[Mn+1AXn]

Hi, I was wondering if this mechanism would be possible (S_N2'?)and if so would it be the major product produced when NaOEt is added?
Thanks

[pgk]

The mechanism is correct but this does not automatically mean that the posted product would be the major one. The (selective) formation of a reaction product depends on a wide range of factors, e.g. thermodynamic, kinetic, nature of leaving group, formed ring strain, hard/soft-acidity/basicity of reactants, solvent effect, etc.
Indicatively, using of NaH instead of EtONa will minimize debromination side products and it will avoid the formation of any 4-(2-hydroxyethyl)-2-cyclohexenyl ethyl ether.

[Mn+1AXn]

Thank you for answering. Please can you tell me why using NaH would reduce debromination side products compared NaOEt?
Thanks

[pgk]

Because NaH is a soft base and it cannot react with alkyl or allyl hydrogens that are hard acids, as being covalently bonded hydrogens; in contrast to hydroxides and alkoxides that are hard bases and thus, they can react with such hydrogens.
However, hydrides can favor dehalogenation under radical conditions, due to the formation of hydride radical. Therefore, small amounts of dehalogenated impurities may be formed during NaH promoted, Williamson synthesis of ethers. Note that daylight and/or the fluorescent lamps in the lab, can weakly initiate radical reactions.