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Thread: E,Z Notation

  1. #1 E,Z Notation 
    Moderator Moderator AlexP's Avatar
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    I'm going through an organic chemistry book before I take the course in the fall, and I'm caught up on this one example of E,Z notation. The molecule is...

    CH<sub>3</sub>CH<sub>2</sub>C(CH<sub>3</sub>)=C(Cl)CH<sub>2</sub>CH<sub>3</sub>

    and is given the name (E)-3-chloro-4-methyl-3-hexene. I simply can't figure out why it's E and not Z, considering the methyl and chlorine substituents are on the same side of the double bond. And I'd think that it is the methyl and chlorine that are considered the substituents because the rest of it is clearly the 3-hexene part of the molecule. Hopefully from what I gave you can figure out the structural formula of the molecule. Any help would be appreciated, I really want to understand this.


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    There is the problem that the structure you have tried to reproduce is ambiguous. It can be either the cis or trans isomer, or represent either one. We cannot tell. Try retyping it as something like:

    CH3CH2C=CCH2CH3
    | |
    CH3 Cl

    or maybe it is the other isomer. In fact, maybe you can identify the above structure as either E or Z, regardless of whether it is the isomer in your question. (Recall that a chlorine atom has higher priority than a carbon atom.)

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    Let's see if this'll do it... There's actually a nice program that I have called ChemSketch, and it'd be easy to make the structural formula in it. Only problem is, I'm not sure if I can put an image from that into a post. So anyways...

    CH<sub>3</sub>CH<sub>2</sub>............CH<sub>2</sub>CH<sub>3</sub>
    ............\........../
    ..............C.=.C
    ............/..........\
    ......CH<sub>3</sub>.............Cl
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    OK, now we have it. This is the (E) isomer.

    On the left side of the double bond, the CH3CH2- has priority over the -CH3. I assume you get that. On the right side the Cl has priority over the -CH2CH3 because a chlorine atom has a higher atomic weight than a carbon atom. Since the two higher priority groups are on opposite sides (trans) of the double bond, the isomer is (E).

    See the Wiki discussion of E,Z notation over here:

    http://en.wikibooks.org/wiki/Organic...es#EZ_Notation.

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  6. #5  
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    Quote Originally Posted by SteveF
    On the left side of the double bond, the CH3CH2- has priority over the -CH3. I assume you get that.
    Not a good assumption. I didn't know that...I'm going to have to go through that section now and see if it's in there, or how I missed it. If I'd known that, I wouldn't have had the problem. So in case I still don't get it from the book, why does the ethyl get priority over the methyl?
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    This is from the Wiki article referred to in my previous post. The author also gives an example using a methyl and an ethyl group:

    "The priority is based on the atomic number of the substituents...

    "On the right side, we have carbon substituents on both the top and bottom, so we go out to the next bond. On to the top, there's another carbon*, but on the bottom, a hydrogen**. So the top gets high priority and the bottom gets low priority."

    *There's our ethyl group.
    ** And now methyl.

    Review the Wiki article and I think you'll catch on quickly to E,Z notation.

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  8. #7  
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    Ok, I've got it now, but let me know if this is a good way to look at it...

    Obviously, the Cl has priority. So from there, we want to examine the substituents on the same and opposite side of the double bond, namely the methyl group and the upper-left ethyl group. Looking at them, the element bonded directly to the C = C in both cases is carbon, so they're of equal priority. So, we next look at the next element bonded to those carbons, which in the case of the methyl group is hydrogen, but in the case of the ethyl group, another carbon, so carbon takes priority over hydrogen and the two highest-priority groups are Cl and CH<sub>3</sub>CH<sub>2</sub>.
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    Yep, you got it now.

    And, as you can see, it's all quite simple.

    Good luck with the rest of your course.

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  10. #9  
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    Thanks. I'll probably end up with more questions as I go through the book, and I'll be sure to post them on here, as you seem to know what you're doing. Thanks again.
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