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Sn2/E2 exercise correction

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xshadow:
Hi, can you help me with this exercise:

https://i.imgur.com/sqB2YQV.jpg



Are the products written corrected?

In the first one can an E2 occours forming  ALSO the secondary product HC=C=H2??

Because  I have a 1° allylic halide : its stability is similar to a secondary alkyl halide...and in 2° alkyl halide usually E2 and Sn2 are in competition (with strong base/nucleohpilic,of course)

Thanks!!

sharbeldam:
following

kriggy:
You are likely going to get more products, especially in the second case, due to the Sn1´ or Sn2´ substitutions. Sn reactions on secondary halides can go both mechanisms and allylic halides make it even more complex.

https://en.wikipedia.org/wiki/Allylic_rearrangement

for more information

xshadow:
Hi

And what about elimination?

Can E2 occour in a secondary  allylic halides? (I 'll get a conjugated diene)
In the second molecule can I get a mixture of E2 and Sn2 products? (Sn1/E1 product low because I'm using a strong base and nucleophile)
Thanks

hollytara:
There are two important variables that aren't given: temperature and solvent. 

The first reaction - 3-chloropropene or allyl chloride - is only going to react by SN2, unless very high temperatures are used.  The allene product is high energy and requires fairly severe conditions.

The second reaction 3-chloro-1-butene is trickier.  If it is set up for SN2 (polar aprotic solvent, low T) you will only get the 3-ethoxy-1-butene.  If you try for SN1 (low concentration of ethoxide, refluxing ethanol) you will get the SN1 products - 3-ethoxy-1-butene and 1-ethoxy-2-butene (probably as a mixture of E and Z) with some elimination product (1,3-butadiene).  If you go for E2 conditions (high concentration of ethoxide, refluxing ethanol) you will get essentially only the 1,3-butadiene product. 

A nice summary of this can be found in the "MAximum Success, Minimum Effort" book by Fredlos

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