Methyl bromide and Ethyl bromide is mixed with equal proportions and the mixture is treated with metallic sodium. Which of the following is not expected to form
a)
Ethane
b)
Propane
c)
Butane
d)
Pentane
Correct answer is option 'D'. Can you explain this answer?

Question

Methyl bromide and Ethyl bromide is mixed with equal proportions and the mixture is treated with metallic sodium. Which of the following is not expected to form

a)
Ethane
b)
Propane
c)
Butane
d)
Pentane

Correct answer is option 'D'. Can you explain this answer?

Answer

The reaction is wurtz when symmetrical alkyl halide react with sodium in presence of dryether form only one product.but unsymmetrical alkyl halide react so different product formation occur.

Answer

WURTZ REACTION CH3Br +C2H5Br in the presence of Na nd dryether CH3-CH3(ethane)+C2H5-C2H5(butane)+C2H5-CH3(propane).Here we get mixture of alkanes  so ans. is D

Answer

Methyl bromide and Ethyl bromide are both alkyl halides, which are organic compounds containing a halogen atom (in this case, bromine) bonded to an alkyl group. When these two compounds are mixed in equal proportions and treated with metallic sodium, a reaction called a nucleophilic substitution reaction takes place.

Nucleophilic substitution reactions involve the replacement of a leaving group (in this case, the bromine atom) by a nucleophile (in this case, the metallic sodium). The reaction proceeds through a series of steps, including the formation of a carbanion intermediate.

Let's analyze the expected products for each option:

a) Ethane:
In the presence of metallic sodium, methyl bromide and ethyl bromide can undergo nucleophilic substitution reactions to form ethane. This is because the carbanion intermediate formed during the reaction can easily be protonated by water or an alcohol, leading to the formation of the corresponding alkane. So, ethane is expected to form.

b) Propane:
Similarly, the nucleophilic substitution reactions of methyl bromide and ethyl bromide can also lead to the formation of propane. The carbanion intermediate can be protonated to yield propane. Therefore, propane is expected to form.

c) Butane:
Again, the reaction of methyl bromide and ethyl bromide can result in the formation of butane. The carbanion intermediate can undergo protonation to give butane. Thus, butane is expected to form.

d) Pentane:
Pentane, which is a five-carbon alkane, would not be expected to form in this reaction. This is because the reactants, methyl bromide and ethyl bromide, both have a maximum of two carbon atoms. Since the reaction involves the substitution of a bromine atom with a nucleophile, the product will only contain a maximum of two carbon atoms. Therefore, pentane is not expected to form.

In summary, when methyl bromide and ethyl bromide are mixed in equal proportions and treated with metallic sodium, ethane, propane, and butane can be formed, but pentane is not expected to form.

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