McMurry (kap.17,3 + 17,4)

Question 1

The C-C double bond can be described in two ways:

1) in valence bond language

2) in molecular bond language

describe the difference?

Answer 1

In valence bond language, the carbons are SP2 hybridized, and have 3 equivilant hybrid orbitals that lie in a plane at angles of 120 degrees to one another.

The carbons form a (sigma) bond by head on head overlap of SP2 orbitals, and a (pi) bond by sideways overlap of unhybridized p orbitals (see figure 1.15 on page 187 for describtion)

In molecular orbital language - interaction between p orbitals leads to one bonding and one antibonding (pi) molecular orbital.

The (pi) bonding has no node between nuclei. The (pi) antibonding has one node beween nuclei.

Question 2

Is there free rotation around the double bonded C-C?

Answer 2

No, for rotation to occur, the double bond (Pi) bond must break and re-form.

Question 3

Can the substitutions on the disubstituted carbons interchange place?

Disubstituted carbon = two substituents other than hydrogen are attached to the double bonds carbon.

Answer 3

No, because bond rotation cant occur, the two (or more) substituents cant interconvert - they are distinct isolable compounds.

Question 4

The lack of rotation around the double bond has important chemical consequences, what?

Answer 4

They can have cis-trans sterioisomers.

The cis isomer = has two substituents on the same side (both carbons have 1 substituent and 1 hydrogen, but the substituents are on the same side (up or down)).

The trans isomer = the substituents are on the opposite side.

Question 5

Is cis-trans isomerism limited to disubstituted alkenes?

Answer 5

No, it can occur whenever both double bond carbons are attached to two different groups:

however if one carbon is attached to two identical groups, then the cis-trans isomerism is not possible.

(further down you can see that YES, cis and trans are only limited to disubstituted alkenes. With trisubstituted or even tetrasubstituted alkenes, the E-Z configuration is designated).

Question 6

What are the rules for the E-Z configuration?

Answer 6

(firstly you have to have a tri or tetra substituted C=C)

1) Look at the two attached substituents on each C and rank them from atomic number - the one with higher atomic number gets the higher rank, and one with lower gets lower rank.

2) If desicion cant be reached from the first atom of the substituents, look at the second, third, fourth etc.. (until a difference is found).

3) Multiple bonds atoms are equivilant to the same number of single bonded atoms.

For examples, look at page 189.