Answer:but-1-ene
Explanation:This is an E2 elimination reaction .
Kindly refer the attachment for complete reaction and products.
Sodium tert-butoxide is a bulky base and hence cannot approach the substrate 2-chlorobutane from the more substituted end and hence major product formed here would not be following zaitsev rule of elimination reaction.
Sodium tert-butoxide would approach from the less hindered side that is through the primary centre and hence would lead to the formation of 1-butene .The major product formed in this reaction would be 1-butene .
As the mechanism of the reaction is E-2 so it will be a concerted mechanism and as sodium tert-butoxide will start abstracting the primary hydrogen through the less hindered side simultaneously chlorine will start leaving. As the steric repulsion in this case is less hence the transition state is relatively stabilised and leads to the formation of a kinetic product 1-butene.
Kinetic product are formed when reactions are dependent upon rate and not on thermodynamical stability.
2-butene is more thermodynamically6 stable as compared to 1-butene
The major product formed does not follow the zaitsev rule of forming a more substituted alkene as sodium tert-butoxide cannot approach to abstract the secondary proton due to steric hindrance.
Answer:
im pretty sure its groundwater just read this and see the answer
Explanation:
The ocean holds about 97 percent of the Earth's water; the remaining three percent is found in glaciers and ice, below the ground, in rivers and lakes. Of the world's total water supply of about 332 million cubic miles of water, about 97 percent is found in the ocean.
Answer:
A chemical equation consists of the chemical formulas of the reactants (on the left) and the products (on the right). The two are separated by an arrow symbol (“→” usually read aloud as “yields”). ... The equation also identifies that all the compounds are in the gaseous state.
Explanation:
The symbol is U and its atomic number is 92
First, we will get the "n", the number of half lives, it is the elapsed time over the half life. In the problem, the time is measured in days, so we have
6/2 = 3
to get the ending amount of radioactive sample, we have
32g x (1/2)³ = 4 grams of radioactive sample ⇒ the amount left after 6 days