What products would you expect from the reaction of 1-bromopropane with each of the following? (

What products would you expect from the reaction of...

Key Concepts

Carbon–Carbon Bond Formation via Nucleophilic Substitution

Using nucleophiles that introduce carbon-containing groups, such as cyanide or acetylide ions, allows for the formation of new carbon–carbon bonds. These reactions typically proceed through an SN2 mechanism, which is particularly effective with primary alkyl halides.

Finkelstein Reaction

The Finkelstein reaction is a halide exchange process, typically involving the substitution of one halogen (usually bromide or chloride) for another like iodide via an SN2 mechanism. The reaction is often driven by the precipitation of a by-product, which shifts the equilibrium toward product formation.

Formation and Application of Grignard Reagents

Organometallic reagents created by reacting alkyl halides with magnesium are known as Grignard reagents. They serve as powerful nucleophiles in forming carbon–carbon bonds and can be further protonated to yield hydrocarbons, making them a cornerstone reagent in synthesis.

SN2 Nucleophilic Substitution

This reaction type occurs when a nucleophile directly displaces a leaving group on an alkyl halide in a one?step, concerted mechanism. It is most favorable with primary (and some secondary) alkyl halides where steric hindrance is minimal, resulting in inversion of configuration at the reaction center.

E2 Elimination Reactions

E2 eliminations take place when a strong base removes a ?-proton from an alkyl halide in a single concerted step, leading to the formation of a double bond (alkene). The reaction outcome is influenced by the strength and steric bulk of the base, as bulky bases tend to favor elimination over substitution.

Transcript

00:01 So we have one bromopropane and we want to know what products we would get when reacting with the following reagents.

00:08 So for example, a, we have sodium amit.

00:12 So we can go ahead and draw our bromopropane here, one bromopropine.

00:22 And we know that sodium amid is a very strong base.

00:26 It's also a very strong nucleophile.

00:28 However, a strong base would predominate here in an e2 reaction, u2 elimination reaction.

00:39 So we have our sodium amid here.

00:41 And it is likely to undergo v2 elimination because it is a strong base.

00:48 So you can go ahead and show that.

00:51 It's going to deprotonate at the beta hydrogen here.

00:59 This bond is going to get moved down here, and the bromide is going to leave.

01:06 And that is going to give us the following.

01:10 Alken.

01:12 However, we will get a little bit of minor product as well with our substitution reaction, so we can go ahead and draw that real quick.

01:23 Since sodium amids also a very strong nucleophile, we can show it doing an sn2 type reaction.

01:38 All right.

01:40 It's going to attack, backside attack.

01:42 Gromide's going to leave.

01:44 And our product is going to look like the following, where we have our propulamine.

01:53 However, the alkene will be our major product, and the propulamine will be our minor product.

02:06 And for our intergrantic products, we are going to get ammonium and sodium bromide.

02:25 Now, for example, b, we have, looks like potassium terapitoxide.

02:33 And we want to know what happens when we react that with our one bromo propane.

02:37 So let's go ahead and draw that.

02:42 So we know that it is a very bulky base, very strong base.

02:45 So it is likely that it is going to undergo an e2 elimination reaction.

02:54 So let's go ahead and draw the product for that.

02:58 So for the aeropushing mechanism, we're going to go ahead and have the deprotonation here.

03:08 And bromine, or sorry, these electrons are going to get pushed down.

03:13 Bromine is going to leave.

03:15 And we're going to get the following product.

03:17 That'll be the only organic product that we get, because it will only undergo e2.

03:28 And for our inorganic products, we are going to get potassium bromide.

03:40 But we'll also get a little bit of terputal alcohol as well, which is a...

03:48 All right.

03:49 So for example, c, we have sodium iodide.

03:52 This is a classic sn2.

03:54 Iodide's a very strong nucleophile.

03:56 We have a primary alkaliad.

03:58 So let's go ahead and draw what that's going to look like.

04:03 So we have that...

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