Chemistry Olympiad Training Program Sec 4/2010

Chemistry Olympiad Training Program Sec 4/2010

Lesson 6

Isomerism in alkanes (continued)

ACTIVITY 1

Newman Projection Practice

(a)  Make a model of 2,3,4-trimethylpentane. Consider the bond along C2-C3. Draw Newman projections for the six conformations corresponding to energy maxima and minima.

(b)  State and explain which conformation is the most stable and which is the least stable.

(c)  Starting with the least stable conformer, sketch the energy profile to illustrate the relative stability of the different conformers of 2,3,4-trimethylpentane when rotated along C2-C3.

(a) Newman projections:

I / II / III
IV / V / VI

(b) Most stable conformation: IV (staggered)

Reasons:

(1) Staggered conformation is more stable than eclipsed conformation.

(2) There are only two pairs of gauche interaction (look up this term!) between alkyl groups in IV but three pairs between alkyl groups in II and VI Þ less steric strain for VI.

Least stable conformation: I (eclipsed)

Reasons:

(1)  -CH(CH3)2 is a larger group than -CH3 and thus the eclipsed interaction between -CH(CH3)2 and -CH3 gives rise to the greater steric repulsion. This interaction is absent in V.

(2)  Stability of I < III due to presence of CH3-CH3 eclipsed interaction which is absent in III.

(c)

ACTIVITY 2

Boat and chair conformations of cyclohexane

(a)  Make a model of the chair and boat form of cyclohexane.

Chair conformation / Boat conformation

(b)  For each conformation, draw a Newman projection when viewed along C1-C2 and C5-C4 Which conformation is more stable? Give two reasons for your answer.

Chair conformation / Boat conformation

More stable conformation: Chair conformation

Reasons:

(1) Presence of additional strain in the boat form due to four pairs of eclipsed C-H bonds.

(2) Steric repulsion between flagpole hydrogens in boat configuration (H attached to carbon 3 and 6)

ACTIVITY 3

Ring-flip in cyclohexanes – conformational mobility

Make a model for each of the following molecules A and B.

A: / B:

For each molecule:

(a)  Flip the model into one of the possible chair conformations and draw it. Label the substituent positions as axial or equatorial.

(b)  Flip the model into the other chair conformation and draw it. Label the substituent positions as axial or equatorial.

(c)  State which is the more stable conformation.

(d)  State what happens to the substituent in the axial and equatorial positions as a result of the ring flip.

(a),(b),(c) Molecule A:

I / II

More stable conformation: II

Molecule B:

III / IV

More stable conformation: III

(d) The axial and equatorial positions interconvert as a result of ring flip.

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