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Finger
1 month ago
11

Draw a structure for the product of nucleophilic substitution obtained on solvolysis of tert−butyl bromide in methanol, and arra

nge the correct mechanism for its formation. be sure to answer all part

Chemistry
2 answers:
Alekssandra [3K]1 month ago
5 0
The mechanism will include a) A slow step, which is the rate-determining step In this phase, the leaving group (chloride ion) departs from the substrate, forming carbocation b) A fast step: The nucleophile will interact with the cation, resulting in the substituted product. This mechanism is depicted as follows:
Alekssandra [3K]1 month ago
4 0
The final product formed from the solvolysis of tert-butyl bromide in methanol is 2-methoxy-2-methylpropane. Further Explanation: Nucleophilic substitution is when a nucleophile with electron excess replaces the leaving group on the substrate. Such reactions play a significant role in converting one functional group into another. There are two main types of nucleophilic substitutions: 1. A two-step substitution, where the leaving group is first removed to create a carbocation intermediate, followed by the nucleophile attacking the carbocation to create the final product. 2. A single-step substitution, in which the nucleophile attacks the leaving group from the opposite side of the electrophilic center, resulting in an inversion of configuration. The mechanism for the solvolysis of tert-butyl bromide in methanol follows this type (refer to the accompanying image), hence: Step 1: a carbocation intermediate forms from tert-butyl bromide upon the departure of the bromide ion. Step 2: Methanol functions as the nucleophile, attacking the carbocation intermediate. Step 3: The final step involves deprotonation, yielding the end product, 2-methoxy-2-methylpropane.
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In a titration experiment, H2O2(aq) reacts with aqueous MnO4-(aq) as represented by the equation above. The dark purple KMnO4 so
Alekssandra [3086]

Respuesta:

El oxígeno en H2O2 es la especie que se reduce a H2O y se oxida a O2.

Explicación:

5 H2O2(aq) + 2 MnO4-(aq) + 6 H+(aq) → 2 Mn2+(aq) + 8 H2O(l) + 5 O2(g)

La oxidación se define como la pérdida de electrones. La oxidación provoca un aumento en el número de oxidación de un elemento.

Si se descompone esta reacción en sus mitades de reducción y oxidación

Se observa que, de los reactivos mencionados anteriormente,

H202 se convierte en H2O y O2

MnO4- + H+ se convierte en Mn2+ y H2O

El número de oxidación de Mn cambia de +7 en MnO4- a +2 en Mn2+ (lo que indica evidentemente una reducción)

El oxígeno en MnO4- no cambia su número de oxidación, ya que se mantiene en -2

El número de oxidación del oxígeno cambia de -1 en H2O2 a -2 en H2O y 0 en O2

El hidrógeno en H2O2 no cambia su número de oxidación, y su número de oxidación se mantiene en +1 tanto en H2O2 como en H2O.

Esto indica que H2O2 sufre tanto oxidación como reducción; más específicamente, el oxígeno en H2O2 es la especie que se reduce a H2O y se oxida a O2.

Espero que esto ayude

7 0
2 months ago
A 20.0–milliliter sample of 0.200–molar K2CO3 so­lution is added to 30.0 milliliters of 0.400–mo­lar Ba(NO3)2 solution. Barium c
KiRa [2933]

Respuesta:

0.16 M

Explicación:

Teniendo en cuenta:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

O sea,

Moles =Molarity \times {Volume\ of\ the\ solution}

Dado que:

Para K_2CO_3 :

Molaridad = 0.200 M

Volumen = 20.0 mL

Convierte mL a L:

1 mL = 10⁻³ L

Entonces, volumen = 20.0×10⁻³ L

Los moles de K_2CO_3 son:

Moles=0.200 \times {20.0\times 10^{-3}}\ moles

Moles de K_2CO_3 = 0.004 moles

Para Ba(NO_3)_2 :

Molaridad = 0.400 M

Volumen = 30.0 mL

Convertimos mL a L:

1 mL = 10⁻³ L

Volumen = 30.0×10⁻³ L

Entonces, los moles de Ba(NO_3)_2 son:

Moles=0.400 \times {30.0\times 10^{-3}}\ moles

Moles de Ba(NO_3)_2 = 0.012 moles

Según la reacción:

Ba(NO_3)_2 + K_2CO_3\rightarrow BaCO_3 + 2KNO_3

1 mol de Ba(NO_3)_2 reacciona con 1 mol de K_2CO_3

Por lo tanto,

0.012 mol de Ba(NO_3)_2 reacciona con 0.012 mol de K_2CO_3

Moles disponibles de K_2CO_3 = 0.004 mol

El reactivo limitante es el que está en menor cantidad, entonces K_2CO_3 es el limitante (0.004 < 0.012).

La formación del producto depende del reactivo limitante, así que,

1 mol de K_2CO_3 reacciona con 1 mol de Ba(NO_3)_2 y produce 1 mol de BaCO_3

0.004 mol de K_2CO_3 reacciona con 0.004 mol de Ba(NO_3)_2 y genera 0.004 mol de BaCO_3

Los moles restantes de Ba(NO_3)_2 son: 0.012 - 0.004 = 0.008 mol

El volumen total es 20 + 30 mL = 50 mL = 0.050 L

Por lo que la concentración del ion bario, Ba^{2+}, después de la reacción es:

Molarity=\frac{0.008}{0.050}\ M = 0.16\ M

3 0
3 months ago
How many grams of calcium cyanide (Ca(CN)2) are contained in 0.79 mol of calcium cyanide?
alisha [2963]
I am not sure, but do you know da wae brudda?


3 0
1 month ago
In a group assignment, students are required to fill 10 beakers with 0.720 M CaCl2. If the molar mass of CaCl2 is 110.98 g/mol a
alisha [2963]
The result is 200 g. Given that the molar mass of CaCl2 is 110.98 g/mol, this indicates that there are 110.98 g in 1 L of a 1 M solution. Let's calculate the amount of CaCl2 in 0.720 M. Using the proportion 110.98 g: 1 M = x: 0.720 M, we find x to be 79.90 g. Therefore, in 1 L of a 0.720 M solution, there is 79.90 g. Next, we need to create ten beakers with 250 mL each, totaling 10 * 250 mL = 2500 mL or 2.5 L. Then, using the equation 79.90 g: 1 L = x: 2.5 L, we calculate x = 79.90 g * 2.5 L: 1 L, resulting in x = 199.75 g, approximately 200 g.
8 0
1 month ago
Read 2 more answers
The normal boiling point of c2cl3f3 is 47.6°c and its molar enthalpy of vaporization is 27.49 kj/mol. what is the change in entr
eduard [2782]
Based on the equation:

ΔG = ΔH - TΔS = 0

It follows that ΔS = ΔH/T

So, ΔS = n*ΔHVap / Tvap

- where n represents the number of moles calculated as mass/molar mass

For a mass of 24.1 g

and a molar mass of 187.3764 g/mol

substituting gives:

∴ n = 24.1 / 187.3764g/mol

      = 0.129 moles

The molar enthalpy of vaporization, ΔHvap, is 27.49 kJ/mol

The temperature in Kelvin, Tvap = 47.6 + 273 = 320.6 K

After substitution, we compute ΔS, the change in entropy:

∴ΔS = 0.129 mol * 27490 J/mol / 320.6 K

      = 11 J/K
7 0
2 months ago
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