Is this a destroyer roadmap error?

[Ryltar]

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2-bromo-2-methylbutane is reacted with CN- and heat to produce 3-methyl-2-butene. HCN is a weak acid, thus CN- is a strong conjugate base. Because CN- is a strong base, I would think this reaction would undergo E2; however, in destroyer it states E1 product.

This is in destroyer roadmap 5, and is right above Simmons Smith rxn.

 

[JMC117]

Could it be that there is an E1 and E2 product but they just wanted to show the E1 product (that's why they labeled it)? Just a guess

 

[Ryltar]

Very little would be E1 product though. E1 would only occur in this case under neutral conditions (e.g. using an alcohol and heat) which would result in an SN1 major product and an E1 minor product. In this case the product would be almost all E2 (97%+), with the remainder being SN2 (~3%).

 

[Badgerfaithful]

When I was going over that road map I was wondering the same thing. I kind of just ignored that reaction. I'm pretty sure it's an error, because everything you've said is correct.

 

[Dotoday]

Destroyer is correct, since 2-Bromo, 2-methyl butane is a tertiary Halide, it would give a stable carbocation. CN- being a base can stabilize that by solvolysis, therefore it is an E1 reaction.

 

[Ryltar]

One would assume that because CN- comes from a weak acid it's a strong base, which is true, as it is a decent base; however, still about 1/100 the strength of bicarbonate ion. I suppose cyanide just isn't quite strong enough for E2 to occur, where you need a base strong base such as OH- or RO-.

My textbook conflicts with this assumption, as it states that with tertiary halides E2 occurs under basic conditions, and secondary halides require strong bases for E2.

 

[Palmetto914]

This is an E1 reaction. Tertiary + Basic = E1...

I don't know how it has been for your organic classes, but our professor always said tertiary is a red flag for either Sn1 or E1, as both of those mechanism involve a carbocation.

 

[Ryltar]

I think that statement is a bit too general. If you have tertiary compound with leaving group and a strong base, such as OH- or RO-, then the reaction would proceed almost completely E2. In fact, E2 doesn't require as strong a base to occur with increasing substitution, and in this case the conditions are basic and the compound is most substituted. By contrast, a tertiary compound reaction under neutral conditions (heating in pure ethanol) leads to a mixture of products resulting from both SN1 and E1 elimination.

Cyanide ion is definitely not a neutral condition, as it is a fairly strong base.

Anyhow, bottom of the line, they probably wouldn't give one this confusing on the DAT, and if they did, they would at least list a solvent to give some clues.

 

[Ryltar]

Btw, I'm directly referencing my O chem book, p.376-377 Organic Chemistry, John McMurry 6e.

 

[Palmetto914]

You're right, that was too general of a statement. Under basic conditions, pretty much every resource I've checked says E2, however a solvent is generally specified.

 

[Sublimation]

First of all the way destroyer named it is technically wrong. it should be 2-methyl but-2-ene. here is the mechanism guys. It can undergo both. Br is a really good leaving group and its on a tertiary carbon.....and cyanide is small planar and will readily abstract a proton. cyanide will abstract protons and force a E2 mechanism before the halide can leave to form the carbocation.