Why did mendeleev switch iodine and tellurium?
1 answer:
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Answer:
b
Explanation:
The reaction that is not a displacement reaction from all the options is
In a displacement reaction, a part of one of the reactants is replaced by another reactant. In single displacement reactions, one of the reactants completely displaces and replaces part of another reactant. In double displacement reaction, cations and anions in the reactants switch partners to form products.
<em>Options a, c, d, and e involves the displacement of a part of one of the reactants by another reactant while option b does not.</em>
Correct option = b.
The complete reaction along with intermediates is given below, with each step highlighted in different color.
Step 1:
In this step an acidic proton at alpha position is abstracted from lactone moiety and corresponding enolate is formed, which is resonance stabilized. Both structures are shown. In this case LDA (<span>Lithium diisopropylamide) acts as a base.
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Step 2:
The enolate formed attacks on Methyl Iodide, as Iodide being greater in size is a good leaving group and alpha methyl moiety is generated.
Step 3:
This step is acid catalyzed Bromination. Bromine is added at alpha position.
Step 4:
This is elimination reaction and is according to <span>Hofmann's Rule. Here less substituted alkene is generated.</span>
Step 1:
In this step an acidic proton at alpha position is abstracted from lactone moiety and corresponding enolate is formed, which is resonance stabilized. Both structures are shown. In this case LDA (<span>Lithium diisopropylamide) acts as a base.
</span>
Step 2:
The enolate formed attacks on Methyl Iodide, as Iodide being greater in size is a good leaving group and alpha methyl moiety is generated.
Step 3:
This step is acid catalyzed Bromination. Bromine is added at alpha position.
Step 4:
This is elimination reaction and is according to <span>Hofmann's Rule. Here less substituted alkene is generated.</span>