Asked by Perry
Hello!
I have a question that leaving me stumped. So E means opposite and Z means together. In order to do this alternate cis/trans arrangement you have to order molecules a priority either 1 and 2. You have to go along either side of double bond and determine priority. I'm so lost with these molecules.
here is what I came up with
First one is E
Second: don't know how to differientiate with one F and the Cl's
Third: don't know if this is even a E, Z
Fourth: Definitely Z
Sixth: Unsure between the F on one side and C on the other. Does this mean Z.
here are the molecules.
h t t p://img242.imageshack.us/img242
/933/picture2jt7.png
you'll have to copy paste it.
Please HElp!!!
Thanks!!
I have a question that leaving me stumped. So E means opposite and Z means together. In order to do this alternate cis/trans arrangement you have to order molecules a priority either 1 and 2. You have to go along either side of double bond and determine priority. I'm so lost with these molecules.
here is what I came up with
First one is E
Second: don't know how to differientiate with one F and the Cl's
Third: don't know if this is even a E, Z
Fourth: Definitely Z
Sixth: Unsure between the F on one side and C on the other. Does this mean Z.
here are the molecules.
h t t p://img242.imageshack.us/img242
/933/picture2jt7.png
you'll have to copy paste it.
Please HElp!!!
Thanks!!
Answers
Answered by
GK
Your URL is not properly typed. It should have been:
http://img242.imageshack.us/img242/933/picture2jt7.png
(Typed, not pasted. Your URL has spaces and carriage returns that should not be there)
The high priority groups on the 2nd formula are:
C3H5Cl2 on the left side of the double bond and F on the right side. It is an E isomer.
The four groups on the 3rd formula are: CH3 and CH3 on the left side of the double bond. H and F on the right. You assign priorities and decide the type of isomer.
The four groups on the last formula are:
C5H11 and C3H7 on the left. H and F on the right. You assign priorities and decide isomer type.
Your first and fourth formulas are labeled correctly.
http://img242.imageshack.us/img242/933/picture2jt7.png
(Typed, not pasted. Your URL has spaces and carriage returns that should not be there)
The high priority groups on the 2nd formula are:
C3H5Cl2 on the left side of the double bond and F on the right side. It is an E isomer.
The four groups on the 3rd formula are: CH3 and CH3 on the left side of the double bond. H and F on the right. You assign priorities and decide the type of isomer.
The four groups on the last formula are:
C5H11 and C3H7 on the left. H and F on the right. You assign priorities and decide isomer type.
Your first and fourth formulas are labeled correctly.
Answered by
Perry
Hi GK!
Does this mean that there are 4 molecules with the E configuration? All but molecules four are E configurations?
Thanks So Much For Your HElp!
Does this mean that there are 4 molecules with the E configuration? All but molecules four are E configurations?
Thanks So Much For Your HElp!
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