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Chapter 18: Problem 411

Write a reasonable mechanism for the reaction of hydrogen chloride and methanol with formaldehyde to give methy1 chloromethy1 ether that is consistent with the fact that the reaction occurs under conditions where neither methylene chloride nor methy1 chloride are formed.

Short Answer

Expert verified
The reaction mechanism for the formation of methyl chloromethyl ether without generating methylene chloride or methyl chloride involves the following steps: 1. Protonation of the carbonyl oxygen in formaldehyde. 2. Nucleophilic attack of methanol on protonated formaldehyde, forming an oxonium ion intermediate. 3. Deprotonation of the oxonium ion intermediate by a chloride ion. 4. Chlorination of the hydroxyl group in the product, forming the final methyl chloromethyl ether product. This mechanism prevents the formation of undesired side products while leading to the desired product.

Step by step solution

01

Identify the reactive sites of the molecules involved

We need to recognize the reactive sites in the reactants: formaldehyde has an electrophilic carbonyl carbon, methanol has a nucleophilic oxygen with a lone pair of electrons, and hydrogen chloride has an acidic hydrogen bound to a nucleophilic chloride.
02

Formation of protonated formaldehyde

The first step in the reaction would involve the protonation of the carbonyl oxygen in formaldehyde, making the carbonyl carbon more electrophilic: \[CH_2O + HCl \rightarrow CH_2(OH)^+ + Cl^-\]
03

Nucleophilic attack of methanol on protonated formaldehyde

In the second step, the oxygen atom with the lone pair in methanol can attack the electrophilic carbonyl carbon of the protonated formaldehyde: \[CH_2(OH)^+ + CH_3OH \rightarrow CH_3OCH_2OH^+ + H^+\] Here, we form an oxonium ion intermediate which contains the carbon-oxygen bonds required for our desired product.
04

Deprotonation of the oxonium ion intermediate

In the next step, a chloride ion (Cl⁻) would act as a base and remove a proton from the oxonium ion intermediate, leading to the formation of the desired product, methyl chloromethyl ether: \[CH_3OCH_2OH^+ + Cl^- \rightarrow CH_3OCH_2OH + HCl\]
05

Chlorination of the hydroxyl group

Finally, the hydroxyl group in the product formed in step 4 reacts with another molecule of hydrogen chloride, leading to the formation of the final product, methyl chloromethyl ether: \[CH_3OCH_2OH + HCl \rightarrow CH_3OCH_2Cl + H_2O\] This proposed mechanism allows for the formation of the desired product without generating any methylene chloride or methyl chloride as side products.

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Most popular questions from this chapter

Find all possible structures of the ketone that reacts with hydroxylamine to form an oxime which contains \(13.9 \%\) nitrogen.

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