5. Instead of redrawing the molecule to make the lowest priority group point back, alternatively, the lowest priority group can be swapped with the group that is currently pointing back. This is now the enantiomer of the original compound. The procedure is described below. i. Determine priorities of the groups attached to the stereocenter. ii. Swap the lowest priority group with the group that is pointing back, and redraw (this is the enantiomer). iii. Determine the absolute configuration (R or S) of the enantiomer. iv. Determine the absolute configuration (R or S) of the original compound. Answer the following questions. Completely fill in the circle in front of your chosen answer. (a) Assign priorities of the groups at this stereocenter (1 = highest priority; 2 = second priority; 3= third priority; 4 = lowest priority). Each group must have a different priority: CH3 Br H3CH₂C H CH3 O 1 O 2 O 3 04 Br O 1 O 2 O 3 4 (b) The absolute configuration of this compound is: OR OS H O 1 O 2 O 3 O t CH₂CH3 O 1 O 2 O 3 O 4

Organic Chemistry: A Guided Inquiry
2nd Edition
ISBN:9780618974122
Author:Andrei Straumanis
Publisher:Andrei Straumanis
Chapter6: Alkanes & Alkenes
Section: Chapter Questions
Problem 21E
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5. Instead of redrawing the molecule to make the lowest priority group point back, alternatively, the
lowest priority group can be swapped with the group that is currently pointing back. This is now the
enantiomer of the original compound. The procedure is described below.
Determine priorities of the groups attached to the stereocenter.
Swap the lowest priority group with the group that is pointing back, and redraw (this is the
enantiomer).
iii.
Determine the absolute configuration (R or S) of the enantiomer.
iv. Determine the absolute configuration (R or S) of the original compound.
i.
ii.
Answer the following questions. Completely fill in the circle in front of your chosen answer.
(a) Assign priorities of the groups at this stereocenter (1 = highest priority; 2 = second priority;
3 = third priority; 4 = lowest priority). Each group must have a different priority:
CH3
Br
H
CH3
*Br
H3CH₂C H
3
O 4
(b) The absolute configuration of this compound is:
OR
OS
CH₂CH3
1
02
Transcribed Image Text:5. Instead of redrawing the molecule to make the lowest priority group point back, alternatively, the lowest priority group can be swapped with the group that is currently pointing back. This is now the enantiomer of the original compound. The procedure is described below. Determine priorities of the groups attached to the stereocenter. Swap the lowest priority group with the group that is pointing back, and redraw (this is the enantiomer). iii. Determine the absolute configuration (R or S) of the enantiomer. iv. Determine the absolute configuration (R or S) of the original compound. i. ii. Answer the following questions. Completely fill in the circle in front of your chosen answer. (a) Assign priorities of the groups at this stereocenter (1 = highest priority; 2 = second priority; 3 = third priority; 4 = lowest priority). Each group must have a different priority: CH3 Br H CH3 *Br H3CH₂C H 3 O 4 (b) The absolute configuration of this compound is: OR OS CH₂CH3 1 02
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