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:
You need 8,324 g of CaCl₂ yo make this solution
Explanation:
Molarity is a way to express concentration in a solution, in units of moles of solute per liter of solution.
To know the grams of CaCl₂ it is necessary to know, first, the moles of this substance with the desired volume and concentration , thus:
0,1500 L × = 0,075 CaCl₂ moles
Now, with the molar mass of CaCl₂ you will obtain the necessary grams, thus:
0,075 CaCl₂ moles × = 8,324 g of CaCl₂
So, you need <em>8,324 g of CaCl₂</em> to make 150,0 mL of a 0,500M solution
I hope it helps!