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Organic Chemistry : E-Z Isomerism
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OmarEdExcel
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#1
Hello TSR,
I hope you are all alright. Recently, I have encountered a problem regarding organic chemistry ; specifically the orientation of alkenes with the display of E-Z isomerism.
Furthermore, the problem is not about when a compound is named E or Z, but the problem is why some compounds cannot form this type of display?
A decent example would be that but-2-ene forms E-Z isomerism. However, but-1-ene does not, why so?
This is how a but-1-ene looks like.
http://prntscr.com/lxbt0n
This is how a but-2-ene looks like :
http://prntscr.com/lxbuzu
Now, why but-1-ene doesn't form E-Z isomerism? This is how I tried to make but-1-ene form an E-Z isomerism :
http://prntscr.com/lxbwmx
In fact, But-1-ene is not the only compound that doesn't form isomerism, 2-methylpentane and methylprop-1-ene do not as well.
Regardless, kindly explain why they do not.
Thank you in advance, your cooperation will be highly appreciated.
Warm regards,
OmarEdExcel.
I hope you are all alright. Recently, I have encountered a problem regarding organic chemistry ; specifically the orientation of alkenes with the display of E-Z isomerism.
Furthermore, the problem is not about when a compound is named E or Z, but the problem is why some compounds cannot form this type of display?
A decent example would be that but-2-ene forms E-Z isomerism. However, but-1-ene does not, why so?
This is how a but-1-ene looks like.
http://prntscr.com/lxbt0n
This is how a but-2-ene looks like :
http://prntscr.com/lxbuzu
Now, why but-1-ene doesn't form E-Z isomerism? This is how I tried to make but-1-ene form an E-Z isomerism :
http://prntscr.com/lxbwmx
In fact, But-1-ene is not the only compound that doesn't form isomerism, 2-methylpentane and methylprop-1-ene do not as well.
Regardless, kindly explain why they do not.
Thank you in advance, your cooperation will be highly appreciated.
Warm regards,
OmarEdExcel.
Last edited by OmarEdExcel; 3 years ago
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username3249896
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#2
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#2
One carbon at the double bond in but-1-ene is bonded to 2 identical groups.
This is E-but-2-ene.
But-1-ene does not show E-Z isomerism
2-methylpentane cannot exhibit E-Z isomerism since it has no double bond.
methylprop-1-ene does not exhibit E-Z isomerism since one carbon at the double bond is bonded to 2 identical groups. In fact both carbon atom of the double bond are bonded to identical groups (2CH3 on one, 2H on the other)
Now, why but-1-ene doesn't form E-Z isomerism? This is how I tried to make but-1-ene form an E-Z isomerism :
http://prntscr.com/lxbwmx
http://prntscr.com/lxbwmx
But-1-ene does not show E-Z isomerism
2-methylpentane cannot exhibit E-Z isomerism since it has no double bond.
methylprop-1-ene does not exhibit E-Z isomerism since one carbon at the double bond is bonded to 2 identical groups. In fact both carbon atom of the double bond are bonded to identical groups (2CH3 on one, 2H on the other)
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OmarEdExcel
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#3
Hello BobbJo, thank you once again for your activity.
If I am to draw a but-1-ene, why it can't be as follows? :
http://prntscr.com/lxc2jo
If I am to draw a but-1-ene, why it can't be as follows? :
http://prntscr.com/lxc2jo
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username3249896
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#4
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#4
(Original post by OmarEdExcel)
Hello BobbJo, thank you once again for your activity.
If I am to draw a but-1-ene, why it can't be as follows? :
http://prntscr.com/lxc2jo
Hello BobbJo, thank you once again for your activity.
If I am to draw a but-1-ene, why it can't be as follows? :
http://prntscr.com/lxc2jo
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maddiesmith
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#5
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#5
Draw out 4 carbons in a row and put the double bond between the first and second, that's but-1-ene. You will see how the first carbon has 2 H attached and the second has a H and the rest of the chain. As carbon number 1 doesn't have any distinctive groups other than H, you can't say it is E or Z
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OmarEdExcel
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#6
Oh. That's interesting.
I did assume that it is butene because of the fact of having 4 carbon atoms overall. How can I distinguish between a methyl-containing carbon and another carbon atom?
I am aware that a methyl is CH3, so whenever it is there, we always consider it a methyl and not the name of the compound? Just like what you have said earlier, it is dimethylpropene and not butene, probably for the fact of having CH3? kindly answer this.
Now, the fact that but-1-ene can't show E-Z isomerism is because if we tried to attach a CH3, it will be another compound - dimethylpropene just like what happened earlier.
And therefore, the only way to do it so to attach, by max, CH2 once on each side and that won't work because butene is supposed to have the molecular formula of C4H8?
That would truly sum it up for now.
I did assume that it is butene because of the fact of having 4 carbon atoms overall. How can I distinguish between a methyl-containing carbon and another carbon atom?
