What is the product of each of the following reactions?
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Question

What is the product of each of the following reactions?

c.

d. Chemical reaction diagram showing an amine and a carbonyl compound with HCl, indicating a conjugate addition process.

Accepted Answer

Welcome back, everyone. We need to draw the product of our reaction shown below. Observing our first reactant we have within our structure, a carbonyl group characterized by a carbon double bonded to oxygen in which that Carboni is surrounded by two alkyl chains or R groups. Now, we can also observe that this carbon in our carbon is our alpha carbon. And our carbons in our alkyl chains are our beta carbons. Where in our first alkyl chain, we have the presence of a pi bond which indicates a lower ratio of hydrogen to carbon atoms, creating an unsaturated alpha beta. And actually, that should be listed the other way around. So in alpha beta unsaturated ketone, because recall that a ketone is characterized by a carbon between two our groups as we have in our structure. Our second reactant we can observe is a amine in a ring which is a five carbon ring. Now, we observe that it is an amine because we see a nitrogen bonded to hydrogen within our ring and this nitrogen is surrounded by two carbon atoms. So thus, we have a secondary amine and we also have an acid catalyst being hydrochloric acid. Now, we're going to begin by thinking of the possible reactions that we can have occur for this type of reaction. Our two possibilities would be one, either a direct one, comma two addition or we can have option two, which is a one comma four conjugate addition. In order to determine which reaction we would follow, we're going to need to consider our nuclear file that is present. So if our nuclear file is a strong nuclear file, we would have either something like a hydroxide ion within our nucleic structure or a grid reagent as our nuclear file, which we recalls an organic organo metallic compound. And having either or of these options as our nucleic structure would result in a strong leuco phyle corresponding to a direct addition. So we'll just say number one instead of writing things out. And then our second option for our nuclear file would be if we have a weak nu file, which could be considered the water molecule alcohols orth primary or secondary amines or our cyanide anion C N one minus, this would then thus correspond to a weak nu file corresponding to our second option for our reaction being a conjugate 14 conjugate addition. Now, as we stated, we have a secondary, I mean, and so we're going to go with our reaction mechanism for number two being the conjugate addition. Since again, we have a secondary, I mean, as our nu file. So as a result, we're going to draw in that lone pair on nitrogen within our, I mean, because we're called that nitrogen has a bonding preference of having three bonds and one lone parent itself. So we can draw that implied lone parent. And we're going to begin by observing that in our alpha beta unsaturated ketone structure, we have resonance which can can occur with our pi bonds. So our pi bond from our first al chain on the left portion of our structure can shift so that we create another shift to the PIB bo in our Carboni resulting in a lone pair on oxygen, which would contribute a minus charge on oxygen because it would have directly attached seven valence electrons from a set of three lone pairs on itself. And then we would also observe that we now have a partially positive charge at carbon number four in our unsaturated ketone structure. And so thus, since we have technically a partially positive charge at carbon number four, due to resonance, we're going to make an attack from our nu file to that carbon there, that is partially electro positives, that is thus going to cause this pie bond between our electro positives carbon and the carbon that is bonded to to shift to the next position in our alch chain, which will thus cause our pie bond in our carbon to shift as lone pairs on oxygen. So the resultant structure would appear as follows where we have our amine, which again is in a five carbon ring. So our mean is now bonded to our alkyl chain. We have that remaining methyl group. And then the rest of our alch chain where we have now a double bond between our alpha carbon from our Carboni originally and our carbon from the alkyl chain. And then now we no longer have a carbon. We have just a single bond between our alpha carbon and oxygen, which now has three lone pairs on itself and a minus one charge. And then of course, we have the methyl group bonded to our alpha carbon. Still notice that the hydrogen from our amine is not attached anymore. And that is due to the fact that recall that hydrochloric acid is a strong acid and is going to dissociate fully so that we have protons as well as chloride anions in solution. And so those chloride anions are going to de proin our amine. So technically, we have two loan pairs on our mean and we would draw in our implied loan pair in our current structure so far. Now, we're going to again, an apologies. This arrow should be a resonance arrow here, we're then again going to shift our electrons to show resonance so that our one set of our lone pairs on oxygen shifts to form a pi bond to reform our Carboni. And because we have our strong acid in solution in solution H C L, we can't have five bonds on this carbon alpha carbon in our carbon. And so we're going to need to use this pi bond to attack and form a bond to our proton in H C L, which is then going to kick off our bond that hydrogen has to chlorine, giving us another spectator ion of chlorine. Now, at this step, you can also show the formation of an en which can to, to get to our final step in which we have our equilibrium arrow now to our final product where we have our amine ring can draw on the lone para nitrogen. Our mean ring is bonded to our alkyl chain bonded to our methyl and then bonded to the rest of our alkyl chain in which we have now our reformed carbonyl with of course, methyl bonded to the alpha carbon in the carbonyl. And notice that our bond that we made from our pi bond to hydrogen from hydrochloric acid is implied here. So we just didn't draw it in but recall that carbon should have four bonds. So it just implied. And so this final structure is going to be our final answer from our reaction between our alpha beta unsaturated ketone and our secondary immune with hydrochloric acid. So I hope that this makes sense and let us know if you have any questions.

What is the product of each of the following reactions?
c.
d.

Key Concepts

Conjugate Addition

Carbonyl Compounds

Amine Reactivity

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