Functional Group Answers

 Most of the answers are shown on the structure of EPOTHILONE A to the right.

(a) Important functional groups are in blue. Note that the N is not an amine since it is part of a double bond. Probably look for the heteroatoms (O,N,S etc) and the multiple bonds. The alkane components has been ignored.

(b) The hybridization of the C atoms asked are indicated in green. Did you remember to account for the CH bonds that aren't shown ?

(c) C14-C15 > C15-C17 > C17-C18.  The hybridization of the C atoms involved in the various bonds affects the bonds lengths. The greater the % s character in the hybrids (sp3 = 25% s, 75% p : sp2 = 33% s, 67 % p : sp = 50% s, 50% p), the shorter the bond will be. C14-C15 is sp3-sp3, C15-C17 is sp3-sp2 and C17-C18 is sp2-sp2.

(d) The bonding MOLECULAR orbitals are indicated in red.
 
 

Epothilone A

Isomers Answers

Ans 1 : The molecular formula of C6H14 contains the maximum number of H atoms for 6C (CnH2n+2) so it can cannot contain double bonds or rings. Therefore the only functional group that can be present is an alkane.  When you draw them it is a good idea to have a system whereby you start with the longest chain, then move to shorter chains with an increasing number of substitutents, moving them to different positions on the chain as you go. Watch that you don't repeat structures.

Constitutional isomers of C6H14

Ans 2 : The molecular formula of C5H10 contains 2 less H atoms than pentane (C5H12) so that it must contain either a double bond or one ring. Therefore in terms of functional groups we need to consider alkenes and cycloalkanes.

Constitutional isomers of C5H10


Ans 3 : The molecular formula of C3H6O contains 2 less H atoms than propane (C3H8) so that it must contain either a double bond or one ring.
The combination of the double bond, ring and oxygen atoms means that the following functional groups are possible: aldehyde, alcohol, alkene, cycloalkane, epoxide, ether, ketone.

Constitutional isomers of C3H8O


Thermodynamics Answers
 
Ans 4 : In order to answer the question, it is a very good idea to draw a diagram that relates the following three equations that can be use to create the Hess's Law  "triangle" that relates the alkanes, their component elements and their combustion products.

Hess's Law cycle for C8H18 combustion


Reaction
DH 
Eqn
C8H18 ----------> 8 CO2 + 9 H2O Combustion (known), Hc(alkane)
1
8 C + 9 H2 -----> 8 CO2 + 9 H2O Combustion (calculate), Hc(elements)
2
8 C + 9 H2 ----------> C8H18 Formation (required) Hf(alkane)
3

From Hess's Law we get that Hf (alkane) = Hc(e) - Hc(alkane)

Hc(e)= 8 x Hc(C) + 9 x Hc(H2) = 8 x -94.05 + 9 x -57.8 = -1272.6 kcal/mol.

So for 2,2-dimethylhexane, I, Hf = -1272.6 - - 1304.6 = 32 kcal/mol

and for 2,2,3,3-tetramethylbutane, II,  Hf = -1272.6 - -1303.0 = 30.4 kcal/mol

Thus 2,2,3,3-tetramethylbutane is the more stable since it has the more exothermic DHf. (care ! both are calculated to be endothermic).  This is also evident in that II has the less exothermic Hc, and is the most branched of the two hydrocarbons. The relative stability is most clearly demonstrated and appreciated by the diagram, otherwise be very cautious with the signs of your calculations.