Draw the structure of an eight-carbon alkene that would yield the following compound (and no others) after treatment with ozone followed by dimethyl sulfide.

The Correct Answer and Explanation is:

Problem Restatement:

You are given a compound (not shown here, but presumably a specific product of ozonolysis), and you want to draw the structure of an eight-carbon alkene that, upon ozonolysis followed by treatment with dimethyl sulfide (Me₂S), would yield only that compound and no other products.

Step 1: Understanding Ozonolysis

Ozonolysis cleaves the double bond (C=C) of an alkene, breaking it into two carbonyl compounds.
Treatment with dimethyl sulfide (Me₂S) is a reducing workup, which prevents further oxidation, so the products are aldehydes or ketones rather than acids.
The carbon-carbon double bond is split, and each alkene carbon becomes the carbonyl carbon of either an aldehyde or ketone.

Step 2: What does “yield the following compound and no others” mean?

The product of ozonolysis will be one specific compound.
This means the starting alkene must be symmetrical, or at least have the double bond positioned such that ozonolysis cleaves to yield only that single product.
If the alkene is unsymmetrical or has multiple double bonds, ozonolysis would yield more than one product.

Step 3: Given the product (hypothetical, as the product structure is not given here)

Since the product isn’t shown in your message, I will assume a common problem type:

Example: The product is a single compound such as two molecules of butanal (CH3-CH2-CH2-CHO).

Step 4: Constructing the alkene

Ozonolysis of an alkene that yields two molecules of the same aldehyde or ketone suggests the alkene is symmetrical and has 8 carbons total.
Since butanal has 4 carbons, two of them mean total 8 carbons.
The alkene must have a double bond in the middle so that ozonolysis splits the molecule into two identical 4-carbon fragments.

Step 5: Drawing the correct alkene

The eight-carbon alkene that gives two butanal molecules upon ozonolysis is 1,2-octene, but it must be symmetrical.
The symmetrical eight-carbon alkene is 4-octene or more precisely 4-octene (cis or trans):

CH3-CH2-CH2-CH=CH-CH2-CH2-CH3

Ozonolysis of this compound cleaves the double bond between C4 and C5, resulting in:

CH3-CH2-CH2-CHO (butanal)
CH3-CH2-CH2-CHO (butanal)

Thus, only butanal forms, no other products.

Final answer:

4-octene (CH3-CH2-CH2-CH=CH-CH2-CH2-CH3)

Ozonolysis is a reaction that cleaves alkenes at the carbon-carbon double bond, converting it into carbonyl compounds — aldehydes or ketones depending on substitution. When treated with ozone followed by dimethyl sulfide, the cleavage results in aldehydes or ketones rather than carboxylic acids. If we want the ozonolysis of an eight-carbon alkene to yield a single specific product, the alkene must be carefully chosen so that cleavage produces only that product, with no alternative fragments.

This scenario strongly suggests the alkene is symmetrical about its double bond because symmetrical alkenes give identical carbonyl products upon cleavage. If the alkene were unsymmetrical, ozonolysis would yield two different carbonyl compounds, increasing the product number.

For an eight-carbon alkene, one natural candidate is 4-octene. This molecule has the double bond between carbons 4 and 5, effectively splitting the chain into two equal four-carbon fragments. Upon ozonolysis, the double bond is cleaved, producing two molecules of butanal (four-carbon aldehyde). Thus, the reaction cleanly yields a single product type: butanal, with no mixtures.

The structure of 4-octene is:
CH3–CH2–CH2–CH=CH–CH2–CH2–CH3

This molecule’s symmetry at the double bond ensures that the ozonolysis cleavage is at the center, producing identical fragments. There is no other site for cleavage, nor alternative double bonds, so no other products form.

Hence, the key to this problem is understanding the connection between the location of the double bond in the starting alkene, symmetry of the molecule, and the products formed after ozonolysis. By placing the double bond centrally in an eight-carbon chain, we guarantee the formation of a single aldehyde product without any others, satisfying the problem’s requirement.