I am aware that a methyl is CH3, so whenever it is there, we always consider it a methyl and not the name of the compound? Just like what you have said earlier, it is dimethylpropene and not butene, probably for the fact of having CH3? kindly answer this.
Now, the fact that but-1-ene can't show E-Z isomerism is because if we tried to attach a CH3, it will be another compound - dimethylpropene just like what happened earlier.
And therefore, the only way to do it so to attach, by max, CH2 once on each side and that won't work because butene is supposed to have the molecular formula of C4H8?
That would truly sum it up for now.
Last edited by OmarEdExcel; 3 years ago
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username3249896
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#7
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#7
(Original post by OmarEdExcel)
Oh. That's interesting.
I did assume that it is butene because of the fact of having 4 carbon atoms overall. How can I distinguish between a methyl-containing carbon and another carbon atom?
I am aware that a methyl is CH3, so whenever it is there, we always consider it a methyl and not the name of the compound? Just like what you have said earlier, it is methylpropene and not butene, probably for the fact of having CH3? kindly answer this.
Now, the fact that but-1-ene can't show E-Z isomerism is because if we tried to attach a CH3, it will be another compound - methylpropene just like what happened earlier.
And therefore, the only way to do it so to attach, by max, CH2 on either sides and that won't work because butene is supposed to have the molecular formula of C4H8?
That would truly sum it up for now.
Oh. That's interesting.
I did assume that it is butene because of the fact of having 4 carbon atoms overall. How can I distinguish between a methyl-containing carbon and another carbon atom?
I am aware that a methyl is CH3, so whenever it is there, we always consider it a methyl and not the name of the compound? Just like what you have said earlier, it is methylpropene and not butene, probably for the fact of having CH3? kindly answer this.
Now, the fact that but-1-ene can't show E-Z isomerism is because if we tried to attach a CH3, it will be another compound - methylpropene just like what happened earlier.
And therefore, the only way to do it so to attach, by max, CH2 on either sides and that won't work because butene is supposed to have the molecular formula of C4H8?
That would truly sum it up for now.
it has an alkene group so propene
it has a methyl group on second carbon so 2-methylpropene
locant is not needed so methylpropene is fine too (i.e name is methylpropene or 2-methylpropene)
It can't show E-Z isomerism because at least one carbon atom at the double bond is bonded to identical groups.
(Original post by abdulkureci)
you're drawing of but-1-ene is incorrect. the C atom on the left hand side is bonded to 2 Hydrogen atoms, and the second C atom is bonded to 1 Hydrogen and 1 C2H5, what you have drawn is not but-1-ene because the double C=C bond is not between the first and second carbon, in your case it is between the second and third carbon
you're drawing of but-1-ene is incorrect. the C atom on the left hand side is bonded to 2 Hydrogen atoms, and the second C atom is bonded to 1 Hydrogen and 1 C2H5, what you have drawn is not but-1-ene because the double C=C bond is not between the first and second carbon, in your case it is between the second and third carbon
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maddiesmith
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#8
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#8
(Original post by OmarEdExcel)
Oh. That's interesting.
I did assume that it is butene because of the fact of having 4 carbon atoms overall. How can I distinguish between a methyl-containing carbon and another carbon atom?
I am aware that a methyl is CH3, so whenever it is there, we always consider it a methyl and not the name of the compound? Just like what you have said earlier, it is methylpropene and not butene, probably for the fact of having CH3? kindly answer this.
Now, the fact that but-1-ene can't show E-Z isomerism is because if we tried to attach a CH3, it will be another compound - methylpropene just like what happened earlier.
And therefore, the only way to do it so to attach, by max, CH2 on either sides and that won't work because butene is supposed to have the molecular formula of C4H8?
That would truly sum it up for now.
Oh. That's interesting.
I did assume that it is butene because of the fact of having 4 carbon atoms overall. How can I distinguish between a methyl-containing carbon and another carbon atom?
I am aware that a methyl is CH3, so whenever it is there, we always consider it a methyl and not the name of the compound? Just like what you have said earlier, it is methylpropene and not butene, probably for the fact of having CH3? kindly answer this.
Now, the fact that but-1-ene can't show E-Z isomerism is because if we tried to attach a CH3, it will be another compound - methylpropene just like what happened earlier.
And therefore, the only way to do it so to attach, by max, CH2 on either sides and that won't work because butene is supposed to have the molecular formula of C4H8?
That would truly sum it up for now.
The way to figure out if a CH3 is a side group (Methyl) or part of the chain is by counting how many carbons are in a row, out of the molecule how long can you count the chain? If a carbon has 2 Methyl groups, it's either or isn't it? Because by going down one side of the chain that's the counting done and you don't go back and count the other Methyl too. You always aim to find the LONGEST CHAIN POSSIBLE. this is very important as this is what the name of the molecule reflects!
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