Why isn’t the answer A? Which molecule is not produced when ethane reacts with bromine in the presence of ultraviolet light? A C2H4Br2 B HBr C H2 D C4H10
The question states ‘not produced’ C - H2 is not produced, it cannot be produced - initiation produce Br radicals, first propagation produces HBr and a radical (C2H5, C2H4Br, C2H3Br2…), second propagation step produces Br radical and a product.
Radical substitution can continue until all H are substituted to C2Br6. Termination step - colliding of 2 radicals E.g C2H5 radical from ( C2H6 + Br. -> C2H5. + HBr) could collide with another C2H5., terminating, instead of Br2.
Hi! I am a bit confused about why a ethanolic solvent is used in a SN2 reaction of a halogenoalkane with potassium cyanide. Will the cyanide ion react with the ethanolic solvent instead of the halogenoalkane?
Hi! I am a bit confused about why a ethanolic solvent is used in a SN2 reaction of a halogenoalkane with potassium cyanide. Will the cyanide ion react with the ethanolic solvent instead of the halogenoalkane?
Ethanol solvent is used instead of aqueous, water. Because water dissociates to OH- , which can undergo nucleophilic substitution with haloalkane, to produce an alcohol.
Hi! I am a bit confused about why a ethanolic solvent is used in a SN2 reaction of a halogenoalkane with potassium cyanide. Will the cyanide ion react with the ethanolic solvent instead of the halogenoalkane?
Why didn't you just start your own thread, rather than piggybacking on someone else's?
Yours is a perfectly decent Q that deserved to be the star of the conversation.
Ethanol solvent is used instead of aqueous, water. Because water dissociates to OH- , which can undergo nucleophilic substitution with haloalkane, to produce an alcohol.
Why didn't you just start your own thread, rather than piggybacking on someone else's? Yours is a perfectly decent Q that deserved to be the star of the conversation.
Sorry, I wasn't sure how to. I will do that next time. Thank you!