[tortoise]

Please arrage these ions in desceding order of stability:
a) CH₃⁺  b) CH₂R⁺  c) CHR₂⁺  d) CR₃⁺

[Yggdrasil]

Please read the Forum Rules[/url] before posting.  We will not help you answer question that you have not shown any attempts to answer on your own.  However, I will offer one bit of help.  A very helpful tutorial on carbocations can be found at the following site:

http://web.chem.ucla.edu/%7Eharding/tutorials/cc.pdf

[tortoise]

sorry, but I really don't know the answer... a solution given by my friend is a - d - b - c. I don't know why it is.

oh thank you for your link. it helps a lot. yeah the order is (page 5)

CH₃⁺ (least stable) < RCH₂⁺ < R₂CH⁺ < R₃C⁺ (most stable)

[Yggdrasil]

Yup, that is the correct order.  Now, the real question is do you understand why?  (if you don't I can try to explain it)

[tortoise]

actually.... I don't know     Help me explain it pls.

[Yggdrasil]

In one word the answer is hyperconjugation.  Basically, since carbocations have empty p-orbitals, they gain stability by filling their empty p-orbital with electrons from neighboring sources.  One way that this can happen is through overlap with neighboring sigma bonds.  For example, lets say you have a methyl group bonded to a carbocation.  When the C-H bond of the methyl group is aligned with the empty p-orbital of the carbocation, some of the electron density in the C-H sigma bond can delocalize to help fill and stabilize the empty p-orbital.  This phenomenon is known as hyperconjugation.  Hydrogens which are directly bonded to the carbocation cannot stabilize the empty p-orbital because the hydrogen's electrons lie in the C⁺-H bond which is orthogonal to the empty p-orbital and therefore cannot donate electron density.

Does this make sense?

[tortoise]

sure! it's surely simplier than the lecture you've given. thanks a lot!