A good way to visualize sn1, e1, e2, sn2

[tyjacobs]

So I think sn1, e1, e2, sn2 is a big topic on ochem dat. When I was studying this I created a flow chart to figure out which one of the 4 is favored and it may be helpful to you guys. This has worked for me and idk if this works 100% of the time you may find this useful


1) determine if you have primary, secondary, tertiary alkyl halide

2a) Primary R-X:
Look to see if base is bulky:
If it is then you will do E2
If it is not, look to see if it's strong:
If it is then you will do sn2.
If it is NOT, no reaction.

2b) Secondary R-X:
Look to see if base is bulky:
If it is then you will do E2
If it is not bulky, then look to see if the base is strong:
If it is, you will do Sn2 + E2 (E2 is FAVORED)
If it is not strong, look to see if the base is a good nucleophile:
If it is, you will do Sn2 + E2 (Sn2 is FAVORED)
If it is not a good nucleophile, you will do Sn1 and E1

2c) Tertiary R-X:
Look to see if base is strong:
If it is you will do E2
If it is not strong, you will do Sn1 + E1


What is a "good nucleophile"? For me, whenever i saw a negatively charged sulfur base like Na+ Ch3Ch2S- I would be thinking thats a weak base but a good nucleophile.

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[prosthaholicDDS]

there is one thing that is confusing me ( Dat destroyer road map number 3 ) . Let say you have a Tertiary halide .. and in the first case we use (CH)3ONa, and in the second case we used C2H5OK .. how are those two different .. I know they give different products I just can't understand it ..

 

[tyjacobs]

When u see (Ch3)3ONa, do E2 elimination and create the LESS SUBSTITUTED double bond.

When you see c2h5oNa, do E2 elimination but you create the MORE SUBSTITUTED double bond.

 

[DentalLonghorn2014]

You are pretty awesome. This is one of my weakest areas. Thanks a lot, friend.

 

[DAT Destroyer]

You are pretty awesome. This is one of my weakest areas. Thanks a lot, friend.

Hey DentalLonghorn!

Always a pleasure to see you pop in! Glad TyJacobs came to the rescue, he's a smart dude! His DAT score proves it!

Nancy

 

[Jay20155555]

For the secondary alkyl halide, why is it that when you have a Strong Nucleophile and Weak base you get a racemic mixture of SN2 and E2.

Wouldn't E2 be eliminated since the base is weak? Unless you have excess of that weak base?

I was thinking it would be a mix of SN2 and E1 favoring SN2, here is the example i was working out and got confused

My thought process.....

Okay we got a secondary alkyl halide, nothing eliminated yet. Then look at nucleophile. NaCN is ionic and dissociates into Na+ and CN-. CN- is a strong nucleophile (eliminating SN1) and a weak base (eliminating E2). Now all we have is E1 and SN2 (favoring SN2), which I thought was the answer.

K just looked back at it and noticed the solvent is aprotic meaning it can be E1, but if we ignore the solvent, following your rules I still dont get why E2 happens

 

[tyjacobs]

if a question asks for a product, you give the single product with the highest yield. Therefore in this situation, you would give the Sn2 product since there will be a higher yield of that

 

[Jay20155555]

I think what I'm confused about is when you said "If it is, you will do Sn2 + E2 (Sn2 is FAVORED)"

2b) Secondary R-X:
Look to see if base is bulky:
If it is then you will do E2
If it is not bulky, then look to see if the base is strong:
If it is, you will do Sn2 + E2 (E2 is FAVORED)
If it is not strong, look to see if the base is a good nucleophile:
If it is, you will do Sn2 + E2 (Sn2 is FAVORED) <-------
If it is not a good nucleophile, you will do Sn1 and E1

If you have a weak base that is a strong nucleophile how can you have a minor product of E2. I thought E2 cant be done with a weak base. Wouldnt it be a minor product of E1 with a major product of Sn2?

 

[yoh95]

When u see (Ch3)3ONa, do E2 elimination and create the LESS SUBSTITUTED double bond.

When you see c2h5oNa, do E2 elimination but you create the MORE SUBSTITUTED double bond.

