Answer:
(2S,3S)-2-Bromo-3-phenylbutane + C₂H₅ONa → E-2-phenyl-2-butene + NaBr + C₂H₅OH
C₁₀H₁₃Br + C₂H₅ONa → C₁₀H₁₂ + NaBr + C₂H₅OH
Explanation:
When we add sodium ethoxide in solvent its ionic character increases Na⁺ ⁻OCH2CH3.
Upon adding (2S,3S)-2-Bromo-3-phenylbutane in the solution, Oxygen (O⁻) take the proton from position 3 of the 2-Bromo-3-phenylbutane. By extracting this proton the electrons at carbon 3 make bond with the carbon 2 having the bromine atom.
As bromine is a good leaving group so it easily get detached from carbon 2 making a double bond between carbon 2 and 3.
Stereo-chemical outcome
As a result of this elimination reaction E-2-phenyl-2-butene produces.
Why we call it as E
As we can see in the attached structure that both preority groups are in the same side or we can say that they are cis to each other.
When both priority groups are on same side or cis to each othe we call it as E if they are opposite or trans we call call them Z.
How to select priority groups
We select the priority groups by their atomic number. We give higher priority to that group which has higher atomic number.
