Why is this E1 reaction?

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wizi

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Hi everyone,

I am going over the Orgo Destroyer Roadmap 5.

Why is this reaction E1?

Isnt CN- a strong nucleophile and weak base --> it should be SN2 right?
but if its SN2, the Carbon attach to Br is tertiary carbon --> its not right either?


orgoe1roadmap.png
 
CN- is a strong nucleophile and weak base at the same time, and since heat is added you can easily assume the reaction will go as an elimination reaction. Since the halide is tertiary, the reaction will be either E1 or S1, and if you remember E1 or S1 reaction favors weak base and weak nucleophile respectively. In order for E2 to occur, the base needs to be big (steric hinderance) and strong. Hope this helps =)
 
CN- is a strong nucleophile and weak base at the same time, and since heat is added you can easily assume the reaction will go as an elimination reaction. Since the halide is tertiary, the reaction will be either E1 or S1, and if you remember E1 or S1 reaction favors weak base and weak nucleophile respectively. In order for E2 to occur, the base needs to be big (steric hinderance) and strong. Hope this helps =)

This is great advice. I just want to add that for E2 the base does need to be strong, but not necessarily big (though a classic E2 base is big: t-butoxide).
 
This is great advice. I just want to add that for E2 the base does need to be strong, but not necessarily big (though a classic E2 base is big: t-butoxide).

Also keep in mind that when there is heat under the reaction arrow it is usually going to be an E reaction.
 
Not so sure I agree with some of the above. SN2 occurs frequently with secondary substrates and a good nucleophile.
 
Last edited:
Not so sure I agree with some of the above. SN2 occurs frequently with secondary substrates and a good nucleophile.

It occurs frequently depending on how sterically hindered the molecule is. It's in competition with E2 and SN2.

In regards to the question, E1 because it's a tertiary molecule so you can't do an SN2 on it because of steric hindrance. Halogens are good leaving groups so it leaves and makes a tertiary carbocation, instead of a secondary, and a cyanide comes in and plucks a hydrogen off and forms a double bond.
 
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