Review of Nitriles - Online Tutor, Practice Problems & Exam Prep

Nitriles can be synthesized through three primary methods: SN2 reactions, cyanohydrin formation, and amide dehydration. In SN2 reactions, a cyanide ion displaces a halide from an alkyl halide. Cyanohydrins form when cyanide adds to aldehydes or ketones, creating a chiral center. Amide dehydration involves using dehydrating agents like diphosphorus pentoxide to convert primary amides into nitriles. Nitriles can undergo hydrolysis to yield carboxylic acids, reduction to form primary amines, or conversion to ketones via Grignard reagents followed by hydrolysis.

Review of Nitriles Concept 1

Hey everyone. So in this video, let's take a look and review the functional group of nitriles. Now with this reviewing, we're going to talk about the synthesis of nitriles. Nitriles can be synthesized through 3 different methods. The first method deals with SN2 reaction.

We're going to have the reaction of an alkyl halide, so RX, with some type of nucleophile, in this case, our cyanide ion. Secondly, we can do a cyanohydrin formation. Remember, this is a nucleophilic addition of your cyanide ion to either an aldehyde or a ketone. And then finally, we have amide dehydration. Here we have a reaction of an amide with a dehydrating agent.

Typically, we're using Diphosphorus Pentoxide, so P₂O₅ or we're using Phosphorus Oxychloride which is SOCl₂. So these represent our 3 major approaches to synthesizing nitriles.

Review of Nitriles Concept 2

Hey, everyone. In our discussion of the synthesis of nitriles, let's take a look at SN2 reactions. Now here we're going to say that the cyanide ion can displace halides to yield nitriles. In this example, we're going to create a nitrile via an SN2 reaction. We're going to say that the carbon connected to our electronegative bromine is partially positive, the bromine itself partially negative.

Because of this, our cyanide ion will come in through SN2, hitting this carbon and kicking out the bromine. Now, remember with an SN2 reaction, we get inversion of our original configuration. So the bromine had a wedged bond, but with the incoming cyanide ion, it now changes into a dashed bond. So we've gone from an alkyl bromide as a reactant to our nitrile as a final product, with our bromide ion just laying out there to the side. Alright.

So just remember, this is an easy type of way to make a nitrile starting from an alkyl halide.

Review of Nitriles Concept 3

Now, a second way to create a nitrile is through cyanohydrin formation. Cyanohydrins are formed when an aldehyde or ketone undergoes nucleophilic addition with a cyanide ion. If we take a look here, we have our ketone. Remember with our ketone, oxygen is partially negative, and the carbon connected to it is partially positive. This KCN, it's ionic, so it's made up of K⁺ and CN^-.

The CN^- comes in, hits this carbonyl carbon, breaking the bond and kicking the double bond to the oxygen. So initially, what we would get is that CN getting attached to that carbonyl carbon, and this oxygen will be negative with K⁺. H₃O⁺ at the end is just there to protonate that negative oxygen. So at the end, we would have an OH group still with our CN group present. Now, here's the thing, that carbon is now a chiral center because this carbon that was our carbonyl carbon has four different groups attached to it.

So technically, we don't know the stereochemistry of the OH group and the CN group. To be as accurate as possible, we should show them as both wavy lines because the OH could be wedged, and the CN could be dashed, or vice versa. Those lines illustrate that the configuration of those bonds could be either wedged or dashed. And again, this happens because that carbon center, that carbonyl carbon center, has become a chiral center. Alright.

So just remember, cyanohydrin formation, which is a nucleophilic addition of a cyanide ion to an aldehyde or a ketone, is yet another way we can make a nitrile.

Review of Nitriles Example 2

Hey everyone. Let's take a look at the following example question. Here it says suggest reagents for each step of the following synthesis. If we take a look at the first step, we have an alkene and we're going to an alcohol. Now both of these alkene carbons, there's nothing that differentiates them from one another.

