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OCR A2 Chemistry Question

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    I don't understand how they got the answers to Q5 cii and fii, so could anyone explain them to me?

    For Q5cii, why are there three stereoisomers? There isn't a carbon-carbon double bond so there aren't any E/Z isomers. For optical isomers, the compound is symmetrical about the carbon-carbon double bond so there should only be two optical isomers?

    For Q5fii, I get the correct answer if I use the concentration of sodium hydroxide as the concentration of the conjugate base [A-] but I don't understand why.

    Thanks in advance!
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    Q5c(ii) For every 'chiral carbon' there are 2 stereoisomers possible, since there are 2 chiral carbons there is a maximum of four possible. Because of the symmetry in the molecule 2 out of those 4 are the same as each other - leaving 3 different stereoisomers. Try drawing them out to convince yourself of this (even build a model!) :yep:

    Q5f(ii) The sodium hydroxide reacts with the acid: HA + NaOH ---> H2O + NaA so the number of moles of NaOH is the same as the conjugate base, once reacted
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    (Original post by EierVonSatan)
    Q5c(ii) For every 'chiral carbon' there are 2 stereoisomers possible, since there are 2 chiral carbons there is a maximum of four possible. Because of the symmetry in the molecule 2 out of those 4 are the same as each other - leaving 3 different stereoisomers. Try drawing them out to convince yourself of this (even build a model!) :yep:

    Q5f(ii) The sodium hydroxide reacts with the acid: HA + NaOH ---> H2O + NaA so the number of moles of NaOH is the same as the conjugate base, once reacted
    I understand that there are two optical isomers for each chiral carbon.
    If we consider just one of the chircal carbon first then the four groups are COOH, H, I and A (where A is just another group with COOH, H and I attached to another carbon). This compound has two optical isomers but if we consider the other chiral carbon, we get the same four groups because the compound is symmetrical about the carbon-carbon bond so the optical isomers of this chiral carbon are both identical to the first chiral carbon's optical isomers.

    I don't understand where I'm going wrong :/

    Thanks for the explanation for Q5fii, it makes perfect sense (it's so simple now I don't understand how I didn't see that :/)
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    (Original post by Tamagotchi)
    I understand that there are two optical isomers for each chiral carbon.
    If we consider just one of the chircal carbon first then the four groups are COOH, H, I and A (where A is just another group with COOH, H and I attached to another carbon). This compound has two optical isomers but if we consider the other chiral carbon, we get the same four groups because the compound is symmetrical about the carbon-carbon bond so the optical isomers of this chiral carbon are both identical to the first chiral carbon's optical isomers.

    I don't understand where I'm going wrong :/
    Yeah, I understand why you're having problems - this isn't the easiest thing you'll come across :nah:

    Don't simplify the other group to just ''A'' as it's stereochemistry alters when it's reflected! I've attached a pdf that will help (hopefully!)

    I've drawn out one isomer and called it A. The dotted line is a mirror and drawn A's reflection, then rotated it so it looks more like A (for comparison), you'll see that this is not the same isomer, called it B.

    Next row down, I altered the top carbons stereochemistry (swapped the H and COOH) and called it C. C is then reflected and rotated and it looks like a new stereoisomer D. However, if you simply turn it upside down it is in fact C as well, and this is where the symmetry comes into play :yes:
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  2. File Type: pdf meso-stereo.pdf (129.8 KB, 16 views)
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    (Original post by EierVonSatan)
    Yeah, I understand why you're having problems - this isn't the easiest thing you'll come across :nah:

    Don't simplify the other group to just ''A'' as it's stereochemistry alters when it's reflected! I've attached a pdf that will help (hopefully!)

    I've drawn out one isomer and called it A. The dotted line is a mirror and drawn A's reflection, then rotated it so it looks more like A (for comparison), you'll see that this is not the same isomer, called it B.

    Next row down, I altered the top carbons stereochemistry (swapped the H and COOH) and called it C. C is then reflected and rotated and it looks like a new stereoisomer D. However, if you simply turn it upside down it is in fact C as well, and this is where the symmetry comes into play :yes:
    Ohh, so I have to take into account the stereochemistry of the carbon (and its groups) which is attached to the chiral carbon? Because that's what you did in A and C isnt it?
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    (Original post by Tamagotchi)
    Ohh, so I have to take into account the stereochemistry of the carbon (and its groups) which is attached to the chiral carbon? Because that's what you did in A and C isnt it?
    Yeah :yep:
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    (Original post by EierVonSatan)
    Yeah :yep:
    I get it now Thank you!

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