A chloro compound A, (i) Forms aldehydes on ozonolysis followed by the hydrolysis. (ii) When vaporis

Q: Why do we use 22400 mL for the molar volume at STP?

At Standard Temperature and Pressure (STP), 1 mole of any ideal gas occupies a volume of 22.4 liters, which is equivalent to 22400 milliliters. This is a fundamental constant derived from the Ideal Gas Law ($PV = nRT$).

Q: How do we know the compound is an alkene?

The problem states that the compound undergoes ozonolysis to form aldehydes. Ozonolysis is a classic reaction that cleaves carbon-carbon double bonds. Therefore, the compound must be an alkene.

Q: Why did we assume the formula $C_nH_{2n-1}Cl$?

The compound is a "chloro compound" and an alkene. An alkane has the formula $C_nH_{2n+2}$. An alkene has $C_nH_{2n}$. Replacing one hydrogen atom with a chlorine atom gives the monochloroalkene formula $C_nH_{2n-1}Cl$.

Q: What if the double bond was terminal?

If the double bond were terminal (e.g., $CH_2=CH-CH_2Cl$), ozonolysis would produce formaldehyde ($HCHO$), which is an aldehyde, but the other fragment might be a ketone or another aldehyde depending on the structure. The problem says it "forms aldehydes" (plural), which is consistent with an internal double bond like $CH_3-CH=CH-Cl$.

Q: Can we solve this without knowing the exact structure?

Yes! The exact position of the double bond or the chlorine atom isn't strictly necessary to find the number of carbon atoms. The general formula $C_nH_{2n-1}Cl$ is sufficient to set up the algebraic equation for the molar mass.

Question

A chloro compound A,
(i) Forms aldehydes on ozonolysis followed by the hydrolysis.
(ii) When vaporised completely, 1.53 g of A gives 448 mL of vapour at STP.
The number of carbon atoms in a molecule of compound A is ……… .

Sigma Percentile · Accepted Answer

The journey to solving this problem is a beautiful blend of physical chemistry and organic chemistry. It requires us to first determine the macroscopic properties of the gas and then dive into the microscopic structure of the molecule. Let's break it down step-by-step.

### Phase 1: The Macroscopic World (Mole Concept)

We are given a mysterious chloro compound, let's call it **Compound A**. We don't know its formula yet, but we do know its physical behavior. At Standard Temperature and Pressure (STP), $448 	ext{ mL}$ of its vapor weighs $1.53 	ext{ g}$.

To identify any compound, the first crucial piece of information we need is its **molar mass**. The molar mass is simply the mass of one mole of the substance.

Recall Avogadro's hypothesis and the ideal gas law: at STP, one mole of any ideal gas occupies a volume of exactly $22400 	ext{ mL}$ (or $22.4 	ext{ L}$). This is our golden key! If we know the mass of $448 	ext{ mL}$, we can easily find the mass of $22400 	ext{ mL}$ using the unitary method.

$$M = \frac{1.53 	ext{ g}}{448 	ext{ mL}} 	imes 22400 	ext{ mL mol}^{-1}$$

Let's do the math. Dividing $22400$ by $448$ gives exactly $50$.

$$M = 1.53 	imes 50 = 76.5 	ext{ g mol}^{-1}$$

So, the molar mass of Compound A is **$76.5 	ext{ g mol}^{-1}$**. We have successfully completed the first phase!

### Phase 2: The Microscopic World (Organic Structure)

Now, we put on our organic chemistry hats. The problem states that Compound A forms **aldehydes** upon ozonolysis followed by hydrolysis.

What does this tell us? Ozonolysis is a reaction that acts like a pair of molecular scissors, specifically cutting carbon-carbon double bonds ($C=C$). Because it forms aldehydes, we know for sure that the compound is an **alkene**. Furthermore, since it's a "chloro compound", it must be a **monochloroalkene**.

Let's construct its general formula. An alkane has the formula $C_nH_{2n+2}$. An alkene has $C_nH_{2n}$. If we replace one hydrogen atom with a chlorine atom, we get the general formula for a monochloroalkene:

$$	ext{General Formula} = C_nH_{2n-1}Cl$$

We know the molar mass of this compound is $76.5 	ext{ g mol}^{-1}$. Let's express the molar mass algebraically using the atomic masses of Carbon ($12$), Hydrogen ($1$), and Chlorine ($35.5$).

$$M = 12(n) + 1(2n - 1) + 35.5$$

$$M = 12n + 2n - 1 + 35.5$$

$$M = 14n + 34.5$$

Now, we equate this algebraic expression to the numerical molar mass we found in Phase 1:

$$14n + 34.5 = 76.5$$

This is a simple linear equation. Let's solve for $n$:

$$14n = 76.5 - 34.5$$

$$14n = 42$$

$$n = \frac{42}{14} = 3$$

The value of $n$ represents the number of carbon atoms in the molecule. Therefore, Compound A contains exactly **$3$ carbon atoms**.

### The Way Forward: Visualizing the Reaction

For the curious minds, what exactly is this compound? With $3$ carbon atoms, one double bond, and one chlorine atom, the formula is $C_3H_5Cl$. Given that it produces *aldehydes* (plural) on ozonolysis, the double bond must be internal to yield two aldehyde fragments.

The structure is **1-chloropropene** ($CH_3-CH=CH-Cl$). When treated with ozone ($O_3$) followed by zinc and water ($Zn/H_2O$), the double bond cleaves perfectly down the middle:

$$CH_3-CH=CH-Cl \xrightarrow{O_3, Zn/H_2O} CH_3-CHO + OHC-Cl$$

This yields acetaldehyde and formyl chloride, perfectly satisfying all the conditions given in the problem. Chemistry is truly a beautiful puzzle!

Animated Solution for Chemistry - Basic Concepts in Chemistry: A chloro compound A, (i) Forms aldehydes on ozonolysis followed by the hydrolysis. (ii) When vaporised completely, 1.53 g of A gives 448 mL of vapour at STP. The number of carbon atoms in a molecule of compound A is ……… .

Enter Numerical Value:

Visualized Solution

$V i s u a l A n c h or$

$L o g i c B r i d g e$

$R a w S e t u p$

$A t o mi c C o m p u t e$

$G e n er a l F or m u l a$

$M o l a r M a ss E q u a t i o n$

$S o l v in g f or n$

$F ina l A n s w er$

$T h e W a y F or w a r d$

The Sigma Insight: Molecular Mass, Mole Concept and Concentration

Phase 1

Phase 2

The Way Forward

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Animated Solution for Chemistry - Basic Concepts in Chemistry: A chloro compound A, (i) Forms aldehydes on ozonolysis followed by the hydrolysis. (ii) When vaporised completely, 1.53 g of A gives 448 mL of vapour at STP. The number of carbon atoms in a molecule of compound A is ……… .

Enter Numerical Value:

Visualized Solution

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