# What are the products obtained after ozonolysis of the following compounds: 2,4,4-trimethylpent-2-ene; 2-methylbut-1-ene; 1-methylclohexene?

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Jul 13, 2017

A mixture of aldehydes and ketones

#### Explanation:

Let's start with drawing out the structure for 2,4,4 - trimethylpent - 2 ene.

• 5 carbons due to "pent" prefix
• 3 methyl groups: 1 group at carbon 2; 2 groups at carbon 4
• a double bond starting at carbon 2 due to -2-ene

Since we want to do ozonlysis, the reagent should be $1. {O}_{3}$ followed by 2. Reducing agent.
For the sake of clarity, I'll use DMSO as the reducing agent, but keep in mind there many more such as DMS, ${H}_{2} {O}_{2}$, etc..

Anyways, it is important that you put the 1. And 2. In front of the reactants for ozonlysis and some other reactions that you study since it shows the reagents being added separately, not stimutaneously.

To draw the product, cut the double bond in half and put oxygen's where you cut the double bond.

I've seen students and textbooks leave the product as is without redrawing it as two separate products.

Technically, you could do that if you want, but if you separate the products, you will notice that you produced a ketone and an aldehyde.

Now let's proceed to 2-methylbut-1-ene .

• 4 carbon chain due to but- prefix
• methyl group at carbon 2
• double bond starting at carbon 1

As before, add the ozonlysis Reagents and cut the double bond in half. After cutting the double bond in half, place 2 oxygens where the cut was.

Lastly, let's solve 1-methylcyclohexene

• 6 carbon ring due to cyclohex- prefix
• methyl group at carbon 1
• double bond starting at carbon 1, although not specified.. it's safe to assume.

Do the same procedure as before: add the ozonlysis reagents, cut double bond in half, add 2 oxygens where the cut was.

However, there is a twist with this one.

You cannot leave it as it is. You have to draw It in a chain.

Start by numbering the ring then drawing it in a chain. Do not worry about the oxygens just yet.

In this case. It's just a six carbon chain.

Add the oxygens and their double bonds where they match. We should have the oxygen groups at carbon 1 and carbon 6.

Okay, I hope this write up helps ! Sorry about the length, but hey, we did solve three problems :P

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Jul 13, 2017

The products depend on the nature of the workup.

Reducing workup

This involves reaction with ozone, then workup with $\text{Zn/acetic acid}$ or ("CH"_3)_2"S".

The process cleaves the alkene into aldehydes and/or ketones.

(a) 2,4,4-Trimethylpent-2-ene

$\left(\text{CH"_3)_3"CH=C"("CH"_3)_2 → underbrace(("CH"_3)_3"CH=O")_color(red)("2,2-dimethylpropanal") + underbrace("O=C"("CH"_3)_2)_color(red)("propanone}\right)$

(b) 2-Methylbut-1-ene

"CH"_3"CH"_2"C"("CH"_3)"=CH"_2 → underbrace("CH"_3"CH"_2"C"("CH"_3)"=O")_color(red)("butanone") + underbrace("O=CH"_2)_color(red)("methanal")

(c) 1-Methylcyclohexene

Oxidative workup

This involves reaction with ozone, followed usually by workup with ${\text{H"_2"O}}_{2}$.

The process converts any potential aldehydes into carboxylic acids.

(a) 2,4,4-Trimethylpent-2-ene

$\left(\text{CH"_3)_3"CH=C"("CH"_3)_2 → underbrace(("CH"_3)_3"C(OH)=O")_color(red)("2,2-dimethylpropanoic acid") + underbrace("O=C"("CH"_3)_2)_color(red)("propanone}\right)$

(b) 2-Methylbut-1-ene

"CH"_3"CH"_2"C"("CH"_3)"=CH"_2 → underbrace("CH"_3"CH"_2"C"("CH"_3)"=O")_color(red)("butanone") + underbrace("O=C(OH)H")_color(red)("methanoic acid")

(c) 1-Methylcyclohexene

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anor277 Share
Jul 13, 2017

Ketones, and aldehydes, depending on workup......

#### Explanation:

${R}_{1} {R}_{2} C = C {R}_{3} {R}_{4} \stackrel{{O}_{3}}{\rightarrow} {R}_{1} {R}_{2} C = O + O = C {R}_{3} {R}_{4}$.

While aldehydes are unstable with respect to oxidation, you should be able to chop up an olefin to give a series of aldehydes and ketones. If you can individually identify each oxo-compound, you could evidence the structure of the original olefin by wet chemical methods.

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