# Reactivity order of electrophilic addition

Find the order of the ease of electrophilic addition of the following:

1. $\ce{H3C-O-CH=CH2}$
2. $\ce{ F-CH=CH2}$
3. $\ce{Cl-CH=CH2}$
4. $\ce{NO2-CH=CH2}$

According to me, it should be 1 > 4 > 2 > 3 but in my textbook, it says 1 > 2 > 3 > 4.

If we take $\ce{F-CH=CH2}$ and $\ce{NO2-CH=CH2}$, the result after electrophilic addition would be $\ce{F-CH+-CH2E}$ and $\ce{NO2-CHE-CH2+}$ where E is the electrophile. Since both F and $\ce{NO2}$ are -I groups, further the $\ce{C+}$ carbocation is situated, more stable is the compound. The latter should be more reactive than the former. So why is my textbook going against this logic?

• "Since both F and NO2 are -I groups, further the C+ carbocation is situated, more stable is the compound." This is wrong. F is also a +M effect group, which NO2 is not. Thus the former is more stabilized than the latter, which is also what your textbook says. Mr pea's answer below is correct. What is the cause of the reluctance you have in accepting that answer? – Gaurang Tandon Mar 2 '18 at 11:32
• I verified with my professors and they say I am correct. Reactivity among halide and other groups is made only on the bases of -I effect and not +M. Only the orientation is made on the bases of +M not reactivity. @GaurangTandon – dr.drizzy Mar 2 '18 at 14:01
• What do you mean by "orientation"? – Gaurang Tandon Mar 2 '18 at 14:05
• Meaning, where the carbocation will form after the electrophile just attacks. @GaurangTandon – dr.drizzy Mar 2 '18 at 15:15
• Which book stated that order? – Avyansh Katiyar Mar 2 '18 at 23:32

For the electrophilic addition to carbondouble bonds, the $\pi$- electron cloud double bonds first attack the electrophile and then a carbocation is created adjacent to the carbon where the electrophile is added. Then that carbocation is attacked by the nucleophile to form the addition product. $$\ce{R-CH=CH2 + E -Nu -> R-CH^+-CH2-E -> R-CH(Nu)-CH2E}$$The above path is the general electrophillic addition reaction in the case of alkenes. The carbocation position may vary depending upon the $\ce{R}$ group.

1. When the $\ce{R}$ is $\ce{CH3-O}$, there is a positive mesomeric effect (+M or +R effect) of the group due to which when the carbocation is formed in the adjacent carbon, it gets high stabilisation due to the +R effect, which in turn decreases the energy of the transition state of carbocation heavily, and thus the reaction rate becomes very high. That's why, the rate of electrophilic addition is very high in this case is very high, i.e it is the most easy.
2. When the $\ce{R}$ is $\ce{F}$, there is a significantly stronger +R effect due to $\ce{2p\pi -2p\pi}$ overlap. So, the carbocation gets a slight stabilisation due to this +R effect but there is also a -I effect in case of $\ce{F}$ which is the factor of destabilisation of carbocation. That's why, ease is relatively lower due to significant +R but stronger -I effect.
3. When the $\ce{R}$ is $\ce{Cl}$, there is a very weak +R effect (vinyl chloride has a weak effect of +R, which is observed from the lesser dipole moment due to lesser charge separation than ethyl chloride, where there is no +R), but dominant -I effect. So, in this case carbocation is more destabilised. So, the ease of addition is decreased even further.
4. When the $\ce{R}$ is $\ce{NO2}$, there is -I effect and also a dominant -R effect or -M effect. So the $\pi$ -electron cloud gets delocalised with the nitro group and thus due to decreased $\pi$-electron cloud density, it becomes difficult for it to even attack the electrophile. So, fractions of molecules forming the carbocation is extremely less, which actually increases the activation energy of the reaction. Thus, the reaction becomes very difficult with nitroethylene. Consideering the overall effects described above, the ease of electrophilic addition should be,$$\ce{CH3 -O - CH=CH2 > F - CH=CH2 > Cl - CH=CH2 > NO2 - CH=CH2 }$$

• "significantly stronger +R effect" stronger than what? – Gaurang Tandon Mar 8 '18 at 17:44
• @GaurangTandon I think it is as compared to chlorine. Would have made more sense if he answered fluorine after chlorine, or mentioned that resonance is less significant with chlorine, in the 3rd part. – Mr_Pea Mar 9 '18 at 11:11

There will be +R(or +M) in the cases involving F and Cl, +R being better with F than Cl due to better orbital overlap with C.

Fluorine donates a lone pair to carbon, leading to a positive charge on F. Although F has a positive charge, this resonating form has special stability since this resonance form has more number of bonds, and since octet of all atoms and duplet of $\ce{-H}$ are complete.*

There is no lone pair on N in $\ce{-NO_2}$ so there is no +R effect in this case.

• Yes, your assumption is correct, I edited my question so it's easier to understand. I know the cases involving F and Cl. My question is, which would be more reactive between the halogen group(II and III) or nitro group(IV)? Could you explain that? Thank you. – dr.drizzy Feb 25 '18 at 19:56
• @raajsuriya both II and III will be more reactive than IV due to more stability of the carbonation intermediate. -I effect of nitro group is much stronger compared to that of fluorine. Effect at second carbon of nitro group will be comparable to that of fluorine group at first carbon(not completely sure about them being comparable, but there is a considerable difference in their inductive effects) – Mr_Pea Feb 25 '18 at 20:52
• Since F and Cl are both deactivators (in the context of EAS) with F being more deactivating than Cl, it would thus make sense to say that F must be more electron-withdrawing from the pi bond than Cl, making the bond less nucleophilic. The + M/R effect is less significant than -I effect, as evidenced by the rates of reaction in EAS of F-substituted arenes and Cl-substituted arenes. Do you have any references to support what you are saying? – Tan Yong Boon Feb 25 '18 at 23:01
• @TanYongBoon I’m sorry but I don’t have any reference to support what I am saying. Although what I am saying is what my professor has said at least a couple of times and according to him, in this case the extent of orbital overlap between F and C is the deciding factor. – Mr_Pea Feb 26 '18 at 13:21
• @TanYongBoon I guess Mr_Pea is right...I've heard the same thing from my teacher a decent number of times. +M of F>+M of Cl and -I of F<-I of Cl – Carrick Mar 5 '18 at 14:14

In electrophilic addition of a hydrogen halide ($\ce{HX}$), the $\ce{H+}$ attacks first. Reactivity in this case is measured by the stability of the carbocation. The reason for this is more stable the intermediate/carbocation, greater the time for the subsequent attack by the anion.

Pertaining to the question in hand we will be forming a two degree carbocation in all the cases as the only other option is a very unstable three degree carbocation. Let's analyse the carbocations one by one.

First let's rank the cases according to the mesomeric effect.

1. Oxygen(+M): Good overlap due to matching orbital size.
2. Fluorine(+M): Good overlap due to matching orbital size
3. Chlorine(+M): Poor overlap due to a larger orbital size than carbon.
4. Nitro (-M): Actively destabilizes the intermediate

From this we can deduce the order to be:

$$\ce{O} \overset{?}{<=>} \ce{F > Cl > NO2}$$

The deciding factor between oxygen and fluorine is the $\ce{-I}$ effect. Fluorine has a greater pulling tendency therefore it destabilizes the carbocation more than oxygen. Thus, the final order is
$$\ce{ O > F > Cl > NO2}$$
which is also the order given by your book.