Answer:
See explanation
Explanation:
The reaction that we are considering here is quite a knotty reaction. It is difficult to decide if the mechanism is actually E1 or E2 since both are equally probable based on the mass of scientific evidence regarding this reaction. However, we can easily assume that the methylenecyclohexane was formed by an E1 mechanism.
Looking at the products, one could convincingly assert that the reaction leading to the formation of the two main products proceeds via an E1 mechanism with the formation of a carbocation intermediate as has been shown in mechanism attached to this answer. Possible rearrangement of the carbocation yields the 3-methylcyclohexene product.
I’m pretty sure the answer is C
Answer:
C.)One electron in each p orbital
Explanation:
In a P-sublevel with 3 electrons, they should be arranged with one electron going into each p-orbitals.
This is in accordance with the Hund's rule of maximum multiplicity.
The rule states that "electrons go into degenerate orbitals or sub-levels(p,d and f) singly before paring up".
Since the p-orbital is 3-fold degenerate with a capacity to accommodate a maximum number of 6 electrons, given 3 electrons, they will follow the Hund's rule in order to fill the orbitals.
So one electron will go in each p - orbitals easily.
The molarity of solution made by dissolving 15.20g of i2 in 1.33 mol of diethyl ether (CH3CH2)2O is =0.6M
calculation
molarity =moles of solute/ Kg of the solvent
mole of the solute (i2) = mass /molar mass
the molar mass of i2 = 126.9 x2 = 253.8 g/mol
moles is therefore= 15.2 g/253.8 g/mol = 0.06 moles
calculate the Kg of solvent (CH3CH2)2O
mass = moles x molar mass
molar mass of (CH3CH2)2O= 74 g/mol
mass is therefore = 1.33 moles x 74 g/mol = 98.42 grams
in Kg = 98.42 /1000 =0.09842 Kg
molarity is therefore = 0.06/0.09842 = 0.6 M
Answer:
<span>In the addition of hbr to 1-butyne the electrophile in the first step of the mechanism is <u>Hydrogen atom of HBr</u>.
Explanation:
In this reaction first of all HBr approaches the triple bond. A Pi Complex (weak inter-molecular interactions) is formed between the two molecules. And the triple bond attacks the partial positive hydrogen atom creating a negative charge on Bromine along with positive charge on itself (Sigma Complex). In second step the negative Bromide attacks the positive carbon of Butyne.</span>
