Answer:
E = h v = 6.626 x x 97.20 x
= 6.44 x
J = 4.01 x
eV
n = Energy Level = / E = -13.6 / 4.01 x
= 3.39 x
= 1.84 x
energy level...
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.
1 answer:
Answer:
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.
You might be interested in