So here we could use different approaches. We could do oxymercuration demercuration, we could do acid-catalyzed hydration, or we could do hydroboration oxidation. Here, I'm going to do hydroboration oxidation. So BH₃, our borane with THF, and that's going to be over our Hydrogen Peroxide and Hydroxide Ion. So that's a way of us going from an alkene to this alcohol.

Next, we're going from this secondary alcohol to a ketone. Now remember, we can go from a secondary alcohol to a ketone through, that's right, Oxidation. So, because it's secondary in nature, we could do it either through weak oxidation or strong oxidation, it really doesn't matter. Here, I'll do it through strong oxidation. So here, we're going to use Potassium dichromate which is a strong oxidizing agent.

So K₂Cr₂O₇ over sulfuric acid. So we've oxidized our secondary alcohol to a ketone. Now, finally, we need to change this ketone into this structure. Things we should notice here is 1, we have an introduction of a nitrile or cyanide group originally, and we have an ether that was also made. Now, this one's a bit tricky.

So this is what happened. We have our ketone. We're going to say that the carbonyl carbon is partially positive and oxygen is partially negative. We're going to react this with KCN, so K+ CN-. Through nucleophilic addition, this cyanide ion hits the carbonyl carbon and kicks the double bond to the oxygen.

And this is how we got that CN group there. And right now, we have a negative oxygen. Typically, at this point, if we wanted to make a cyanohydrin, we would just protonate that negative oxygen. But if we look at our product, it's not an OH group there, it's an OCH₃. To add an OCH₃ there, we would introduce a methyl halide.

So let's say CH₃I, iodine's great leaving group. This negative oxygen could do a nucleophilic attack hitting this methyl carbon and by SN2 kick out this iodine. And that's how we got that CH₃ connected to that oxygen. So here we would say KCN for the first step and for the second step, we do methyl iodide. So, these will be the necessary reagents to take us from our initial reactant all the way to our final product.

Review of Nitriles Example 3

Hey, everyone. So in this example question, it says, select the best set of reagents to achieve the following transformation. Initially, we're starting out with methylbenzene or toluene, and we're obtaining a nitrile group as the final product. First of all, we could say that methyl is an ortho para director, but our nitrile group represents meta directors. There's no way these meta directors would be ortho para to each other.

If we look at our options, since this is an ortho para director, if I added anything, it would go first to the para position. But, as we observe our product, there is nothing here in the para position. That tells us that, commonly, we block the para position. This happens through Sulfonation.

We're going to have our toluene. We do Sulfonation with SO3 over H2SO4. That's going to add my sulfonic acid group to the para position since the methyl group is an ortho para director. So, I've blocked that para position. Now, we have an ortho para, sulfonic acid as a meta director.

To add a nitrile group to benzene, we do it through nitration. Remember, in order deciding, ortho para directors, which are activating, always determine where the incoming group will go over meta directors who are deactivating. So, we use nitric acid over sulfuric acid; this is nitration. The methyl group being more activated decides where the nitrile group will go, which is ortho to it. Now, we finally have our methyl group and our nitrile group ortho to one another.

Since our product doesn’t have anything in the para position, we must do desulfonation. We accomplish this with H3O⁺. That just removes the sulfonic acid group altogether. Now, how do I transform an alkyl methyl into a nitrile group?

Next, we do side chain oxidation with a strong oxidizing agent in the form of potassium permanganate. Remember that the carbon directly connected to benzene is a benzylic carbon. As long as it possesses at least 1 hydrogen, a strong oxidizing agent will change the entire alkyl chain into a carboxylic acid. So, we're going to have that methyl group become a carboxylic acid. Next, we do NH3 over DCC.

Remember, when reacting a carboxylic acid with an amine, we're trying to make an amide. So now, we have a primary amide. And remember, we can dehydrate primary amides into nitriles. We can do that with diphosphorus pentoxide. We're going to lose water, and that will transform my primary amide into a nitrile.

