[guybrush]

Hi,

Im stuck for ideas on how to solve a couple of organic chemistry questions which have me stumped.

1. On the basis of charge separation due to electronegativity differences and electrostatics, predict which atoms in the following questions is likely to react with the reagent shown. a) Bromoethane with O atom of OH-     b) propanal with H+

2. I have four molecules and i need to determine, in order, which has a higher boiling point than the others and rank them in increasing boiling point. Im not sure how to work out. Is there a way to work out boiling points for molecules? All the molecules end in -ane.

Could you point me in the right direction please?

Thanks

G

[Markovnikov]

The boiling points increases depending on the intermolecular forces between the molecules. One could list the molecules in if they have possibilities to do hydrogenbonding and by how large they are (vdw-forces and dipole moment).

[guybrush]

The boiling points increases depending on the intermolecular forces between the molecules. One could list the molecules in if they have possibilities to do hydrogenbonding and by how large they are (vdw-forces and dipole moment).

Thankyou for your reply. Two of the molecules i have are: Heptane, and 2,2,3-tri-methyl-butane

Which would have the higher boiling point and why? Im still a bit fuzzy on it but if i can understand it with an example i reckon i would be able to work the rest of that question out myself.

[stewie griffin]

In general the strength of the van dar Waals forces depends on the area of contact between molecules. So longer molecules have more contact than shorter molecules (thus heptane boils higher than propane). Furthermore, branching in alkanes reduces the total area of contact, thus reducing the boiling point.

[sjb]

In terms of q1, how is the molecule (as opposed to the ion) polarised?

[guybrush]

In general the strength of the van dar Waals forces depends on the area of contact between molecules. So longer molecules have more contact than shorter molecules (thus heptane boils higher than propane). Furthermore, branching in alkanes reduces the total area of contact, thus reducing the boiling point.

Thanks for that, appreciated.

[guybrush]

In terms of q1, how is the molecule (as opposed to the ion) polarised?

Hi,

I think Bromine is the + as opposed to the H-.

[sjb]

In terms of q1, how is the molecule (as opposed to the ion) polarised?

Hi,

I think Bromine is the + as opposed to the H-.

Where is the H- of which you speak? Do you recognise any polar bonds in the molecules bromoethane and propanal?

[guybrush]

In terms of q1, how is the molecule (as opposed to the ion) polarised?

Hi,

I think Bromine is the + as opposed to the H-.

Where is the H- of which you speak? Do you recognise any polar bonds in the molecules bromoethane and propanal?

Apologies for the late delay in replying.

The H- is on a separate molecule with Oxygen.

I think the Oxygen atom will react with the Bromine of Bromoethane due to Oxygen being electrophilic and Bromine being Nucleophilic. Looking at the periodic table the polar bonds will be: Bromine will be Br- and the Oxygen will be O+.

The second one i think the O atom of Propanal with reaction with the H+ atom due to the difference in polar bonds. O will be - in this case and more likely to gain an electron from H+.

Does this seem correct?