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Size:  17.7 KB Hi I am uncertain about why the first molecule has a secondary carbocation but not the second molecule. The carbocation in 1 is bonded to 3 other carbon, so why is this not tertiary carbocation ? Whereas the the carbocation formed from 2, the carbocation is only attached to 2 carbon only, so why is this not the secondary carbocation?

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    The first is tertiary and the second is secondary.
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    (Original post by Pigster)
    The first is tertiary and the second is secondary.
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Size:  47.4 KB I am basically stuck on this question, this is what I am confused with. So i have deduced the molecule to be a carboxylic acid but I am struggling to work out whether it would be butanoic acid or methylpropanoic acid. I thought it would be butanoic acid as this will form a secondary cabocation but the answer is methylpropanoic acid...

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    A peak at 43 should scream C3H7+ to you. That will either be CH3CH2CH2+ OR CH3CH+CH3. The first is primary, the second is secondary.
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    (Original post by Pigster)
    A peak at 43 should scream C3H7+ to you. That will either be CH3CH2CH2+ OR CH3CH+CH3. The first is primary, the second is secondary.
    CH3CH2CH2+ -> why is this not a secondary carbocation as the second carbon atom has 2 carbon atoms on either sides..

    CH3CH+CH3 When you identify the carbocation , how do you know that the positive charge could be on the second carbon not the third carbon as either way it still produces a C3H7 + ?

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    Hi hope this helps. The positive charge is not due the loss of an atom (H). The fragment will still contain all atoms of the original part of the molecule it is take from. The positive charge comes from loss of an electron. The electron will come from part of the bond that will be broken when the fragment is formed. So positive charge will be on the carbon which was attached to the bond you removed. If you still are not sure this link explains it in more detail: http://www.chemguide.co.uk/analysis/...works.html#top[/QUOTE]
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    (Original post by coconut64)
    Name:  uiogl.png
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Size:  47.4 KB I am basically stuck on this question, this is what I am confused with. So i have deduced the molecule to be a carboxylic acid but I am struggling to work out whether it would be butanoic acid or methylpropanoic acid. I thought it would be butanoic acid as this will form a secondary cabocation but the answer is methylpropanoic acid...

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    Work out the empirical formula first. Also the molecule is an ester not a carboxylic acid. The peak is at 1740, use your wavenumber table.
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    (Original post by RevisionGuide)
    Work out the empirical formula first. Also the molecule is an ester not a carboxylic acid. The peak is at 1740, use your wavenumber table.
    The molecular ion peak is at 88, bare that in mind.
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    (Original post by RevisionGuide)
    Work out the empirical formula first. Also the molecule is an ester not a carboxylic acid. The peak is at 1740, use your wavenumber table.
    (broad) peak at 2500–3300 is O–H
    peak at 1630-1820 C=O so it is a carboxylic acid not an ester
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    I get it now thanks for the help
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    (Original post by A.K57)
    (broad) peak at 2500–3300 is O–H
    peak at 1630-1820 C=O so it is a carboxylic acid not an ester
    No you're wrong lol. A C=O bond can make the compound an aldehyde, ketone, ester amide ester etc. The peak is between 1735-1750 which is the wave number of an ester (search it). Also the peak at 3000 shows that it is an alkene (3000-3100) containing a C-H bond not O-H. This can be proven by the mass spectrometer, which does not show any peak at M/Z 17 (OH has an ar of 17). Therefore, there is no carboxylic acid or alcohol present in the compound
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    (Original post by RevisionGuide)
    No you're wrong lol. A C=O bond can make the compound an aldehyde, ketone, ester amide ester etc. The peak is between 1735-1750 which is the wave number of an ester (search it). Also the peak at 3000 shows that it is an alkene (3000-3100) containing a C-H bond not O-H. This can be proven by the mass spectrometer, which does not show any peak at M/Z 17 (OH has an ar of 17). Therefore, there is no carboxylic acid or alcohol present in the compound
    Hey Mr revision guide. I do not know what is your exam board but I actually sat this exam last year and it is ocr chemistry A June 2016 and I got this right. You dont use the mass spectrometer to find the O-H bond you use it to find the type of carbocation in the compound. Some bonds will not show up, just how it works and anyway you usually just need to look at biggest peak and molecular ion peak. Here is the mark scheme Answer

