Electrogeek
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Hello everyone,

I'm doing this worksheet (in attachment) and I'm having trouble at coming up with the name of the last four chemicals - I know whether they're E or Z isomers, but I don't know how to do the rest of their names...

Can anyone help me with a couple please?
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RMNDK
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(Original post by Electrogeek)
Hello everyone,

I'm doing this worksheet (in attachment) and I'm having trouble at coming up with the name of the last four chemicals - I know whether they're E or Z isomers, but I don't know how to do the rest of their names...

Can anyone help me with a couple please?
O.oer these are a bit mean.

I'll go through 4 and see if you can do the rest.

One way to name it is to name is based off the alkene. This is because alcohols have the prefix: hydroxy- whereas alkenes have no prefix.

To get the lowest numbering, you'd want to start at the carbon with the alcohol group.
Therefore this molecule is 1-hydroxy-2-methylpent-2-ene.

But perhaps the more correct way (which you might not know) is:
It's both an alcohol and an alkene, which comes under a new group called enols

The reason why this is more correct is that the alcohol functional group (-OH) has a higher priority than an alkene functional group and so it should be the suffix of the name. So we use both suffixes.
Spoiler:
Show
Alkenes are one of the lowest in priority.
We still start numbering at the carbon with the alcohol group as it takes priority.
To name this molecule would be 2-methyl-2-penten-1-ol

Here's a video if you're unsure: https://www.youtube.com/watch?v=NrjxBCwxalQ

5 is like 4 but you've got a chloro group.

In number 6, the main functional group, if you didn't know, is an aldehyde and it also has a higher priority than the alkene functional group (I don't think it even has a prefix anyway). If alcohol/alkene hybrids are called enols, what could aldehyde/alkene hybrids be called?

Apply similar reasoning to number 7.

If you get it, great! If not don't worry too much.


I hope someone else responds so that I'm not wrong xD
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Electrogeek
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Thank you so much! So.. For number 5, do halides go first in the name, making it 3-chloro-2-methyl-2-penten-1-ol?
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charco
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(Original post by RMNDK)
O.oer these are a bit mean.

I'll go through 4 and see if you can do the rest.

One way to name it is to name is based off the alkene. This is because alcohols have the prefix: hydroxy- whereas alkenes have no prefix.

To get the lowest numbering, you'd want to start at the carbon with the alcohol group.
Therefore this molecule is 1-hydroxy-2-methylpent-2-ene.

But perhaps the more correct way (which you might not know) is:
It's both an alcohol and an alkene, which comes under a new group called enols

The reason why this is more correct is that the alcohol functional group (-OH) has a higher priority than an alkene functional group and so it should be the suffix of the name. So we use both suffixes.
Spoiler:
Show
Alkenes are one of the lowest in priority.
We still start numbering at the carbon with the alcohol group as it takes priority.
To name this molecule would be 2-methyl-2-penten-1-ol

Here's a video if you're unsure: https://www.youtube.com/watch?v=NrjxBCwxalQ

5 is like 4 but you've got a chloro group.

In number 6, the main functional group, if you didn't know, is an aldehyde and it also has a higher priority than the alkene functional group (I don't think it even has a prefix anyway). If alcohol/alkene hybrids are called enols, what could aldehyde/alkene hybrids be called?

Apply similar reasoning to number 7.

If you get it, great! If not don't worry too much.


I hope someone else responds so that I'm not wrong xD
Here's a priorities table to help:

Name:  nomenclature-priority.png
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Electrogeek
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(Original post by charco)
Here's a priorities table to help:

Name:  nomenclature-priority.png
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Thanks for the table. For number 7 am I right in saying that there's an ester group connected to the chain (I have absolutly no idea for number 7)?

Also, why is there a aldehyde group in number 6 - wouldn't they all be single bonds?
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RMNDK
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(Original post by Electrogeek)
Thanks for the table. For number 7 am I right in saying that there's an ester group connected to the chain (I have absolutly no idea for number 7)?

