By studying and learning by using your brain to accomplish new goals. :)
(if that makes sense)
Hope this helps :3
Answer:
Case 1:
X = Any element from Group I
i) H
ii) Li
iii) Na
iv) K
v) Rb
vi) Cs
Y = 1
Case 2:
X = Any element from Group II
i) Be
ii) Mg
iii) Ca
iv) Sr
v) Ba
vi) Ra
Y = 2
Case 3:
X = Any element from Group III
i) B
ii) Al
iii) Ga
iv) In
v) Ti
Y = 3
Explanation:
The general formula given is as follow,
XCly
So, if X has +1 oxidation state, then it will require only one Cl atom with oxidation number -1 to form a neutral compound, therefore, y = 1.
If X has +2 oxidation state, then it will require two Cl atoms with oxidation number -1 to form a neutral compound, therefore, y = 2.
If X has +3 oxidation state, then it will require three Cl atoms with oxidation number -1 to form a neutral compound, therefore, y = 3.
Well.. if ur going down a ramp, friction beat gravity, right....??? I guess, I I have an idea hahha
The question is incomplete, the complete question is;
Why is a terminal alkyne favored when sodium amide (NaNH2) is used in an elimination reaction with 2,3-dichlorohexane? product. A) The terminal alkyne is more stable than the internal alkyne and is naturally the favored B) The terminal alkyne is not favored in this reaction. C) The resonance favors the formation of the terminal rather than internal alkyne. D) The strong base deprotonates the terminal alkyne and removes it from the equilibrium.
E) The positions of the Cl atoms induce the net formation of the terminal alkyne.
Answer:
E) The positions of the Cl atoms induce the net formation of the terminal alkyne.
Explanation:
In this reaction, sterric hindrance plays a very important role. We know that sodamide is a strong base, it tends to attack at the most accessible position.
The first deprotonation yields an alkene. The strong base attacks at the terminal position again and yields the terminal alkyne. Thus the structure of the dihalide makes the terminal hydrogen atoms most accessible to the base. Hence the answer.
