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abruzzese [7]
4 years ago
9

Compound a has molecular formula c5h10. hydroboration-oxidation of compound a produces 2-methylbutan-1-ol. draw the structure of

compound a:

Chemistry
2 answers:
maw [93]4 years ago
8 0
To help you answer this question, you can search a sample reaction in the internet for hydroboration-oxidation. Take a look at the attached picture. The sample reaction is shown at the top. We can deduce that the original compound must be an alkene. Following the pattern, compound a must be 2-methylbutene as shown in the picture.

Ugo [173]4 years ago
5 0

The structure of compound that has general formula {{\mathbf{C}}_{\mathbf{5}}}{{\mathbf{H}}_{{\mathbf{10}}}} and leads to the formation of 2-methylbutan-1-ol is attached in the image.

Further explanation:

Hydroboration-Oxidation is a two-step pathway used to produce alcohol. The reaction proceeds in a manner, where the hydrogen attaches to carbon that is more substituted and the boron attach to the least substituted carbon in the alkene double bond. The borane serves as a lewis acid by accepting two electrons from an alkene. This process completes octet of boron. Also,this process does not require any activation by a catalyst. The hydroboration oxidation mechanism has the elements of both hydrogenation and electrophilic addition.

Hydroboration oxidation reaction involves two steps given as follows:

1. In the first step, borane is added to alkene to form alkyl borane. This step is called as hydroboration of alkene.

2. In the next step, alkyl borane reacts with hydrogen peroxide and sodium hydroxide to form alcohol. This step is known as oxidation.

Given, 2-methylbutan-1-ol is formed by alkene having general formula {{\text{C}}_{\text{5}}}{{\text{H}}_{{\text{10}}}}. The given alcohol is formed by alkene through hydroboration oxidation reaction. The alkene having general formula {{\text{C}}_{\text{5}}}{{\text{H}}_{{\text{10}}}} and that leads to the formation of 2-methylbutan-1-ol must be 2-methylbutene.

The mechanism of the formation of 2-methylbutan-1-ol is attached in the image.

Learn more:

1. What coefficients are needed to balance the equation for the complete combustion of methane? brainly.com/question/1971314

2. Calculate energy required to break the hydrogen bond?brainly.com/question/10278354

Answer details:

Grade: Senior School

Subject: 2.Chemistry

Chapter: Hydroboration oxidation reaction

Keywords: Borane, hydroboration oxidation reaction, hydrogen peroxide, sodium hydroxide, alkyl borane, C5H10, 2-methylbutene and 2-methylbutan-1-ol.

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Answer:

4.) 9, 1, and 4    5.) 4, 1, and 4

Explanation:

I am not quite sure about this because I cannot remember if the coefficient (the number before the elements) is applied to every element in the compound. If it is then your number of atoms are as follows: CORRECTION: you do not have to apply the coefficient to every element only the one that is after it. So when you back and fix the error your number of atoms will be as follows:

number 4

H: 9

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you can calculate the number of atoms present in this compound by multiplying the coefficient and the subscripts of each atom.

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3 years ago
A sample of a pure compound that weighs 60.3 g contains 20.7 g Sb (antimony) and 39.6 g F (fluorine). What is the percent compos
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Answer:

The percent composition of fluorine is 65.67%

Explanation:

Percent Composition is a measure of the amount of mass an element occupies in a compound. It is measured in percentage of mass.

That is, the percentage composition is the percentage by mass of each of the elements present in a compound.

The calculation of the percentage composition of an element is made by:

percent composition element A=\frac{total mass of element A}{mass of compound} *100

In this case, the percent composition of fluorine is:

percent composition of fluorine=\frac{39.6 g}{60.3 g} *100

percent composition of fluorine= 65.67%

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