Which of the following Newman projections that represent the most stable conformation of 2,3-dimethylbutane.
Which of the following Newman projections that represent the most stable conformation of 2,3-dimethylbutane.
Recall that these are CONFORMERS, meaning that they can take different positions in order to favor the most stable position.
Analysis will be done on carbns --> C2-C3 respectively
C2 --> front carbon, bonded species --> to H, CH3, CH3
C3 --> back carbon, bonded species --> H, CH3, H
Note that
there are two main types of position; they are either:
- STAGGERED --> will have 60° separation between all groups, therefore, it is much more stable, the groups are the farthest away possible from all
- ECLIPSED --> Half of the Groups are 0° away from each other, the other half is at 120°.. This requires higher energy since it takes more energy to get together at the 0° configuration.
Also...
keep in mind that bulky groups... Bulky groups are those groups which are "larger" considered to other groups
The least bulky group/species is -H atom... since it is small.
Note that there are also energy considerations depending on the type of configuration:
Bulky - Bulky --> most energy
Bulky - nonbulky --> medium energy requirement
nonbulky - nonbulky --> least energy
Now, knowing this....
Number your species from top to bottom as ;
A,B,C,D,E
ignore all eclipsed configurations, that is, A and E
B,C,D are left
2,3-dimethylbutane. --> total carbons = 2+4 = 6
recall that 2 carbons are not shown explicitly in the newman projeciton, so we must see 6-2 = 4 carbons in the substitents
ignore B since it has only 3
Only C and D are left
note that
D has all carbons on the 50% side of the molecule ( equivalent of Southest side)
therefore, this is loaded to this side, implies high energy
this is not recommended.
The most stable configuration has 50% - 50% in each load
therefore
C is the best configuration
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