[deutdeut]

Consider the carbocation in which one carbon atom in cyclohexane carries a positive charge, I know it's  a secondary carbocation, as I found 2 carbons attaching the postively-charged carbon, but can anyone tell me where is the 2 alkyl groups that exerts positive inductive effect in the structure?

[Mitch]

The cation obviously has a positive charge, the next nearest carbons by induction will have a smaller magnitude partial negative charge, the carbons attached to those will have a smaller magnitude partial negative charge and so on.

[deutdeut]

But I want to know is how to determine that there are 2 alkyl groups attching the positively charged carbon in the example above. Can you explain a bit?

[Mitch]

Its cyclohexane, every carbon is attached to 2 other carbons.

[deutdeut]

I know, but now one of the carbon atoms in postively charged, how I know its secondary carbocation based on the fact that it has two alkyl groups? What are the 2 alkyl groups in the structure?

[Mitch]

The 2 alkyl groups is the cyclohexane ring its attached too. What makes sense, that its a primary carbocation, or a tertiary carbocation, no, secondary is the only thing that makes rational sense.

[deutdeut]

Can you explain a little more of the statement : The 2 alkyl groups is the cyclohexane ring its attached too.

[english]

The following is a side view of what cyclohexane actually looks like in its most stable conformation.  It is not actually planar.  If you rotate this image by rotating it down, eventually you would see the famous planar hexagon.

From the side, the real geometry can be revealed in three dimensions.


Sigma bonds stabilize cationic charges on carbons.  Generally this means the hydrogen-carbon sigma bonds adjacent to the carbocation, and somewhat farther out at least one carbon-carbon sigma bond can stabilize the same carbocation.  The sigma bonds (single bonds) closest to the carbocation are the most important.

Pick any one of the carbons in this structure and pretend that one hydrogen is removed to depict your carbocation.  This carbon will now have an empty p orbital.  The carbons next to your carbocation have sigma bonds with their repsective hydrogens.  The bonding electrons from these bonds donate electric charge to the carbocation thorugh hyperconjugation, a phenomenon much like the overlap of p orbitals to form pi bonds.

[deutdeut]

That means, in if a carbocation is formed from cycloalkanes, no separate alkyl groups can be identified, like the above example, since the carbon atoms are joined together in ring form?

[english]

When we have a cyclic structure, the first alkyl group starts from the carbon to the left of the chosen carbon and all of the other carbons, let's say, counterclockwise up to your chosen carbon.

The second starts to the right of the chosen carbon and extends clockwise all across the ring back to the chosen carbon.


You're letting the cyclic aspect confuse you.  The alkyl groups depend on what direction you move.

You're thinking that it is the same no matter what direction, but for this purpose just focus on where the donating bonds are coming from.


We typically only focus on the two neighboring carbons, because only they tend to matter.  But if you move around the ring from those neighboring carbons you'll get two alkyl groups, depending on what direction you go.

There's no easy way to explain this.   


Think of it like this. 

If we choose one of those six carbons, we can say that it is bonded to two —CH₂CH₂CH₂CH₂CH₂ groups, one to the left, and one to the right.

That they are all connected is throwing you.

[deutdeut]

But the so-called two alkyl groups are in fact the same,  isn't it ridiculous to recount the same thing twice? And also, from the above example, the positively-charged carbon is connected to C5H10, is it an alkyl group? Isn't we call C5H11 an alkyl group?

[english]

You can see the significance of this if you see a disubstituted cyclohexane.

The carbon bonded to the bromide, Br, has two alkyl groups, —CH₂CH(CH₃)CH₂CH₂CH₂—  and   
—CH₂CH₂CH₂CH(CH₃)CH₂—

Same principle I mentioned earlier.  We're moving clockwise and counterclockwise with respect to our chosen carbon, which is bonded to bromine.

You see the significance now that the two directions result in different groups? 


With cyclohexane, you have two alkyl groups, they're just the same. 

[deutdeut]

But returning to my example, the positively-charged carbon has 2 alkyl groups as you've mentioned, but is C5H10 an alkyl group. Isnlt the pentyl group is C5H11?

[english]

You can't really say that the whole ring is one alkyl group.


And even if you did, since it is cyclic it would be C₅H₁₀.

 

[Mitch]

How big of a chain would it have to be until you feel comfortable calling it a separate alkyl group?