Increasing the concentration of a reactant shifts the position of a chemical equilibrium towards formation of more products. wha
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Answer:
See explanation and picture below
Explanation:
First, in the case of methyloxirane (Also known as propilene oxide) the mechanism that is taking place there is something similar to a Sn2 mechanism. Although a Sn2 mechanism is a bimolecular substitution taking place in only step, the mechanism followed here is pretty similar after the first step.
In both cases, the H atom of the HBr goes to the oxygen in the molecule. You'll have a OH⁺ in both. However, in the case of methyloxirane the next step is a Sn2 mechanism step, the bromide ion will go to the less substitued carbon, because the methyl group is exerting a steric hindrance. Not a big one but it has a little effect there, that's why the bromide will rather go to the carbon with more hydrogens. and the final product is formed.
In the case of phenyloxirane, once the OH⁺ is formed, the next step is a Sn1 mechanism. In this case, the bond C - OH⁺ is opened on the side of the phenyl to stabilize the OH. This is because that carbon is more stable than the carbon with no phenyl. (A 3° carbon is more stable than a 2° carbon). Therefore, when this bond opens, the bromide will go there in the next step, and the final product is formed. See picture below for mechanism and products.
The given molarity of sodium hydroxide solution = 2.0 M
The required concentration of sodium hydroxide is 65 mL of 0.6 M NaOH
Converting 65 mL to L:
Calculating the moles of NaOH in the final solution:
Finding out the volume of 2.0 M solution taken to prepare the final solution:
Therefore, 19.5 mL of 2.0 M NaOH solution and make it up to 65 mL to prepare 0.6 M NaOH solution.