Why is E2 favored for C2H5ONa? I thought it was also strong nucleophile since it has a negative charge on it so I was assuming SN2

 

[deleted678432]

Why is E2 favored for C2H5ONa? I thought it was also strong nucleophile since it has a negative charge on it so I was assuming SN2

Depends on the Alkyl halide.

 

[yoh95]

Depends on the Alkyl halide.

Could you explain more? All the destroyer organic chemistry they do E2 elimination for all NaC2H5O I am very confused. Tyjacobs also said this so am I missing something?

 

[deleted678432]

Could you explain more? All the destroyer organic chemistry they do E2 elimination for all NaC2H5O I am very confused. Tyjacobs also said this so am I missing something?

So just as Tyjacobs said, there are a few major components into choosing the mechanism for a reaction. Two main ones are the basicity/nucleophilicity of the reagent, and the placement of the alkyl halide.

If there is a primary alkyl halide (say Et-I), and you react it with NaC2H5O, then you are reacting a PRIMARY alkyl halide with a STRONG nucleophile! This would favor SN2 because the nucleophile can easily kick the strong leaving group out of it's unhindered "Home". There might be a very very tiny amount of E2 since Ethyl oxide is a great base, however, it wouldn't be an amount substantial enough to be a correct answer in a problem.

Now, if you were to react NaC2H5O with, say, a secondary iodo alkyl halide, then you would get a different product. Here you have a SECONDARY alkyl halide with a STRONG nucleophile! You'd think that the nucleophile would then just come and kick off the Iodine and reside happily on the secondary carbon. Wrong. Due to SN2 being a backside-attack mechanism, it has to squeeze behind the iodine to kick it off. It is easy to do in the first option, where a carbon and a few hydrogens are "blocking" the iodine, but here, there are TWO carbon groups blocking the iodine and will make it tough for the nucleophile to come in and attack the iodine. So now instead of kicking off the iodine directly, the base will grab one of the adjacent hydrogens, kicking down its electrons to form a pi-bond, thereby kicking off the iodine via E2 mechanism. Make sense??

 

[yoh95]

So just as Tyjacobs said, there are a few major components into choosing the mechanism for a reaction. Two main ones are the basicity/nucleophilicity of the reagent, and the placement of the alkyl halide.

If there is a primary alkyl halide (say Et-I), and you react it with NaC2H5O, then you are reacting a PRIMARY alkyl halide with a STRONG nucleophile! This would favor SN2 because the nucleophile can easily kick the strong leaving group out of it's unhindered "Home". There might be a very very tiny amount of E2 since Ethyl oxide is a great base, however, it wouldn't be an amount substantial enough to be a correct answer in a problem.

Now, if you were to react NaC2H5O with, say, a secondary iodo alkyl halide, then you would get a different product. Here you have a SECONDARY alkyl halide with a STRONG nucleophile! You'd think that the nucleophile would then just come and kick off the Iodine and reside happily on the secondary carbon. Wrong. Due to SN2 being a backside-attack mechanism, it has to squeeze behind the iodine to kick it off. It is easy to do in the first option, where a carbon and a few hydrogens are "blocking" the iodine, but here, there are TWO carbon groups blocking the iodine and will make it tough for the nucleophile to come in and attack the iodine. So now instead of kicking off the iodine directly, the base will grab one of the adjacent hydrogens, kicking down its electrons to form a pi-bond, thereby kicking off the iodine via E2 mechanism. Make sense??

So for any secondary halide with strong base/strong nucleophile the major product is the E2?

 

[deleted678432]

So for any secondary halide with strong base/strong nucleophile the major product is the E2?

Yes with some SN2 minor product as well.


Sent from my iPhone using SDN mobile

 

[yoh95]

Yes with some SN2 minor product as well.


Sent from my iPhone using SDN mobile

Okay thank you. Not sure how I didn't catch that during chads videos since his chart just says products are SN2+E2

 

[deleted678432]

Okay thank you. Not sure how I didn't catch that during chads videos since his chart just says products are SN2+E2

I'd just memorize what Ty posted above! It explains it all pretty well.