3. A. Draw the Newman Projection of pentane looking downthe C2-C3 bond in a staggered confirmation.B. Draw an energy diagram showing the conformationalanalysis using the Newman Projection above (showing a 360degree total rotation, 60 degrees at a time). Hint – see Figure4.6, 4.9 and 4.11 for examples. Estimate the potential energy ofeach 60 degree turn as you do not have data to calculate thevalues exactly. Label each Newman Projection asstaggered/eclipsed or gauche/anti as appropriate.

Answered: Image /qna-images/answer/cbae0672-a0a8-41c5-a13a-b6490b057c20.jpg

Answered: 3. A. Draw the Newman Projection of…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Please answer the following questions A-D. Label everything and circle answer. A.) (4.) Draw the R and S enantiomer for 2-bromobutane. Using the model of 2-bromobutane representing the R enantiomer from 4., switch any two groups on the chiral carbon. Compare this to your S enantiomer from 4. B.) Are they the same or different? ______ C.) Are they the same or enantiomers? ________________ Now switch any pair of groups around the chiral carbon of the “switched” model again! Compare the “double-switched” model with the other unmodified S molecule from 4. D.) Are they enantiomers or are they identical? ______
6. Did you really think I would forget to include Newman projections? Draw the given wedge- dash structure in a Newman projection (as drawn; do NOT rotate it yet). Then perform the following rotations, indicating if the resulting conformation is anti staggered, staggered, eclipsed, gauche, or syn eclipsed. Use abbreviated substituent names for Newman projections and full structures for wedge-dash projections. Finally, rank the conformers in order of stability (using > or =) Newman as drawn: A conformation: back carbon front carbon rotate back carbon 60 degrees counterclockwise Newman after rotation: C rotate front Newman after rotation: B carbon 180 degrees conformation: conformation: rotate back carbon 120 degrees clockwise Newman after rotation: D Wedge-dash redraw as wedge-dash conformation: Most stable: (least)
Complete the question on paper and snap a photo and upload it as your submission in jpg or pdf format. CI IO H OH a. Generate a Newman projection of the molecule shown above. The red arrow indicated the direction from which your eyes are looking at the molecule. b. Also draw a Newman projection of a conformation that is less stable than the one that you have draw in a. There may be more than one correct option to generate a less stable Newman projection - you need only generate one. You can rotate around the C-C single bond to go between eclipsed and/or staggered conformations. c. Describe why the conformation in b. is less stable than a (name the interactions that make the conformation less stable). Point out where on your structure those interactions are (circling is fine).
Consider the sugar below, one of several sugars in the antibiotic streptomycin. Draw in all of the hydrogens and label them as axial or equatorial. Reverse, reverse!Draw the molecule with traditional wedge/dash notation. Draw a Newman projection looking down the indicated bond. Practice self- care by abbreviating the ring when doing so.
Give detailed Solution with explanation needed with structure. Don't give Handwritten answer. Give correct solution
a. Use strain energy increments in the OWL Table Reference (see References button, Strain Energy Increments) to calculate the energy difference between the two chair conformations of the compound b. Specify substituent positions (axial or equatorial) in the more stable chair. c. Estimate the percent of the more stable chair at equilibrium at 25°C. (To determine the percent of the more stable chair at equilibrium, first calculate Keq, and then use this value to find the percentage.) CH3 CH3 Answers: a. The energy difference is b. In the more stable chair: • The phenyl group is in the • The isopropyl group is in the | kJ/mol. position. position. c. At 25°C the equilibrium percent of the more stable chair conformation is approximately [
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.
The S isomer of 2-chlorobutame (below) can react with chlorine to make 5 isomers of dichlorobutane. Draw a Fisher projection each isomer, label all of the chiral centers as R or S, and designate the enantiomer and diastereomer relationships. (This is a reaction you have never seen before, it is capable of removing any hydrogen in the molecule and replacing it with a chlorine. It is a 2,step reaction with a free radical intermediate) CH, -CI + Н-с-н Ch C4H3 Ch + HCI CH,
3. Consider the following compound, which is a derivative of cholesterol. HO Identify all asymmetric carbons and assign their absolute configuration. Write the configuration directly next to each asymmetric carbon in the above image.
Several compounds are shown below. For each, make two models: one model of the compound as shown, and the other where each tetrahedral stereocenter is inverted. Compare the two models side-by-side to see if the model with the inverted stereocenter(s) is the enantiomer, or if it is identical to the original. Remember, you can rotate around o-bonds to place the molecule in the conformation with the highest symmetry. Is each compound meso or not? OH OH OH OH DOKOOK OH OH Compound 1 Compound 2 Compound 1's mirror image is identical to Compound 1 Compound 1's mirror image is the enantiomer of Compound 1 Compound 1 is meso Compound 2's mirror image is identical to Compound 2 Compound 2's mirror image is the enantiomer of Compound 2 Compound 2 is meso Compound 3's mirror image is identical to Compound 3 Compound 3's mirror image is the enantiomer of Compound 3 Compound 3 is meso Compound 4's mirror image is identical to Compound 4 Compound 4's mirror image is the enantiomer of Compound 4…

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