
How many compounds can be obtained in yield by Wurtz reaction?
(a)- 6
(b)- 3
(c)- 4
(d)- 5
Answer
525.9k+ views
Hint:Wurtz reaction is a preparation method in which we can prepare symmetrical alkanes, the yield of a symmetrical alkane is good, but unsymmetrical alkanes are not easily prepared in this reaction. In this process, two moles of alkyl halide are treated with sodium and we get alkane.
Complete step-by-step answer:Wurtz reaction is a preparation method in which we can prepare symmetrical alkanes, the yield of a symmetrical alkane is good, but unsymmetrical alkanes are not easily prepared in this reaction. In this process, two moles of alkyl halide are treated with sodium and we get alkane.
The general reaction for this method will be:
$2R-X+2Na\to R-R+2NaX$
Let us study all the options given above:
This compound will be made from a secondary halide having the formula $C{{H}_{3}}-CH(C{{H}_{3}})-X$. The reaction is given below:
This compound is a symmetrical alkane, hence it will have a good yield.
This compound will be made from primary halides having formula $C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}-X$ and $C{{H}_{3}}-C{{H}_{2}}-X$. The reaction is given below:
This compound is an unsymmetrical alkane, hence it will not have a good yield.
This compound will be made from primary halide having formula $C{{H}_{3}}-CH(C{{H}_{3}})-C{{H}_{2}}-X$. The reaction is given below:
This compound is a symmetrical alkane, hence it will have a good yield.
This compound will be made from primary and secondary halides having formula $C{{H}_{3}}-CH(C{{H}_{3}})-X$ and $C{{H}_{3}}-C{{H}_{2}}-X$. The reaction is given below:
This compound is an unsymmetrical alkane, hence it will not have a good yield.
This compound will be made from a primary halide having the formula $C{{H}_{3}}-C{{H}_{2}}-X$. The reaction is given below:
This compound is a symmetrical alkane, hence it will have a good yield.
$C{{H}_{4}}$ has only one carbon atom, hence it cannot be prepared from the Wurtz reaction.
Therefore, the correct answer is an option (b)- 3.
Note:The Wurtz reaction follows the bimolecular Nucleophilic substitution reaction, i.e., ${{S}_{N}}2$ mechanism. The smallest alkane that can be prepared from this method will be ethane.
Complete step-by-step answer:Wurtz reaction is a preparation method in which we can prepare symmetrical alkanes, the yield of a symmetrical alkane is good, but unsymmetrical alkanes are not easily prepared in this reaction. In this process, two moles of alkyl halide are treated with sodium and we get alkane.
The general reaction for this method will be:
$2R-X+2Na\to R-R+2NaX$
Let us study all the options given above:
This compound will be made from a secondary halide having the formula $C{{H}_{3}}-CH(C{{H}_{3}})-X$. The reaction is given below:
This compound is a symmetrical alkane, hence it will have a good yield.
This compound will be made from primary halides having formula $C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}-X$ and $C{{H}_{3}}-C{{H}_{2}}-X$. The reaction is given below:
This compound is an unsymmetrical alkane, hence it will not have a good yield.
This compound will be made from primary halide having formula $C{{H}_{3}}-CH(C{{H}_{3}})-C{{H}_{2}}-X$. The reaction is given below:
This compound is a symmetrical alkane, hence it will have a good yield.
This compound will be made from primary and secondary halides having formula $C{{H}_{3}}-CH(C{{H}_{3}})-X$ and $C{{H}_{3}}-C{{H}_{2}}-X$. The reaction is given below:
This compound is an unsymmetrical alkane, hence it will not have a good yield.
This compound will be made from a primary halide having the formula $C{{H}_{3}}-C{{H}_{2}}-X$. The reaction is given below:
This compound is a symmetrical alkane, hence it will have a good yield.
$C{{H}_{4}}$ has only one carbon atom, hence it cannot be prepared from the Wurtz reaction.
Therefore, the correct answer is an option (b)- 3.
Note:The Wurtz reaction follows the bimolecular Nucleophilic substitution reaction, i.e., ${{S}_{N}}2$ mechanism. The smallest alkane that can be prepared from this method will be ethane.
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