    1. Molecular formula Element % mass Ar moles ratio C 54.5/12 = 4.54
    2 H 9.1/1= 9.1 4
    O 36.4/16=2.28
     empirical formula = C2H4O
     molecular ion peak m/z or Mr = 88
     molecular formula = C4H8O2

    2. Infrared spectrum  peak at 2500–3500 (cm–1 ) is O–H
     peak at 1630-1820 (cm–1 ) is C=O
     C is a carboxylic acid ALLOW stated values within the ranges above IGNORE references to C–O peaks 3. Identifying the carboxylic acid

     (CH3CH2CH2COOH OR (CH3)2CHCOOH)
     Mass spectrum peak at m/z = 43 = C3H7( + )
     secondary carbocation: CH3C +HCH3
     compound C: (CH3)2CHCOOH IGNORE name of carboxylic acid if structure given

    Now I hope you are convinced, if not I will upload the actual question and answer for your satisfaction. I expect you to acknowledge you were wrong lol.
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    (Original post by A.K57)
    Hey Mr revision guide. I do not know what is your exam board but I actually sat this exam last year and it is ocr chemistry A June 2016 and I got this right. You dont use the mass spectrometer to find the O-H bond you use it to find the type of carbocation in the compound. Some bonds will not show up, just how it works and anyway you usually just need to look at biggest peak and molecular ion peak. Here is the mark scheme Answer

    1. Molecular formula Element % mass Ar moles ratio C 54.5/12 = 4.54
    2 H 9.1/1= 9.1 4
    O 36.4/16=2.28
     empirical formula = C2H4O
     molecular ion peak m/z or Mr = 88
     molecular formula = C4H8O2

    2. Infrared spectrum  peak at 2500–3500 (cm–1 ) is O–H
     peak at 1630-1820 (cm–1 ) is C=O
     C is a carboxylic acid ALLOW stated values within the ranges above IGNORE references to C–O peaks 3. Identifying the carboxylic acid

     (CH3CH2CH2COOH OR (CH3)2CHCOOH)
     Mass spectrum peak at m/z = 43 = C3H7( + )
     secondary carbocation: CH3C +HCH3
     compound C: (CH3)2CHCOOH IGNORE name of carboxylic acid if structure given

    Now I hope you are convinced, if not I will upload the actual question and answer for your satisfaction. I expect you to acknowledge you were wrong lol.
    As I said before, a peak between 1630 and 1820 is C=O. Many different homologous series can have a C=O bond, and not just a 'carboxylic acid' as you stated. The wave numbers of the homologous series are narrowed down between 1630 and 1820. Go ahead and upload the actual question for my 'satisfaction', so i can accept that I'm wrong lmao
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    (Original post by coconut64)
    CH3CH2CH2+ -> why is this not a secondary carbocation as the second carbon atom has 2 carbon atoms on either sides..
    You have to look at the carbon atom with the + charge on it. CH3CH2CH2+ has the + on the end carbon, which has an ethyl group attached to it, i.e. only one C directly attached, i.e. primary.

    In the case of CH3CH+CH3, the + is on the central carbon, which has two methyl groups attached, i.e secondary.

