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Hi all, <o:p></o:p>

<o:p></o:p>

I have some difficulties on carbanion which I desperately need help with, <o:p></o:p>

<o:p></o:p>

Methane has a full outer shell of electrons. Each hydrogen atom contributes 1 electron (represented by 'o') and the carbon contributes 4 electrons ( represented by x) [ignoring the C inner-shell non-bonding electrons. <o:p></o:p>

H<o:p></o:p>

xo<o:p></o:p>

Methane: H x o H<o:p></o:p>

o C x<o:p></o:p>

xo<o:p></o:p>

H<o:p></o:p>

<o:p></o:p>

I have to draw the structure as above of the carbon species formed by the following reactions: <o:p></o:p>

<o:p></o:p>

1. CH4 -------> -CH3 [Carbanion] <o:p></o:p>

<o:p></o:p>

I think the structure would be : <o:p></o:p>

<o:p></o:p>

H <o:p></o:p>

xo<o:p></o:p>

Methane: H x <o:p></o:p>

o C <o:p></o:p>

xo<o:p></o:p>

H<o:p></o:p>

<o:p></o:p>

I'm not very sure on this, can you confirm this with me?

-H.

CH4 ------------> .CH3 <o:p></o:p>

<o:p></o:p>

what will be the structure of this? [Radical] <o:p></o:p>

<o:p></o:p>

3.) <o:p></o:p>

-H -<o:p></o:p>

CH4 ----------> +CH3 <o:p></o:p>

<o:p></o:p>

and the structure of this? [Carbocation] <o:p></o:p>

<o:p></o:p>

And what will be the 3D structures and the angles between the C-H bonds in -CH3, .CH3 and +CH3 ? <o:p></o:p>

<o:p></o:p>

4. How to draw the resonance tautomers of the carbanions obtained from<o:p></o:p>

<o:p></o:p>

a ) CH3NO2<o:p></o:p>

b ) CH3CN<o:p></o:p>

c ) CH2=CHCH3 <o:p></o:p>

<o:p></o:p>

I will be test on this topic on friday,I have spent some time on it, but really don't understand them.

hope somebody can kindly answer my questions. <o:p></o:p>

<o:p></o:p>

Thanks in advance! <o:p></o:p>

<o:p></o:p>

Isabel

The previous message had some problem. Hope this message comes out alright.

I have some difficulties on carbanion questions which I desperately need help with,

Methane has a full outer shell of electrons. Each hydrogen atom contributes 1 electron (represented by 'o') and the carbon contributes 4 electrons (represented by 'x') <ignoring the C inner-shell non-bonding electrons.

I have to draw the structure as above of the carbon species formed by the following reactions:

CH4 ----------> -CH3

I think the structure would be:

H x C

I'm not very sure , can you confirm this with me?

What will be the structure of it? (Radical)

CH4 -----------> +CH3

and the structure of this? (Carbocation)

And what will be the 3D structures and angles between the C-H bonds in -CH3, .CH3 and +CH3 ?

4. how to draw the resonance tautomers of the carbanions obtained from

I will be test on this topic on friday, spent some time on it, but really dont understand them.

hope somebody can kindly answer my questions.

Hi Isabel, remember that carbanions will have a lone pair on a carbon atom and that carbon will have a formal charge of -1. For #1 we have

CH4 --> CH3^- + H⁺

For #2, you are removing a hydrogen radical (atom) with its one electron, leaving the other electron on the carbon. Or,

CH4 --> CH3�?nbsp; + H�?/FONT>

where CH3�?has the structure

This is the methyl radical, with the formal charge on the carbon now zero.

In #3 you are removing a hydrogen atom along with its bonding electron pair, that is, as the hydride ion H: ^-

CH4 --> CH3⁺ + H: ^-

This time we are forming the methyl carbocation, which has a formal charge of +1 on the carbon:

In #4, remove an H⁺ from the CH3 group of each molecule, to leave a lone pair (and a formal -1 charge) on that carbon. The second one (acetonitrile) is easier to type so I'll try it:

-H⁺ .. ..
CH3--C

N: ----> CH2--C

CH2=C=N:

In the first resonance structure the formal charge is on the first carbon, and in the second the formal charge is on the nitrogen. Same idea for the other two molecules! Remember to never have more than 8 electrons around row two elements like C, N, and O. That's why we can't have a triple bond on the CH2 carbon, it would have 10 electrons around it, or five bonds, a no-no. In resonance structures we are just rearranging the electrons; the atoms themselves do not change positions. Finally, you want to be comfortable with calculating the formal charge on any atom. It is also clearer to use curved arrows to show the movement of electrons in going from one resonance structure to the next, but unfortunately I couldn't show that here.

Thank you for your explaination! I understand the things that you explained to me,and start to get the idea of these stuff

but I still have some questions on this carbanion topic....hoope you don't mind answering them 1)if the compound is not a methane, there are more than one carbon e.g. CH3CH2CH3, which carbon atom of the compound will loose the H+ on treatment with base?and... 3)if there are 3 compounds, CH3CH2CH3, CH3CH2Ph and CH3CH2OCH3... which one will form the most stable carbanion? can you please explain why to me? and 3) why the compound CH3COCH2COCH3 does not undergo cyclisation on treatment with base? Sorry that i'm asking so many questions and taking your time!!!!!

The trend in alkyl carbanion stability is basically opposite that of carbocation stability. Groups that can donate electron density towards the positive carbon of a carbocation, such as alkyl groups, stabilize the carbocation by allowing the +1 charge to spread out, even if only partially. By the same token, electron-withdrawing groups tend to stabilize a negative charge, while electron-donating groups will destabilize a negative charge. Thus, the general order of alkyl carbanion stability is 1° > 2° > 3°. Note that these carbanions are not prepared by reacting an alkane with a strong base, because there are practically no bases strong enough to react with alkanes directly. Rather, they can be prepared as organomatallic compounds from the reaction of alkyl halides with an active metal such as lithium, forming alkyllithium compounds RLi.

Carbanions and carbocations can also be stabilized by resonance. For instance, the benzyl (PhCH2) carbanion (and carbocation) is stabilized by resonance, which again allows the charge to "spread out" a little onto the ring. Thus, a benzyl carbanion such as the one in your three compound question will be more stable than a comparable alkyl carbanion.

The ether CH3CH2OCH3 is tricky. The oxygen is inductively an electron-withdrawing group, and as such, should favor the formation of a carbanion of the type R-O-CH2^-. However, oxygen can also donate electron density with one of its lone pairs, which would have the opposite effect on the carbanion stability. The overall effect depends on which effect dominates, and this is generally the electron-donating effect by a lone pair of the oxygen. Here is an opposite example using carbocations: the carbocation CH3CH2-O-CH2⁺ is easier to form (in SN1 reactions) than CH3CH2CH2CH2⁺, implying that the ether carbocation is more stable. Thus, I would expect an opposite trend in the carbanion stabilities, namely, that the ether carbanion would be less stable than the alkyl carbanion.

The most acidic hydrogens in the last compound, 2,4-pentanedione, are the central CH2 ones (pKa = 9), and therefore they will react most easily with a base. When you remove one of these as H⁺, the resulting carbanion has three resonance structures. The CH3 hydrogens have much less acidic character (pKa = about 19), and the resulting carbanion that results from removal of one of these has only two resonance structures. But removal of one of the CH3 hydrogens is necessary for an intramolecular aldol condensation to occur (giving the cyclic product).

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