[davidenarb]

Hi all,

When we want to predict the products of a an electrocyclic reaction, we use either a conrotatory or disrotatory fashion of the outermost lobes of a given conjugated pi bond. It is, therefore, important to know the MOs of that compound, or at least, it is important to know the HOMO of that compound.

Is there any method to draw the HOMO of a given conjugated compound?

The HOMOs of butadiene, hexatriene are well known, but what about, for example, the HOMO of octatetraene and the rest of the conjugated pi bond?

Thank you.

[davidenarb]

Although it is bizarre and fun to respond to my question, the answer that I found will maybe help other students.

The answer of that question is undemanding; I found it while I was practicing several problems. we can indirectly infer the signs of the outermost lobes, which are the important lobes involved in electrocyclic reactions since the Woodward-Hoffman rules will by implication oblige a specific signs of the outermost lobes. Therefore, we don't need to know how to draw the MOs of conjugated pi compounds when realizing such reactions.

[Irlanur]

you can use Hückel Theory.

[clarkstill]

...the Woodward-Hoffman rules will by implication oblige a specific signs of the outermost lobes. Therefore, we don't need to know how to draw the MOs of conjugated pi compounds when realizing such reactions.

I don't think this is true, is it? The Woodward-Hoffman rules merely predict whether a given pericyclic reaction is 'allowed' or not, as determined by the conservation of orbital symmetry. It's not really constructive to think of it in terms of the phases or orbitals in the HOMO anyway - although it gives the right answer, this is more coincidence that anything else.  In an electrocyclic reaction there is no reason to believe the overlap of the HOMO with itself alone is the main determining factor in stereochemistry, since all the pi orbitals change during the reaction. 

You need to use Huckel theory to determine the orbitals involved (or the simplified "sine" construction, described here: https://www.google.co.uk/url?sa=t&rct=j&q=&esrc=s&source=web&cd=5&cad=rja&uact=8&ved=0CDYQFjAE&url=http%3A%2F%2Fchemweb.bham.ac.uk%2F~preecjaz%2F2010_CHM3A2%2F01.%2520Part%25201.ppt&ei=SpK6U6XcAamp7AbPjoH4CQ&usg=AFQjCNE7zHlYUCxoB4GAggQGSXw5bzknRQ&sig2=vnFJ0P-qI8eYkc8FmJckeA).

The roots of the W-H rules can only really be understood if you consider the orbital and state correlation diagrams for the particular reaction in question.  The Fleming book "Molecular Orbitals and Organic Chemical Reactions" has a pretty good description of this.

[davidenarb]

...the Woodward-Hoffman rules will by implication oblige a specific signs of the outermost lobes. Therefore, we don't need to know how to draw the MOs of conjugated pi compounds when realizing such reactions.

I don't think this is true, is it? The Woodward-Hoffman rules merely predict whether a given pericyclic reaction is 'allowed' or not, as determined by the conservation of orbital symmetry. It's not really constructive to think of it in terms of the phases or orbitals in the HOMO anyway - although it gives the right answer, this is more coincidence that anything else.  In an electrocyclic reaction there is no reason to believe the overlap of the HOMO with itself alone is the main determining factor in stereochemistry, since all the pi orbitals change during the reaction. 

You need to use Huckel theory to determine the orbitals involved (or the simplified "sine" construction, described here: https://www.google.co.uk/url?sa=t&rct=j&q=&esrc=s&source=web&cd=5&cad=rja&uact=8&ved=0CDYQFjAE&url=http%3A%2F%2Fchemweb.bham.ac.uk%2F~preecjaz%2F2010_CHM3A2%2F01.%2520Part%25201.ppt&ei=SpK6U6XcAamp7AbPjoH4CQ&usg=AFQjCNE7zHlYUCxoB4GAggQGSXw5bzknRQ&sig2=vnFJ0P-qI8eYkc8FmJckeA).

The roots of the W-H rules can only really be understood if you consider the orbital and state correlation diagrams for the particular reaction in question.  The Fleming book "Molecular Orbitals and Organic Chemical Reactions" has a pretty good description of this.

Thank you very much for the powerpoint.