Nancy Said:

Can anyone answer these questions?

We Answered:

Argh...I answered all of these, and Yahoo decided to mess up, so its all gone.

Sorry mate, i dont want to repeat it all again :(


Simply put:


1) The C-C double bond is an area of high electron density (basically, VERY negative) and it causes one Bromine to temporarily be [delta] positive.



2) The C-C double bond does not allow bonds to rotate, so geometric isomers occur. (search for cis-trans in google)



3) Its Nucleophilic Substitution. The NH3 acts as a nucleophile, as it has one lone pair of electrons.

CH3Br + 2NH3 ----> CH3NH2 + NH4Br


4) Yeah it is.
Initiation is producing the Free-radicals.
Propogation is continuing producing them:
CH3BR + BR2 ----> CH3BR2 + BR. (Free radical)

Termination = ending all free radicals



b) Youre kinda right.
As far as I remmeber, it was something like:
CH4 + BR2 = CBR + HBR

Its a free-radical mechanism.
Or Nucleophilic substitution if you want.
(But state its Free-radical substitution!!)




c) Seems right. But use INCREASE in exposure to UV light




5) Dunno about the colour change, but Electrophilic Addition occurs. (C-C double bond has a high electron density etc etc)

Now, I know if you Add HBR to an alkene, i will react.
But bromine solution = H-BR or BR-BR?

If its H-BR, The H does the curly arrow thingy to give the electrons to the BR, and the H then forms a bond with the C-C double bond (which just broke).

The BR is now -ve charged (due to the addition of electrons) and it attacks the CarboCation produced when the C-C double bond breaks.
(The hydrogen will have bonded with ONE Carbon, leaving the other carbon positively charged).
+ve carbon atom = carbocation