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vaieri [72.5K]
3 years ago
11

Devise a 6‑step synthesis of the starting material from the product. The cyclopentene is a key intermediate.

Chemistry
1 answer:
Grace [21]3 years ago
7 0

Cyclopentene is industrially produced in large quantities. It is used as a monomer for the synthesis of plastics, in addition to other chemical syntheses.

<h3>The 6‑step synthesis of the starting material from the product</h3>

Reaction 1 - Preparation of keto-aldehyde

Reaction 2 - Preparation of cyclopentene carbaldehyde

Reaction 3 -Preparation of α,β-epoxy-carbaldehyde cyclopentane

Reaction 4 - Preparation of cyclopentene alcohol

Reaction 5 -Preparation of α,β-epoxy-cyclopentane alcohol

Reaction 6 -Preparation of α,β-epoxy-carbaldehyde cyclopentane

With this information we can conclude that cyclopentene is used as a monomer for the synthesis of plastics.

Learn more about synthesis of plastics in brainly.com/question/13833034

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3 years ago
Identify the outer electron configurations for the (a) alkali metals, (b) alkaline earth metals, (c) halogens, (d) noble gases.
Doss [256]
a) Alkali metals

=> group 1

=> Li:  1s2 2s  => 1s
     Na: [Ne] 3s => 3s
     K: [Ar] 4s => 4s
     Rb: [Kr] 5s => 5s
     Cs: [Xe] 6s => 6s
     Fr: [Rn] 7s => 7s

=> outer electron configuration is ns, where n is the main energy level: 1, 2, 3, 4, 5, 6,7.
     
b) Alkaline earth metals

=> group 2 => you have to add 1 electron to the alkaly metal of the same row.

=> Be: [He] 2s2 => 2s2
     Mg: [Ne] 3s2 => 3s2
     Ca: [Ar] 4s2 => 4s2
     Sr: [Kr] 5s2 => 5s2
     Ba: [Xe] 6s2 => 6s2
     Ra: [Rn[ 7s2 => 7s2

=>outer electron configuration is n s2, where n is the main energy level: 1, 2, 3, 4, 5, 6, 7

c) halogens

=> group 7

=> F: [He] 2s2 2p5 => 2s2 2p5
     Cl: [Ne] 3s2 3p5 => 3s2 3p5
     Br: [Ar] 3d10 4s2 4p5 => 4s2 4p5
     I: [Kr] 4d10 5s2 5p6 => 5s2 5p5
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=> outer electron configuration is ns2 np5, where n is the main energy level 1, 2, 3, 4, 5, 6, 7

d) Noble gases

=> group 8

I will show only the outer shell which is what is requested

=> He: 1s2
     Ne: ... 2s2 2p6
     Ar: ... 3s2 3p6
     Kr: ... 4s2 4p6
     Xe: ... 5s2 5p6
     Rn: ... 6s2 6p6

=> the outer electron configuration is ns2 np6, except for He for which it is 1s2
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4 years ago
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At 46°C a sample of ammonia gas exerts a pressure of 5.3 atm. What is the pressure when the volume of the gas is reduced to one-
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Hello there!

As you may know, pressure and volume have an inverse relationship. This means that as pressure increases, volume decreases (and as volume increases, pressure decreases).

In this problem, the volume of the gas is reduced to 0.250 its original value. This means that the pressure should be higher than it was originally.

We can use Boyle's Law to solve this problem, seeing that temperature is kept constant.

P1V1 = P2V2

Let's rearrange this equation to isolate the variable we need, P2.

P1V1

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 V2

That's better! Now that we have that down, let's plug in the numbers we were given.

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We can get rid of V1, as the number would be equal to 1 (anything divided by itself is equal to 1).

5.3atm

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0.25L

Now, while the number is significantly higher, this is quite reasonable.

Let's plug it in and see if we get what we should:

5.3(V1) = (0.25V1)(21.2)

--------

0.25V1

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Seems right to me!

Feel free to comment if you need any more help. :)

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