    CH3CH+CH3 When you identify the carbocation , how do you know that the positive charge could be on the second carbon not the third carbon as either way it still produces a C3H7 + ?
    The Q states that the peak is due to a secondary fragment ion.
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    (Original post by RevisionGuide)
    As I said before, a peak between 1630 and 1820 is C=O. Many different homologous series can have a C=O bond, and not just a 'carboxylic acid' as you stated. The wave numbers of the homologous series are narrowed down between 1630 and 1820. Go ahead and upload the actual question for my 'satisfaction', so i can accept that I'm wrong lmao
    Why do you have to be so stubborn you need to accept when you are wrong and learn to do it correctly rather than waste everyone's time by trying to prove you are right and everyone else isn't. The answer was from the mark scheme I clearly copied it but no you were so entrenched in your preconceived notions that you did not even try to read what I said. I would suggest revising this part again as some of your concepts are not up to scratch.
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    (Original post by A.K57)
    Why do you have to be so stubborn you need to accept when you are wrong and learn to do it correctly rather than waste everyone's time by trying to prove you are right and everyone else isn't. The answer was from the mark scheme I clearly copied it but no you were so entrenched in your preconceived notions that you did not even try to read what I said. I would suggest revising this part again as some of your concepts are not up to scratch.
    Firstly, no need to be a rude prick. Secondly, I do OCR B Salters, and you mark scheme says that 'a peak on the IR spec between 1680 and 1800 is a carboxylic acid.' We learn that that a peak between 1600 and 1800 is not just a carboxylic acid. If can be either one of these: http://www.ocr.org.uk/Images/247606-...t-specimen.pdf
    (second page).
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    (Original post by RevisionGuide)
    Firstly, no need to be a rude prick. Secondly, I do OCR B Salters, and you mark scheme says that 'a peak on the IR spec between 1680 and 1800 is a carboxylic acid.' We learn that that a peak between 1600 and 1800 is not just a carboxylic acid. If can be either one of these: http://www.ocr.org.uk/Images/247606-...t-specimen.pdf
    (second page).
    Well if the person did what you told him to he would not have gotten the marks in the exams as he does OCR A not B. You could have asked what exam board it was, but no you had to be cocky and pretend like MR knows everything. Even when I told you I sat this exam you still did not agree. Its frustrating when you have done one question like three times and some one insists they know it even when they are clearly doing it wrong according to the spec it will be marked in. Look again on page 2 of your insert go to OH and see broad peak at 2500-3300- it clearly says it is for carboxylic acid. So your different spec logic doesn't work here even. You are clearly wrong you can see the mark scheme, so you need to admit or can you not even hold on to your word? Revise the spec once more I suggest as well. LOL
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    (Original post by A.K57)
    Well if the person did what you told him to he would not have gotten the marks in the exams as he does OCR A not B. You could have asked what exam board it was, but no you had to be cocky and pretend like MR knows everything. Even when I told you I sat this exam you still did not agree. Its frustrating when you have done one question like three times and some one insists they know it even when they are clearly doing it wrong according to the spec it will be marked in. Look again on page 2 of your insert go to OH and see broad peak at 2500-3300- it clearly says it is for carboxylic acid. So your different spec logic doesn't work here even. You are clearly wrong you can see the mark scheme, so you need to admit or can you not even hold on to your word? Revise the spec once more I suggest as well. LOL
    Are you a frickin retard or something? The peak was at 3000-3100 in my eyes. Look at the insert, it says that a peak between 3000-3100 is an alkene fml. You're calling me cocky, but you're coming out like a know it all with some mad ego.
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    (Original post by RevisionGuide)
    Are you a frickin retard or something? The peak was at 3000-3100 in my eyes. Look at the insert, it says that a peak between 3000-3100 is an alkene fml. You're calling me cocky, but you're coming out like a know it all with some mad ego.
    Its not ego its the frustration, its the frustration and your unwillingness to accept that you were wrong. Well perhaps I was too annoyed and my language should not have been that harsh, but you did insist on a wrong idea for a very long time. To be fair to you some of my friends made the same mistake in the real thing. Even though there is a more rapid dip in region you specified, but peak starts before that point. Now that you have seen your mistake, can you acknowledge it as well in clear words?
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    (Original post by RevisionGuide)
    Are you a frickin retard or something?
    There's no need for that.

    The peak was at 3000-3100 in my eyes. Look at the insert, it says that a peak between 3000-3100 is an alkene fml. You're calling me cocky, but you're coming out like a know it all with some mad ego.
    The shapes of peaks for alkenes are different to that for O-H bonds. In general you can identify an O-H band by how wide it is (about 1000 wavenumbers) compared to that for an alkene (about 250 wavenumbers: see ethylene http://webbook.nist.gov/cgi/cbook.cg...R-SPEC&Index=0).
 
 
 
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