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<em>Calculate the molar fraction of NaClO in a commercial bleach solution that contains 3.62% by mass of NaClO in water</em>
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<h3>Further explanation</h3>convert mass to mol
mol NaClO (MW=74.5 g/mol) :
mol H₂O (MW= 18 g/mol) :
Total mol :
0.0486 + 5.354 = 5.4026
mol fraction of NaClO :
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>