Provide the structures of the fragments that result when the molecular ion of 2-heptanone undergoes fragmentation via McLafferty
rearrangement. Include charges and single electrons.
1 answer:
Answer:
See explanation
Explanation:
We have to start, remembering the mechanism behind the <u>McLafferty rearrangement</u>. The hydrogen in the g<u>amma carbon</u> (in this case, carbon 5) would be removed by a <u>heterolytic rupture</u> due to the <u>cation-radical</u> placed in the oxygen of the <u>carbonyl group</u>. Then we will have several heterolytic ruptures. Between <u>carbons alpha and beta</u> (in this case, 4 and 3) and a rupture in the <u>carbonyl group</u>. Due to these ruptures, <u>two double bonds would be formed</u>. One double bond in the alcohol cation-radical and the other one in the alkene.
See figure 1
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Answer:
Explanation:
The cell reaction properly written is shown below:
Cu|Cu²⁺ || Ag⁺
| Ag
From this cell reaction, to get the net ionic equation, we have to split the reaction into their proper oxidation and reduction halves. This way, we can know that is happening at the electrodes and derive the overall net equation.
Oxidation half:
Cu ⇄ Cu²⁺
+ 2e⁻
At the anode, oxidation occurs.
Reduction half:
Ag⁺ + 2e⁻ ⇄ Ag
At the cathode, reduction occurs.
To derive the overall reaction, we must balance the atoms and charges:
Cu ⇄ Cu²⁺
+ 2e⁻
Ag⁺ + e⁻ ⇄ Ag
we multiply the second reaction by 2 to balance up:
2Ag⁺ + 2e⁻ ⇄ 2Ag
The net reaction equation:
Cu + 2Ag⁺
+ 2e⁻⇄ Cu²⁺
+ 2e⁻ + 2Ag
We then cancel out the electrons from both sides since they appear on both the reactant and product side:
Cu + 2Ag⁺
⇄ Cu²⁺
+ 2Ag