The Resonance (Mesomeric) and Electromeric Effects

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. In phenol, an O H group sits on a benzene ring, and oxygen is more electronegative than carbon. So O H pulls electrons out of the ring; it is a minus I group, remember? But in fact, O H activates the ring, making it richer in electrons. So how can one group pull and push at the same time? The answer is the resonance effect. It comes up often in JEE, so let's begin.

0:30The resonance (mesomeric) effect

So, where does the push come from? Pi, the Greek letter on screen, means a pi bond, like the second bond of a double bond. Sometimes two pi bonds interact, or a pi bond and a lone pair on the next atom. With me? The polarity this produces is the resonance, or mesomeric, effect: R or M for short, okay? Now, the inductive effect travels along sigma bonds, sigma being the Greek letter in the question. So what path does this one take? Through the pi system, yes, exactly! And like the inductive effect, it is permanent, always present in the molecule. Keep that word in mind. Clear?

1:13+R and -R groups: look at the atom attached

Now, groups come in two kinds. Plus R, a plus sign and a capital R, means the group donates electrons to the conjugated system. The clue is a lone pair on the atom attached. Okay? The short dash in front is the bond to the rest of the molecule. O minus, where the minus sign written high is a negative charge; N H two, where the two written low is a subscript; NHR and N R two, where R is an alkyl group; O H, O R, NHCOCH three, OCOCH three, and the halogens. Each attached atom has a lone pair. Right? Now, for minus R, the attached atom carries a multiple bond to a more electronegative atom, which pulls the pi electrons onto itself, so the group withdraws electrons. Got it? The minus R list: N O two; C triple bond N, the three short lines being a triple bond; CHO, COR, COOH, COOR, and S O three H, okay? Each has a multiple bond from the attached atom to oxygen or nitrogen. With me?

2:20On a benzene ring: ortho and para feel it most

Now put such a group on a benzene ring. Ortho, small o, means the carbons next to the group; para, small p, the carbon straight across. Plus R group: it pushes its lone pair in, and the resonance structures put the extra negative charge mostly at ortho and para. Okay? So the ring is activated towards electrophiles, the electron-seeking reagents, and they attack at ortho and para. That is the push in our puzzle. Clear? Now let us flip it. Minus R group: it pulls electrons out, again mostly from ortho and para, where the resonance structures now put a positive charge. With me? So an electrophile now goes to meta, the carbons between ortho and para. Meta must have extra electrons, right? No, no, careful, that is exactly the trap! The whole ring has lost electron density, so it is deactivated. Meta, small m, gained nothing; it simply lost the least, so it is the least deprived position, so the electrophile attacks there. Got it?

3:25Halogens, and the tug of war with -I

Now, the halogens carry lone pairs, so they are on the plus R list. So they activate the ring, just like O H does, right? Oh no, careful here, it is not so simple! Halogens are also minus I groups, a minus sign and a capital I for inductive, pulling through the sigma bonds. And for them, minus I is the stronger effect. Okay? So halogens deactivate the ring. But their plus R effect still puts more electron density at ortho and para than at meta, so they still direct an electrophile there. Deactivating, yet ortho and para directing. Right? And now, back to phenol. For O H, O R and N H two, the plus R effect outweighs the minus I effect. So O H does pull through the sigma bonds, but its push through the pi system wins, and that solves our puzzle. Clear? Now, here is one last rule. Resonance moves charge only to the ortho and para carbons, so from the meta position there is no resonance effect, only the inductive effect. Got it?

4:32The electromeric effect: only when a reagent attacks

Now, a different effect, the electromeric effect. Here the shared pi pair of a multiple bond is transferred completely to one of its two atoms. Okay? Remember the word permanent? This one is not! It happens only at the demand of an attacking reagent, so it is temporary: remove the reagent, and it vanishes. With me? Now the sign, with a capital E for electromeric. In plus E, the electrons move to the atom the reagent attaches to. When H plus, the plus sign written high, adds to an alkene, the pi pair moves to the carbon where the H plus attaches. Right? Now you try this one. C N minus, the cyanide ion, attacks the carbon of a carbonyl group, and the pi electrons move onto the oxygen. Is that plus E or minus E? It is minus E, yes, exactly! The reagent attaches to carbon, but the electrons move away from it, to oxygen. Clear?

5:32Example 1: classify OH, NO2, NH2, CHO, OCH3, CN as +R or -R

Alright, example one, a warm up. Classify as plus R or minus R: O H, N O two, N H two, CHO, OCH three, and C N, six groups in all. Pause the video and try it first. Done? Okay, let's check. Look at the atom attached: the oxygen of O H, the nitrogen of N H two and the oxygen of OCH three all have lone pairs, so all three are plus R groups. Got it? Now, the nitrogen in C N has a lone pair too, so C N is a plus R group as well, right? No, no, wait, which atom is attached? It is the carbon, and it has a triple bond to the more electronegative nitrogen, so C N withdraws electrons. N O two and CHO have a double bond to oxygen. So all three are minus R groups. Did you get all six?

6:25Example 2: rank aniline, anisole, bromobenzene by reactivity

Now example two, a step harder. Arrange in order of decreasing reactivity towards an electrophile: benzene, aniline, benzaldehyde, anisole and bromobenzene. Pause the video and try it first. Done? Okay, let's check. Aniline carries N H two, and anisole OCH three, the groups in brackets. Both activate the ring by plus R, just like O H in our puzzle. And nitrogen donates better than oxygen, so aniline comes first. With me? Now look at bromobenzene. Bromine has lone pairs, so it is plus R as well. Does it go above benzene, or below it? It goes below benzene, very good, remember the halogens! For bromine, minus I is greater than plus R, see the greater than sign, so it deactivates the ring, but only weakly. Okay? Benzaldehyde carries CHO, which pulls by both effects, minus I and minus R, with nothing pushing back. So it deactivates the ring strongly and comes last. Clear? So the order is aniline, greater than anisole, greater than benzene, greater than bromobenzene, greater than benzaldehyde. See how simple that was?

7:39Quick revision: +R and -R, ortho para, halogens, +E and -E

So, quick revision. Plus R groups have a lone pair on the atom attached; minus R groups, a multiple bond to a more electronegative atom. On a ring, the resonance effect, capital R, acts at ortho and para; for halogens, minus I wins, but they direct to ortho and para. The electromeric effect is temporary, at a reagent's demand: plus E towards its atom, minus E away. Now go and try the practice problems on this page. All the best!