As long as each of the sp2carbons of an alkene is bonded to only one substituent, we can use the terms cis and trans to designate the structure of the alkene: if the hydrogens are on the same side of the double bond, it is the cis isomer; if they are on opposite sides of the double bond, it is the trans isomer. But how do we designate the isomers of a compound such as 1-bromo-2-chloropropene?
1. CH3CH 2. CH3CH
CHCH2CH3 CCH3 CH3
4. CH3CH2CH CH2 3. CH3CH CHCH3
The relative priorities of the two groups bonded to an sp2carbon are determined using the following rules:
1. The relative priorities of the two groups depend on the atomic numbers of the two atoms bonded directly to the sp2carbon. The greater the atomic number, the higher the priority. For example, in the following compounds, one of the sp2 carbons is bonded to a Br and to an H:
If the atoms attached to an sp2carbon are the same, the atoms attached to the “tied”
atoms are compared; the one with the greatest atomic number belongs to the group with the higher priority.
Br has a greater atomic number than H, so Br has a higher priority than H. The other sp2carbon is bonded to a Cl and to a C. Chlorine has the greater atomic num- ber, so Cl has a higher priority than C. (Notice that you use the atomic number of C, not the mass of the CH3group, because the priorities are based on the atomic numbers of atoms, not on the masses of groups.) The isomer on the left has the high-priority groups (Br and Cl) on the same side of the double bond, so it is the Z isomer. (Zee groups are on Zee Zame Zide.) The isomer on the right has the high- priority groups on opposite sides of the double bond, so it is the E isomer.
2. If the two groups bonded to an sp2carbon start with the same atom (there is a tie), you then move outward from the point of attachment and consider the atom- ic numbers of the atoms that are attached to the “tied” atoms. For example, in the following compounds, both atoms bonded the sp2carbon on the left are C’s (in a CH2Cl group and a CH2CH2Cl group), so there is a tie.
If an atom is doubly bonded to another atom, treat it as if it were singly bonded to two of those atoms.
If an atom is triply bonded to another atom, treat it as if it were singly bonded to three of those atoms.
CH3 CHCH3
CH2OH
the Z isomer the E isomer
C C ClCH2CH2
CH3 CHCH3
CH2OH C C ClCH2CH2
ClCH2
ClCH2
the Z isomer the E isomer
CH CH2
CH2CH3
C CCH2 HC
C
HC CCH2
C
HOCH2CH2 CH2CH3
C
CH CH2 HOCH2CH2
The C of the CH2Cl group is bonded to Cl, H, H, and the C of the CH2CH2Cl group is bonded to C, H, H. Cl has a greater atomic number than C, so the CH2Cl group has the higher priority. Both atoms bonded to the other sp2carbon are C’s (from a CH2OH group and a CH(CH3)2group), so there is a tie on that side as well. The C of the CH2OH group is bonded to O, H, and H, and the C of the CH(CH3)2group is bonded to C, C, and H. Of these six atoms, O has the greatest atomic number, so CH2OH has a higher priority than CH(CH3)2. (No- tice that you do not add the atomic numbers; you consider the single atom with the greatest atomic number.) The E and Z isomers are as shown above.
3. If an atom is doubly bonded to another atom, the priority system treats it as if it were singly bonded to two of those atoms. If an atom is triply bonded to another atom, the priority system treats it as if it were singly bonded to three of those atoms. For example, one of the sp2carbons in the following pair of isomers is bonded to a CH2CH2OH group and to a CH2C‚CHgroup:
Because the atoms immediately bonded to the sp2carbon on the left are both bonded to C, H, H, we ignore them and turn our attention to the groups attached to them. One of these is CH2OH and the other is The triple-bonded C is considered to be bonded to C, C, and C; the other C is bonded to O, H, and H.
Of the six atoms, O has the greatest atomic number, so CH2OH has a higher pri- ority than Both atoms bonded to the other sp2carbon are C’s, so they are tied. The first carbon of the CH2CH3group is bonded to C, H, and H. The first carbon of the group is bonded to an H and doubly bonded to a C. Therefore, it is considered to be bonded to H, C, and C. One C cancels in each of the two groups, leaving H and H in the CH2CH3group and C and H in the group. C has a greater atomic number than H, so has a higher priority than CH2CH3.
CH“CH2 CH“CH2
CH“CH2
C‚CH.
C‚CH.
Mechanistic Tutorial:
E and Z Nomenclature
CH3 H C H3C
C BrCH2 BrCH2C
CH3 CHCH3
a. b.
PROBLEM 7♦
Assign relative priorities to each set of substituents:
a.
b.
PROBLEM 8
Draw and label the E and Z isomers for each of the following:
ơCH2CH2OH, ơOH, ơCH2Cl, ơCH“CH2
ơCH3
ơOH,
ơI,
ơBr,
a.
b.
c.
CH3CH2C CH3CH2CH2CH2
CCH2Cl CH3CHCH3 CH3CH2C CHCH2CH3
Cl
CH3CH2CH CHCH3
PROBLEM 9
Indicate whether each of the following is an E isomer or a Z isomer (black carbon, white = hydrogen, green = chlorine, and red = oxygen):
=
PROBLEM-SOLVING STRATEGY
Draw the structure of (E)-1-bromo-2-methyl-2-butene.
First draw the compound without specifying the isomer so you can see what groups are bonded to the sp2carbons. Now determine the relative priorities of the two groups on each of the sp2carbons.
The sp2carbon on the right is attached to a CH3and to an H; CH3has the higher priority. The other sp2carbon is attached to a CH3and to CH2Br; CH2Br has the higher priority. To get the E isomer, draw the compound with the two high-priority groups on opposite sides of the double bond.
Now continue on to Problem 10.
most stable
least stable C C >
R R
R R
C C
R R
R H
> C C
R H
R H
> C C
R H
H H
relative stabilities of alkyl-substituted alkenes
CH2CH3 CH2CH3
CH2CH3 CH2CH3
CH2CH3 CH2CH3
C C
H H H
H H
H
C H C
H
cis-2-butene
C C
H H
H C H
H
H H
C H
trans-2-butene
the trans isomer does not have steric strain the cis isomer has steric strain
The fewer hydrogens bonded to the sp2 carbons of an alkene, the more stable it is.
PROBLEM 10♦
Draw the structure of (Z)-3-isopropyl-2-heptene.