How many Significant Figures are in 45.000?
1 answer:
You might be interested in
Answer:
See explanation
Explanation:
In order to do this, we need to use 3 reagents to get the final product.
The first one, and logic is the halogenation of the alkene. Doing this, with Br2/CCl4, we'll get an alkane with two bromines, one in carbon 2 and the other in carbon 3.
Then, the next step is to eliminate one bromine of the reactant. The best way to do this, is using sodium ethoxide in ethanol. This is because sodium ethoxide is a relatively strong base, and it will promove the product of elimination in major proportions rather than the sustitution product. If we use NaOH is a really strong base, and it will form another product.
When the sodium ethoxide react, it will form a double bond between carbon 1 and 2 (The carbon where one bromine was with the methyl, changes priority and it's now carbon 3).
The final step, is now use acid medium, such H3O+/H2O or H2SO4/H2O. You can use any of them. This will form an carbocation in carbon 2 (it's a secondary carbocation, so it's more stable that in carbon 1), and then, the water molecule will add to this carbon to form the alcohol.
See the attached picture for the mechanism of this.
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
