Br H₂C H H₂C Br Br. A D G CH₂ H,C. Br B CH₂ E Br CH₂ H Br. H₂C .. CH₂ F Br Br CH₂ To answer it correctly: (a) Draw both chair conformations. (b) Determine which chair conformation is more stable. Make sure to keep track of the highlighted (c) Draw two connected Newman projections for each highlighted bond for each conformation. (d) Fill in the blanks below: The answer for the more stable chair conformation: Hint: More stable conformation means there are more equatorial substituents than axial. The answer for the less stable chair conformation:

Br H₂C H H₂C Br Br. A D G CH₂ H,C. Br B CH₂ E Br CH₂ H Br. H₂C .. CH₂ F Br Br CH₂ To answer it correctly: (a) Draw both chair conformations. (b) Determine which chair conformation is more stable. Make sure to keep track of the highlighted (c) Draw two connected Newman projections for each highlighted bond for each conformation. (d) Fill in the blanks below: The answer for the more stable chair conformation: Hint: More stable conformation means there are more equatorial substituents than axial. The answer for the less stable chair conformation:

Organic Chemistry: A Guided Inquiry

2nd Edition

ISBN:9780618974122

Author:Andrei Straumanis

Publisher:Andrei Straumanis

Chapter6: Alkanes & Alkenes

Section: Chapter Questions

Problem 10CTQ: Consider the Newman projection below. a. Draw a full Lewis structure of this molecule with...

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Consider the molecule 1-bromo-2-methylbutane. C3 and C4 should be drawn as Et as in theexample. This group is called an ethyl group and can be considered a sphere about twice the sizeof a methyl group. Draw the following Newman projections sighting down the C1C2 bond... a. The lowest potential energy conformation. b. The highest potential energy staggered conformation.
Consider the Newman projection below. a. Draw a full Lewis structure of this molecule with R1=Me,R2=Et , and R3=iPr . b. Given the sizes of these R groups (R3R2R1) , does the Newman projection above show thelowest potential energy conformation of this bond? If not, draw a Newman projectionshowing the lowest P.E. conformation (sighting down this same bond). c. To draw a Newman projection in the lowest P.E. conformation, the following rule of thumbusually applies: Place the largest group on the front carbon anti to the largest group on theback carbon. Is your answer to the previous question consistent with this rule of thumb?
Imagine that this worksheet has sucked you the plane of the page, and you are now facing the indicated bond of the drawn molecule. a. Draw the six main Newman projections for the indicated bond: three eclipsed, three staggered. b. Label which conformers are eclipsed ands taggered.c. Identify the most and least stable conformer, and explain your reasoning.
Following is a staggered conformation for one of the enantiomers of 2-butanol Q. viewed along the bond between carbons 2 and 3, draw a Newman projection for this staggered conformation.
Then draw the most stable and least stable Newman projection conformation from the C4-C5 bond in the molecule above
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Please do parts A and B Consider (2R,3R)-2,3-diidobutane: (a) Draw a Fischer Projection of this structure with C1 on top and C4 on bottom. (b) Draw a Newman Projection of of the meso compound of 2,3-dibromobutane looking down the C2-C3 bond with C2 in front, and label your asymmetric carbons as S or R.
Draw each compound in its most stable conformation(s). Then draw it in its most symmetric conformation, and determine whether it is chiral.(a) 2-methylbutane