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Shtirlitz [24]
3 years ago
7

Using the reagents below, list in order (by letter, no period) those necessary to transform 1-chlorobutane into 1-butyne. Note:

Not all entry-fields provided may be needed. Choose "na" in any space where you have no reagent .
a. H2, Pt:
b. H2, Lindlar's Catalyst
c. H2, Ni2B{P-2)
d. Br2, CCI4
e. NaNH2, Mineral oil, heat
f. NH4CI G.
g. t -BuOK, t -BuOH, heat
h. CH3 I
i. CH3CH2 I
Chemistry
1 answer:
Andru [333]3 years ago
3 0

Answer:

gde

Explanation:

We are attempting to synthesize 1-butyne from 1-chlorobutane. Since 1-chlorobutane is a primary alkyl halide, 1-butene is formed when 1-chlorobutane is reacted with a bulky base such as t -BuOK or t -BuOH in presence of strong heat. This is an E2 reaction.

Secondly, the 1-butene is reacted with bromine in carbon tetrachloride. The vicinal dihalide (1,2-dibromobutane) is formed. This can now undergo further elimination reactions in the presence of sodamide and strong heat to yield 1-butyne which is the desired product. These reactions involve the elimination of the first HBr molecule to give an alkenyl bromide. A second elimination now gives the terminal alkyne.

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Why are relative atomic masses on the periodic tables not whole numbers?
Kobotan [32]

Answer:

They are averages.

Explanation:

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6 0
3 years ago
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The compound Xe(CF3)2 decomposes in afirst-order reaction to elemental Xe with a half-life of 30. min.If you place 7.50 mg of Xe
Irina-Kira [14]

Answer:

t=147.24\ min

Explanation:

Given that:

Half life = 30 min

t_{1/2}=\frac{\ln2}{k}

Where, k is rate constant

So,  

k=\frac{\ln2}{t_{1/2}}

k=\frac{\ln2}{30}\ min^{-1}

The rate constant, k = 0.0231 min⁻¹

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

Given that:

The rate constant, k = 0.0231 min⁻¹

Initial concentration [A_0] = 7.50 mg

Final concentration [A_t] = 0.25 mg

Time = ?

Applying in the above equation, we get that:-

0.25=7.50e^{-0.0231\times t}

750e^{-0.0231t}=25

750e^{-0.0231t}=25

x=\frac{\ln \left(30\right)}{0.0231}

t=147.24\ min

6 0
3 years ago
If an experiment calls for 0.200mole acetic acid (Hc2H3O2)how many grams of glacial acetic acid do we need?
bija089 [108]
<h3>Molar mass:-</h3>

\\ \sf\longmapsto HC_2H_3O_2

\\ \sf\longmapsto 1u+2(12u)+3(1u)+2(16u)

\\ \sf\longmapsto 1u+24u+3u+48u

\\ \sf\longmapsto 28u+48u

\\ \sf\longmapsto 76u

\\ \sf\longmapsto 76g/mol

  • No of moles=0.2mol
  • Given mass=?

\\ \sf\longmapsto No\:of\;moles=\dfrac{Given\:Mass}{Molar\:Mass}

\\ \sf\longmapsto 0.2=\dfrac{Given\:mass}{76}

\\ \sf\longmapsto Given\:Mass=0.2\times 76

\\ \sf\longmapsto Given\:Mass=1.52g

7 0
3 years ago
the heat of fusion of acetone is 5.7 kJ/mol. Calculate to two significant figures the entropy change when 6.3 mol of acetone mel
shtirl [24]

<u>Answer:</u> The entropy change of the process is 2.0\times 10^2J/K

<u>Explanation:</u>

To calculate the entropy change for different phase at same temperature, we use the equation:

\Delta S=n\times \frac{\Delta H_{f}}{T}

where,  

\Delta S = Entropy change

n = moles of acetone = 6.3 moles

\Delta H_{f} = enthalpy of fusion = 5.7 kJ/mol = 5700 J/mol    (Conversion factor:  1 kJ = 1000 J)

T = temperature of the system = -94.7^oC=[273-94.7]=178.3K

Putting values in above equation, we get:

\Delta S=\frac{6.3mol\times 5700J/mol}{178.3K}\\\\\Delta S=201.4J/K=2.0\times 10^2J/K

Hence, the entropy change of the process is 2.0\times 10^2J/K

4 0
3 years ago
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