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Topics: Molecular orbital, Chemical bonding, Lewis structure Pages: 23 (1986 words) Published: June 5, 2013
Chemistry S-20

Week 1

Alkenes: Isomers and Nomenclature
1. There are 6 unique alkene isomers of the hydrocarbon C5H10. Draw each of these isomers, and provide a systematic name for each.

1-pentene

(E)-2-pentene

(Z)-2-pentene

2-methyl-1-butene

2-methyl-2-butene

3-methyl-1-butene

2. For the three alkenes above which are various isomers of pentene, rank them in order of stability. Explain your ranking.

most stable (more subst. double bond)

less stable (cis alkene is slightly less stable)

least stable (less substituted double bond)

3. Provide a systematic name for the following alkene.

(Z)-3-propyl-2-heptene
32

Chemistry S-20

Week 2

Bicyclic Compounds and Bredt's Rule
1. Draw each of the following bicyclic compounds in a good "perspective" drawing.

CH3

CH3

trans-decalin: two chairs

CH3

CH3

CH3

norbornane skeleton:
CH3 H3C CH3

2. The molecule trans-cyclooctene is known to exist. (It is chiral, by the way). Why is the analagous molecule trans-cyclohexene unstable? Far too strained to have a trans-alkene in a six-membered ring. H H H

=
H

H H

trans-cyclooctene (planar representation)

trans-cyclooctene (perspective representation)

trans-cyclohexene????

3. Draw each of the following bicyclic alkenes in a good "perspective" representation. Only one of these three compounds actually exists. Which one, and why?

BAD!

In these two, the alkenes can't be planar. Highly strained. (Try to model!)

BAD!

Ok! ** Bredt's Rule: Can't have sp2 carbon at a bridgehead. (No alkenes, no carbocations!)

Or, notice the trans-alkenes in small rings.

60

Chemistry S-20

Week 1

Carbocation Stability
1. Draw the structure of the methyl cation, CH3+. Describe the hybridization, geometry, and orbitals in this species.

H H

H

Carbon is sp2-hybridized, with 120° bond angles and a vacant p-orbital.

2. Each of the following carbocations is significantly more stable than the methyl cation. Explain why each of these species is especially stable. You should use both molecular orbital and resonance arguments in your explanations. H2 C CH2

H3C

H2C

C H

CH2

H3C

O

CH2

MO: hyperconjugation: overlap of C-H s-bond with vacant p-orbital

MO: overlap of p bonding orbital with vacant p orbital

MO: overlap of oxygen lone-pair orbital with vacant p orbital (Because of overlap, lone pair will be in p-orbital)

H H3C H C

H2 H C H

H2C H

C

H2 H C H

H3C

O

H2 H C H

There isn't a particularly satisfying resonance explanation...

Resonance: there is a second, equally stable structure:

Resonance: there is a second, more stable structure:

H2C

C H

CH2

H3C

O

CH2

3. Rank the above cations in terms of stability. Again, use both MO and resonance arguments to explain the relative order of stability. They increase in stability from left (least stable) to right (most stable). MO: look at the energies of the donor orbitals. The donor orbital becomes higher in energy (thus a better donor) as you go from C–H to C=C to the nonbonding lone pair on O. Higher energy donor leads to better overlap and more stabilization of the carbocation. Carbocation p-orbital (nonbonding) Oxygen lone pair (nonbonding) C=C C–H

Best overlap comes from highest energy donor (HOMO) because it is closest in energy to the acceptor (carbocation)

Resonance: The alternate resonance structures become more stable as you go from left to right. (Why is the structure with the positive formal charge on the O more stable? Because it is the only Lewis structure in which all the atoms have filled octets. ) 33

Chemistry S-20

Week 1

Carbocation Formation and Rearrangement
1. Each of the following alkenes will generate a carbocation by protonation. Draw a curved arrow mechanism for the single step of carbocation formation, and show the carbocation product.

H+

H+

H+

2. Each of the following alkenes will...
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