Assign relative priorities to the groups or atoms in each of t...

Question

Assign relative priorities to the groups or atoms in each of the following sets:

c. $-$ CH(CH₃)₂

d. $-$ CH $=$ CH₂

Accepted Answer

Hello everyone. Let's do this problem. It says rank the groups or atoms in each of the following sets according to their relative priorities. For set A, we have the groups C C H 33 C H two C H two C L B R N C H two C H two C H two C L in set B, we have C H double bonded to C H C H three C H two C H two, C H three, CH two CH 3 and C C H three double bond C H two. So when the question asks us to rank these groups according to their relative priorities, we're talking about the priorities mentioned in the K OL pre log nomenclature rules. So this relates to assigning R or S configuration or absolute configuration. So we're going to imagine that these four groups are attached to a chiral center and then we'll assign their priorities based on atomic mass. So the higher the atomic mass specifically of the first atom attached to that Carl center, then the higher the priority of that group. OK. And what if there's a tie between that first atom in two groups or multiple groups, then we'll simply compare their adjacent atoms using the atomic mass priority rule. OK. So let's start with set A, I'll draw a center carbon with four bonds and starting from the top center and going clockwise, I will write these groups in order and I'll write them in their expanded formula so that we can see the groups attached to each atom. So in the first position, the first group is C CH three CH 3 ECH three Group two. On the right, we have CH two C H two C L group three at the bottom. At six o'clock, we have a Bromine Adam And group four on the left, we have CH two C H two C H two C L. OK. So we're first going to look at the atom that is directly attached to the center carbon. So we have carbon carbon bromine carbon. Well, the bromine has a higher atomic mass than carbon. So this bromine atom is going to be the highest priority group. OK. Now, we have to compare these three carbon atoms and their adjacent atoms since they are all equivalent, right? They're all carbon atoms. So let's look at the three next atoms attached to these carbons. So group number one, we have three carbons group, number two, we have hydrogen, hydrogen carbon group four, we have hydrogen, hydrogen carbon. Well, since group one has 3 carbons and the other groups have two hydrogens and a carbon. That first group with the three methyl is going to be the second highest priority group. OK. Now up to this point groups, two and four have the same atoms. So we need to keep comparing their adjacent atoms. So moving on to the three atoms attached to the second carbon on carbon two, we have hydrogen chlorine hydrogen. On group # four, we have hydrogen carbon hydrogen. While chlorine has a higher atomic mass than carbon. So that second group will be third priority and the fourth group will be fourth priority. So listing the names of these groups out in their descending priority order, We have bromine highest priority. Then CCH- 33, then C H two C H two C L And lastly lowest priority we have CH two, CH 2 CH two CL. All right, that's it. For set A, let's do all this again for set B drawing my center carbon with four bonds. OK. First group in my list is carbon with a hydrogen double bonded two C H C H three C H C H three. My next group I have CH two CH 2 CH three. My 3rd group, I have CH two CH And my fourth group, I have CCH three double bonded two C H two. All right. So let's compare the initial atom in every group connected to that chiral center, they are all carbon atoms. So we have to move on to adjacent atoms. So the first group and the fourth group have double bonds. So that's something to note. So in group one, we have hydrogen and then we'll count that carbon double bond twice. So hydrogen, carbon carbon group two, we have hydrogen, hydrogen, carbon group three, we have hydrogen, hydrogen carbon and group four, we have carbon carbon carbon. So group four has the most carbon atoms attached to that first carbon. So that's going to be our highest priority group. Group number four. Well, group # four is our first priority group. OK. And our first group will be the second priority, right? The group here at the top that has hydrogen carbon carbon. All right, the C H double bond C H C H three group and now groups two and three both have a hydrogen, hydrogen carbon bonded to that first carbon. So we need to look one step further at those adjacent atoms. So group two has hydrogen hydrogen carbon and group three has three hydrogens. So those three hydrogens are going to make this group the lowest priority group. Number four for group number three, rather group number three will be the lowest priority and group number two will be third priority. So listing those from highest to lowest priority, we have C C H three double bonded to C H two. Then we have CH double bonded to CHCH three. Then we have CH two, CH 2, CH three and our lowest priority group is CH two CH 3. So that is it for set B which completes the answers for this entire problem. That's it for this one. I hope this video is helpful and thanks for watching.

Assign relative priorities to the groups or atoms in each of the following sets:
c. $-$ CH(CH₃)₂
d. $-$ CH $=$ CH₂

Key Concepts

Cahn-Ingold-Prelog Priority Rules

Chirality

Substituent Effects

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