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PolarNik [594]
3 years ago
7

Draw the major organic product formed when the compound shown below undergoes a reaction with two equivalents of ch3ch2mgbr and

then is treated with water.

Chemistry
1 answer:
alekssr [168]3 years ago
3 0
I have attached an image showing the starting material required for the questions, which is a molecule with an ester functionality. Esters are very reactive towards Grignard reagents and the reagent will add twice to the carbonyl of the ester, as is implied by the question saying that 2 equivalents of the Grignard reacts.

The first Grignard reacts with the ester and adds the ethyl group to the carbonyl carbon of the ester, to undergo an addition-elimination reaction to form the ketone. However, ketones are also very reactive to Grignard reagents, and the carbon of the cabronyl will react once more with the second Grignard to add a second ethyl group. At this point the alkoxide is formed, which is why water is added as a work-up step, to protonate the alkoxide to isolate the final alcohol product.

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A volume of 120 mL of H2O is initially at room temperature (22.00 ∘C ). A chilled steel rod at 2.00 ∘C ∘C is placed in the water
aliina [53]

Answer:

\large \boxed{\text{28.5 g}}

Explanation:

There are two heat flows in this process and, since energy (heat) can neither be destroyed nor created, the energy change for the system must equal zero.

Data:

For Fe,    m₁ = ?;         C₁ = 0.452 J°C⁻¹g⁻¹; Ti =   2.00 °C; T_f = 21.50 °C

For H₂O, m₂ = 120 g; C₂ = 4.18    J°C⁻¹g⁻¹; Ti = 22.00 °C; T_f = 21.50 °C

Calculations:

1. Temperature changes

ΔT₁ = T_f - Ti = 21.50 °C -   2.00 °C = 19.50 °C

ΔT₂ = T_f - Ti = 21.50 °C - 22.00 °C = -0.50 °C

2. Mass of steel rod

\begin{array}{ccl}\text{Heat  gained by steel rod + heat lost by water} & = & 0\\m_{1}C_{1} \Delta T_{1} + m_{2}C_{2} \Delta T_{2}& = & 0\\m_{1} \times 0.452 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}\times 19.50 \, ^{\circ}\text{C} + \text{120 g} \times 4.18\text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times (-0.50)\, ^{\circ}\text{C}& = & 0\\8.814m_{1}\text{ g}^{-1} - 250.8 & = &0\\\end{array}\\

\begin{array}{ccl}8.814m_{1}\text{ g}^{-1} & = &250.8\\m_{1} & = & \dfrac{250.8}{\text{8.814 g}^{-1}}\\\\& = & \textbf{28.5 g}\\\end{array}\\\text{The mass of the steel rod is $\large \boxed{\textbf{28.5 g}}$}

6 0
3 years ago
Disease caused by virus are more dangerous.Why?
nlexa [21]

Answer:

Explanation:

1. Virus's are hard to detect because of their simple construction.

2. Some mutate very easily.

3. It is hard to isolate the virus and kill it without doing damage to the host.

7 0
3 years ago
What is the si unit for the quantity of particles in a sample?
kaheart [24]
The SI unit for the amount or quantity of small particles in a sample is referred to as the mole.

This tells how much in terms of atoms or molecules are present for a particular mass.
7 0
3 years ago
What is the balance equation for I2 (s) dissolving in water
hjlf

Answer:

I₂ + H₂O --------→ HI + HIO

Explanation:

Iodine is not fully dissolve in water but only dissolve up to some extent.

When iodine was mixed with water it give mixture of two products.

  • Hydrogen iodide or hydroiodic acid (HI)
  • Hypoiodous acid (HIO)

So the balance reaction is as under

        I₂ + H₂O --------→ HI + HIO

All the reactant are one mole and one one mole of each product produced in the shown reaction.

This is a reversible reaction and only 0.05% of Iodine molecule react with water and the other  Iodine molecule remain in water unreacted.

3 0
4 years ago
Gaseous hydrogen and oxygen can be prepared in the laboratory from the decomposition of gaseous water. The equation for the reac
nata0808 [166]

Answer:

m_{O_2}=87.2gO_2

Explanation:

Hello.

In this case, given the chemical reaction, we can compute the grams of oxygen by using the 98.2 g of water via the 2:1 mole ratio between them, the molar mass of water that is 18.02 g/mol, the molar mass of gaseous oxygen that is 32.00 g/mol and the following stoichiometric procedure relating the given information:

m_{O_2}=98.2gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{1molO_2}{2molH_2O}*\frac{32.00gO_2}{1molO_2}   \\\\m_{O_2}=87.2gO_2

In which the result is displayed with three significant figures because the given mass of water 98.2 g, has three significant figures too.

Best regards!

6 0
3 years ago
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