Answer:
4.602 g H20 was used.
14.634 g H20 was left over
Explanation:
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1 answer:
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Answer:
Explanation:
According to this. Let's analize the possible products a, b and c.
First, the problem states that we have 2 eq. of base, For this case, let's assume it's KOH. Now, As we are doing a reaction with base, means that this reaction can only take places under conditions of SN2 and E2, a fast reaction that is taking place in only 1 step.
With this in mind, let's analyze product a. This states that it has two sp hybridized carbon, in other words, a triple bond between two carbons. So the product is with no doubt, an alkyne.
Product b has only one sp hybridized carbon, which means that this carbon should cannot be an alkyne because we need two carbon atoms. The only way to have one atom of C sp hybridized, is with two double bonds, so product b would have to a alkene with two double bonds.
Product c do not have sp hybridized carbon, therefore, it only has two double bonds in two different Carbon atoms, so it's another alkene with two double bonds, but in two different atoms of carbon.
Picture attached show the product a, b and c. Hope this can help
