Question:
Identify the major product formed from the following reaction
Identify the major product formed from the following reaction
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Wurtz reaction couples two alkyl halide molecules using sodium in dry ether to form a higher alkane.
Updated On: Jun 15, 2026
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The Correct Option is D
Solution and Explanation
Step 1: Identify the starting compound.
The given alcohol is propan-1-ol:
\[ \mathrm{CH_3CH_2CH_2OH} \]
Step 2: Reaction with HCl/ZnCl\(_2\).
HCl in the presence of ZnCl\(_2\) (Lucas reagent) converts alcohol into alkyl chloride.
Thus, propan-1-ol forms 1-chloropropane:
\[ \mathrm{CH_3CH_2CH_2OH \rightarrow CH_3CH_2CH_2Cl} \]
Step 3: Reaction with NaI in dry acetone.
This is the Finkelstein reaction.
Alkyl chloride is converted into alkyl iodide:
\[ \mathrm{CH_3CH_2CH_2Cl \rightarrow CH_3CH_2CH_2I} \]
Thus, 1-iodopropane is formed.
Step 4: Reaction with Na in dry ether.
Alkyl iodide reacts with sodium metal in dry ether through the Wurtz reaction.
Two propyl groups combine together:
\[ 2\mathrm{CH_3CH_2CH_2I}+2\mathrm{Na} \rightarrow \mathrm{CH_3CH_2CH_2CH_2CH_2CH_3}+2\mathrm{NaI} \]
Step 5: Identify the final product.
The product formed contains six carbon atoms in a straight chain.
Hence, the final product is hexane.
Step 6: Final conclusion.
Therefore, the major product formed is
\[ \boxed{\text{Hexane}} \]
The given alcohol is propan-1-ol:
\[ \mathrm{CH_3CH_2CH_2OH} \]
Step 2: Reaction with HCl/ZnCl\(_2\).
HCl in the presence of ZnCl\(_2\) (Lucas reagent) converts alcohol into alkyl chloride.
Thus, propan-1-ol forms 1-chloropropane:
\[ \mathrm{CH_3CH_2CH_2OH \rightarrow CH_3CH_2CH_2Cl} \]
Step 3: Reaction with NaI in dry acetone.
This is the Finkelstein reaction.
Alkyl chloride is converted into alkyl iodide:
\[ \mathrm{CH_3CH_2CH_2Cl \rightarrow CH_3CH_2CH_2I} \]
Thus, 1-iodopropane is formed.
Step 4: Reaction with Na in dry ether.
Alkyl iodide reacts with sodium metal in dry ether through the Wurtz reaction.
Two propyl groups combine together:
\[ 2\mathrm{CH_3CH_2CH_2I}+2\mathrm{Na} \rightarrow \mathrm{CH_3CH_2CH_2CH_2CH_2CH_3}+2\mathrm{NaI} \]
Step 5: Identify the final product.
The product formed contains six carbon atoms in a straight chain.
Hence, the final product is hexane.
Step 6: Final conclusion.
Therefore, the major product formed is
\[ \boxed{\text{Hexane}} \]
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