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Serga [27]
3 years ago
14

What compound is formed when methyloxirane (1,2-epoxypropane) is reacted with ethylmagnesium bromide followed by treatment with

aqueous acid

Chemistry
1 answer:
Pavlova-9 [17]3 years ago
5 0

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> (H^+)  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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The table below shows the content in mg of some chemical elements found in 100g of milk:
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According to the law of conservation of matter, the number of ________ is not changed by a chemical reaction. A. molecules B. at
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2 years ago
Read 2 more answers
Be sure to answer all parts. Hydrogen iodide decomposes according to the reaction 2 HI(g) ⇌ H2(g) + I2(g) A sealed 1.50−L contai
Zarrin [17]

Answer : The concentration of HI and I_2 at equilibrium is, 0.0158 M and 0.00302 M respectively.

Explanation :

First we have to calculate the concentration of H_2, I_2\text{ and }HI

\text{Concentration of }H_2=\frac{\text{Moles of }H_2}{\text{Volume of solution}}=\frac{0.00623mol}{1.50L}=0.00415M

\text{Concentration of }I_2=\frac{\text{Moles of }I_2}{\text{Volume of solution}}=\frac{0.00414mol}{1.50L}=0.00276M

\text{Concentration of }HI=\frac{\text{Moles of }HI}{\text{Volume of solution}}=\frac{0.0244mol}{1.50L}=0.0163M

Now we have to calculate the value of equilibrium constant (K).

The given chemical reaction is:

                            2HI(g)\rightleftharpoons H_2(g)+I_2(g)

Initial conc.      0.0163     0.00415      0.00276

At eqm.        (0.0163-2x) (0.00415+x)  (0.00276+x)

As we are given:

Concentration of H_2 at equilibrium = 0.00467 M

That means,

(0.00415+x) = 0.00467

x = 0.00026 M

Concentration of HI at equilibrium = (0.0163-2x) = (0.0163-2(0.00026)) = 0.0158 M

Concentration of I_2 at equilibrium = (0.00276+x) = (0.00276+0.00026) = 0.00302 M

8 0
3 years ago
What is the concentration of a solution in which 15 grams of sugar is dissolved in 0.2 L of water?
Margaret [11]

Answer:

0.2 M

Explanation:

Step 1: Given data

  • Mass of sugar (sucrose): 15 g
  • Volume of water: 0.2 L (we will assume it is the volume of the solution)

There are different ways to express the concentration of a solution. We will calculate molarity, which is one of the most used.

Step 2: Calculate the moles of sucrose

The molar mass of sucrose is 342.3 g/mol.

15 g × 1 mol/342.3 g = 0.044 mol

Step 3: Calculate the molarity of the solution

Molarity is equal to the moles of solute divided by the liters of solution.

M = 0.044 mol/0.2 L = 0.2 M

7 0
3 years ago
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