When 2-bromo-2-methylbutane is treated with a base, a mixture of 2-methyl-2-butene and 2-methyl-1-butene is produced when potass
ium hydroxide is the base, 2-methyl-1-butene accounts for 45% of the product mixture. However, when potassium tert-butoxide is the base, 2-methyl-1-butene accounts for 70% of the product mixture. What percent of 2-methyl-1-butene would be in the mixture if potassium propoxide were the base?
1 answer:
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Answer:
The energy for 1 mole of these photons is E = 31 ×
Explanation:
Given data
Wavelength 639.8 ×
m
Plank constant h = 6.626 × J sec
Speed of light c = 3 × meter per second
We know that Energy of a photon is given by
Value of ( h c ) = 6.626 × × 3 ×
= 19.878 ×
639.8 ×
m
Now Energy of a photon
E = 0.031 × Joule per mole
E = 31 ×
Therefore the energy for 1 mole of these photons is E = 31 ×
Answer:
Following are the solution to this question:
Explanation:
The octapeptides are:
Its structure includes free glycine amino, and alanine residues free carboxylic. Its peptides are customarily extracted from the left side free NH2 residues as well as the right side free -COOH residues only in the same direction, the sequence is provided.
please find the structure in the attached file.
