Rank the following acetophenone derivatives on the basis of th...

Question

Rank the following acetophenone derivatives on the basis of the carbonyl stretching frequency (1 = highest ; 4 = lowest ).

(a) Chemical structure of para-methoxyacetophenone, showing a benzene ring with methoxy and carbonyl groups.

(b) Chemical structure of 4-chloroacetophenone, showing a benzene ring with a chlorine atom and a carbonyl group attached.

(c) Chemical structure of acetophenone, showing a benzene ring attached to a carbonyl group.

(d) Chemical structure of a nitro-substituted acetophenone derivative with a carbonyl group.

Accepted Answer

All right. Hi, everyone. So this question says when positioned correctly, substituents can transmit electronic data through the benzene ring. This idea can be discovered and used in Ir spectroscopy sort. The following derivatives of Benzal Hyde according to the frequency of carbonel stretching, one being the highest and three being the lowest. So compound A is for hydroxy Benzal Hyde. Compound B is for fluor Robens aldehyde and compound C is for nitro Benz aldehyde. So recall that Ir spectroscopy is primarily used to find what functional groups may be present on a molecule, right, because each distinct IR band on A given Ir spectrum has two characteristics, the intensity and the wave number. Now the wave number is directly proportional to the frequency at which that particular bond vibrates at. And there are a few factors that are going to overall affect the position of the vibrational band because there are few factors that determine the frequency and therefore the wave number, right. Number one is going to be the reduced mass of both both atoms that are connected together. But in this case, right, for this example, what we're doing here is comparing the three car bottles of A B and C. So because it's the same type of bond, right, the reduced mass isn't going to play a factor because we do have the same atoms being connected together. However, right, what will change? And we'll discuss why later on is the strength of the bond relative to the others. In this case, why? Because as soon as you increase the strength of the bond overall, you're also going to increase the wave number at which it vibrates and therefore increase the frequency. So let's talk about what will affect the strength of the bond. And before we continue, let's just notice here that for compounds A B and C, we all have benzene rings in conjugation with our carbon. Now what I mean by conjugation, right, is it the pi bond of the car bal and the p system of the benzene ring is separated by one signal bond in between. And that single bond in between is what we mean when we say that the system is conjugated. Now, the reason why this is so important, right is that conjugation is going to overall affect the strength of a bond because it determined whether or not that bond can participate in resonance. So in this case, right, because the car bono is in conjugation with the benzene ring, our car bals here are going to participate in resonance with the benzene rings. Now, the reason for this right is as a result of resonance are all the hides here are going to have a resonance form in which the carbonel becomes his witter. So what that means right is that in the resonance form, there will be a single bond in between carbon and oxygen with a negative charge on oxygen and a positive charge on carbon. And by introducing this resonance form, right, what we're doing is we're reducing the double bond character of the carbonel. And recall also that double bonds are stronger than single bonds. So by reducing the double bond character and introducing more single bond character, what that does is actually weaken the car bal as a result, which will therefore decrease the frequency at which it's going to vibrate. But here, right, because all three car bals participate in conjugation, we're going to see the same effect. So to compare them even further, right, we can consider how other substituents on the ring affect this conjugation. So if we consider first and foremost our four hydroxy Benzal Hyde, the hydroxy group, particularly the oxygen is going to have the effect of actually extending the conjugated system that involves the ring and the car bono itself. So because you're increasing the conjugation, right, what you're doing is increasing the number of resonance forms. So because the hydroxy group is going to donate electrons towards the ring, what that does is actually reduce or stabilize the positive charge density on the carbon carbon. So when you stabilize this Twitter ionic resonance form in which carbon has a positive charge. What you do is increase the single bond character, which therefore makes the bond even weaker. So in this case, four hydroxy Benzal, the hede is actually going to stretch at the lowest frequency. Now, by contrast, right, for nitro Benzal Hyde, of course, has a nitro group and recall that nitro groups are very powerful electron withdrawing groups. And because the nitro group is so electron with drawing what it's actually going to do is a decrease the conjugation of the pie system, which will therefore reduce the positive charge density on the carbon carbon. Which means that this resonance form here in which our car bono is a wi or ion is going to be less prominent. So our single bond character is reduced. And when the single bond character is reduced, the bond overall becomes stronger, which means that for nitro Benz aldehyde compound C is actually going to vibrate at the highest frequency because of the car bole of four nitro Benzal thede is going to be the strongest due to the fact that it has more double wand character because of the decrease in conjugation. And so compound B right or fluoro bez aldehyde, it's going to have an effect on the ring in the sense that the fluorine being electron is going to reduce the electron density of the ring by pulling electrons towards itself. However, this effect is going to be less than the one that the nitro group causes because the nitro group is a stronger electron was drop. So if were setting up everything to answer the question, right? With number one being the highest and number three being the lowest, number one is going to be compound C two is going to be compound B and three, which is the lowest is going to be compound A and there you have it. So with that being said, thank you so very much for watching and I hope you found this helpful.

Rank the following acetophenone derivatives on the basis of the carbonyl stretching frequency (1 = highest ; 4 = lowest ).
(a)
(b)
(c)
(d)

Key Concepts

IR Spectroscopy

Electronic Effects of Substituents

Carbonyl Stretching Frequency

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