Elimination Answers

Qu 1:	When heated with H₂SO₄, 2^o alcohols undergo dehydration via an E1 mechanism.
	The major product is the tri-substituted alkene, methylcyclohexene

Qu 2:	When heated with strong bases such as NaOEt, alkyl bromides undergo E2 elimination.
	The outcome of E2 reactions is dependent on the antiperiplanar arrangement of the C-H and C-LG bonds. For substituted cyclohexanes this requires that the LG be axial.
	For the cis-isomer with the -Br axial, the more highly susbtituted alkene can be formed by removal of the H adjacent to the methyl group.
	For the trans-isomer, when the -Br is axial the methyl group is also axial. Therefore the elimination must occur from the C3-H bond giving the anti-Zaitsev product. The reactive conformation is an unfavorable diaxial conformer, therefore the reaction will be slower than that of the cis-isomer.

Qu 3:

These are elimination reactions :

(a) First the E2 reaction of an alkyl halide with a strong base. There two possible H atoms at C3 that can be removed to give 2-butene, look at each in turn:


In this staggered conformation with the methyl groups anti, the H is anti to the Br leaving group giving trans-2-butene.	With the H anti to the Br, the methyl groups are a less stable gauche relationship. This leads to the cis-alkene

(b) Now the E1 reaction of an alcohol with a strong acid. Again there are two possible H atoms at C3 to consider:


With H⁺lined up as shown to allow formation of the p bond, the methyl groups end up trans in the alkene.	With H⁺lined up as shown to allow formation of the p bond, the methyl groups are in close proximity and end up cis in the alkene.

Implications: The steric interactions in the product forming steps control the stereoselectivity favoring trans-2-butene.

Qu 4:	The lowest energy conformation of menthyl chloride has the chlorine atom in a equatorial postion.
	In this position there is no antiperiplanar H , ring flip is difficult as it would require to formation of a triaxial conformer. In contrast, in neomenthyl chloride, the lowest energy confromation has the chlorine atom axial with 2 H in the correct orientation to give the products. The major product is the more highly substituted alkene.