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alekssr [168]
2 years ago
11

Draw the structure with the molecular formula C6H14O that most likely produced the below IR and 1H NMR. The 13C NMR data is 78.3

ppm, 43.9 ppm, 27.3 ppm, 13.3 ppm.

Chemistry
1 answer:
spayn [35]2 years ago
8 0

Answer:

                   The answer is 2,3-dimethylbutan-2-ol and the structure is attached below.

Explanation:

                     Although we are not provided with ¹H-NMR spectrum and IR spectrum but still we can elucidate the ¹³C-NMR data and finalize a plausible structure.

                      First of all we look at the molecular formula, we can conclude from the formula that the structure given is saturated in nature because the hydrogen deficiency index of this formula is zero. Hence, we can say that there is no double bond either between Carbon atoms or between carbon and oxygen atom. This can also be proved by the absence of peaks in downfield as unsaturated compounds and carbonyl compounds give value above 100 and 200 ppm respectively.

                      Secondly, we can also conclude that among the six carbon atom two pairs of them are having same electronic environment because we are having only 4 signals hence we can conclude that two pairs have same chemical shift values.

                       Also, after making every possinble isomer of given molecular formula the structure of 2,3-dimethylbutan-2-ol was found to be the most accurate structure.

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Complexes containing metals with d10 electron configurations are typically colorless because ________. Complexes containing meta
algol [13]

Answer:

there is no d electron that can be promoted via the absorption of visible light

Explanation:

One of the properties of transition elements is the possession of incompletely filled d orbitals. This property accounts for their unique colours.

The colours of transition metal compounds stem from d-d transition of electrons due to the presence of vacant d orbitals of appropriate energy to which electrons could be promoted.

For elements whose atoms have a d10 configuration, such vacant orbitals does not exist hence their compounds are not colored.

Sometimes, the colour of transition metal compounds stem from ligand to metal charge transfer(LMCT) for instance in KMnO4.

8 0
2 years ago
Question 6 pls I need help this is due in 2 minutes!!..
MAVERICK [17]

Answer:

c

Explanation:

7 0
2 years ago
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A mysterious white powder could be powdered sugar (C12H22O11), cocaine (C17H21NO4), codeine (C18H21NO3), norfenefrine (C8H11NO2)
rodikova [14]

Norfenefrine (C₈H₁₁NO₂).

<h3>Further explanation</h3>

We will solve a case related to one of the colligative properties, namely freezing point depression.

The freezing point of the solution is the temperature at which the solution begins to freeze. The difference between the freezing point of the solvent and the freezing point of the solution is called freezing point depression.

\boxed{ \ \Delta T_f = T_f(solvent) - T_f(solution) \ } \rightarrow \boxed{ \ \Delta T_f = K_f \times molality \ }

<u>Given:</u>

A mysterious white powder could be,

  • powdered sugar (C₁₂H₂₂O₁₁) with a molar mass of 342.30 g/moles,
  • cocaine (C₁₇H₂₁NO₄) with a molar mass of 303.35 g/moles,
  • codeine (C₁₈H₂₁NO₃) with a molar mass of 299.36 g/moles,
  • norfenefrine (C₈H₁₁NO₂) with a molar mass of 153.18 g/moles, or
  • fructose (C₆H₁₂O₆) with a molar mass of 180.16 g/moles.

When 82 mg of the powder is dissolved in 1.50 mL of ethanol (density = 0.789 g/cm³, normal freezing point −114.6°C, Kf = 1.99°C/m), the freezing point is lowered to −115.5°C.

<u>Question: </u>What is the identity of the white powder?

<u>The Process:</u>

Let us identify the solute, the solvent, initial, and final temperatures.

  • The solute = the powder
  • The solvent = ethanol
  • The freezing point of the solvent = −114.6°C
  • The freezing point of the solution = −115.5°C

Prepare masses of solutes and solvents.

  • Mass of solute = 82 mg = 0.082 g
  • Mass of solvent = density x volume, i.e., \boxed{ \ 0.789 \ \frac{g}{cm^3} \times 1.50 \ cm^3 = 1.1835 \ g = 0.00118 \ kg  \ }

We must prepare the solvent mass unit in kg because the unit of molality is the mole of the solute divided by the mass of the solvent in kg.

The molality formula is as follows:

\boxed{ \ m = \frac{moles \ of \ solute}{kg \ of \ solvent} \ } \rightarrow \boxed{ \ m = \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }

Now we combine it with the formula of freezing point depression.

\boxed{ \ \Delta T_f =  K_f \times \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }

It is clear that we will determine the molar mass of the solute (denoted by Mr).

We enter all data into the formula.

\boxed{ \ -114.6^0C - (-115.5^0C) = 1.99 \frac{^0C}{m} \times \frac{0.082 \ g}{Mr \times 0.00118 \ kg} \ }

\boxed{ \ 0.9 = \frac{1.99 \times 0.082}{Mr \times 0.00118} \ }

\boxed{ \ Mr = \frac{0.16318}{0.9 \times 0.00118} \ }

We get \boxed{ \ Mr = 153.65 \ }

These results are very close to the molar mass of norfenefrine which is 153.18 g/mol. Thus the white powder is norfenefrine.

<h3>Learn more</h3>
  1. The molality and mole fraction of water brainly.com/question/10861444
  2. About the mass and density of ethylene glycol as an  antifreeze brainly.com/question/4053884
  3. About the solution as a homogeneous mixture  brainly.com/question/637791

Keywords: a mysterious white powder, sugar, cocaine, codeine, norfenefrine, fructose, the solute, the solvent, dissolved, ethanol, normal freezing point, the freezing point depression, the identity

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3 years ago
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Which title is most appropriate for this table?
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a.uses of functional groups involving oxygen

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3 years ago
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Is 61 degrees considered cold?
IrinaVladis [17]
Yes because my great grandfather froze at 60 degrees
7 0
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