This ether can, in principle, be synthesized by two different combinations of haloalkane and metal alkoxide. Draw the combination of alkyl chloride and potassium alkoxide that forms the higher yield of ether. You do not have to consider stereochemistry. You do not have to explicitly draw H atoms. Include counter-ions, e. g., Na , I-, in your submission, but draw them in their own separate sketcher. Separate multiple reactants using the sign from the drop-down menu.

The 2 different combinations are given in the attached figure. The 2nd combination has a higher yield due to less hindrance by the alkyl halide.

Explanation:

The first reaction is between an alkyl halide and metal alkoxide. In this case, the alkyl chloride would be a secondary component and thus will cause greater hindrance to the yield.  The reaction is as given in the attached figure

The second reaction is between a benzene ring containing halide and metal alkoxide. Now as the reaction is via alkyl chloride being the primary agent, there is less hindrance and thus greater yield in this case.