Question

The empirical formula weight of 'Z' in the given reaction sequence is
n-propyl bromide \( \xrightarrow{\text{Na}} \) X \( \xrightarrow{\text{V}_2\text{O}_5, 773 \, \text{K}} \) Y \( \xrightarrow{\text{Cl}_2, \, \text{UV}, 500 \, \text{K}} \) Z
Dry ether, 20 atm

• A. \(47.5 \)
• B. \(54.5 \)
• C. \(84.5 \)
• D. \(48.5 \)

Hint:

When faced with a reaction sequence leading to a specific empirical formula weight: 1. Identify standard reactions: Wurtz reaction for alkyl halides is common. 2. Consider reaction conditions: High temperature and catalysts like V\(_2\)O\(_5\) can indicate severe oxidation or even cracking, leading to smaller fragments, especially if the final product's molecular weight is very low. 3. Work backward from empirical formula weight: If the options are empirical formula weights, calculate the empirical formula from the atomic weights to identify potential compounds. An empirical formula weight of 48.5 strongly suggests $\operatorname{CHCl}$ (e.g., from $\operatorname{C}_2\operatorname{H}_2\operatorname{Cl}_2$). 4. Connect steps: Check if the identified intermediate (Y) can plausibly lead to the final product (Z) under the given conditions.

Answer 1

The Correct Option is D

Let's analyze the given reaction sequence step by step:
Step 1: Reaction of n-propyl bromide with Na in dry ether to form X.
This is a Wurtz reaction. In a Wurtz reaction, two molecules of an alkyl halide couple to form a higher alkane.
n-propyl bromide is \( \operatorname{CH}_3\operatorname{CH}_2\operatorname{CH}_2\operatorname{Br} \).
\[ 2\operatorname{CH}_3\operatorname{CH}_2\operatorname{CH}_2\operatorname{Br} + 2\operatorname{Na} \xrightarrow{\text{Dry ether}} \operatorname{CH}_3\operatorname{CH}_2\operatorname{CH}_2\operatorname{CH}_2\operatorname{CH}_2\operatorname{CH}_3 + 2\operatorname{NaBr} \]
So, X is n-hexane (\( \operatorname{C}_6\operatorname{H}_{14} \)).
Step 2: Reaction of X (n-hexane) with \( \text{V}_2\text{O}_5 \) at 773 K and 20 atm to form Y.
This step represents a catalytic oxidation of n-hexane under harsh conditions. While common industrial oxidation of n-hexane using \( \operatorname{V}_2\operatorname{O}_5 \) at high temperatures and pressures can lead to adipic acid (\( \operatorname{C}_6\operatorname{H}_{10}\operatorname{O}_4 \)), the low empirical formula weight of Z (48.5) suggests that Y must be a much smaller molecule, implying significant cleavage of the carbon chain during this oxidation step.
Under severe conditions of catalytic oxidation (high temperature, high pressure), alkanes can undergo oxidative cleavage to produce smaller, more stable fragments. Considering the subsequent chlorination to form Z with an empirical formula weight of 48.5 (which corresponds to an empirical formula of \( \operatorname{CHCl} \)), it is plausible that n-hexane is completely fragmented and oxidized to ethyne (acetylene, \( \operatorname{C}_2\operatorname{H}_2 \)), which is a highly unsaturated small hydrocarbon.
So, Y is likely ethyne (\( \operatorname{C}_2\operatorname{H}_2 \)).
Step 3: Reaction of Y (ethyne) with \( \text{Cl}_2 \) under UV light at 500 K to form Z.
Ethyne (\( \operatorname{C}_2\operatorname{H}_2 \)) undergoes addition reactions with halogens. Under UV light, free radical addition/substitution can occur. When ethyne reacts with chlorine, it can form 1,2-dichloroethene.
\[ \operatorname{HC}\equiv \operatorname{CH} + \operatorname{Cl}_2 \xrightarrow{\text{UV}, 500 \, \text{K}} \operatorname{ClHC}=\operatorname{CHCl} \]
So, Z is 1,2-dichloroethene (\( \operatorname{C}_2\operatorname{H}_2\operatorname{Cl}_2 \)).
Step 4: Calculate the empirical formula weight of Z.
The molecular formula of Z is \( \operatorname{C}_2\operatorname{H}_2\operatorname{Cl}_2 \).
To find the empirical formula, divide the subscripts by the greatest common divisor (which is 2 in this case):
Empirical formula = \( \operatorname{C}_{(2/2)}\operatorname{H}_{(2/2)}\operatorname{Cl}_{(2/2)} = \operatorname{CHCl} \).
Now, calculate the empirical formula weight:
Atomic weight of C \( \approx \) 12.01
Atomic weight of H \( \approx \) 1.01
Atomic weight of Cl \( \approx \) 35.45
Empirical formula weight of \( \operatorname{CHCl} \) = 12.01 + 1.01 + 35.45 = 48.47 g/mol.
Rounding to one decimal place, the empirical formula weight is 48.5 g/mol.
This matches option (4).
The final answer is \( \boxed{48.5} \).