCCCZEBr−>Figure 11.3
Reaction coordinate diagram for
the E2 reaction of 2-bromopentane
and ethoxide ion.3-D Molecule:
3-Bromo-2,2,3-
trimethylpentaneThe more stable alkene has the more stable transition state and therefore is formed
more rapidly (Figure 11.3). In Section 4.11, we saw that the alkene with the
bulkiest groups on the sameside of the double bond is less stable because the
electron clouds of the large substituents can interfere with each other, causing
steric strain.Elimination of HBr from 3-bromo-2,2,3-trimethylpentane leads predominantly to
the Eisomer because this stereoisomer has the methyl group, the bulkiest group on
one carbon, opposite the tert-butyl group, the bulkiest group on the other
carbon.If the from which a hydrogen is to be removed is bonded to only one hy-
drogen, there is only one conformer in which the groups to be eliminated are anti.
Therefore, only one alkene product can be formed. The particular isomer that is
formed depends on the configuration of the reactant. For example, anti elimination of
HBr from (2S, 3 S)-2-bromo-3-phenylbutane forms the Eisomer, whereas anti elimina-
tion of HBr from (2S, 3 R)-2-bromo-3-phenylbutane forms the Zisomer. Notice that the
groups that are not eliminated retain their relative positions. (If you have troubleb-carbonCH 3 CH 2 C C CH 3Br CH 3
3-bromo-2,2,3-trimethyl-
pentaneCH 3 CH 3
CH 3 CH 2 O−
CH 3 CH 2 OH(E)-3,4,4-trimethyl-
2-pentene
major productCC CH 3H 3 C CH 3+H CH 3 C CH 3
(Z)-3,4,4-trimethyl-
2-pentene
minor productCC CH 3H 3 C CH 3
H 3 C CH CH 3sp^2 sp^2H 3 CCH 2 CH 3HHCCH 3 C CH 2 CH 3H H(E)-2-pentene (Z)-2-penteneinteracting electron
clouds cause steric strainCCTutorial:
E2 StereochemistryBRUI11-400_436r3 26-03-2003 10:20 AM Page 415