What compound is formed when methyloxirane (1,2-epoxypropane) is reacted with ethylmagnesium bromide followed by treatment with
aqueous acid
1 answer:
Answer:
Pentan-2-ol
Explanation:
On this reaction, we have a <u>Grignard reagent</u> (ethylmagnesium bromide), therefore we will have the production of a <u>carbanion</u> (step 1). Then this carbanion can <u>attack the least substituted carbon</u> in the epoxide in this case carbon 1 (step 2). In this step, the epoxide is open and a negative charge is generated in the oxygen. The next step, is the <u>treatment with aqueous acid</u>, when we add acid the <u>hydronium ion</u> () would be produced, so in the reaction mechanism, we can put the hydronium ion. This ion would be <u>attacked by the negative charge</u> produced in the second step to produce the final molecule: <u>"Pentan-2-ol".</u>
See figure 1
I hope it helps!
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Answer:
The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol
Explanation:
The ∆H (heat of reaction) of the combustion reaction is the heat that accompanies the entire reaction. For its calculation you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient (number of molecules of each compound that participates in the reaction) and finally subtract them:
Enthalpy of the reaction= ΔH = ∑Hproducts - ∑Hreactants
In this case, you have: 2 NOCl(g) → 2 NO(g) + Cl₂(g)
So, ΔH=
Knowing:
- ΔH= 75.5 kJ/mol
= 90.25 kJ/mol
= 0 (For the formation of one mole of a pure element the heat of formation is 0, in this caseyou have as a pure compound the chlorine Cl₂)
=?
Replacing:
75.5 kJ/mol=2* 90.25 kJ/mol + 0 -
Solving
-=75.5 kJ/mol - 2*90.25 kJ/mol
-=-105 kJ/mol
=105 kJ/mol
<u><em>The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol</em></u>
First, we have to remember the molarity formula:
Part 1:
In this case, our solute is sodium nitrate (NaNO3), and we have the mass dissolved in water, then we have to convert grams to moles. For that, we need the molecular weight:
Then, we calculate the moles present in the solution:
Now, we have the necessary data to calculate the molarity (with the solution volume of 200 mL):
The molarity of this solution equals 0.2339 M.
Part 2:
In this case, we have the same amount (in moles and mass) of sodium nitrate, but a different volume of solution, then we only have to change it:
So, the molarity of this solution is 0.1701 M.