Destroyer Orgo question 2010

[mx41]

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#52

In my classnotes, I have that E2 mechanisms are for primary compounds, and only primary compouds because carbocations do not form in E2 reactions. So I am wondering if this question is a mistake but it says the most reactive are 3*>2*>1* so the answer is C.
But I wouldve said
E>D>B>C>A

Can someone explain this please?

 

[lex1489]

That question got me too. You want always want to form the most stable product, right? For SN2, product stability is largely unaffected by how substituted the C atom is-- primary, secondary, tertiary carbons only affect the ability of the nucleophile to attack the carbon atom.

For elimination reactions, the stability of the final product (an alkene) IS affected by the level of substitution on the carbon atoms. 2,4-dimethyl-2-butene is more stable than 2-methyl-2-butene which is more stable than 2-butene which is more stable than propene which is more stable than ethene.

If you draw those out you'll see we go from a quad-substituted alkene to a non-substituted alkene, losing stability (and increasing in heat of hydrogenation) in the process.

Since a substituted alkene is more stable, 3* is the best alklyl halide for elimination.

 

[mx41]

That question got me too. You want always want to form the most stable product, right? For SN2, product stability is largely unaffected by how substituted the C atom is-- primary, secondary, tertiary carbons only affect the ability of the nucleophile to attack the carbon atom.

For elimination reactions, the stability of the final product (an alkene) IS affected by the level of substitution on the carbon atoms. 2,4-dimethyl-2-butene is more stable than 2-methyl-2-butene which is more stable than 2-butene which is more stable than propene which is more stable than ethene.

If you draw those out you'll see we go from a quad-substituted alkene to a non-substituted alkene, losing stability (and increasing in heat of hydrogenation) in the process.

Since a substituted alkene is more stable, 3* is the best alklyl halide for elimination.

Oh okay, that makes sense. Thank you!!

I have another question, #67:

After adding the bromine to the compound, you add a strong base. This is an elimination mechanism, so wouldnt the double bond be formed in the ring? Zeitsevs rule says you form the double bond to the most substituted carbon. But CH3 is not more substituted than CH2 carbon, so I am not understanding the second step.

 

[lex1489]

You're using a bulky base, which can only attack the unhindered H, thus giving you the anti-zaitsev (Hoffman) product.

And the hoffman elimination: http://en.wikipedia.org/wiki/Hofmann_elimination

 

[DATkiller]

What would be the product if we substitude HBR and ROOR with NBS in the last step in #67? Is that gonna give us answer choice e?

 

[lex1489]

I believe NBS also gives the anti-markonikov product