[Dougal]

Okay, so I'm having some trouble with identifying chiral carbons. My textbook says that for a carbon to be chiral, it must have 4 different groups bonded to it. However, in several questions there are structures in which a carbon is bonded to more than one of a certain group, and is still considered chiral. What's that about? For example:



I have marked the molecules that have chiral carbons with a tick, and the molecule that does not with a cross. Now, I'm sure you see my  problem. To me, at least, the molecules that are supposed to have chiral carbons don't seem to have 4 different groups bonded to them. Maybe I am mistaken though. If I'm not, then I guess there is some more detailed explanation of what constitutes a chiral carbon out there, and I would be grateful if anyone could help me.

On a somewhat unrelated note, I was wondering if I someone could explain the basics of the inductive effect to me? I know that it has to do with the dispersal of charge, but I'm still a bit hazy on its implications and mechanisms.

A big thank you to anyone who helps!

[macman104]

Hey Dougal,

Let's start at the top left and work our way around clockwise:

1)  We see that there are two chiral carbons here.  The carbon with the -OH and the carbon with the -CH₃.  Now, it might be a little confusing since they are in a ring, which is fine.  If we look at the carbon with the -OH.  We see that it has an OH, a H and then two possible paths around the ring.  If we look at those two paths, we see that they are indeed different.  If we go around the ring to the left, we encounter a CH2-CH2-CH2-CH(CH3), but if we go around the right, we encounter CH(CH3)-CH2-CH2-CH2.  Even though it is in a ring, we see that the environments as we go around the ring are different depending on the direction, hence two "different" groups, in addition to the H and OH.  This same process can be carried out for the carbon with the -CH3 on it.

2)  This is much the same thing.  The only difference is that we only have 1 chiral center, namely the carbon with the -CH3.  See if after the previous explanation this one makes more sense.

3)  This one is analogous to the first one, only with a carbon cut out of the middle

4)  This one is a little different.  We initially suspect that the carbon with the CH3 and OH groups might be chiral.  However, if we trace the path around the ring for that carbon, we'll notice that it is actually symmetrical.  Because of this symmetry, the carbon lacks chirality, because it lacks 4 distinct substituents.

Inductive effects are the electronegative effects that adjacent atoms have on each other.  For example, if we have chloroform, CHCl₃, we can see that the chlorines have an inductive effect on the molecule as they pull electron density towards themselves.  For a more applicable (albeit slightly more advanced example), this trend is probably most noticeable when looking at pKa values.  If we look at acetic acid CH₃COOH, we see that it has a pKa of about 4.7, indicating it is fairly acidic.  However, if we look at something like Trichloroacetic acid, CCl₃COOH, we see that it has a pKa of only 0.7.  This can be explained by the strong inductive (electron withdrawing) that the chlorines have on the molecule.  The chlorines pull electron density away from the rest of the molecule making the hydrogen even more acidic.

[sjb]

Which molecules are you having difficulties in deciding whether they have chiral forms or not? Are you considering all of the substituent, not just the adjacent atoms?

Consider for instance 3-methylcyclohexanone (top right). Here the chiral centre is the one with the methyl group attached. It's true that at "atom distance 1" you have a methyl, a hydrogen and 2 methylenes, and so seemingly not 4 different groups, but if you look beyond the methylenes you see that one of them is bonded to a carbonyl and the other is bonded to a third methylene, so different..

[Dougal]

That was a fantastic answer, macman. The pKa example was great too, as that seems to be the context most of our questions are in.

Thanks!

[Borek]

CH₃COOH pKa = 4.75
CH₂ClCOOH pKa = 2.87
CHCl₂COOH pKa = 1.48
CCl₃COOH pKa = 0.7

[macman104]

That was a fantastic answer, macman. The pKa example was great too, as that seems to be the context most of our questions are in.Thanks!

Glad it helped, I wasn't sure what your level of knowledge was so I was a little worried that it might be overload.

As Borek posted, pKa's are usually a great way to see inductive effects in action.  Granted there can be other effects at work, but for the acetic acid and halogen analogues it works great.

I also tend to be verbose so....*shrug*

[AWK]

4)  This one is a little different.  We initially suspect that the carbon with the CH3 and OH groups might be chiral.  However, if we trace the path around the ring for that carbon, we'll notice that it is actually symmetrical.  Because of this symmetry, the carbon lacks chirality, because it lacks 4 distinct substituents.

I disagree with explanation of macman104 for this compound.
Very often hydrogen atoms in formulas of  cycloalkanes are omitted then you have two carbon atoms in cyclopentane substituted with 4 different substituents. If two different external subtituents are at different carbon atoms then such a cyclopentane derivative can be optically active.

[macman104]

I disagree with explanation of macman104 for this compound.
Very often hydrogen atoms in formulas of  cycloalkanes are omitted then you have two carbon atoms in cyclopentane substituted with 4 different substituents. If two different external subtituents are at different carbon atoms then such a cyclopentane derivative can be optically active.

I'm not sure I understand your post.  The bolded part seems to be talking about a carbon with 2 hydrogens and then 2 different substituents, but then you say 4 different substituents.  We both know what chirality is, so I'm sure it's just me not comprehending your post correctly...can you try and rephrase?  It's possible I phrased my explanation poorly, but I'm a little confused by your post.

[AWK]

 I do hope the picture is sufficient

[Borek]

macman104: you have numbered compounds

1 2
4 3

and AWK thinks

1 2
3 4

Or at least thats what I think about the problem source

[AWK]

Yes my nunbering was
1 2
3 4

[adam]

Borek are right, it's a numbering confusion.