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A-level chemistry help

Hi,
I was wondering if anyone could help me name the attached compound below - I'm just not sure how to go about it when there is a chain attached to a ring.
Thank you :smile:

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Reply 1
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Lydia.Mx
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IMG_7923.JPG
You don't seem to have attached anything?
Original post by Lydia.Mx
Hi,
I was wondering if anyone could help me name the attached compound below - I'm just not sure how to go about it when there is a chain attached to a ring.
Thank you :smile:
Reply 3
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Lydia.Mx
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Sorry it wouldn't load so had to do it separately - should be done now!
Original post by AmazingPhysicist
You don't seem to have attached anything?
Benzyl ethanoic acid
Ok, so you've got this compound. Focus on the straight chain first, it's got two carbons and an aldehyde functional group. So part of the name is ethanal. There is a benzene ring attached to the second carbon, so this is referred to as a phenyl group because it isn't the main part. So the name is 2 phenyl ethanal.
Original post by Lydia.Mx
IMG_7923.JPG
Phenylacetaldehyde
Yes I can confirm this name is another alternative
Original post by ethancruise15
Phenylacetaldehyde
Reply 8
Avatar for kaiju78
kaiju78
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1-phenylethanal is the IUPAC name
Original post by Lydia.Mx
Hi,
I was wondering if anyone could help me name the attached compound below - I'm just not sure how to go about it when there is a chain attached to a ring.
Thank you :smile:
Reply 9
Avatar for Lydia.Mx
Lydia.Mx
OP
14
Original post by AmazingPhysicist
Ok, so you've got this compound. Focus on the straight chain first, it's got two carbons and an aldehyde functional group. So part of the name is ethanal. There is a benzene ring attached to the second carbon, so this is referred to as a phenyl group because it isn't the main part. So the name is 2 phenyl ethanal.

Okay, so is there a reason for it being called a phenyl group? Sorry I've just started Yr 13 chemistry and we've started to look at carbonyls first so am a bit confused. The rest of the name makes sense!

*I've just looked on chem guide and I understand now so because the chain is attached to the benzene ring it becomes a phenyl group as a hydrogen is effectively being 'lost', well not lost you've just removed one of the hydrogens to become C6H5 instead of C6H6?
(edited 4 years ago)
Reply 10
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Lydia.Mx
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Thank you :smile:
Original post by ethancruise15
Phenylacetaldehyde
Reply 11
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Lydia.Mx
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Great, thank you :smile:
Original post by kaiju78
1-phenylethanal is the IUPAC name
Yes that it correct
Original post by Lydia.Mx
Okay, so is there a reason for it being called a phenyl group? Sorry I've just started Yr 13 chemistry and we've started to look at carbonyls first so am a bit confused. The rest of the name makes sense!

*I've just looked on chem guide and I understand now so because the chain is attached to the benzene ring it becomes a phenyl group as a hydrogen is effectively being 'lost', well not lost you've just removed one of the hydrogens to become C6H5 instead of C6H6?
Reply 13
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Lydia.Mx
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Great, thanks so much for your help, I really appreciate it :smile:
Original post by AmazingPhysicist
Yes that it correct
You're very welcome, if you need any other help just let me know.
Original post by Lydia.Mx
Great, thanks so much for your help, I really appreciate it :smile:
Reply 15
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Lydia.Mx
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Original post by AmazingPhysicist
You're very welcome, if you need any other help just let me know.

I have another question that I'm rather stuck on - would you mind having a look for me? I'm so sorry!IMG_7925.jpg
It's 16c - ii) I've got i but just don't know what it would make in ii.
I believe the OH directly attached to the benzene ring would be converted to O-Na+
Original post by Lydia.Mx
I have another question that I'm rather stuck on - would you mind having a look for me? I'm so sorry!IMG_7925.jpg
It's 16c - ii) I've got i but just don't know what it would make in ii.
What did you guys get for part 1? :smile: ( I'm not too sure myself)
What do you mean part 1?
Original post by hoiler2000
What did you guys get for part 1? :smile: ( I'm not too sure myself)
c part one or is that your answer ^^ with the O-NA
Original post by AmazingPhysicist
What do you mean part 1?

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