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sergeinik [125]
3 years ago
8

Compound a, c8h10, undergoes nitration to give 2 products. The 1h nmr spectrum of a shows a complex five-proton multiplet at 7.2

7-7.40 δ, a two-proton quartet at 2.60 δ, and a three-proton triplet at 1.25 δ. What is the structure of a?

Chemistry
1 answer:
alekssr [168]3 years ago
8 0
<h3>Answer:</h3>

             Compound A is Ethylbenzene.

<h3>Explanation:</h3>

                   The structure of Ethylbenzene was elucidated and finalized based on following facts:

1)    The presence of multiplet signal ranging from δ  7.27 to 7.40 ppm confirms the presence of aromatic protons. Also the statement says that this signal is for 5 protons indicating that the benzene ring is mono substituted.

2)   The substituted group on benzene is Ethyl because the quartet for two protons shows that two protons (-CH₂-) are coupled with a CH₃ group. Hence, the ethyl group gives one quartet and one triplet signal.

3)    The chemical reaction stated in statement is Nitration reaction. Also, the alkyl group (ethyl) is ortho-para directing in nature. Hence, it will give two products i.e. 1-Ethyl-2-nitrobenzene and 1-Ethyl-4-nitrobenzene respectively.

Attached figure gives the calculated NMR spectrum of Ethylbenzene and the chemical shifts calculated are almost near to the given values. Also, the two products formed are shown.

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Lunna [17]
The answer is in the picture which is given below:

8 0
3 years ago
For each set of values, calculate the missing variable using the ideal gas law.
allsm [11]

Answer:

1. n = 0.174mol

2. T= 26.8K

3. P = 1.02atm

4. V = 126.88L

Explanation:

1. P= 2.61atm

V = 1.69L

T = 36.1 °C = 36.1 + 273= 309.1K

R = 0.082atm.L/mol /K

n =?

n = PV / RT = (2.61x1.69)/(0.082x309.1)

n = 0.174mol

2. P = 302 kPa = 302000Pa

101325Pa = 1atm

302000Pa = 302000/101325 = 2.98atm

V = 2382 mL = 2.382L

T =?

n = 3.23 mol

R = 0.082atm.L/mol /K

T= PV /nR = (2.98x2.382)/(3.23x0.082) = 26.8K

3. P =?

V = 0.0250 m³ = 25L

T = 288K

n = 1.08mol

R = 0.082atm.L/mol /K

P = nRT/V = (1.08x0.082x288)/25 = 1.02atm

4. P = 782 torr

760Torr = 1 atm

782 torr = 782/760 = 1.03atm

V =?

T = 303K

n = 5.26 mol

R = 0.082atm.L/mol /K

V = nRT/P

V = (5.26x0.082x303)/1.03 = 126.88L

8 0
4 years ago
18 An important environmental consideration is the appropriate disposal of cleaning solvents. An environmental waste treatment c
Katyanochek1 [597]

Answer:

a) Percentage by mass of carbon: 18.3%

   Percentage by mass of hydrogen: 0.77%

b)  Percentage by mass of chlorine: 80.37%

c) Molecular formula: C_{2} H Cl_{3}

Explanation:

Firstly, the mass of carbon must be determined by using a conversion factor:

0.872g CO _{2} *\frac{12g C}{44g CO_{2} } = 0.238g CO_{2}

The same process is used to calculate the amount of hydrogen:

0.089g H_{2}O*\frac{2g H}{18g H_{2}O }  = 0.010g H

The percentage by mass of carbon and hydrogen are calculated as follows:

%C\frac{0.238g}{1.3g} *100%= 18.3%

%H\frac{0.010g}{1.3g} *100%=0.77%

From the precipation data it is possible obtain the amount of chlorine present in the compound:

1.75 AgCl*\frac{35.45g Cl}{143.45g AgCl}= 0.43g AgCl

Let's calculate the percentage by mass of chlorine:

%Cl=\frac{0.43g}{0.535g} * 100%= 80.37%

Assuming that we have 100g of the compound, it is possible to determine the number of moles of each element in the compound:

18.3g C*\frac{1mol C}{12g C} = 1.52mol C

0.77g H*\frac{1mol H}{1g H} = 0.77mol H

80.37gCl*\frac{1molCl}{35.45g Cl} = 2.27mol Cl

Dividing each of the quantities above by the smallest (0.77mol), the  subscripts in a tentative formula would be

C=\frac{1.52}{0.77} = 1.97 ≈ 2

H = \frac{0.77}{0.77} = 1

Cl =\frac{2.27}{0.77}=2.94≈3

The empirical formula for the compound is:

C_{2} H Cl_{3}

The mass of this empirical formula is:

mass of C + mass of H + mass of Cl= 24g +1+ 106.35 =131.35g

This mass matches with the molar mass, which means that the supscript in the molecular formula are the same of the empirical one.

5 0
3 years ago
What is the molality of a solution of 10 g NaOH in 500 g water?<br>Molar mass NaOH = 40 g​
umka2103 [35]

Explanation:

Moles of NaOH = 10g / (40g/mol) = 0.25mol.

0.25mol / 500g = 0.50mol / 1000g = 0.50mol/dm³.

The molarity is 0.50mol/dm³.

4 0
3 years ago
A sample of table sugar (sucrose, C12H22O11) has a mass of 1.202 g.
Archy [21]

Answer:

a) 0.003512 moles

b) Moles C= 0.04214 moles carbon

Moles H = 0.07726 moles hydrogen

Moles O = 0.03863 moles of oxygen

c) C atoms = 2.54 *10^22 carbon atoms

H atoms = 4.65 *10^22 hydrogen atoms

O atoms = 2.33 *10^22 oxygen atom

Explanation:

Step 1: Data given

Mass of sucrose = 1.202 grams

Molar mass of sucrose = 342.3 g/mol

Step 2: Calculate moles of sucrose

Moles sucrose = Mass sucrose / molar mass sucrose

Moles sucrose = 1.202 grams / 342.3 g/mol

Moles sucrose = 0.003512 moles

Step 3: Calculate moles of each element

For 1 mol of C12H22O11 we have 12 moles of carbon, 22 moles of hydrogen and 11 moles of oxygen

Moles C: 12*0.003512 = 0.04214 moles carbon

Moles H: 22* 0.003512 = 0.07726 moles hydrogen

Moles O: 11* 0.003512 = 0.03863 moles of oxygen

Step 4: Calculate the number of atoms

C atoms = 6.022 *10^23 / mol * 0.04214 moles = 2.54 *10^22 atoms carbon

H atoms = 6.022 * 10^23 / mol * 0.07726 moles = 4.65 *10^22 atoms H

O atoms = 6.022 * 10^23 / mol * 0.03863 moles = 2.33 *10^22 atoms O

0.003512 moles of sucrose containse 6.022 *10^23 * 0.003512 = 2.11 * 10^21 sucrose molecules

7 0
3 years ago
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