HELP ME ! im on DEATH ROW T___T

teraexa SG

2) Toluene is the common name for methylbenzene.

So for the 2nd qn Toluene + Bromine:

C6H5CH3 + Br2 -------FeBr3 catalyst----------> C6H4BrCH3 + HBr

In this case, depends on the reagent and conditions. If there is uv light present, then free radical substitution will occur on the methyl side chain of toluene.

And just in case you were wondering since it's bromine water, why do we still need an iron(III) bromide catalyst and no tri-substitution occurs, remember that only on phenols and phenylamine rings will tri-substitution occur with bromine water (aka aqueous bromine). If you were to react phenols and phenylamine with bromine in CCl4, then mono-substitution occurs.

3) For cyclohexane, it is just basically free radical substitution. So the equation should ideally be:

C6H12 + Br2 -> C6H11Br + HBr
C6H11Br + Br2 -> C6H10Br2 + HBr

and so on and so forth...

teraexa SG

Oh and to add on for the benefit of those taking H3 Chemistry or even H2 Chemistry.

In the above mentioned reaction, since there's a ring and a different substituent on different carbons, some might invariably wonder if cis-trans isomerism may occur with this compound since the Br atoms can be on the same or different side of the ring plane and the rotation of the C-C bond is restricted by the rigid ring.

Ok, generally, if just given this compound (ie 1,2-dibromocyclohexane), yes such cis-trans (or E-Z if you are really fussy) isomerism can be present. However, if the compound is formed through the above reaction, no geometric isomers can be found. This is because all the products formed through the above reaction will always be of the trans (or E) isomer.

The exact mechanism is due to the formation of the bromonium ion. When the first Br atom, acting as an electrophile, is added across the double bond, it overlaps with the 2 carbon centers (due to its large electron cloud) to form a bromonium bridge. Hence it prevents the second Br atom from approaching from the same side of the double bond and the poor 2nd Br atom has to approach from the other side of the double bond and hence forming the trans isomer exclusively. This also happens when I2 is added to a double bond.

Do correct me if I am wrong. I didn't take uni level organic chemistry =p

And yes, this is just FYI. Ignore it if you do not understand.

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