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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:

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Explanation:

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The moles of Ca²⁺ can be found using the molar relationship between Ca(NO₃)₂ and Ca²⁺

(0.100mol Ca(NO₃)₂) (Ca²⁺ /Ca(NO₃)₂) = 0.100 mol Ca²⁺

The concentration of Ca²⁺  is then:

[Ca²⁺] = n/V = (0.100mol)/(100.0mL) x (1000ml)/(1L) = 1M

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(0.100mol Ca(NO₃)₂) (2NO₃⁻/Ca(NO₃)₂) = 0.200 mol NO₃⁻

The concentration of NO₃⁻ is then:

[NO₃⁻] = (0.200mol)/(100.0mL) x (1000ml)/(1L) = 2M

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