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Galina-37 [17]
3 years ago
10

The 1H-NMR of a compound with molecular formula C6H12O2 consists of four signals: 1.1 (triplet, integrating to 3 Hydrogens), 1.2

(doublet, integrating to 6 Hydrogens), 2.3 (quartet, integrating to 2 Hydrogens) and 5.0 (septet, integrating to 1 Hydrogen). Propose a structural formula for this compound consistent with this information.

Chemistry
1 answer:
Tamiku [17]3 years ago
4 0

Answer:

Isopropyl propionate  

Explanation:

1. Information from formula

The formula is C₆H₁₂O₂. A six-carbon alkane would have the formula C₆H₁₄. The deficiency of two H atoms indicates the presence of either a ring or a double bond.

2. Information from the spectrum

(a) Triplet-quartet

A 3H triplet and a 2H quartet is the classic pattern for a CH₃CH₂- (ethyl) group

(b) Septet-doublet

A 1H septet and a 6H doublet is the classic pattern for a -CH(CH₃)₂ (isopropyl) group

(c) The rest of the molecule

The ethyl and isopropyl groups together add up to C₇H₁₂.

The rest of the molecule must have the formula CO₂ and one unit of unsaturation. That must be a C=O group.

The compound is either

CH₃CH₂-COO-CH(CH₃)₂ or (CH₃)₂CH-COO-CH₂CH₃.

(d) Well, which is it?

The O atom of the ester function should have the greatest effect on the H atom on the adjacent carbon atom.

The CH of an isopropyl is normally at 1.7. The adjacent O atom should pull it down perhaps 3.2 units to 4.9.  

The CH₂ of an ethyl group is normally at 1.2. The adjacent O atom should pull it down to about 4.4.

We see a signal at 5.0 but none near 4.4. The compound is isopropyl propionate.

3. Summary

My peak assignments are shown in the diagram below.

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A sample of hydrated copper (II) sulfate (CuSO4•nH2O) is heated to 150°C and produces 103.74 g anhydrous copper (II) sulfate and
wariber [46]

Answer:

5

Explanation:

Firstly, we convert what we have to percentage compositions.

There are two parts in the molecule, the sulphate part and the water part.

The percentage compositions is as follows:

Sulphate- (103.74)/(103.74 + 58.55) × 100% = apprx 64%

The water part = 100 - 64 = 36%

Now, we divide the percentages by the molar masses.

For the CuSO4 molar mass is 64 + 32 + 4(16) = 160g/mol

For the H2O = 2(1) + 16 = 18g/mol

Now we divide the percentages by these masses

Sulphate = 64/160 = 0.4

Water = 36/18 = 2

The ratio is thus 0.4:2 = 1:5

Hence, there are 5 water molecules.

3 0
3 years ago
A satellite and a hockey puck are alike because they illustrate Newton's first law of motion.
lutik1710 [3]

Answer:

It would be True

Explanation:

Because they both have the same push of gravity. Gravity affects all objects equally. If you drop an egg and a watermelon at the same time they would both collide with the floor at the same time.

8 0
3 years ago
What Went Wrong – Balancing Chemical Equations
OLEGan [10]
1-It has to be 3 Fe and not Fe3.
2-The oxygens aren't balanced

Balanced equation:
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4 0
3 years ago
This question deals with waste disposal in the Solutions and Spectroscopy experiment. What should be done to waste solutions con
Korvikt [17]

Answer:

b. It should be dumped in a beaker labeled "waste copper" on one's bench during the experiment.

d. It should be disposed of in the bottle for waste copper ion when work is completed.

Explanation:

Solutions containing copper ion should never be disposed of by dumping them in a sink or in common trash cans, because this will cause pollution in rivers, lakes and seas, being a contaminating agent to both human beings and animals. They should be placed in appropriate compatible containers that can be hermetically sealed. The sealed containers must be labeled with the name and class of hazardous substance they contain and the date they were generated.

It never should be returned to the bottle containing the solution, since it can contaminate the solution of the bottle.

In the Solutions and Spectroscopy experiments there is always wastes.

3 0
3 years ago
The following reaction shows the products when sulfuric acid and aluminum hydroxide react.
Elden [556K]

The correct answer is approximately 11.73 grams of sulfuric acid.

The theoretical yield of water from Al(OH)3 is lower than that of H₂SO₄. As a consequence, Al(OH)3 is the limiting reactant, H₂SO₄ is in excess.

The balanced equation is:

2Al(OH)₃ + 3H₂SO₄ ⇒ Al₂(SO₄)₃ + 6H₂O

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The molar mass of H₂SO₄ is 98.706 g/mol. The mass of 0.28845 moles of H₂SO₄ is 0.28845 × 98.706 = 28.289 g

40 grams of sulfuric acid is available, out of which 28.289 grams is consumed. The remaining 40-28.289 = 11.711 g is in excess, which is closest to the first option, that is, 11.73 grams of H₂SO₄.

6 0
3 years ago
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