Answer:
(2R,3S)-2-ethoxy-3-methylpentane
and
(2S,3S)-2-ethoxy-3-methylpentane
Explanation:
For this case, we will have as nucleophile. Also, this compound is also in excess. So, we will have as solvent
a protic solvent. Therefore the Sn1 reaction would be favored.
The first step would be the carbocation formation followed by the attack of the nucleophile. In this case both isomers would be produced: R and S (see figure).
The mass of sodium sulfite that was used will be 1,890 grams.
<h3>Stoichiometric problems</h3>First, the equation of the reaction:
The mole ratio of SO2 produced and sodium sulfite that reacted is 1:1.
Mole of 960 grams SO2 = 960/64 = 15 moles
Equivalent mole of sodium sulfite that reacted = 15 moles
Mass of 15 moles sodium sulfite = 15 x 126 = 1,890 grams
More on stoichiometric problems can be found here: brainly.com/question/14465605
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