Identify the structure of a minor product formed from the radical chlorination of propane, which has molecular formula C3H6Cl2 a
nd exhibits the given 1H NMR spectrum.
1 answer:
Answer:
As we are not provided with ¹H-NMR spectrum therefore we can not conclude a single result for the minor chlorination of propane with Molecular formula C₃H₆Cl₂.
Hence, we will draw all the possible isomers with molecular formula C₃H₆Cl₂ and predict the ¹H-NMR spectrum for each of them separately.
Below is the structures of possible minor products and the NMR spectrum respectively.
Possible Product A:
If the minor product formed is 2,2-Dichloropropane then it will give only one singlet peak in NMR spectrum.
Possible Product B:
If the minor product is 1,1-Dichloropropane then it will exhibit two triplets and one multiplet as shown in figure.
Possible Product C:
If the minor product is 1,2-Dichloropropane then it will give four peaks in NMR. The extra peak is formed due to two enantiomers about the chiral carbon.
Possible Product D:
If the minor product is 1,3-Dichloropropane then it will exhibit one triplet and one quintet signal as shown in figure.
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Answer :
The equilibrium concentration of CO is, 0.016 M
The equilibrium concentration of Cl₂ is, 0.034 M
The equilibrium concentration of COCl₂ is, 0.139 M
Explanation :
The given chemical reaction is:
Initial conc. 0.1550 0.173 0
At eqm. (0.1550-x) (0.173-x) x
As we are given:
The expression for equilibrium constant is:
Now put all the given values in this expression, we get:
x = 0.139 and x = 0.193
We are neglecting value of x = 0.193 because equilibrium concentration can not be more than initial concentration.
Thus, we are taking value of x = 0.139
The equilibrium concentration of CO = (0.1550-x) = (0.1550-0.139) = 0.016 M
The equilibrium concentration of Cl₂ = (0.173-x) = (0.173-0.139) = 0.034 M
The equilibrium concentration of COCl₂ = x = 0.139 M