Which of the following represents a propagation step in the monochlorination of methylene chloride (CH2Cl2)?a. CHCl3 + Cl. Right
arrow .CCl3 + HCl.b. CHCl2 + Cl2 right arrow CHCl3 + Cl..c. CH2Cl + Cl2 right arrow CH2Cl2 + Cl..d. CHCl2 + Cl. Right arrow CHCl3
1 answer:
Answer:
B = CHCl2 + Cl2 --> CHCl3 + Cl
Explanation:
Free radical halogenation is a chlorination reaction on Alkane hydrocarbons. This involves the splitting of molecules into radicals/ unstable molecules in the presence of sunlight/ U.V light which ensures bonding of the molecules.
Free radical chlorination is divided into 3 steps which are:
The initiation step
The propagation step
The termination step
So in reference to the question, propagation step involves two steps.
The first step is where the molecule in this case the methylene chloride(CH2Cl2) loses a hydrogen atom and then bond with a chlorine atom radical to give a nethylwnw chloride radical and HCl.
The second step involves the reaction of this methylene chloride got in the first step with chlorine molecule to form trichloride methane and a chlorine radical.
You would find in the attachment the 2 step mechanism.
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Answer:
2-ethoxy-2-methylpropan-1-ol
Explanation:
On this reaction, we have an "<u>epoxide"</u> (2-methyl-1,2-epoxypropane). Additionally, we have <u>acid medium</u> (due to the sulfuric acid ). The acid medium will produce the <u>hydronium ion</u> (
). This ion would be attacked by the oxygen of the epoxide. Then a <u>carbocation</u> would be produced, in this case, the most stable carbocation is the <u>tertiary one</u>. Then an <u>ethanol</u> molecule acts as a nucleophile and will attack the carbocation. Finally, a <u>deprotonation </u>step takes place to produce <u>2-ethoxy-2-methylpropan-1-ol</u>.
See figure 1
I hope it helps!
