A chiral amine a having the r configuration undergoes hofmann elimination to form an alkene b as the major product. b is oxidati
vely cleaved with ozone, followed by ch3sch3, to form ch2═o and ch3ch2ch2cho. what are the structures of a and b? indicate stereochemistry where appropriate.
1 answer:
Hoffman Product is always a less substituted alkene. In given scenario Structure A (Given Below) was taken as a starting chiral Amine. For the sake of elimination the amine was taken tertiary so that on methylation it becomes a good leaving group. This chiral amina (A) when treated with Methyl Iodide gives a quarternary amine which on treatment with Silver oxide yields less substituted Alkene (B) as shown Below.
Alkene B on Ozonolysis give two aldehydes i.e. Formaldehyde and Butyraldehyde.
Alkene B on Ozonolysis give two aldehydes i.e. Formaldehyde and Butyraldehyde.
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Answer:
Explanation:
Hello,
In this case, the described chemical reaction is:
Thus, for the given reacting masses, we must identify the limiting reactant for us to determine the maximum mass of water that could be produced, therefore, we proceed to compute the available moles of ethane:
Next, we compute the moles of ethane consumed by 13.0 grams of oxygen by using the 1:7/2 molar ratio between them:
Thus, we notice there are less available moles of ethane, for that reason, it is the limiting reactant, thereby, the maximum amount of water is computed by considering the 1:3 molar ratio between ethane and water:
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