Answer:
see explanation below
Explanation:
You are missing the reaction scheme, but in picture 1, I found a question very similar to this, and after look into some other pages, I found the same scheme reaction, so I'm gonna work on this one, to show you how to solve it. Hopefully it will be the one you are asking.
According to the reaction scheme, in the first step we have NaNH2/NH3(l). This reactant is used to substract the most acidic hydrogen in the alkine there. In this case, it will substract the hydrogen from the carbon in the triple bond leaving something like this:
R: cyclopentane
R - C ≡ C (-)
Now, in the second step, this new product will experiment a SN2 reaction, and will attack to the CH3 - I forming another alkine as follow:
R - C ≡ C - CH3
Finally in the last step, Na in NH3 are reactants to promvove the hydrogenation of alkines. In this case, it will undergo hydrogenation in the triple bond and will form an alkene:
R - CH = CH - CH3
In picture 2, you have the reaction and mechanism.
They are too small to see with the naked eye
Answer:
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Answer:
Therefore, the oxidation state of N in NaNO₂ is +3
Explanation:
Problem: calculating the oxidation state of nitrogen N in NaNO₂
let us denote the oxidation number of Nitrogen as N:
We know from the periodic table that Na has an oxidation number of +1 i.e it will readily want to lose 1 electron so as to complete its octet.
Oxygen is known to have an oxidation number of -2
Summation of the oxidation number of each atoms is 0 for neutral compound.
Therefore to calculate the oxidation state of Nitrogen in NaNO₂, we express as:
+1 + N + (-2 x 2) = 0
1 + N = 4
N = 4-1 = +3
Therefore, the oxidation state of N in NaNO₂ is +3