Bootcamp organic 5.

[Faux]

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Question 3 and 16 are a bit unclear for me.

http://postimg.org/image/c1zfpqm5z/



I understand anticoplaner. If you have a wedge as a bromine, you want a dash hydrogen. With that mindset, I picked B. Why does this give the E isomer over the Z isomer? Is it because the groups that are untouched after the reaction are a dash hydrogen and a wedge methyl?



http://postimg.org/image/wne7hn3qv/

For 16, I understand that this is anti addition. Would it be right to assume that every cis alkene that undergoes anti addition will give you a trans product? And that every trans alkene will give you a cis product when you do anti addition? I narrowd the question down to C and D and didn't know what to do from there.

 

[LazyP]

For 3, I think you are forgetting that they want an Z product. B will give E product. C will give Z because once you rotate the between the c-c bond to get the H anti planar to Br, the product will become Z.

16) essentially yes and thats why D is correct. If its hard to see, check the R/S configuration after the anti-addition.

 

[Faux]

For 3, I think you are forgetting that they want an Z product. B will give E product. C will give Z because once you rotate the between the c-c bond to get the H anti planar to Br, the product will become Z.

16) essentially yes and thats why D is correct. If its hard to see, check the R/S configuration after the anti-addition.

Thanks for confirming 16 🙂. Is there anyway else to look at 3 without rotating bonds in my head? Usually for organic chemistry, theres always another way to look at it.


Also, last question for 5.


http://postimg.org/image/mvov3zunr/

Explanation mentions that its an sn2 attack and inversion takes place. No inversion actually takes place in this reaction. Is it because the epoxide didn't show stereochemistry? I picked E.

 

[LazyP]

for anti planar stuff or E2, if the original alkane has both are dashed (basically if hydrogen is anti planar without rotation around that bond) then the alkene will be as is (the subs will stay where they are, so could be Z or E). If one is solid dash and another just dashed (no H in anti planar as is, meaning you would have to rotate the C-C bond), then the alkene will have one side rotated always either forming Z or E, opposite of what the original would have been if H was already anti planar (Z or E). So what I am saying is probably confusing, may be flawed (others please confirm this), but thats what I used and worked always.

5) It is SN2, the acid protonated the O then the EtOH attacks the more substituted side, causing an inversion. This is fine example of always checking the chiral center config (R or S). Your choice didn't invert, check the config of the epoxide (R) then that of the answer B (S). In E it still is R.