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IrinaVladis [17]
3 years ago
13

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.

Chemistry
1 answer:
ozzi3 years ago
4 0
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.

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Glucose, C6H12O6, is used as an energy source by the human body. The overall reaction in the body is described by the equation C
Crank

Answer:

70.3824 grams of carbondioxide is produced.

Explanation:

C_6H_{12}O_6(aq)+6O_2(g)⟶6CO_2(g)+6H_2O(l)

Moles of glucose =  \frac{48.0 g}{180 g/mol}=0.2666 mol

According to reaction 1 mole of glucose reacts with 6 mole of oxygen gas.

Then 0.2666 mol of gluocse will reactwith :

\frac{6}{1}\times 0.2666 mol =1.5996 mol of oxygen

Mass of 1.5996 moles of oxygen gas:

1.5996 mol\times 32 g/mol = 51.1872 g

51.1872 grams of oxygen are required to convert 48.0 grams of glucose to CO_2 and H_2O.

According to reaction, 1 mol, of glucose gives 6 moles of carbon dioxide.

Then 0.2666 moles of glucose will give:

\frac{6}{1}\times 0.2666 mol =1.5996 mol of carbon dioxide

Mass of 1.5996 moles of carbon dioxide gas:

1.5996 mol\times 44 g/mol = 70.3824 g

70.3824 grams of carbondioxide is produced.

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Is a eagle a producer?
Galina-37 [17]
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What type of evidence is a warm cup of coffee?
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C. Transient evidence

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1.34 milligrams is the same as _______kg and ______g
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There are 1000 mg in 1 g
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