An organic compound (C8H10O2) rotates plane-polarized light. It produces pink color with neutral FeC

Q: Why can't the compound be an ether like anisole?

While an ether would fit the molecular formula, the positive neutral $FeCl_3$ test strictly requires a free phenolic $-OH$ group directly attached to the benzene ring.

Q: Could the side chain be $-CH_2CH_2OH$?

No. Although $-CH_2CH_2OH$ fits the remaining atom count ($C_2H_5O$), it does not have a chiral center. The problem states the compound rotates plane-polarized light, so it must be chiral.

Q: Why do we multiply the 3 structural isomers by 2?

Each of the 3 positional isomers (ortho, meta, para) contains exactly one chiral center. A molecule with one chiral center exists as a pair of non-superimposable mirror images (enantiomers), usually denoted as R and S. Thus, 3 structures $\times$ 2 enantiomers = 6 total isomers.

Q: What if there were two different substituents on the ring instead of one side chain?

If we split the remaining $C_2H_5O$ into two groups (like $-CH_3$ and $-OCH_3$, or $-CH_3$ and $-CH_2OH$), none of the resulting arrangements would possess a chiral center. The only way to achieve chirality with these atoms is the 1-hydroxyethyl group ($-CH(OH)CH_3$).

Q: Does the position of the OH group on the chiral center matter for the total count?

Yes, the spatial arrangement (wedge vs. dash) of the OH group on the chiral carbon is exactly what creates the two enantiomers (R and S) for each positional isomer.

Question

An organic compound ($	ext{C}_8	ext{H}_{10}	ext{O}_2$) rotates plane-polarized light. It produces pink color with neutral $	ext{FeCl}_3$ solution. What is the total number of all the possible isomers for this compound?

Sigma Percentile · Accepted Answer

### Decoding the Molecular Formula

Imagine you are a detective, and the molecular formula is your first clue at the crime scene. We are given the compound $C_8H_{10}O_2$.

The very first step in any structural identification problem is to calculate the **Degree of Unsaturation (DU)**.

Using the formula $DU = C + 1 - \frac{H}{2}$, we plug in our values to get $DU = 8 + 1 - 5 = 4$.

Whenever you see a DU of 4, an alarm should go off in your head: this strongly suggests the presence of a **benzene ring**!

### The Telltale Chemical Test

Now, let's look at the chemical behavior. The problem states that our mystery compound produces a pink color with a neutral $FeCl_3$ solution.

This is a classic, high-yield concept for JEE. The neutral $FeCl_3$ test is a specific confirmatory test for **phenols**.

This tells us something crucial: our benzene ring has at least one hydroxyl group ($-OH$) directly attached to it. We have just locked in the core of our molecule as a phenol derivative, $C_6H_4(OH)-$.

### Deducing the Side Chain

If we know the core is a phenol, what is the rest of the molecule? Let's do some simple chemical arithmetic.

We take our total formula $C_8H_{10}O_2$ and subtract the phenol portion, $C_6H_5O$.

This leaves us with exactly $C_2H_5O$. This remaining cluster of atoms must form the side chain attached to our benzene ring.

### The Requirement of Chirality

Here is where the problem gets really interesting. The question mentions that the compound **rotates plane-polarized light**.

This is the physical manifestation of **optical activity**. For a molecule to be optically active, it must possess a **chiral center**—a carbon atom bonded to four different groups.

So, our side chain $C_2H_5O$ cannot just be any random arrangement; it must be structured in a way that creates a chiral center.

### Constructing the Chiral Center

Let's explore the possibilities for $C_2H_5O$.

Could it be $-CH_2CH_2OH$? No, that has no chiral carbon.

Could it be an ether linkage like $-OCH_2CH_3$? Again, no chiral center.

The only valid arrangement that gives us a chiral carbon is the 1-hydroxyethyl group: $-CH(OH)CH_3$.

Look closely at that first carbon: it is attached to a methyl group, a hydroxyl group, a hydrogen atom, and the massive phenol ring. Four different groups! We have found our chiral side chain.

### Assembling the Final Isomers

Now we have our two building blocks: a phenol ring and a chiral $-CH(OH)CH_3$ side chain.

Where can we attach this side chain? Relative to the $-OH$ group on the ring, we can place it at the **ortho**, **meta**, or **para** positions.

This gives us exactly 3 distinct structural isomers.

But wait, we aren't done! Because each of these 3 structures contains exactly one chiral center, each one will exist as a pair of non-superimposable mirror images (enantiomers).

For every position, we have an **R-configuration** and an **S-configuration**.

Therefore, our final calculation is beautifully simple: 3 structural isomers $	imes$ 2 enantiomers = **6 total isomers**.

Animated Solution for Chemistry - Organic Chemistry: An organic compound ( $C_{8} H_{10} O_{2}$ ) rotates plane-polarized light. It produces pink color with neutral $FeCl_{3}$ solution. What is the total number of all the possible isomers for this compound?

Enter Numerical Value:

Visualized Solution

$D eco d in g t h e M o l ec u l a r F or m u l a$

$N e u t r a l F e C l_{3} T es t$

$D e d u c in g t h e S i d e C hain$

$O pt i c a l A c t i v i t y & C hi r a l i t y$

$I d e n t i f y in g t h e C hi r a l S i d e C hain$

$P os i t i o na l I so m er s$

$T o t a l S t er eo i so m er s$

The Sigma Insight: Alcohols, Phenols, Ethers

Decoding the Molecular Formula

The Telltale Chemical Test

Deducing the Side Chain

The Requirement of Chirality

Constructing the Chiral Center

Assembling the Final Isomers

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } DecodingtheMolecularFormula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } NeutralFeCl3​Test

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } DeducingtheSideChain

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OpticalActivity&Chirality

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } IdentifyingtheChiralSideChain

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PositionalIsomers

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } TotalStereoisomers