[orgo814]

How would I arrange these in order of decreasing acidity?
a) CH3CH=CH2, CH3CH2CH3, CH3C(triple bond)CH
b) CH3CH2CH2OH, CH3CH2CO2H, CH3CHClCO2H
c) CH3CH2OH, CH3CH2OH2+, CH3OCH3

The only thing the textbook tells me to do is look at pKa values. Lower pKa= stronger acid, weaker base. Larger pKa= vice versa

However, I'm not given pKa values so there has to be some conceptual way to figure this out without resorting to a table of pKa values. For a few of these, I can find the pKa values in the book but some I can't.

Any insight would be appreciated.

[Hunter2]

What do you know about bondings, halogenatoms in organic molecules, Oxygen in organic molecules?

[Babcock_Hall]

In part C what is the relationship between the first two entries?  Hint: fill in the blanks:  The second entry is the blank blank of the first entry.

[IsotopeBill]

Think in terms of the definition of a Bronsted acid.  Which compound would be more apt to donate a proton, and which would be more apt to accept a proton.  What structural aspects of a compound affect the proton donating and accepting abilities?

[orgopete]

Many textbooks explain that as electron withdrawing groups, sp>sp2>sp3.
As electron withdrawing, F>O>N>C.

What can you deduce about the the order by applying these observations?

[PhDoc]

You don't need to know pKa's for this. Simply rationalize.

S-orbitals are very electronegative; hydrogens bound to an sp-hybridized orbital (50% s) are more acidic than those bound to an sp2-hybridized orbital (33% s), as well as those bound to an sp3-hybridized orbital (25% s).

Relative pKa:  sp < sp2 < sp3

You're then comparing n-propanol with propanoic acid and 2-chloropropanoic acid. Carboxylic ACIDS are more acidic than alcohols (pKa 5 vs 16). Carbo acids with electron withdrawing groups are more acidic yet.

A protonated alcohol is a strong acid, just as strong as H3O+, then comes ethanol, and finally dimethyl ether, which isn't very acidic.

It's crucial that you rationalize these things without having to consult pKa tables.

Good luck!

[orgo814]

Thank you, that last explanation helped a ton!