Just to be clear, is this not an Aldehyde?

[ipodtouch]

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I was wondering why this wasn't an aldehyde.
http://imgur.com/lnQ3m


I was working on this problem:
http://imgur.com/av5w6

And I originally chose 2 isomers because we couldn't use Aldehydes. In the solutions, it even says that there is a presence of Ester bond, but NOT aldehydes or Ketones. But apparently I was wrong as it uses 4 isomers, including the compound in the first picture.
I understand that it's also an ester. Is there some kind of functional group hierarchy? where if it's an ester it cancels out the aldehyde identity?

 

[milski]

It's an ester. Ether/ester/aldehyde/ketone are mutually exclusive - you cannot call an ester aldehyde or an ether.

 

[MedPR]

I was wondering why this wasn't an aldehyde.
http://imgur.com/lnQ3m


I was working on this problem:
http://imgur.com/av5w6

And I originally chose 2 isomers because we couldn't use Aldehydes. In the solutions, it even says that there is a presence of Ester bond, but NOT aldehydes or Ketones. But apparently I was wrong as it uses 4 isomers, including the compound in the first picture.
I understand that it's also an ester. Is there some kind of functional group hierarchy? where if it's an ester it cancels out the aldehyde identity?

Not really sure how to answer your question. An ester is an ester and an aldehyde is an aldehyde. I"m not sure what you mean by "cancels out"

Aldehyde = carbonyl bonded to an alkyl group and a hydrogen
Ester = carbonyl bonded to an alkyl and a alkoxy

In other words, in that first picture, if you were to erase that oxygen (the non-carbonyl oxygen) you would have an aldehyde.

For your second picture, the 1750 peak means it is an ester, as indicated by the solution. The four isomers are four different esters. An aldehyde is not an option here because there is not a peak indicative of an aldehyde. You might need to go back and review some definitions. "4 isomers" does not mean 1 ester, 1 ether, 1 ketone, 1 aldehyde.. In this case it means 4 different structures all with formula C4H8O2. The functional group can be determined from the IR data.

 

[ipodtouch]

Oh sorry

I thought an Aldehyde was just a carbonyl bonded to an H on one side.

 

[SaintJude]

Sorry for what?! 🙂 Regarding your initial hunch about hierachy. It's not "technically correct" but I understand what you are saying. And you're right. It's the same with carboxylic acid, you can see an OH but it's part of larger, common functional group--so you identify it as a COOH. So here it's an ester that "overtake" the aldehyde identity.

Ipodtouch, how did they find the 4 structural isomers? Most questions should take like 60-70 seconds to answer and I can't answer this that quickly + I also got it wrong.

 

[MedPR]

Sorry for what?! 🙂 Regarding your initial hunch about hierachy. It's not "technically correct" but I understand what you are saying. And you're right. It's the same with carboxylic acid, you can see an OH but it's part of larger, common functional group--so you identify it as a COOH. So here it's an ester that "overtake" the aldehyde identity.

Ipodtouch, how did they find the 4 structural isomers? Most questions should take like 60-70 seconds to answer and I can't answer this that quickly + I also got it wrong.

Make the ester functional group, then rearrange the alkyl chain.

So they are:

HCOOprop
HCOOisoprop
EthCOOmet
MetCOOEth

 

[Godric]

Not really sure how to answer your question. An ester is an ester and an aldehyde is an aldehyde. I"m not sure what you mean by "cancels out"

Aldehyde = carbonyl bonded to an alkyl group and a hydrogen
Ester = carbonyl bonded to an alkyl and a alkoxy

In other words, in that first picture, if you were to erase that oxygen (the non-carbonyl oxygen) you would have an aldehyde.

For your second picture, the 1750 peak means it is an ester, as indicated by the solution. The four isomers are four different esters. An aldehyde is not an option here because there is not a peak indicative of an aldehyde. You might need to go back and review some definitions. "4 isomers" does not mean 1 ester, 1 ether, 1 ketone, 1 aldehyde.. In this case it means 4 different structures all with formula C4H8O2. The functional group can be determined from the IR data.

@MedPR bulls-eye! He/She is right, I remember that chapter. Its the isomers, and I believe this was a in chapter example problem. And they told you certain spectra features, and none of them indicated a aldehyde. Make sure you dont forget to incorporate spectra data, it can save valuable time (and makes the difference in getting the problem right or wrong)