Predict the products of the following S_N2 reactions...

Question

Predict the products of the following S_N2 reactions.

(e)

(f)

Accepted Answer

welcome back everyone. We need to give the substitution products formed in each of the sn two reactions given below. Beginning with part a. We have the following solvent which we see contains an amine group where our nitrogen is surrounded by two carbon atoms. So this is specifically a secondary amine. Now due to the presence of this lone pair, although this is considered neutral because nitrogen is bonded by its bonding preference of three bonds and two lone pairs. That lone pair attributes negative character. And so this would be acting as our nuclear file. It's specifically going to be a moderate nuclear file since it's neutral. Now moving on to our substrate, we have an alcohol. Hey lead where roman is attached to a carbon surrounded by one other carbon atoms. So this is specifically a primary alcohol. He lied and we want to recall that roaming located across period four on the periodic table is a pretty large atom. So even though it's electro negative, it's able to easily stabilize extra electrons over itself because it's so large. And so we would consider to be a good leaving group since it's an algal hey lead and a weak base. So we would understand that from all of these details. We've outlined that this is going to follow an sN two mechanism. We want to recall that this is specifically considered the calculation of an amine, where we should recall that in this type of reaction we have the following intermediate phase occur, which would include an acid base reaction. And this would keep going up until the point where we form a tertiary amine, which would be the most eric lee hindered and therefore the least reactive, so least nuclear filic. So going back to our reaction, we have our backside nuclear filic attack from our nuclear Felix center to the carbon atom that are leaving group is bonded to at the same time kicking out our leaving group which would allow us to form the following product. Where again in our asset based intermediate phase we have the depot donation step occurs. So we would have minus H plus as our shorthand for discrimination occurring. We would form the following structure where we have our ring Now with the leftover bond from the depot donation and then with the addition of our alcohol highlight train and we can see that our nitrogen atom is now surrounded by a total of three carbon atoms. So this is already the formation of a tertiary amine, meaning that this would be our major product and the end of this mechanism also note that we do have HBR that is also formed Since our bromide an ion will depart in eight. So for our first example, this would be our first answer for part A And now let's move on to example. B. So looking at example B, we have another primary calculated structure and we have the second reactant being ammonia, which we recognize as one of our memorized weak bases from general chemistry but in this case we recognize that it has a lone pair on the nitrogen here because it's bonded to three other atoms and it's therefore going to act as our nuclear file, particularly a weak one, but nonetheless a nuclear file. So we follow an sN two mechanism where we again have a backside attack from our nuclear file to the carbon that are leaving group is bonded to king out our leaving group at the same time. And then in the intermediate phase we do have that deep throat nation step occur And our asset-based equilibrium where we would form our first product here, which would be the following change structure now with NH2 since we lost a proton. Now this is actually an even better nuclear file than our ammonia. Originally because we have this electron withdrawing alcohol group. And so we can consider that we would have excess space which we can react this nuclear file to. So we would take another molecule of our actual highlight chain and we would have another nuclear filic attack of the carbon atom that are leaving group is bonded to kicking out the leaving group again and then ultimately another deep rotation step. So we'll have minus H plus here. And so we would form actually another product which is going to be the following structure where we have N. H. And then a second alcohol chain now and sorry, just so everything is visible scoot this over. So we can see that so far we have the formation of a primary amine and now a secondary amine. So just to note this down, primary amine and secondary amine. Now our secondary mean is still pretty nuclear filic. So we would still be able to take another molecule of excess space and react it with our secondary amine. So we would have another backside nuclear filic attack to the carbon atom that are leaving group is bonded to kicking out our leaving group again. We also have another deep throat nation step occur in the acid base equilibrium intermediate phase. And we would form our final major product which is going to have the following structure which we can see is now a nitrogen or a mean sorry, surrounded by a total of three carbon atoms being our tertiary amine and our major product. And we can see that it's pretty hysterically hindered. So it's going to have very low nuclear Felicity and be not as reactive as our prior and mean products. So what's boxed in yellow is try proposal. I mean, which is our second final substitution product for example, be of this reaction. So what's highlighted and boxed in yellow are our two final answers to complete this example. I hope that everything I reviewed was clear. If you have any questions please leave them down below and I'll see everyone in the next practice video

Predict the products of the following S_N2 reactions.
(e)
(f)

Key Concepts

S<sub>N</sub>2 Mechanism

Nucleophiles

Leaving Groups

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