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mario62 [17]
4 years ago
13

Treatment of 2-hexanone with naoch2ch3 followed by ch3br affords compound x (c7h14o) as the major product. x shows a strong abso

rption in the ir spectrum at 1713 cm-1, and its 1h nmr data is given below. what is the structure of x?

Chemistry
1 answer:
VashaNatasha [74]4 years ago
8 0
The given ketone when reacted with base gave enolate, the enolate formed due to loss of methylene proton next to carbonyl group. Enolate when treated with methyle Bromide gave alpha substituted product. 

Strong absorption around 1713 cm⁻¹ in IR spectrum confirms the presence of Carbonyl group.

The product along with ¹H-NMR values is given below,

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Answer:

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Explanation:

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What is an Atom?<br><br> What are the 5 most common elements that make up the human body?.
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6 0
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What is the molarity of 225 grams of Cu(NO2)2 in a total volume of 2.59 L?
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Answer:

The molarity is 0.56\frac{moles}{L}

Explanation:

In a mixture, the chemical present in the greatest amount is called a solvent, while the other components are called solutes. Then, the molarity or molar concentration is the number of moles of solute per liter of solution.

In other words, molarity is the number of moles of solute that are dissolved in a given volume.

The Molarity of a solution is determined by:

Molarity (M)=\frac{number of moles of solute}{Volume}

Molarity is expressed in units (\frac{moles}{liter}).

Then you must know the number of moles of Cu(NO₂)₂. For that it is necessary to know the molar mass. Being:

  • Cu: 63.54 g/mol
  • N: 14 g/mol
  • O: 16 g/mol

the molar mass of Cu(NO₂)₂ is:

Cu(NO₂)₂= 63.54 g/mol + 2*(14 g/mol + 2* 16 g/mol)= 155.54 g/mol

Now the following rule of three applies: if 155.54 g are in 1 mole of the compound, 225 g in how many moles are they?

moles=\frac{225 g*1 mole}{155.54 g}

moles= 1.45

So you know:

  • number of moles of solute= 1.45 moles
  • volume=2.59 L

Replacing in the definition of molarity:

Molarity=\frac{1.45 moles}{2.59 L}

Molarity= 0.56\frac{moles}{L}

<u><em>The molarity is 0.56</em></u>\frac{moles}{L}<u><em></em></u>

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