Answer:
Repeated SN2 reactions occur leading to the formation of a racemic mixture
Explanation:
S-2-iodooctane is a chiral alkyl halide with an asymmetric carbon atom. The presence of an asymmetric carbon atom implies that it can rotate plane polarized light and thus lead to optical isomerism. The two configurations of the compound are R/S according to the Cahn-Prelong-Ingold system.
However, when S-2-iodooctane is treated with sodium iodide in acetone, repeated SN2 reactions occur since the iodide ion is both a good nucleophile and a good leaving group. Hence a racemic modification is formed in the system with time hence we end up with (±)- Iodooctane.
Answer:
56.9 mmoles of acetate are required in this buffer
Explanation:
To solve this, we can think in the Henderson Hasselbach equation:
pH = pKa + log ([CH₃COO⁻] / [CH₃COOH])
To make the buffer we know:
CH₃COOH + H₂O ⇄ CH₃COO⁻ + H₃O⁺ Ka
We know that Ka from acetic acid is: 1.8×10⁻⁵
pKa = - log Ka
pKa = 4.74
We replace data:
5.5 = 4.74 + log ([acetate] / 10 mmol)
5.5 - 4.74 = log ([acetate] / 10 mmol)
0.755 = log ([acetate] / 10 mmol)
10⁰'⁷⁵⁵ = ([acetate] / 10 mmol)
5.69 = ([acetate] / 10 mmol)
5.69 . 10 = [acetate] → 56.9 mmoles
Oxygen is the 8th element in the periodic table. This means that oxygen has 8 protons<span> and 8 electrons. In order to get the number of neutrons you take the atomic weight in this case 15.9999~16 and you subtract it by the number of protons (16-8) (o_O)
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