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Salsk061 [2.6K]
3 years ago
10

Amantadine is effective in preventing infections caused by the influenza A virus and in treating established illnesses. It is th

ought to block a late stage in the assembly of the virus. Amantadine is synthesized by treating 1-bromoadamantane with acetonitrile in sulfuric acid to give N-adamantylacetamide, which is then converted to amantadine.
(a) Propose a mechanism for the transformation in Step 1.
(b) Describe experimental conditions to bring about Step 2.

Chemistry
1 answer:
Sliva [168]3 years ago
7 0

Answer:

See image attached

Explanation:

a)

The full reaction mechanism of step 1 was obtained from Bartleby and attached to this answer. The steps involved in the reaction are:

1) Loss of Br- and formation of a carbocation

2) Attack of CH3CN on the carbocation

4) Formation of a quaternary nitrogen intermediate

5) Attack of water on the quaternary nitrogen intermediate

6) Loss of the water molecule

5) Formation of the amide product

b)

i) sodium hydroxide

ii) HCl

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The volume of 0.555M KNO3 solution would contain 12.5 g of solute iss 223 mL.

<h3>What is the relationship between mass of solute and concentration of solution?</h3>

The mass of solute in a given volume of solution is related by the formula below:

  • Molarity = mass/(molar mass * volume)

Therefore, volume of solution is given by:

Volume = Mass /molarity * molar mass

Molar mass of KNO₃ = 101 g/mol

Volume = 12.5/(0.555 * 101)

Volume = 0.223 L or 223 mL

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State the law of multiple proportions.
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<u>Answer:</u> The percent yield of the 1-bromobutane is 48.65 %

<u>Explanation:</u>

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Given mass of NaBr = 11.1 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

\text{Moles of NaBr}=\frac{11.1g}{103g/mol}=0.108mol

The chemical equation for the reaction of 1-butanol and NaBr is:

\text{1-butanol + NaBr}\rightarrow \text{1-bromobutane}

By Stoichiometry of the reaction

1 mole of NaBr produces 1 mole of 1-bromobutane

So, 0.108 moles of NaBr will produce = \frac{1}{1}\times 0.108=0.108 moles of 1-bromobutane

  • Now, calculating the mass of 1-bromobutane from equation 1, we get:

Molar mass of 1-bromobutane = 137 g/mol

Moles of 1-bromobutane = 0.108 moles

Putting values in equation 1, we get:

0.108mol=\frac{\text{Mass of 1-bromobutane}}{137g/mol}\\\\\text{Mass of 1-bromobutane}=(0.108mol\times 137g/mol)=14.80g

  • To calculate the percentage yield of 1-bromobutane, we use the equation:

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Experimental yield of 1-bromobutane = 7.2 g

Theoretical yield of 1-bromobutane = 14.80 g

Putting values in above equation, we get:

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