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JulsSmile [24]
3 years ago
9

Draw a mechanism for the reaction of methylamine with 2-methylpropanoic acid. Draw any necessary curved arrows. Show the product

s of the reaction. Include any nonzero formal charges and all lone pairs of electrons. Indicate which side of the reaction is favored at equilibrium.

Chemistry
1 answer:
Nitella [24]3 years ago
5 0

Answer:

See figure 1

Explanation:

On this case we have a <u>base</u> (methylamine) and an <u>acid</u> (2-methyl propanoic acid). When we have an acid and a base an <u>acid-base reaction </u>will take place, on this specific case we will produce an <u>ammonium carboxylate salt.</u>

Now the question is: <u>¿These compounds can react by a nucleophile acyl substitution reaction?</u> in other words <u>¿These compounds can produce an amide? </u>

Due to the nature of the compounds (base and acid), <u>the nucleophile</u> (methylamine) <u>doesn't have the ability to attack the carbon</u> of the carbonyl group due to his basicity. The methylamine will react with the acid-<u>producing a positive charge</u> on the nitrogen and with this charge, the methylamine <u>loses all his nucleophilicity.</u>

I hope it helps!

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Answer:

P2= 0.696atm

Explanation:

Applying Boyle's law

P1V1= P2V2

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3 years ago
An electron is the largest part of an atom. True or False?
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Answer: False

Explanation:

Electrons are the smallest of the three particles that make up atoms. Electrons are found in shells or orbitals that surround the nucleus of an atom. Protons and neutrons are found in the nucleus. They group together in the center of the atom.

6 0
3 years ago
The equilibrium concentrations of the reactants and products are [HA]=0.280 M, [H+]=4.00×10−4 M, and [A−]=4.00×10−4 M. Calculate
Tems11 [23]

Answer:

6.24

Explanation:

The following data were obtained from the question:

Concentration of HA, [HA] = 0.280 M,

Concentration of H+, [H+] = 4×10¯⁴ M

Concentration of A-, [A−] = 4×10¯⁴ M

pKa =.?

Next, we shall write the balanced equation for the reaction. This is given below:

HA <===> H+ + A-

Next, we shall determine the equilibrium constant Ka for the reaction. This can be obtained as follow:

Equilibrium constant for a reaction is simply the ratio of concentration of the product raised to their coefficient to the concentration of the reactant raised to their coefficient.

The equilibrium constant for the above equation is given below:

Ka = [H+] [A−] /[HA]

Concentration of HA, [HA] = 0.280 M,

Concentration of H+, [H+] = 4×10¯⁴ M

Concentration of A-, [A−] = 4×10¯⁴ M

Equilibrium constant (Ka) =

Ka = (4×10¯⁴ × 4×10¯⁴) / 0.280

Ka = 1.6×10¯⁷/ 0.280

Ka = 5.71×10¯⁷

Therefore, the equilibrium constant for the reaction is 5.71×10¯⁷

Finally, we shall determine the pka for the reaction as follow:

Equilibrium constant, Ka = 5.71×10¯⁷

pKa =?

pKa = – Log Ka

pKa = – Log 5.71×10¯⁷

pKa = 6.24

Therefore, the pka for the reaction is 6.24.

6 0
3 years ago
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Alja [10]

2.1648 kg of CH4 will generate 119341 KJ of energy.

Explanation:

Write down the values given in the question

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CH4 +2 O2 → CO2 +2 H20

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to produce 119341 KJ of energy, 135.3 mol of CH4 and 270.6 mol of O2 will require

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=2.1648 kg of CH4

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4 0
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Answer:

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