From our steps, option D would have to be the answer. Option A doesn't work because potassium permanganate right off the bat would change this into a carboxylic acid, which is a meta director. So adding doing nitration next would have put the NO₂ group meta to it, which wouldn’t have worked. Option B wouldn't work because, in nitration, if we did nitration here as the first step, it'll go to the para position.

Again, in our products, the nitrile group and the nitriles group are ortho to each other, not para. Option C doesn't work because, yes, we're doing sulfonation here, then we're doing nitration. Okay. We're doing desulfonation. Yes. But then, here's the thing; we're doing NBS. NBS would replace one of my benzylic hydrogens with a bromine. And then if we did KCN, that would hit this carbon and kick out the BR. And if we looked at the end, we wouldn't have a nitrile directly connected to benzene. That nitrile would be connected to CH₂, which is in turn connected to benzene. We have one too many carbons. That’s why C wouldn’t work.

So here, out of all the options, we see that option D, going step by step, is the best method to see that it will give us our final product.

Review of Nitriles Concept 5

Hey everyone. In our continued review of nitriles, now let's take a look at the reaction of nitriles. Here, we're going to say nitrile reactions are similar to other carboxylic acid derivatives. First, we have hydrolysis. We're going to say here, nitriles undergo hydrolysis to produce carboxylic acids.

Second, we have reduction. Here, we're going to say catalytic or hydride reduction produces primary amines. And then finally, we have conversion to ketones. We're going to say that nitriles react with Grignard reagents to yield ketones. So these will be our three primary reactions when it comes to nitriles.

Review of Nitriles Concept 6

Hey, everyone. So in our first reaction of nitriles, we're going to take a look at hydrolysis. We're going to say nitriles produce carboxylic acids in acidic hydrolysis. So if we take a look here, we have our nitrile here, and we're doing acidic hydrolysis, so we're using H₃O⁺, this will transform it into a carboxylic acid and we're going to create a positive amine, so NH₄⁺. And we're going to say we can create a carboxylate anion or carboxylate salts in basic hydrolysis.

So in basic hydrolysis, we still have our CN, but now we're using hydroxide solution. Instead of making a carboxylic acid, we make a carboxylate anion or salt. And then here we're going to have our neutral amine in the form of ammonia. So just remember, hydrolysis is just a way of transforming a nitrile group into either a carboxylic acid or a carboxylate anion or carboxylate salt.

Review of Nitriles Example 4

Everyone, let's examine this example question, which says, "Select the best set of reagents to achieve the following transformation." Observing the problem, we have our initial reactant, and at the end, we have our product. What's the difference? Well, we can see that connected to this carbon now is a carboxylic acid.

As we examine our options, we notice that all choices begin with either catalytic halogenation or radical catalyzed halogenation. Notably, bromination is more controllable, and we prefer it over chlorine due to the potential for a mixture of products. We will start with Br_2 / HBr. With this radical catalyzed bromination, we're likely to replace a tertiary hydrogen with a bromine. Initially, there was a tertiary hydrogen here that got converted into a bromine atom.

This first step created a new structure. Next, we identify the halogen's location as the alpha carbon, while the adjacent ones are beta carbons. The following step involves an E2 elimination process. We could follow either Zaitsev's rule or Hoffman's rule. Considering that the carboxylic acid is added to the terminal carbon, we desire elimination where the double bond is formed nearby. Among our two beta carbons, one has two hydrogens, and the other has three. We will deprotonate the beta carbon that is less substituted, which has more hydrogens, following Hoffman's rule, which utilizes a bulky base.

Examining our choices, options A and B are unsuitable since sodium ethoxide is not a bulky base. Option C is correct because it includes potassium tert-butoxide, our required bulky base. The tert-butoxide will deprotonate the less substituted beta carbon, causing the bond to break and the bromine to be expelled, resulting in a double bond formation.

