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3 years ago

By what mechanism does cyclohexanol react when treated in concentrated sulfuric acid/phosphoric acid and what compound results?

Chemistry

1 answer:

slava [35]3 years ago

6 0

Answer:

The mechanism is E1 and the product is cyclohexene

Explanation:

Attached to this answer is an image of the reaction mechanism of the reaction between cyclohexanol and concentrated sulfuric acid/phosphoric acid. The acid (H3O^+) acts as catalyst.

We can see that the first step in the mechanism is the protonation of the -OH group of the alcohol. This is followed by loss of water molecule which is a good leaving group.

The water now acts as a Lewis base by abstracting a proton from the substrate, forming cyclohexene and regenerating the acid catalyst.

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Gnom [1K]

Answer:

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Explanation:

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3 years ago

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wolverine [178]

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3 years ago

Does a low or a high pressure air mass have a greater atmospheric pressure than the surrounding air masses?

Ghella [55]

Answer:

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Explanation:

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3 years ago

Read 2 more answers

What substance cannot be decomposed by physical or chemical means?

kakasveta [241]

Elements. Elements cannot be further separated as they are the most basic substances.

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4 years ago

Air, with an initial absolute humidity of 0.016 kg water per kg dry air, is to be dehumidified for a drying process by an air co

amid [387]

Answer:

(a) The proportion of dry air bypassing the unit is 14.3%.

(b) The mass of water removed is 1.2 kg per 100 kg of dry air.

Explanation:

We can express the proportion of air that goes trough the air conditioning unit as $p_{d}$ and the proportion of air that is by-passed as $p_{bp}$ , being $p_{d}+p_{bp}=1$ .

The amount of water that goes into the drier inlet has to be 0.004 kg/kg, and can be expressed as:

$0.004 = 0.016*p_{bp}+ 0.002*p_{d}$

Replacing the first equation in the second one we have

$0.004 = 0.016*(1-p_{d})+ 0.002*p_{d}=0.016-0.016*p_{d}+0.002*p_{d}\\0.004 - 0.016 = (-0.016+0.002)*p_{d}\\-0.012 = -0.014*p_{d}\\p_{d}=\frac{-0.012}{-0.014}=0.857\\\\p_{bp}=1-p_{d}=1-0.857=0.143$

(b) Of every kg of dry air feed, 85.7% goes in to the air conditioning unit.

It takes (0.016-0.002)=0.014 kg water per kg dry air feeded.

The water removed of every 100 kg of dry air is

$100 kgDA*0.857*0.014 kgW/kgDA= 1.1998 \approx 1.2 kgW$

It can also be calculated as the difference in humiditiy between the inlet and the outlet: (0.016-0.004=0.012 kgW/kDA) and multypling by the total amount of feed (100 kgDA).

100 * 0.012 = 1.2 kgW

7 0

4 years ago