Product please

It depends highly on where the C14 atom is ?

i gues debosmit is right.
first eto^- will attack c¹⁴ and c-o bond will break rendering a negative charge on o.
now internal nucleophilic rxn. => o^- will attack the carbon and cl^- will leave.

EDIT: ^ c¹⁴ is just for marking the carbon.doesn't affect the reaction.

Your mechanism is right.

But even if OEt attacks dirrectly at cl- attached carbon in an SN2 fasion, the product will be same ?

So why it isn't so?

because OEt will attack from least hindered site . (ie the C14 atom)
hence in the product given by debosmit, the rightmost carbon is C14

Thanks..

My doubt was cleared regarding steric hindrance with 3D viewer..