Answer:
Both reactions share a common intermediate and differ only in the leaving group
Explanation:
The elimination reaction of tertiary alkyl halides usually occur by E1 mechanism. In E1 mechanism, the substrate undergoes ionization leading to the loss of a leaving group and formation of a carbocation.
Loss of a proton from the carbocation completes the reaction mechanism yielding the desired alkene.
In the cases of t-butanol and t-butyl bromide, the mechanism is the same. The both reactions proceed by E1 mechanism. The leaving groups in each case are water and chloride ion respectively.
Answer:
The statement that best describes the trend in first ionization enery of elements on the periodic table is:
It generally decreases down a group because valence electrons are farther from the nucleus.
The first ionization energy measures how difficult is to release an electron from the outermost shell. The higher the ionization energy the more difficult it is to release an electron, the lower the ionication energy the easier to release an electron.
As the atomic number of the atom increases (which is what happens when you go down a group) the furthest the outermost shell of electrons will be (the size of the atoms increases) and so those electrons require less energy to be released, which means that the ionization energy decreases.
Hope it helps!
Molecular geometry about the left carbon atom in CH₃CO₂CH₃ is tetrahedral.
The geometry around left carbon that is CH₃ is tetrahedral.
As the hybridization around left carbon is sp³ that shows its geometry should be tetrahedral and as there are 4 ligands around carbon and there is no lone pair present so the geometry is tetrahedral. So, the molecular geometry about the left carbon atom in CH₃CO₂CH₃ is tetrahedral.
Answer:
13.75
Explanation:
Density is equal to mass divided by volume. 110/8 = 13.75
Answer:
5.50 moles of magnesium oxide is 221.6742 grams
Explanation:
to do this you multiply the number of moles by the molar mass
