Consider the e2 elimination of 3‑bromopentane with hydroxide. the starting material consists of a chiral carbon with an in plane bond to bromine pointing to the upper left, an in plane bond to ethyl pointing to the right, a wedged bond to ethyl pointing to the lower left and a dashed bond to hydrogen pointing to the lower left. this reacts with hydroxide to form the product, water and bromide ion. complete the curved arrow electron-pushing mechanism and draw the structure of the organic product formed.

Scheme is attached

Explanation:

When 3‑bromopentane reacts with hydroxide, (Z)-pent-2-ene will produce through E 2 (Z)-pent-2-ene.

Mechanism

Hydroxide ion (OH⁻ ) is a strong nucleophile so it abstract the proton from carbon next to the carbon attached with bromine.

The the carbon next to carbon of bromine gets -ve charge, mean while it shares its electrons with the carbon having bromine and make a double bond.

As bromine  is a good leaving group so it easily gets detached from carbon, so that carbon comes to its normal state.

As a result (Z)-pent-2-ene will produce. we call it Z because mostly trans products will form.

the correct answer to this question is 2