Boiling point q

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MoooShuuu

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CH3CH2CH2OH has a higher boiling point than CH3CH2CH2SH because of Hbonding right?

An alcohol will always have the higher BP unless there is an carboxylic acid or if another molecule is H bonded to F or N with a longer chain. Is this correct?

My test is Thurs and the more I look at this stuff the more it gets mixed up!!!
 
Another question....

Which compound would not react with sodium metal?

1 CH3OCH3
2 CH3CH2OH
3 CH3COOH
4 C6H5OH
5 CH3I

I don't think it is 2, 3, or 4 because of H bonding and Na reacts violently with water (apparently). So I'm thinking it wouldn't react with the ether, 1. Don't know about 5.

Thoughts?
 
Another question....

Which compound would not react with sodium metal?

1 CH3OCH3
2 CH3CH2OH
3 CH3COOH
4 C6H5OH
5 CH3I

I don't think it is 2, 3, or 4 because of H bonding and Na reacts violently with water (apparently). So I'm thinking it wouldn't react with the ether, 1. Don't know about 5.

Thoughts?

It has nothing to do with H bonding. We are talking reactivity. Na will abstract Hs.
You should know that ethers are for the most part inert, and thus are used as solvents in many organic chemistry reactions. I, much like the Hydrogen of alcohol is a very good leaving group, so it is likely to be abstracted too.
 
CH3CH2CH2OH has a higher boiling point than CH3CH2CH2SH because of Hbonding right?

An alcohol will always have the higher BP unless there is an carboxylic acid or if another molecule is H bonded to F or N with a longer chain. Is this correct?

My test is Thurs and the more I look at this stuff the more it gets mixed up!!!

N, O, F undergo H bonding.. S doesnt.
Since F has higher electronegativity than O and N, a molecule with F will exhibit stronger hydrogen bonding.
N will exhibit the wekeast H bonding.

if the same OH group is present in two molecules, the molecule with more polar subgroups to act as a protein receiver will have the higher boiling point. becuase there are more points for H bonding to occur. This is what is happening with the carboxylic acid because of the ketone group present (acts as proton acceptor in addition to delocalizing electron density from the H of the acid).
 
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