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Prévia do material em texto

CHAPTER 16 593 
 
 
Next, look for exocyclic double bonds. In this case, two 
double bonds (highlighted below) are exocyclic. One of 
the double bonds is exocyclic to two rings giving a total 
of three exocyclic relationships that each add +5 for a 
total of +15. 
 
 
 
Finally, look for a homoannular diene. There appears to 
be no ring that has two double bonds contained wholly 
within it, so no adjustment is required for this compound. 
In summary, the following calculation predicts a max of 
approximately 287 nm: 
 
 Base = 217 
 Additional double bond (1) = +30 
 Auxochromic alkyl groups (5) = +25 
 Exocyclic double bonds (3) = +15 
 Homoannular diene (0) = 0 
 Total = 287 nm 
 
This prediction is in fairly good agreement with the 
experimentally observed value of 283 nm. 
 
 
16.60. 
(a) α-Terpinene reacts with two equivalents of molecular 
hydrogen, so it must have two  bonds. These  bonds 
must be associated with two C=C double bonds (rather 
than being associated with a C≡C triple bond), because 
the carbon skeleton (which does not change during 
hydrogenation) cannot support a triple bond: 
 
 
 
A triple bond could not have been in the ring, because a 
six-membered ring cannot support the linear geometry 
required by the sp hybridized carbon atoms of a triple 
bond. The other C-C bonds (outside the ring) can also 
not support a triple bond (without violating the octet rule 
by giving a carbon atom with five bonds). Therefore, α-
terpinene must have two double bonds. 
 
(b) The ozonolysis products indicate how the molecule 
must have been constructed, because ozonolysis breaks 
C=C bonds into C=O bonds: 
 
 
 
(c) This conjugated system is a homoannular diene: 
 
 
 
In addition, there are four auxochromic alkyl groups, 
highlighted here: 
 
 
 
The calculation for max is shown here: 
 
 
Base = 217 
Additional double bonds = 0 
Auxochromic alkyl groups = +20 
Exocyclic double bonds = 0 
Homoannular diene = +39 
Total = 276 nm 
 
16.61. The molecular formula (C7H10) indicates three 
degrees of unsaturation (see Section 14.16). The 
problem statement indicates that compound A will react 
with two equivalents of molecular hydrogen (H2). 
Therefore, we can conclude that compound A has two  
bonds, which accounts for two of the three degrees of 
unsaturation. The remaining degree of unsaturation must 
be a ring. Ozonolysis yields two products, which 
together account for all seven carbon atoms: 
 
 
Focus on the product with three carbonyl groups. Two 
of them must have been connected to each other in 
compound A (as a C=C bond), and the third carbonyl 
group must have been connected to the carbon atom of 
formaldehyde (CH2O). This gives two possibilities: 
Either C1 was connected to C6 (and C2 was connected to 
C7): 
 
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