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lesya692 [45]
3 years ago
11

Question 096 Propose a three-step synthetic sequence to accomplish the transformation below. 1) HBr, ROOR; 2) t-BuOK; 3) CH3CH2C

CNa 1) NaOEt; 2) HBr, ROOR; 3) CH3CH2CCNa 1) t-BuOK; 2) NaNH2; 3) CH3CH2CCNa 1) CH3CH2CH2CH2Br; 2) NaOEt; 3) HBr, ROOR 1) t-BuOK; 2) HBr, ROOR; 3) CH3CH2CCNa 1) NaOEt; 2) NBS, hν; 3) NaSBu

Chemistry
1 answer:
salantis [7]3 years ago
8 0

Complete Question

Question 096 Propose a three-step synthetic sequence to accomplish the transformation below.

Option 1 =>  1) HBr, ROOR; 2) t-BuOK; 3) CH3CH2CCNa

Option 2 => 1) NaOEt; 2) HBr, ROOR; 3) CH3CH2CCNa

Option 3 => 1) t-BuOK; 2) NaNH2; 3) CH3CH2CCNa

Option 4 => 1) CH3CH2CH2CH2Br; 2) NaOEt; 3) HBr, ROOR

Option 5 => 1) t-BuOK; 2) HBr, ROOR; 3) CH3CH2CCNa

Option 6 => 1) NaOEt; 2) NBS, hν; 3) NaSBu

Answer:

The correct option is option 5

Explanation:

   The mechanism of the reaction is shown on the first uploaded image

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Data:
V_{initial} = 590\:mL
T_{initial} = -55.0^0C
converting to Kelvin
TK = TC + 273
TK = -55.0 + 273 → TK = 218.0 → T_{initial} = 218.0\:K
V_{final} = ? (in\:milliliters)
T_{final} = 30.0^0C
TK = TC + 273
TK = 30.0 + 273 → TK = 303.0 → T_{final} = 303.0\:K

By the first Law of Charles and Gay-Lussac, we have: 
\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }

Solving:
\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }
\frac{ 590 }{ 218.0 } = \frac{ V_{f} }{ 303.0 }
Product of extremes equals product of means:
218.0* V_{f} = 590*303.0
218.0 V_{f} = 178770
V_{f} = \frac{178770}{218.0}
\boxed{\boxed{V_{f} \approx 820.04\:mL}}\end{array}}\qquad\quad\checkmark
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3 years ago
The diagram shows Niels Bohr’s model of an atom.
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Answer:

Energy is absorbed, and an emission line is produced.

Explanation:

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When a copper is drawn into wire the only change that occurs is change in its shape and size no change will take place into its composition that is the wires are still possessing the properties of copper metal. Thus, a physical change takes place when copper is drawn into wire.

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A mixture of CS2(g) and excess O2(g) is placed in a 10 L reaction vessel at 100.0 ∘C and a pressure of 3.10 atm . A spark causes
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Answer:

PCO2  = 0.6 25 atm

PSO2  = 1.2 75 atm

PO2 = 0.6  atm

Explanation:

Step 1: Data given

Volume = 10.0 L

Temperature = 100.0 °C

Pressure = 3.10 °C

After reaction, the temperature returns to 100.0 ∘C, and the mixture of product gases (CO2, SO2, and unreacted O2) is found to have a pressure of 2.50 atm

Step 2: The balanced equation

CS2(g)+3O2(g)→CO2(g)+2SO2(g)

Step 3: Name the reactants and products

a = CS2

b = O2 before reaction

c = CO2

d = SO2

e = nS O2 after reaction with n = the number of moles

Step 4: Calculate moles before reaction

PV = nRT

n = PV/(RT)

(na + nb) = (3.10atm) * (10.0L) / ((0.08206 Latm/moleK) * (373.15K))

(na + nb) = 1.0124

Step 5: Calculate moles after reaction

PV = nRT

n = PV/(RT)

nc + nd + ne) = PV/(RT) = (2.50 atm)*(10.0L) / ((0.08206 Latm/moleK)*(373.15K))

(nc + nd + ne) = 0.816 moles

Step 6: Calculate mol fraction

For  1 mole CS2 we need 3 moles O2  to produce 1 mole of CO2 and 2 moles of SO2

moles O2 remaining = ne = nb - 3na

moles CO2 produced = nc = na

moles SO2 producted = nd = 2na

(nc + nd + ne) = 0.816 moles = nb - 3na + na + 2na = 0.816

nb = 0.816

. (na + nb) = 1.0124

na = 1.0124 moles - 0.816 moles = 0.208

which leads to  

nc = na = 0.208

nd = 2na = 2*0.208 = 0.416

ne = 0.816 - 3*0.208 = 0.192

mole fraction CO2 = 0.208 / (0.208 + 0.416 + 0.192) = 0.25

mole fraction SO2 = 0.416 / (0.208 + 0.416 + 0.192) = 0.5 1

mole fraction O2 = 0.192 /(0.208 + 0.416 + 0.192) = 0.24

Step 6: Calculate partial pressure

PCO2 = 0.25 * 2.50 atm = 0.6 25 atm

PSO2 = 0.51 * 2.50 atm = 1.2 75 atm

PO2 = 0.24 * 2.50 atm = 0.6  atm

Step 7: Control results

now let's verify a couple of things

PV = nRT

P = nRT/V

before rxn

P = (0.208 + 0.816) * (0.08206 L*atm/mole*K) * (373.15K) / (10.0L) ≈ 3.10 atm

after rxn

P = ((0.208 +0.416+0.192) * (0.08206 L*atm/mole*K) * (373.15K) / (10.0L) ≈ 2.50 atm

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