H-bond and BP

According to wikipedia:
Boiling pt of CH3COOH (118.1 ℃)
while for CH3CONH2 (222 ℃)

Hence boiling point of CH3CONH2 should be larger than CH3COOH.

Explanation of the above observation:
Both CH3COOH and CH3CONH2 can form hydrogen bonds with neighbouring molecules. For CH3CONH2, each molecule contains two slightly positive hydrogen atoms and two lone pairs of electrons on the oxygen atom, thus it can form 4 hydrogen bonds at maximum with neighbouring molecules. However, for CH3COOH, only 2 hydrogen bonds are formed between two molecules to produce a dimer. Since the number of hydrogen bonds formed between CH3CONH2 is much larger than CH3COOH. The boiling point of the former should be larger than the latter.

Reference:
http://en.wikipedia.org/wiki/Ethanoic_acid
http://en.wikipedia.org/wiki/Acetamide

2011-05-28 16:17:48 補充：
Why CH3CONH2 form 4 H-bonds?
For a large sample of CH3CONH2 , will it form 1 H-bond? Can I say that although it has 2 N-H bonds, it has only 1 lone pair e ?

amide can form 4 hydrogen for each molecule:
http://www.chemguide.co.uk/organicprops/amides/background.html

2011-05-28 16:20:54 補充：
CH3CONH2 has higher BP since it is a intermolecular H-bond?
CH3COOH has lower BP since it is a dimer?
You have a wrong concept... both molecules have intermolecular hydrogen bonds.

For dimer, plz refer to: http://www.chemguide.co.uk/organicprops/acids/background.html

2011-05-28 16:22:28 補充：
In this question, the focus is the number of hydrogen bonds formed between the molecules but not the polarity of the atoms.

B.P. of RCOOH > B.P. of RCONH2

The hydrogen bond between O and H is stronger than that of between N and H. So more energy is need to overcome the H bond in RCOOH than in RCONH2 to make the molecules moving freely and therefore the B.P. of RCOOH is higher than the B.P. of RCONH2.