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IgorC [24]
3 years ago
6

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 <u>"anti-Markovnikov" reaction</u>. This means that the "OH" will be added at the <em>least substituted carbon of the double bond.</em>

In the case of <u>2-methyl-2-butene</u>, 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 <u>the "OH" will be added to carbon three</u> producing <u>3-methylbutan-2-ol</u>.

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<u> the "OH" will be added to carbon 2</u> producing <u>2-methylcyclohexan-1-ol</u>.

See figure 1

I hope it helps!

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Answer:

(a) color

(b) endothermic

(c)

b The ΔH value would have the same magnitude value but opposite sign.

c The K expression would be inverted.

Explanation:

Let's consider the following reaction at equilibrium.

      A(g) ⇄ 2 B(g)

colorless    dark colored

<em>(a) The cylinder should appear (color or colorless)</em>

At equilibrium, there is a mixture of A and B, so the cylinder should appear colored.

<em>(b) When the system is cooled, the cylinder's appearance becomes very light colored. Therefore, the reaction must be (endothermic or exothermic)</em>

According to Le Chatelier's Principle when a perturbation is made to a system at equilibrium it will react to counteract such effect. When the system is cooled, it will tend to increase the temperature by releasing heat. In this case, the reaction is endothermic so when the reverse reaction is favored, colorless A is favored as well.

<em>Suppose the reaction equation were written as follows: 2 B(g) ⇄ A(g) </em>

<em>(c) Which of these statements would then be true?</em>

<em>a The value of K would not change.</em> FALSE. The new K would be the inverse of the direct K.

<em>b The ΔH value would have the same magnitude value but opposite sign. </em>TRUE. This is stated by Lavoisier-Laplace Law.

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<em>d The color of the cylinder would be darker.</em> FALSE. Changing the way the reaction is expressed has no effect on the equilibrium.

7 0
4 years ago
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leva [86]

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IceJOKER [234]

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14.61 g NaCl

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7 0
2 years ago
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pantera1 [17]

Answer:

<h3>\huge{ \underline{ \underline { \bold{ \sf{ \orange{See \: below}}}}}}</h3>

Explanation:

▪️\underline{ \bold{ \sf{ \blue{Chemical \: reaction}}}}

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▪️\underline{ \bold{ \sf{ \purple{ \: Further \: more \: explanation}}}}

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▪️\underline{ \bold{ \sf{ \purple{For \: example}}}}

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N₂ + 3H₂ ⇒ 2NH₃

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Hope I helped!

Best regards!!

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