Question #27b9d

2 Answers
Sep 30, 2015

Answer:

It is #R#.

Explanation:

enter image source here

You need to assign priorities for the groups atached to the stereocenter by taking a look at the atomic number of the atom that is directly attached to the carbon atom, i.e. to the stereocenter.

So, for the groups you have, the atoms attached to the stereocenter are

  • For #W -># the #"N"# in #"NH"_2#;
  • For #X -># the #"C"# in #"CH"_3#;
  • For #Y -># the chlorine atom, #"Cl"#;
  • For #Z -># the #"C"# in #"CH"_2"NH"_2#.

The idea here is that you assign priorities in order of decreasing atomic number. The atom will the biggest atomic number will get priority #color(red)(1)#, the atom with the second-biggest atomic number will get priority #color(red)(2)#, and so on.

In this case, you have nitrogen, #Z = 7#, chlorine, #Z = 17#, and two carbon atoms, #Z = 6#.

So chlorine will get priority #color(red)(1)# and nitrogen will get priority #color(red)(2)#.

But what do you do for the two carbon atoms? Which one gets priority #color(red)(3)# and which one gets priority #color(red)(4)#?

To decide that, look at the atoms that are attached to these two carbon atoms in their respective groups.

Group #X# has its carbon atom attached to three hydrogen atoms, while group #Z# has its carbon atom attached to two hydrogen atoms and a nitrogen atom.

#"H"_2"C" - "NH"_2#

So, for group #X# you have three hydrogen atoms, #Z = 1#. For group #Z#, you have two hydrogen atoms and one nitrogen atom, so you get

#"For group X: " "H " "H " "H" #
#"For group Z: " "N " "H " "H"#

The first point of difference between these two lists will constitute the tiebreaker. In your case, the tiebreaker is the nitrogen atom, atomic number of #Z = 7#, which beats the hydrogen atom, atomic number #Z=1#.

This means that group #Z# will get priority #color(red)(3)# and group #X# will get priority #color(red)(4)#.

So, this is what you have

enter image source here

Now, in order to assign an #R# or #S# configuration for your stereocenter, you need to have the group with the lowest priority on the dash, i.e. pointing away from the plane of the page.

So basically, you need #color(red)(4)# to be where #color(red)(3)# is.

Now, you could rotate the molecule in your mind to see how the configuration would change if group #X# is placed on the dash, or you could use a little trick to find the #R# or #S# configuration without changing the orientation of the molecule.

Since this answer is already pretty long, I'll show you the quick way.

So, you know that can determine the configuration of the stereocenter by ignoring the group with the lowest priority and checking to see the direction in which the remaining priorities increase

  • if they increase counterclockwise, then you have #S#
  • if they increase clockwise, then you have #R#

Since the group with the lowest priority is not on the dash, you can switch it with the group that is on the dash so that it lands on the dash.

Keep in mind, however, that switching the position of any two groups will change the stereocenter configuration!.

In your case, if you switch #color(red)(4)# with #color(red)(3)#, so that the lowest priority group is now on the dash, you get

enter image source here

The priorities increase from priority #color(red)(1)# to priority #color(red)(3)# counterclockwiseclockwise, so the stereocenter is #S# (I apologise for the awful quality)

enter image source here

However, since you switched two groups to get this configuration, you have to reverse it to get the actual one.

Therefore, the stereocent is not #S#, but #R#, which is what you would get if group #X# was placed on the dash without having to switch it with another group.

Sep 30, 2015

Answer:

The configuration is (R) .

Explanation:

Let us write the structure of your molecule first and assign priorities:
enter image source here

Priorities are assigned based on the atomic number. The rule is the following:
Assign priorities (1, 2, 3, or 4) to the atoms directly bonded to the stereogenic center in order of decreasing atomic number.

Next we look at the lowest priority substituent (#CH_3#), it should be facing backward; which is not in this case!

The rule is very simple, you switch places between the lowest priority substituent (#CH_3#) and the actual substituent facing backward (#CH_2NH_2#), here is what you get:
enter image source here

Label the new configuration after rotation; it is going to be (S) .
Flip the configuration: Which means since it is (S) it will become (R).

You're DONE.

Note: some textbooks mention that you have to rotate the molecule, which means you have to work your imagination, since it is a 3D molecule and you are working on a 2D surface (the paper or computer monitor) sometimes it becomes confusing, this method is way simplier.

Here is a video on labeling stereogenic centres R and S. There is part 1 and part 2.