E2 antiperiplanar Question

[emminent]

Elimination of Br with NaOCH3 from which of the following compounds will give (Z)-3-methyl-2-pentene as the major product?

• B.
  
• C.
  

I chose B but answer is C and I'm getting confused with BC explanation. For antiperiplanar, isn't it suppose to have the H on the wedge if the Br/LG is on the dash?

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[TheEpicFruitCake]

I'm looking at my Chad notes and I'm getting B as well. The hydrogen would need to be a wedge. Also, wouldn't both answer choices make (E)-3-methyl-2-pentene?

 

[flossiraptor863]

The bonds are single bonds, so they can rotate. B would give (E)-3-methyl-2-pentene, since it would stay in that formation in order for the LG to leave antiperiplanar. In C, the Z product would form since the single bond rotates to accommodate for the antiperiplanar loss of Br and H.

If I remember correctly, Chad uses a ring when he displays this antiperiplanar concept. In a ring, the bonds can't rotate so the double bond forms with the anti-Zaitsev product.

Edit: This is easier to understand if you draw the Newman Projection.

 

[TheEpicFruitCake]

The bonds are single bonds, so they can rotate. B would give (E)-3-methyl-2-pentene, since it would stay in that formation in order for the LG to leave antiperiplanar. In C, the Z product would form since the single bond rotates to accommodate for the antiperiplanar loss of Br and H.

If I remember correctly, Chad uses a ring when he displays this antiperiplanar concept. In a ring, the bonds can't rotate so the double bond forms with the anti-Zaitsev product.

Edit: This is easier to understand if you draw the Newman Projection.

Ah i see! this fixes the Z vs E problem while also making the answer C

 

[FOX_ONE]

so the catch here is that single bonds are always in rotation

newman is the way to go then!

this is why its a lot easier with rings