What happens when the haloalkane is primary, secondary or tertiary? Which one reacts faster?
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tertiary carbocations are more stable than secondary carbocations and secondary is more stable than primary.
being more stable means that it requires less energy to create that carbocation. Requiring less energy implies that there is more possibility of the alkane molecule and the halogen molecule colliding so that the collision creates enough energy to form the carbocation intermediate.
Eg. so lets it takes 500kJ of energy to form the primary cation, 400kJ of energy to from the secondary, and 250kJ to from the tertiary (remember these are endothermic reactions)
we know the molecules have to collide for a reaction to happen. so it is more likely that the energy created by the collision will be enough to form the tertiary carbocation -> directly relates to speed of reactions
speed of reactions is mostly about probability.
Since tertiary ones are the most stable and require the least energy, they should react the quickest.
being more stable means that it requires less energy to create that carbocation. Requiring less energy implies that there is more possibility of the alkane molecule and the halogen molecule colliding so that the collision creates enough energy to form the carbocation intermediate.
Eg. so lets it takes 500kJ of energy to form the primary cation, 400kJ of energy to from the secondary, and 250kJ to from the tertiary (remember these are endothermic reactions)
we know the molecules have to collide for a reaction to happen. so it is more likely that the energy created by the collision will be enough to form the tertiary carbocation -> directly relates to speed of reactions
speed of reactions is mostly about probability.
Since tertiary ones are the most stable and require the least energy, they should react the quickest.
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The reactivity depends on many factors. The length of the molecule, the position on the molecule of the halogen, the number of halogen, and even which halogen the tertiary would cause the molecule to be more polar and therefore more reactive.