The next step involves HBr / H_2O_2. Here, HBr typically follows Markovnikov's rule, but with hydrogen peroxide, it undergoes Anti-Markovnikov hydrohalogenation. Thus, bromine will add to the less substituted alkene carbon creating a new structure.

Then, we use KCN / DMSO to facilitate an SN2 reaction wherein the cyanide ion attacks the carbon displacing bromine, adding a nitrile group.

Finally, acidic hydrolysis with H_2O / H^+ converts the nitrile into a carboxylic acid. Through this methodology, we successfully convert our reactant into the desired final product, making option C the best set of reagents for this transformation.

Review of Nitriles Concept 7

Hey, everyone. In this video, let's take a look at the reduction of nitriles. Now nitriles can be converted to primary amines by reduction, and we can do this through lithium aluminum hydride, so LiAlH4, or through H2 over a metal catalyst, typically palladium. So here, we have catalytic hydrogenation. Now, if we take a look here, we have our nitrile to start and we're using Lithium Aluminum Hydride, a strong reducing agent or H2 over Palladium.

This will transform our Nitrile into a primary amine. In essence, this carbon gains 2 hydrogens, becomes CH2, and this nitrogen gains 2 hydrogens to become NH2. So, this is how we're able to go from a triple bond between the carbon and nitrogen to just a single bond. Okay. So again, remember we can reduce a Nitrile through either strong reduction or through catalytic hydrogenation, transforming a Nitrile into a primary amine.

Review of Nitriles Example 5

Hey, everyone. So here it says select the set of reagents that can be used for the following synthesis. If we take a look here, we have benzene to start and, at the end, we have a bromine that's attached to benzene, and then CH₂NH₂. Let's look at our options. If we take a look, we can see that in A, B, and D, we're starting out with bromination.

Remember, a bromine represents an ortho para director. And then if we look, C is starting out with nitration. Nitration would add an NO₂ group to benzene initially. If we look, our 2 substituents in our products are ortho to one another. If I were to add this NO₂ group through nitration, that would mean that the bromine that comes after would have to go to the meta position.

So our products would have our 2 substituents being meta to one another. That's not what our product represents, so it can't be C. So, right now, we're sure that bromination is what we start with because in A, B, and D, they're all starting out with bromination. So let's just straight go to the bromination product. So we have a Br.

Alright. So A, B, and D so far are green. Next, it's saying we should do what? We should do nitration or we should do sulfonation. Now remember, typically, our 2 substituents when we add them to benzene, they're either going to go para to one another or they're going to go meta to each other.

It's hard to have them go ortho to one another because of steric clashing. We can get around this by doing sulfonation, which blocks the para position. So the next thing we're going to do is sulfonation. So that means that B and D are out. I mean, B and D are in, A is out.

So we're going to do sulfonation that's going to block the para position. Again, the sulfonic acid group goes to the para position because bromine is an ortho para director. Alright. So far, B and D are saying the same thing. Next, they're saying we're doing nitration, both of them.

Okay. So let's keep doing that. So here goes NO₂ added in the ortho position because, again, bromine is an ortho para director, so it has the choice, the power to say where the incoming group goes over the meta director. Right. So then next, they're both saying H₃O⁺ is next.

Remember, H₃O⁺ will do desulfonation. It'll remove the sulfonic acid group. So it's going to be gone. So now we just have bromine and NO₂ group. Okay.

Where do they differ now? Now we're saying we see CNOH, not CN. SN, so tin over HCl, or we're going to have NaN0₂ over HCl. Now, if we're looking a little bit down, we can see we have CuCN, which is part of the Sandmeyer reactions. This would change our diazonium salt into a CN group.

What are the steps to create a diazonium salt? Well, first, you have to have ou

Review of Nitriles Concept 8

In this third reaction, we take a look at the conversion to a ketone. The reaction of nitriles with Grignard followed by hydrolysis produces ketones. So we start out first with our nitrile group. Nitrogen is more electronegative, so it's going to be partially negative. Carbon will be partially positive.

