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Size:  502.6 KBCan someone check if I have drawn these optical isomers right please. And does it matter if I put the asterisk on the left side or right side to signify its chiral carbon? Also when I drawing the right side for the mirror image do I have to teverse everything like on question 1 on the left side one pf the functional group is written as ch3ch2 and I wrote it as h2ch3c on the right side is that right?
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    (Original post by Ann_Onymous)
    Name:  image.jpg
Views: 76
Size:  502.6 KBCan someone check if I have drawn these optical isomers right please. And does it matter if I put the asterisk on the left side or right side to signify its chiral carbon? Also when I drawing the right side for the mirror image do I have to teverse everything like on question 1 on the left side one pf the functional group is written as ch3ch2 and I wrote it as h2ch3c on the right side is that right?

    the isomers look right, the asterisk position isn't important as long it is clearly labelling the correct carbon.You shouldn't reverse everything like you have, because C*-H2CH3C looks as though the chiral carbon is connected to a hydrogen which is not the case.

    Instead treat them as CH2 or CH3 units, and reverse the order of the units:

    So CH3CH2 on the left should become CH2CH3, this shows the correct connectivity.
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    (Original post by Ann_Onymous)
    Name:  image.jpg
Views: 76
Size:  502.6 KBCan someone check if I have drawn these optical isomers right please. And does it matter if I put the asterisk on the left side or right side to signify its chiral carbon? Also when I drawing the right side for the mirror image do I have to teverse everything like on question 1 on the left side one pf the functional group is written as ch3ch2 and I wrote it as h2ch3c on the right side is that right?
    Having looked again, be careful with the last compound's drawing (part v), the drawing should make it clear that the chiral carbon is bonded to the carbon atom and not the oxygen! (Assuming this is supposed to be an aldehyde functionality).
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    (Original post by MexicanKeith)
    the isomers look right, the asterisk position isn't important as long it is clearly labelling the correct carbon.You shouldn't reverse everything like you have, because C*-H2CH3C looks as though the chiral carbon is connected to a hydrogen which is not the case.

    Instead treat them as CH2 or CH3 units, and reverse the order of the units:

    So CH3CH2 on the left should become CH2CH3, this shows the correct connectivity.
    I agree, it should be ch2ch3!
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    Name:  image.jpg
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Size:  498.1 KBSo like this? I noticed that they didn't reverse everything in the textbook which is why I asked.
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    (Original post by Ann_Onymous)
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Size:  498.1 KBSo like this? I noticed that they didn't reverse everything in the textbook which is why I asked.
    For part I yes that looks better, for part II though it should be made clear that the chiral carbon is not bonded to the hydrogen on the next carbon, either of the two options on the right of this picture would be acceptable, with the bond bond very obviously going to the carbon or with the C and H in the other order.Name:  WIN_20160918_00_05_15_Pro.jpg
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    (Original post by Ann_Onymous)
    Name:  image.jpg
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Size:  498.1 KBSo like this? I noticed that they didn't reverse everything in the textbook which is why I asked.
    Just notice you've reversed groups that were correct in some other parts too!

    The important thing is that the way you write the answer represents how the atoms are connected to one another! Try drawing the molecules out in full then go along one carbon at a time and think about writing them down! And when you draw any bonds, make sure they actually connect two atoms that are bonded together! (eg in part III the chiral carbon is not bonded to the bromine but that is how you've drawn it on the right hand side!)
 
 
 
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