In a two-step synthesis, C6H11Br is converted into C6H12O. From the structure of the product, molecular formula of the starting
material, and reagents given, provide the structures for the precursors to C6H12O that provide the product in the highest yield.
1 answer:
Answer:
See explanation below
Explanation:
The question is incomplete. However in picture 1, you have the starting materials and the structure of the product, which you miss in this part.
Now, in picture 2, you have the starting reactant and the product, and the mechanism that is taking place here.
First, all what we have here is an acid base reaction. In the first step, we are using the acid medium to convert the reactant into an alcohol. The bromine there, is not leaving the molecule yet, because it's neccesary for the next step. The starting reactant is an alkene, in that way, we can convert the reactant in the first step into a secondary alcohol. In other words, the first reaction is a alkene hydration.
In the second step, we use a strong base. You may say this is a strong nucleophile and will do a Sn2 reaction to form another alcohol there, but it's not the case, because, before any kind of reaction happens, the priority here is always the acid base, so the base will react with the acidic hydrogen. In this case, it will substract an hydrogen from the OH. When this happens, the lone pair will do an auto condensation here, and attacks the bromine in the molecule. In this way, the molecule will become a cyclomolecule, and that way it form the final product.
See picture 2, for mechanism
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Answer:
At equilibrium, the forward and backward reaction rates are equal.
The forward reaction rate would decrease if is removed from the mixture. The reason is that collisions between
molecules and
molecules would become less frequent.
The reaction would not be at equilibrium for a while after was taken out of the mixture.
Explanation:
<h3>Equilibrium</h3>Neither the forward reaction nor the backward reaction would stop when this reversible reaction is at an equilibrium. Rather, the rate of these two reactions would become equal.
Whenever the forward reaction adds one mole of to the system, the backward reaction would have broken down the same amount of
. So is the case for
and
.
Therefore, the concentration of each species would stay the same. There would be no macroscopic change to the mixture when it is at an an equilibrium.
<h3>Collision Theory</h3>In the collision theory, an elementary reaction between two reactants particles takes place whenever two reactant particles collide with the correct orientation and a sufficient amount of energy.
Assume that and
molecules are the two particles that collide in the forward reaction. Because the collision has to be sufficiently energetic to yield
, only a fraction of the reactions will be fruitful.
Assume that molecules were taken out while keeping the temperature of the mixture stays unchanged. The likelihood that a collision would be fruitful should stay mostly the same.
Because fewer molecules would be present in the mixture, there would be fewer collisions (fruitful or not) between
and
molecules in unit time. Even if the percentage of fruitful collisions stays the same, there would fewer fruitful collisions in unit time. It would thus appear that the forward reaction has become slower.
The backward reaction rate is likely going to stay the same right after was taken out of the mixture without changing the temperature or pressure.
The forward and backward reaction rates used to be the same. However, right after the change, the forward reaction would become slower while the backward reaction would proceed at the same rate. Thus, the forward reaction would become slower than the backward reaction in response to the change.
Therefore, this reaction would not be at equilibrium immediately after the change.
As more and more gets converted to
and
, the backward reaction would slow down while the forward reaction would pick up speed. The mixture would once again achieve equilibrium when the two reaction rates become equal again.