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Lecture Questions (Q 21 and 28)
LiveraD
#1 Posted : Thursday, August 26, 2010 1:21:28 AM
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For Q21 I dont quite understand how they worked out the plane of rotation/symmetry for the answer given in the back of the book. Could someone please explain the answer in terms of free rotation of bonds? it seems to me that all bonds for the answer choices given can be rotated since they are all sigma bonds.

Q28 The question refers to the cis isomer of the compound, but the answer seems to refer to the trans isomer (shouldn't both chlorines be on the same side; either equatorial or axial?). Could some one please explain why the answer is not 'A' (since substituent groups prefer the equatorial position) or 'C'. If the answer in the back of the book is correct i guess my question is, Whats the difference between the cis and trans isomers in a ring compound?
mathewchui
#2 Posted : Tuesday, June 07, 2011 5:53:48 PM
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Hello,

Re Q21
About free rotation in the Fischer projection. One way to look at it is to draw the actual 3D geometry of it with hashed lines and wedges and then rotate it that way. Remember that vertical lines point into the page and horizontal lines point out of the page.

It's also helpful to know the "rules" about rotating in a fischer projection. On the terminal carbons, a "legit" rotation is one that changes the position of all 3 substituents. So that (CH3, Br, H) in option B can be rotated to (Br, CH3, H) and (CH3, H, Br) from left-top-right.

Switching the positions of only two substituents changes the configuration about a carbon and is disallowed.

A: you would have to switch H-Br to achieve an internal plane of symmetry
C: you would have to swtch H-CH3
D: you would have to switch H-Br

Re Q28
"Cis" on the ring refers to both substituents pointing "up" or both pointing "down".

Note that in cyclohexane, substituents at positions 1, 3, 5 are "cis" if they are all equatorial or all axial. However, substituents that are 1 carbon removed (positions 1,2) are "cis" if one is equatorial and one is axial.

Put another way, if two substituents which are 1 carbon removed (positions 1,2) are both axial, then they must be trans.

Also worthy to note. If a substituent is pointing "up" while in the axial position, after a ring flip, it will still be pointing "up". Same for the "down" position. Try it out.

Alternatively, go to wikipedia (http://en.wikipedia.org/wiki/Ring_flip) and look up the "Ring Flip" article. Look at the figure. Pay attention to the numbered carbons and the directions their substituents are pointing.

Cheers
mathewchui
#3 Posted : Tuesday, June 07, 2011 5:55:56 PM
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More regarding Q28

Option A: it is impossible for both to be in the "equatorial" if they are cis
Option B: again, impossible
Option C: again, impossible
Option D: this is what my explanation in the previous post shows. Two "cis" substituents that are one carbon removed must be either equ/ax or ax/equ, but not both axial or both equatorial.
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