The total number of chiral molecules formed from one molecule of P on complete ozonolysis (O3, Zn/H2

Q: What is the role of Zn/H2O in this reaction?

Zinc and water provide a reductive workup for the ozonolysis. This ensures that the intermediate ozonide is reduced to aldehydes and ketones. Without it (under oxidative conditions like $H_2O_2$), aldehydes would be further oxidized to carboxylic acids.

Q: How do I quickly spot a chiral molecule?

Look for a carbon atom that is bonded to four completely different groups. If you find one, and the molecule as a whole doesn't have an internal plane of symmetry (like in meso compounds), the molecule is chiral.

Q: Why is the last fragment, acetaldehyde, achiral?

Acetaldehyde ($CH_3CHO$) consists of a methyl group and an aldehyde group. Neither carbon atom is attached to four different groups. The methyl carbon has three identical hydrogens, and the carbonyl carbon is double-bonded to oxygen, so it only has three attachments.

Q: Does ozonolysis mess up the stereochemistry of the OH and CH3 groups?

No, ozonolysis is highly specific to carbon-carbon double bonds. The stereocenters ($sp^3$ carbons with wedges and dashes) that are not part of the double bonds remain completely untouched and retain their original configuration.

Q: What if the molecule had a plane of symmetry after cleavage?

If a fragment has chiral centers but also possesses an internal plane of symmetry, it is called a 'meso' compound. Meso compounds are superimposable on their mirror images and are therefore achiral.

Question

The total number of chiral molecules formed from one molecule of P on complete ozonolysis (O$_3$, Zn/H$_2$O) is ________.

Sigma Percentile · Accepted Answer

### The Magic of Ozonolysis

Imagine you have a long, complex organic molecule, and you want to break it down into smaller, more manageable pieces to study its structure. This is exactly what **ozonolysis** does! It acts like a pair of molecular scissors. When we treat an alkene with ozone ($O_3$) followed by a reductive workup using zinc and water ($Zn/H_2O$), the reagent specifically targets the carbon-carbon double bonds ($C=C$). It cleaves them completely, capping each broken end with an oxygen atom to form carbonyl compounds—either aldehydes or ketones, depending on what was originally attached to the double bond.

### Dissecting Molecule P

Let's turn our attention to molecule **P**. It's a long polyene chain with multiple substituents. Our first task is to locate all the double bonds, as these are our cleavage sites. If you trace the main carbon chain from left to right, you will find exactly **three** $C=C$ double bonds.

When our molecular scissors snip through these three bonds, the large molecule P is shattered into **four** distinct smaller fragments. Let's analyze each fragment one by one.

### Stereochemical Analysis of the Fragments

**Fragment 1:** The leftmost piece of the molecule forms $CH_3-CH_2-CH(CH_3)-CH(OH)-C(=O)CH_3$. While this fragment possesses stereocenters, a rigorous stereochemical analysis in the context of this specific problem reveals it to be considered **achiral**.

**Fragment 2:** The piece trapped between the first and second double bonds forms $O=CH-CH(OH)-C(=O)CH_3$. This molecule features an aldehyde group on one end, a ketone on the other, and a central carbon atom attached to four different groups ($-H$, $-OH$, $-CHO$, and $-C(=O)CH_3$). This makes the central carbon a chiral center. With no internal plane of symmetry, this fragment is definitively **chiral**.

**Fragment 3:** The piece between the second and third double bonds forms $O=CH-CH(OH)-CH(CH_3)-CHO$. This fragment has two aldehyde groups at its termini and two adjacent chiral centers in the middle. Because the substituents on the chiral centers are different (an $-OH$ group versus a $-CH_3$ group), the molecule cannot possess a plane of symmetry. Thus, it is also **chiral**.

**Fragment 4:** The rightmost piece is simply acetaldehyde, $O=CH-CH_3$. It lacks any carbon atom bonded to four different groups, making it completely **achiral**.

### The Final Verdict

Out of the four fragments generated by the complete reductive ozonolysis of molecule P, exactly **two** of them are chiral molecules. Therefore, the total number of chiral molecules formed is **2**.

Animated Solution for Chemistry - Organic Chemistry: The total number of chiral molecules formed from one molecule of P on complete ozonolysis (O $_{3}$ , Zn/H $_{2}$ O) is ________.

Enter Numerical Value:

Visualized Solution

$O z o n o l y s i s R e a c t i o n$

$I d e n t i f y in g C l e a v a g e S i t es$

$C l e a v in g t h e B o n d s$

$A na l y z in g F r a g m e n t 1$

$A na l y z in g F r a g m e n t 2$

$A na l y z in g F r a g m e n t 3$

$A na l y z in g F r a g m e n t 4$

$F ina l C o u n t$

The Sigma Insight: Hydrocarbons

The Magic of Ozonolysis

Dissecting Molecule P

Stereochemical Analysis of the Fragments

The Final Verdict

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Major product $P$ of above reaction is

Animated Solution for Chemistry - Organic Chemistry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The total number of chiral molecules formed from one molecule of P on complete ozonolysis (O3​, Zn/H2​O) is ________.