sgreen Posted February 7, 2013 Posted February 7, 2013 I am confused with IUPAC naming with more functional group -CHO, -COOH. What is the IUPAC naming of the following compound? CH3-CH2-CH(CHO)-CH2-COOH . I think it's 3-alpentanoic acid. Please help me to choose the right one.
Iota Posted February 7, 2013 Posted February 7, 2013 I don't know, myself. But you might consider downloading 'Accelrys Draw', it's a free chemistry tool on which you can draw a molecule, then have it generate IUPAC names, MMRs etc. Takes a while to get used to using but it's very useful, we use it at university.
sgreen Posted February 7, 2013 Author Posted February 7, 2013 (edited) I don't know, myself. But you might consider downloading 'Accelrys Draw', it's a free chemistry tool on which you can draw a molecule, then have it generate IUPAC names, MMRs etc. Takes a while to get used to using but it's very useful, we use it at university. Thanks formyl pentanoic acid Thanks .Is it possible 3-ethyle-4-oxobutanoic acid or 3-alpentanoic acid? Please, please ensure . Edited February 7, 2013 by sgreen
John Cuthber Posted February 7, 2013 Posted February 7, 2013 An oxo pentanoic acid would have a ketone (rather than an aldehyde) group as well as an acid group. Al is only used as a suffix, not a prefix
sgreen Posted February 8, 2013 Author Posted February 8, 2013 An oxo pentanoic acid would have a ketone (rather than an aldehyde) group as well as an acid group. Al is only used as a suffix, not a prefix In case of Aldehyde, what is prefix except formyl- replacing in 3-alpentanoic acid?
hypervalent_iodine Posted February 8, 2013 Posted February 8, 2013 Well, oxo really applies to any group with a =O in it, so were it another type of aldehyde attached to that structure, you would use oxo. For instance, if we had something like this: The carboxylic acid is still the parent functional group, which means it still ends in -oic acid. The aldehyde is at the 6 position and the ethyl group is at position 4, making the full name of the compound, 4-ethyl-6-oxohexanoic acid. Edit: I agree with John that 2-formylpentanoic acid would be the name you would use for your compound, although I also think that 2-(oxomethyl)-pentanoic acid would be correct.
sgreen Posted February 8, 2013 Author Posted February 8, 2013 Well, oxo really applies to any group with a =O in it, so were it another type of aldehyde attached to that structure, you would use oxo. For instance, if we had something like this: The carboxylic acid is still the parent functional group, which means it still ends in -oic acid. The aldehyde is at the 6 position and the ethyl group is at position 4, making the full name of the compound, 4-ethyl-6-oxohexanoic acid. Edit: I agree with John that 2-formylpentanoic acid would be the name you would use for your compound, although I also think that 2-(oxomethyl)-pentanoic acid would be correct. CH3-CH2-CH(CHO)-CH2-COOH.... exactly 3-formylpentanoic acid or 2-formylpentanoic acid. And how it can be 2-(oxomethyl)-pentanoic acid? Would you please elaborate? IF YOU HAVE TIME..PLEASE
hypervalent_iodine Posted February 8, 2013 Posted February 8, 2013 Sorry, I had meant to say 3-formylpentanoic acid / 3-(oxomethyl)-pentanoic acid.
sgreen Posted February 8, 2013 Author Posted February 8, 2013 Sorry, I had meant to say 3-formylpentanoic acid / 3-(oxomethyl)-pentanoic acid. Thanks a lot. I have unterstood 3-formylpentanoic acid. But why 3-(oxomethyl)-pentanoic acid? If you give the answer in more detail, it will be better for me.
hypervalent_iodine Posted February 8, 2013 Posted February 8, 2013 The oxo prefix simply refers to =O. Oxomethyl means you have a =O attached to a methyl group, giving you the -CHO group. Having 3-(oxomethyl) means that the -CHO is attached to the 3 position, as in your compound.
sgreen Posted February 8, 2013 Author Posted February 8, 2013 The oxo prefix simply refers to =O. Oxomethyl means you have a =O attached to a methyl group, giving you the -CHO group. Having 3-(oxomethyl) means that the -CHO is attached to the 3 position, as in your compound. Thanks for quick reply. Now I get the clear understanding.
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