If I come in with my Grignard reagent, this bond here would break. The carbon would hit here, kicking one of the pi bonds to nitrogen here. So what we have now is our nitrogen with its two lone pairs and two bonds, so it's negatively charged. We have MgBr positive, and we have this ring attaching itself to this carbon. Now in this form, this is just our intermediate before we do our hydrolysis.

What do we have? We have a CN, we have a carbon double bonded to a nitrogen. Remember that represents an imine functional group. And here we would say that this represents a deprotonated imine. Then we would follow it up with acid hydrolysis or acidic hydrolysis, and that would just remove the nitrogen entirely and leave behind an oxygen, forming our ketone as the final product.

So again, a great way of transforming a nitrile into a ketone is through the introduction of a Grignard reagent followed by hydrolysis.

Review of Nitriles Example 6

In this example question, it says, propose a synthesis to convert benzene to Benzophenone with at least one step being a nitrile reaction. Again, the beautiful thing about synthesis is that there are a lot of different ways to get to our end result. This is going to be yet one way that I can go from a benzene to our desired product. So, we want to go from benzene to Benzophenone, so that would just be benzene connected to a carbonyl connected to another benzene. So all I have to do is try to add this group to my original benzene.

Now here, I'm giving myself the stipulation that it has to be through a nitrile reaction. Right? So let's just do that. We're going to have benzene, and I know that one way to make a nitrile is through going through the nitration reaction first. So we're going to have our nitric acid over sulfuric acid.

That's going to add my NO₂ group to benzene. Next, I can reduce this NO₂ group to NH₂ through tin over HCl. So now it becomes an NH₂. Next, I can change that NH₂ group into a diazonium salt. I do this by using sodium nitrite over HCl.

Once I get my diazonium salt, I can do a Sandmeyer reaction in the form of CuCN. So there goes my nitrile reaction. So I just did the nitrile reaction portion of it. So all I have to do now is change this nitrile into this ketone because it's a ketone. And we know that a nitrile can be changed into a ketone by reacting it with a Grignard reagent followed by hydrolysis.

Alright. So here I would take in my Grignard reagent. Here's my Grignard reagent here. Okay. This bond here would break, hit this carbon, kick this bond here.

And then so that would be step 1. And step 2 would just be H₃O⁺. Alright. So, again, this portion here would be this benzene part. We'd be double-bonded.

We're connecting here this benzene ring, and then H₃O⁺ on the bottom here is just being used so we add an oxygen at the end. So we just created our ketone. Right? So these would be the steps necessary, first, to add our nitrile group to benzene, and then just remember that a nitrile group reacting with Grignard reagent followed by hydrolysis would transform it into a ketone, giving us our final answer.

Select the best set of reagents for the following synthesis.

• A. 1) Br₂/hν; 2) Mg/ether; 3) CH₃I; 4) Br₂/hν; 5) KOtBu/tBuOH; 6) HBr/H₂O₂; 7) NaCN/DMSO; 8) H₂/Pd
• B. 1) Br₂/hν; 2) CH₃MgBr; 3) Br₂/hν; 4) KOtBu/tBuOH; 5) HBr/H₂O₂; 6) NaCN/DMSO; 7) H₂/Pd
• C. 1) Br₂/hν; 2) Mg/ether; 3) CH₃I; 4) Br₂/hν; 5) NaOEt/EtOH; 6) HBr/H₂O₂; 7) NaCN/DMSO; 8) H₂/Pd
• D. 1) Br₂/hν; 2) CH₃CH₂MgBr; 3) Br₂/hν; 4) NaCN/DMSO; 5) H₂/Pd

Propose a synthetic route for the following transformation.

