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Vladimir79 [104]
3 years ago
5

2-Methyl-2-pentanol can be made starting from two different ketone electrophiles using two different Grignard reagents: one from

a lower molecular mass Grignard reagent and one from a higher molecular mass Grignard reagent. Provide the retrosynthetic analysis for both routes using bromine as the halogen.
Chemistry
1 answer:
elena-s [515]3 years ago
7 0

Answer:

Explanation:

In this case, we can do this by thinking a little.

Both ways use a grignard reagent, the difference between both ways is that one use an electrophile ketone with a low molecular mass, and the other has a high molecular mass.

The grignard reagent is commonly used to reduce carbonyle groups to alcohols. In the first step, a complex with the reagent is formed in the carbonile, and in the second step, the oxygen atom is hidrated in acid or basic medium and form the respective alcohol.

For the first way, we will use a high molecular mass ketone. In this case the 2-pentanone reacting with CH₃MgBr as a grignard reagent.

For the second way, we will use a low molecular mass ketone, in this case Acetone, reacting with CH₃CH₂CH₂MgBr. Both of them, will give the same product of 2 methyl-2-pentanol. See picture below for that

Hope this helps

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Answer:

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Explanation:

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2 years ago
CS2 (s) + 3 O2 (g) → CO2 (g) + 2 SO2 (g)
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Answer:

2.067 L ≅ 2.07 L.

Explanation:

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It is clear that 1.0 mole of CS₂ react with 3.0 mole of O₂ to produce 1.0 mole of CO₂ and 2.0 moles of SO₂.

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It is known that at STP: every 1.0 mol of any gas occupies 22.4 L.

<u><em>using cross multiplication:</em></u>

1.0 mol of O₂ represents → 22.4 L.

??? mol of O₂ represents → 3.1 L.

∴ 3.1 L of O₂ represents = (1.0 mol)(3.1 L)/(22.4 L) = 0.1384 mol.

  • To find the no. of moles of SO₂ produced from 3.1 liters (0.1384 mol) of hydrogen:

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3.0 mol of O₂ produce → 2.0 mol of SO₂, from stichiometry.

0.1384 mol of O₂ produce → ??? mol of SO₂.

∴ The no. of moles of SO₂ = (2.0 mol)(0.1384 mol)/(3.0 mol) = 0.09227 mol.

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