In the following molecules, identify the carbon where the radical is most likely to form in the first propagation step.
(d)
In the following molecules, identify the carbon where the radical is most likely to form in the first propagation step.
(d)
Draw the major product(s) of the reaction of 1-methylcyclohexene with the following reagents, disregarding stereoisomers:
1. NBS/∆/peroxide
2. Br2/CH2Cl2
A student adds NBS to a solution of 1-methylcyclohexene and irradiates the mixture with a sunlamp until all the NBS has reacted. After a careful distillation, the product mixture contains two major products of formula C7H11Br.
(a) Draw the resonance forms of the three possible allylic free radical intermediates.
(b) Rank these three intermediates from most stable to least stable.
(c) Draw the products obtained from each free-radical intermediate.
Predict the products of the following allylic halogenation reactions.
(d)
When N-bromosuccinimide is added to hex-1-ene in CCl4 and a sunlamp is shone on the mixture, three products result.
(a) Give the structures of these three products.
(b) Propose a mechanism that accounts for the formation of these three products
Predict the products of the following allylic halogenation reactions.
(a)
How many stereoisomers are formed from the reaction of cyclohexene with NBS?
Draw the products of the following reactions, including all stereoisomers:
f.
Identify the two different alkenes that undergo allylic halogenation to prepare the compound shown.
How many stereoisomers are formed from the reaction of 3-methylcyclohexene with NBS?
Suggest a mechanism for the following reactions.
(c)
What reagent would you use to brominate the allylic carbon of cyclohexene?
The light-initiated reaction of 2,3-dimethylbut-2-ene with N-bromosuccinimide (NBS) gives two products:
b. The bromination of cyclohexene using NBS gives only one major product, as shown on the previous page. Explain why there is no second product from an allylic shift.
In the following allylic radicals, identify the carbon where the new C–Br bond is most likely to form in the second propagation step.
(a)
Draw the products of the following reactions, including all stereoisomers:
e.