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:
Alternative C would be the correct choice.
Explanation:
- The dual compounds were evaluated on something like a TLC plate through three separate additives in conducting a TLC study of ferrocene versus acetylferrocene.
- The polar as well as nonpolar ferrocene where nonpolar is about 0.63 with the maximum
value, and indeed the polar is somewhere around 0.19 with
- TLC plate (30:1 toluene/ethanol) established with.
The other three choices are not related to the given circumstances. So that option C would be the appropriate choice.
The amount of hydrogen chloride that can be made is 1064 g
Why?
The two reactions are:
2H₂O → 2H₂ + O₂ 75.3 % yield
H₂ + Cl₂ → 2HCl 69.8% yield
We have to apply a big conversion factor to go from grams of water (The limiting reactant), to grams of HCl, the final product. We have to be very careful with the coefficients and percentage yields!
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