1
answer
0
watching
380
views
11 Nov 2019

Synthesis of an Alkyne Part II: Double Elimination Reaction

(triethylene glycol)

Meso-1,2 - dibromo - 1,2, diphenylethane + 2 KOH --------------------------> diphenylacetylene + 2KBr + 2H2O

1) Using orbitals, explain the geometry of the first elimination. (A double elimination reaction occurs between the dibromide and potassium hydroxide which is in alcoholic solution) (The action of potassium hydroxide in alcoholic solution on compounds containing halogen atoms on adjacent atoms (1,2-dihalides) results in the elimination of two molecules of hydrogen halide (double elimination) and the formation of two new π bonds to give a carbon-carbon triple bond. In this experiment you will perform a double elimination on stilbene dibromideto synthesize the alkyne 1,2- diphenylacetylene according to the reaction equation above:)

2) Alkynes are shorter than alkene bonds, are shorter than alkanes. Why?

3) What is the difficulty of using UV to quantify the amount of a product?

4) What conditions could you use to best favour an E2 elimination of 2-bromo-butane rather than an SN2, E1 or SN1?

5) Why would toluene be a bad idea as a solvent for the final recrystallization (assume it would work to give you crystals…it wouldn’t really, but that’s not the problem I’m getting at here).

6) What is a co-spot on a TLC? Why does it help?

For unlimited access to Homework Help, a Homework+ subscription is required.

Jarrod Robel
Jarrod RobelLv2
10 Apr 2019

Unlock all answers

Get 1 free homework help answer.
Already have an account? Log in
Start filling in the gaps now
Log in