DropBoite Posted April 28, 2012 Report Share Posted April 28, 2012 Attached is the image of the desired structure, my question is based on what is pictured as (CH3)2HC- I represented this group as C3H7 which does seem like the logical equivalent, but just being safe here. Am I right? Reply Link to post Share on other sites More sharing options...
Drake Glau Posted April 28, 2012 Report Share Posted April 28, 2012 Yes, those are the same "group" Random note: they're enantiomers which, if you are HL, is all IB will care about Reply Link to post Share on other sites More sharing options...
DropBoite Posted April 28, 2012 Author Report Share Posted April 28, 2012 Yes, those are the same "group" Random note: they're enantiomers which, if you are HL, is all IB will care about Thank you Drake, but 2 other questions on the same line please Referring to the image above I'd like to ask whether the same thing inverted 90 degrees to the right is correct? As for this one It really puzzles me how the IBO started drawing good old branched structures as these 3-D enantiomer like structures and note this image is from a diagram of SN1. Though I'm not denying that it could be chiral.. explanation? Reply Link to post Share on other sites More sharing options...
Drake Glau Posted April 28, 2012 Report Share Posted April 28, 2012 I don't get what you're asking for the first pictureIt's not chiral. Easiest way to tell is that it has 2 H groups. As far as you're concerned, Sn1 does nothing to the stereochemistry of the molecule...until you take organic in college >.> Reply Link to post Share on other sites More sharing options...
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