GS-1 Biological Sciences Question 22

[rubin861644]

I dont even know where to even start on this question. The explanation made me more confused. Any help would be appreciated.

[admin]

The first step in producing an elimination product from a secondary alkyl halide is to remove a hydrogen. Any compound that has the power to break a C-H bond (which is always very stable) must be, by definition, a very strong base (answer choice D).

However, had it been a very strong base like, for example, OH- then it may engage in SN2 producing an alcohol. To avoid a nucleophilic attack, the base must be large and hindered so it is restricted to removing protons from the edge of the secondary alkyl halide rather than attacking the nucleus.

[vpteruna3172]

I have a question about carbanion stability.
The explanation of Question #22 states, "For example, removing a H+ from the methoxy substituent would create a primary carbanion (verrrry unstable) with no real stable options to get rid of the negative charge."

From what I remember, primary carbanion is the most stable compared to secondary and tertiary carbanions. --> 1° > 2° > 3°
Conversely, primary carbocation is the least stable compared to secondary and tertiary carbocation -->
3° > 2° > 1°.

Can somebody explain it to me?

Thanks..

[jellywing_2058]

You are right about the order of stability of carbocations and carbanions. The reason why the explanation says the primary carbanion is very unstable is mostly because it says that it is a metoxy group (CH3-O-R).

The O is partially negative, and cannot stabilize the negative charge added to the C. The explanation also states the charge has nowhere else to go to because the Hydrogens cannot take care of it; they cannot form a double bond.

[admin]

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