Draw all constitutional isomers with molecular formula c3h8o, and rank them in terms of increasing acidity:
1 answer:
There are only <em>three</em> constitutional isomers (see the image below).
Ethyl methyl ether is the least acidic because it has only C-H bonds. However, it should be more acidic than an alkane (pKₐ ≈ 50) because of the electron withdrawing effect of the O atom. I would expect it to have pKₐ ≈ 30.
The alcohols are <em>more acidic</em> because the O in the O- H bond is highly electron withdrawing. Alcohols typically have pKₐ ≈ 17.
Alkyl groups are electron-donating. This increases electron density in the OH bond and makes the acid weaker. Propan-2-ol has two methyl groups on the α-carbon, while propan-1-ol has one ethyl group. Thus propan-2-ol should be a <em>slightly weaker</em> acid than propan-1-ol.
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Answer:
Rb: [Kr] 5s
Step-by-step explanation:
Rb is element 37, the first element in Period 5.
It has one valence electron, so its valence electron configuration is 5s.
The noble gas configuration uses the symbol of the previous noble gas as a shortcut for the electron configurations of the inner electrons.
The preceding noble gas is Kr, so the electron configuration is Rb: [Kr] 5s.
