A. Examine the 3 staggered and 3 eclipsed conformations resulting from rotation

Question

A. Examine the 3 staggered and 3 eclipsed conformations resulting from rotation about the C2/C3 bond in 2,3-dimethylbutane. Using the supplied conformational energies, determine which conformation is most stable and which conformation is least stable. Please explain/show how you got the answer for each part.

H,H eclipsed   1.0 kcal/mol

CH3,H eclipsed 1.4 kcal/mol

CH3,CH3 eclipsed 2.5 kcal/mol

CH3,CH3 gauche 0.9 kcal/mol

B.

Calculate the overall heats of reaction (?H) for each of the transformations below. Guided by these calculations, order the reactions from most likely to least likely to proceed.

Useful Bond Dissociation Energies H-H H-Br H-CI Br-Br (CH3)2CH-H CH32CH-Br CH32 CH-OH 104 kcaVmol 87.5 kcal/mol 119 kcaVmol 46 kcal/mol 95 kcal/mol 68 kcal/mol 92 kcal/mol Br Reaction 1 CH,CH2CH3 + Br2--CHCHCH3 + HBr 9H Br Reaction 2 CH3CHCH3HBr- 3 + HBr CH3CHCH3 + H2O Br Reaction 3 CHsCH2CH3 + HBr- .

Explanation / Answer

A.
Least energy implies more stable.
Most stable is : CH3,CH3 gauche
Least stable: CH3,CH3 eclipsed

B.
For reaction 1:
delta H = Bond dissociation of reactant - bond dissociation of product
     = 95 + 46 - 68 - 87.5
      = -14.5 Kcal /mol

For reaction 2:
delta H = Bond dissociation of reactant - bond dissociation of product
     = 92 + 87.5 - 68 - 0
      = 111.5 Kcal /mol

For reaction 3:
delta H = Bond dissociation of reactant - bond dissociation of product
     = 95 + 87.5 - 68 - 104
      = 10.5 Kcal /mol

Lesser the energy more stable is the product. More stable the product, most likely product forms.
reaction 1 is most likely and reaction 2 is least likely

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