Also, why is there a aldehyde group in number 6 - wouldn't they all be single bonds?
My mistake, I misread 6 and thought it was HOC. In that case it wouldn't be an aldehyde it would only be an alcohol. Makes it easier.

You're right, 7 is an ester (that's a rather tricky compound)
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Electrogeek
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Am I right in saying that number 6 is 5-chloro-3-chloromethyl-2-ethylpent-2-en-1-ol?

And number 7 is 3-chloromethyl-2-hydroxylmethyl-but-2-enyl methanoate?
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RMNDK
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(Original post by Electrogeek)
Am I right in saying that number 6 is 5-chloro-3-chloromethyl-2-ethylpent-2-en-1-ol?

And number 7 is 3-chloromethyl-2-hydroxylmethyl-but-2-enyl methanoate?
I'm no mark scheme, but I agree with you on number 6.

Number 7, I think you've got the naming of esters the wrong way round.
The R-COO- portion is the -oate part (the carboxylate) The alkyl group that's bonded to the oxygen is your beginning part.

In number 7 we can see that it is based off butanoic acid. The alkyl group at the end is a methyl group.

Therefore, it's methyl butanoate. (Of course that's without all the substituent groups.)
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Electrogeek
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So all the substituent groups go on the butanoate part of the name?
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KombatWombat
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Pedant's corner: you haven't got any enols there! Enols have the hydroxy directly attached to the alkene, and it turns out they're pretty unstable and turn into a ketone or aldehyde. You don't need to know this, but if you think about it I bet you've never come across an enol - this is why!

I agree with 6, but I think 7 would be methyl (Z)-4-chloro-2-hydroxymethyl-3-methylbut-2-enoate.
Chloromethyl and methyl groups are both the same chain length, so you put chloromethyl into the main chain.

BUT what I would say is its a bit pointless writing out the full names of these! The question sheet just wants you to practice Cahn–Infold–Prolog rules to decide E or Z, not to work out the really long and obtuse systematic name. You won't use these very long systematic names much if you do chemistry at uni, they're not that useful. Stick to the sort of thing you'd get in an exam!
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Electrogeek
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(Original post by KombatWombat)
Pedant's corner: you haven't got any enols there! Enols have the hydroxy directly attached to the alkene, and it turns out they're pretty unstable and turn into a ketone or aldehyde. You don't need to know this, but if you think about it I bet you've never come across an enol - this is why!

I agree with 6, but I think 7 would be methyl (Z)-4-chloro-2-hydroxymethyl-3-methylbut-2-enoate.
Chloromethyl and methyl groups are both the same chain length, so you put chloromethyl into the main chain.

BUT what I would say is its a bit pointless writing out the full names of these! The question sheet just wants you to practice Cahn–Infold–Prolog rules to decide E or Z, not to work out the really long and obtuse systematic name. You won't use these very long systematic names much if you do chemistry at uni, they're not that useful. Stick to the sort of thing you'd get in an exam!
Our teacher just set the rest of the name as an extension and challenge to practice using the IUPAC priority rules... Thanks for the help though.
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RMNDK
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(Original post by Electrogeek)
So all the substituent groups go on the butanoate part of the name?
Yes the substituent groups go on the butanoate part.
Spoiler:
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Hint: It would actually be called but-2-enoate, most sources, however, say that you should put the number in front so 2-butenoate. But honestly this is far too complicated for any A-Level syllabus and you need to dig deep into the IUPAC rules.
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Electrogeek
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(Original post by RMNDK)
Yes the substituent groups go on the butanoate part.
Spoiler:
Show
Hint: It would actually be called but-2-enoate, most sources, however, say that you should put the number in front so 2-butenoate. But honestly this is far too complicated for any A-Level syllabus and you need to dig deep into the IUPAC rules.
We were told to do but-2-enoate anyway... Thanks for the help though.
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RMNDK
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(Original post by Electrogeek)
We were told to do but-2-enoate anyway... Thanks for the help though.
That's fine, it's still correct. No problem.
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