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3 years ago

What would be the major product obtained from hydroboration–oxidation of the following alkenes?

Chemistry

1 answer:

zmey [24]3 years ago

8 0

Answer:

a. 3-methylbutan-2-ol

b. 2-methylcyclohexan-1-ol

Explanation:

For this reaction, we must remember that the hydroboration is an "anti-Markovnikov" reaction. This means that the "OH" will be added at the least substituted carbon of the double bond.

In the case of 2-methyl-2-butene, the double bond is between carbons 2 and 3. Carbon 2 has two bonds with two methyls and carbon 3 is attached to 1 carbon. Therefore the "OH" will be added to carbon three producing 3-methylbutan-2-ol.

For 1-methylcyclohexene, the double bond is between carbons 1 and 2. Carbon 1 is attached to two carbons (carbons 6 and 7) and carbon 2 is attached to one carbon (carbon 3). Therefore the "OH" will be added to carbon 2 producing 2-methylcyclohexan-1-ol.

See figure 1

I hope it helps!

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A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli

iVinArrow [24]

Answer:

108.25 ºC

Explanation:

The boiling point elevation for a given solute in water is given by the expression:

ΔTb = i Kbm

where ΔTb is the boiling point elevation

i is the van´t Hoff factor

Kb the boiling constant which for water is 0.512 ºC/molal

m is the molality of the solution

The molality of the solution is the number of moles per kilogram of solvent. Here we run into a problem since we are not given the identity of the coolant, but a search in the literature tells you that the most typical are ethylene glycol and propylene glycol. The most common is ethylene glycol and this it the one we will using in this question.

Now the i factor in the equation above is 1 for non ionizable compounds such as ethylene glycol.

Our equation is then:

ΔTb = Kbm

So lets calculate the molality and then ΔTb:

m = moles ethylene glycol / Kg solvent

Converting gal to L

4.90 g x 3.785 L/gal = 18.55 L

in a 50/50 blend by volume we have 9.27 L of ethylene glycol, and 9.27 L of water.

We need to convert this 9.27 l of ethylene glycol to grams assuming a solution density of 1 g/cm³

9.27 L x 1000 cm³ / L = 9273.25 cm³

mass ethylene glycol = 9273.25 cm³ x 1 g/cm³ = 9273.25 g

mol ethylene glycol = 9273.25 g/ M.W ethylene glycol

= 9273.25 g / 62.07 g/mol =149.4mol

molality solution = 149.4 mol / 9.27 Kg H₂O = 16.12 m

( density of water 1 kg/L )

Finally we can calculate ΔTb:

ΔTb = Kbm = 0.512 ºC/molal x 16.12 molal = 8.25 ºC

boiling point = 100 º C +8.25 ºC = 108.25 ºC

( You could try to solve for propylene glycol the other popular coolant which should give around 106.7 ºC )

6 0

3 years ago

What do non-metals do if they need 8 valence electrons and can't find a metal?

kati45 [8]

Answer:

sharing of electron

Explanation:

As seen in carbon. carbon has four electrons in his outermost shell and needs for electron by sharing of electrons it can gain it by another carbon

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3 years ago

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Vanyuwa [196]

Answer:

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4 years ago

If the solubility of a N30 gas is 2.26g/l at 1.26atm of pressure, what is the solubility of

natali 33 [55]

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3 years ago

Some hormones enter cells via

pochemuha

ANSWER: receptor-meditated endocytosis
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3 years ago