Organic chem

2012-08-25 4:17 pm
Why 4-hydrobenzaldehyde can form more intermolecular hydrogen bonds than
2-hydrobenzaldehyde does? Aren't them isomers?
更新1:

Also, why the boiling point of 2,2,3,3-tetramethylbutane is higher than that of decane?

更新2:

I realised that the boiling point of decane is higher than that of 2,2,3,3-tetramethylbutane , but the exercise book claimed that it is wrong. Thz for your answer.

回答 (1)

2012-08-25 7:15 pm
✔ 最佳答案
1.
What? Of course there could be difference in physical and chemical properties between isomers!

In aldehyde group -CH(=O) there's oxygen atom that can donate lone-pair to lone-pair acceptor;
in hydroxy group -OH there's hydrogen attached to highly-electronegative atom (O), that can accept lone-pair from lone-pair donor.

Thus, with these two groups Hydrogen-bond can be formed.

In 2-hydrobenzaldehyde, the hydroxy group and aldehyde group are close to each other, which makes formation of INTRA-molecular H-bond very feasible. This reduces the extent of formation of INTER-molecular H-bond.

In contrast, in 4-hydroxybenzaldehyde, the two groups in same molecule are far away from each other. Intra-molecular H-bond is not feasible, so it'll give more inter-molecular H-bond.


2.
Are you sure?
from what I see from Wikipedia, decane has higher b.p. than 2,2,3,3-tetramethylbutane. Go and check it.

To explain this, compare their:
- molecular size
- length of main carbon chain, hence shape of the molecule
- strength of instantaneous dipole etc.


2012-08-25 23:01:41 補充:
Don't worry, quite often exercise books include incorrect information.
Just in case Wikipedia was wrong, I checked CRC Handbook of Chemistry and Physics, 79th ed. ; and I am correct.

2012-08-25 23:07:41 補充:
But you may notice that decane has lower Melting Point than 2,2,3,3-tetramethylbutane.
Think about packing etc.


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