Hey, everyone. So here in this practice question, it says, propose a synthetic route for the following transformation. Now this synthesis question is a bit challenging, but here's what we need to understand. What type of functional group am I making as a product? That's right. We're making an ester. Right? We have a carbonyl connected to an oxygen connected to a carbon. So, we're making an ester. We know that esters can be created through reactions of what we learned with carboxylic acid derivatives. Typically, to make an ester, we have a carboxylic acid plus an alcohol by Fischer's esterification. This will help make an ester. So, this will help guide us on what to do. Right? So, somehow, I need to have an alcohol on benzene and somehow I need to create a carboxylic acid in some way. Alright. So remember, if we think backwards in terms of this and we cut this bond, what we had at one point is we had this alcohol and we had this carboxylic acid. And due to loss of water, they closed to make this cyclic ester. Alright. So this tells me, okay, I need to have this CH2. I need to get an OH onto this, but at the same time, I need to have a carboxylic acid here that's ortho to it. Alright. So let's start out with our toluene or methylbenzene. Here goes methylbenzene here. Remember it is an ortho-para director, so the incoming group would have to go para. If we're looking here we have an alcohol group and a carboxylic acid group that are ortho to each other. We can't do that unless we block the para position. Typically, to block the para position we do sulfonation. So we do SO3H over H2SO4, so sulfonation. That's going to block the para position. Next, remember we can create a carboxylic acid if we can somehow get a nitrile group onto benzene and then do the process of reactions of nitriles, we could convert that into a carboxylic

Here’s what students ask on this topic:

What are the primary methods for synthesizing nitriles?

Nitriles can be synthesized through three primary methods: SN2 reactions, cyanohydrin formation, and amide dehydration. In SN2 reactions, a cyanide ion (CN^-) displaces a halide (X) from an alkyl halide (R-X). Cyanohydrins form when a cyanide ion adds to aldehydes or ketones, creating a chiral center. Amide dehydration involves using dehydrating agents like diphosphorus pentoxide (P₂O₅) or thionyl chloride (SOCl₂) to convert primary amides into nitriles.

How does the SN2 reaction mechanism work in the synthesis of nitriles?

In the SN2 reaction mechanism for synthesizing nitriles, a cyanide ion (CN^-) acts as a nucleophile and displaces a halide (X) from an alkyl halide (R-X). The carbon attached to the halide is partially positive, attracting the cyanide ion. The cyanide ion attacks the carbon, causing the halide to leave. This results in the formation of a nitrile (R-CN) and a halide ion (X^-). The reaction proceeds with inversion of configuration at the carbon center.

What is cyanohydrin formation and how does it create nitriles?

Cyanohydrin formation occurs when a cyanide ion (CN^-) adds to the carbonyl carbon of an aldehyde or ketone. The carbonyl carbon is partially positive, attracting the cyanide ion, which breaks the double bond to oxygen. This forms a cyanohydrin, where the carbonyl carbon now has a hydroxyl group (OH) and a cyanide group (CN). This process creates a chiral center at the carbonyl carbon, resulting in a mixture of stereoisomers.

What are the main reactions of nitriles?

Nitriles undergo three main types of reactions: hydrolysis, reduction, and conversion to ketones. Hydrolysis of nitriles produces carboxylic acids in acidic conditions or carboxylate anions in basic conditions. Reduction of nitriles, using agents like lithium aluminum hydride (LiAlH₄) or catalytic hydrogenation (H₂ over palladium), yields primary amines. Conversion to ketones involves reacting nitriles with Grignard reagents followed by hydrolysis.

How are nitriles converted to primary amines?

Nitriles can be converted to primary amines through reduction. This can be achieved using lithium aluminum hydride (LiAlH₄) or catalytic hydrogenation (H₂ over a metal catalyst like palladium). In this process, the triple bond between carbon and nitrogen in the nitrile (R-C≡N) is reduced to a single bond, resulting in a primary amine (R-CH₂-NH₂).