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pochemuha
3 years ago
15

Propose a mechanism to account for the formation of a cyclic acetal from 4-hydroxypentanal and one equivalent of methanol. If th

e carbonyl oxygen of 4-hydroxypentanal is enriched with oxygen-18, do you predict that the oxygen label appears in the cyclic acetal or in the water

Chemistry
1 answer:
timofeeve [1]3 years ago
8 0

Answer:

If carbonyl oxygen of 4-hydroxypentanal is enriched with O^{18}, then the oxygen label appears in the water .

Explanation:

  • In the first step, -OH group at C-4 gives intramolecular nucleophilic addition reaction at carbonyl center to produce a cyclic hemiacetal.
  • Then, one equivalent of methanol gives nucleophilic substitiution reaction by substituting -OH group in cyclic hemiacetal to produce cyclic acetal.
  • If carbonyl oxygen of 4-hydroxypentanal is enriched with O^{18}, then the oxygen label appears in the water produced at the end of reaction.
  • Full reaction mechanism has been shown below.

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What chemical reactions occur in a Nelson's cell? ​
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Explanation:

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a sample of 3.00 g of so2 (g)originally in a 5.00 L vesselat 21 degee Celsius is transferred to a 10.0 L vessel at 26 degree Cel
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Answer:

1) The partial pressure of SO₂ gas in the larger container = 0.115 atm.

2) The partial pressure of N₂ gas in the larger container = 0.206 atm.

3) The total pressure in the vessel = 0.321 atm.

Explanation:

  • To calculate the partial pressure of each gas, we can use the general law of ideal gas: PV = nRT.

where, P is the partial pressure of the gas in atm,

V is the volume of the vessel in L,

n is the no. of moles of the gas,

R is the general gas constant (R = 0.082 L.atm/mol.K),

T is the temperature of the gas in K.

<u><em>1) What is the partial pressure of SO₂ gas in the larger container?</em></u>

<em>∵ P = nRT/V.</em>

n = mass/molar mass = (3.0 g)/(64.066 g/mol) = 0.047 mol.

R = 0.082 L.atm/mol.K.

T = 26 °C + 273.15 = 299.15 K.

V = 10.0 L. (The volume of the new container)

∴ P = nRT/V = (0.047 mol)(0.082 L.atm/mol.K)(299.15 K)/(10.0 L) = 0.115 atm.

<u><em>2) What is the partial pressure of N₂ gas in the larger container?</em></u>

<em>∵ P = nRT/V.</em>

n = mass/molar mass = (2.35 g)/(28.0 g/mol) = 0.084 mol.

R = 0.082 L.atm/mol.K.

T = 26 °C + 273.15 = 299.15 K.

V = 10.0 L. (The volume of the new container)

∴ P = nRT/V = (0.084 mol)(0.082 L.atm/mol.K)(299.15 K)/(10.0 L) = 0.206 atm.

<u><em>3) What is the total pressure in the vessel?</em></u>

  • According to Dalton's law the total pressure exerted is equal to the sum of the partial pressures of the individual gases.

<em>∵ The total pressure in the vessel = the partial pressure of SO₂ + the partial pressure of N₂.</em>

∴ The total pressure in the vessel = 0.115 + 0.206 = 0.321 atm.

5 0
3 years ago
if a baloon filled with dry hydrogen weighs 35 gram,but weighs 440 grams when filled with the vapour of an organic compound. cal
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Answer:

1) The vapor density of the organic compound is approximately 12.57

2) The relative molar mass (RMM) of the organic compound is approximately 25.14 grams  

Explanation:

1) The mass of the balloon filled with dry hydrogen = 35 grams

The mass of the balloon filled with vapor of an organic compound = 440 grams

The vapor density = (Weight of a given volume of gas)/(Weight of equal volume of hydrogen)

The vapor density of the organic compound = (440)/(35) ≈ 12.57

The vapor density of the organic compound ≈ 12.57

2) The relative molar mass (RMM) = 2 × vapor density

The relative molar mass (RMM) of the organic compound = 2 × vapor density of the organic compound

The relative molar mass (RMM) of the organic compound ≈ 2 × 12.57 ≈ 25.14 grams  

The relative molar mass (RMM) of the organic compound ≈ 25.14 grams  

6 0
3 years ago
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