Halogenation

Please see photo from spec. How do you show the formation of X+ or Xdelta+-AlX3delta- ? ImageUploadedByStudent Room1460907615.721948.jpg

I've asked so many people this question but no one can answer for some reason!?
Thank you!

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Reply 1

Kallisto

Entertainment Forum Helper

As you know - I hope so - halogenation are reactions where halogens (Chlorine, Bromine etc.) are involved. The letter X stands for these halogens as a variable for itself.

delta + and delta - are the charges of these halogens in this structural formula. In other words: the halogens exist in ionised state. These states are essential to react.

Here is a picture about how a halogenation could be shown: halogenation (example)

In this example delta(+)X-AlX3delta(-) is delta(+)Cl-AlCl3delta(-). In this example X stands for Cl, the halogenation is a chlorination. The same is possible with bromine, iodine and fluorine. Just substitue Cl in this picture with Br, I or F. You will get a bromination, iodination or fluorination.

Reply 2

maths_4_life

Original post by Kallisto

Doesn't delta mean partial charge not fully ionised? Thank you for the example. Are the charges in circles partial or full charges. And I still can't see where the X+ on its own is formed.

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Reply 3

Kallisto

Entertainment Forum Helper

Original post by maths_4_life

Doesn't delta mean partial charge not fully ionised? Thank you for the example. Are the charges in circles partial or full charges. And I still can't see where the X+ on its own is formed.

They are parital not full charges, as the charges lasts till halogenation is completed. So, you are right. My mistake. X+ seems to be another reaction of halogenation. If I am not mistaken, X+ refers to a radical substitution. It is a chain reaction, you know?

Reply 4

maths_4_life

Original post by Kallisto

Thank you!! Ohhhh I forgot to say this is for halogenation of benzene. Does that make any of your answers different? Sorry about that.

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Original post by KaylaB

Actually the mechanism as you have shown it is not as I believe it should be.

(in situ) 3Br-Br + 2Fe --> 2FeBr₃

Then

Br-Br + FeBr₃ --> Br ----- Br---FeBr₃

Effectively this last structure has the form:

Br⁺ -------[FeBr₄]^-

and NOW the positive bromine is the electrophile that reacts with the benzene ring.

Reply 7

KaylaB

Original post by charco

I feel like the bit in bold is pretty much the same as what I have above, except you've included how the catalyst is made also - which is good to know, but for my exam board we just need to know the reaction of the catalyst with the Halogen so I was unaware of that part :h:

Reply 8

charco

Study Forum Helper

Original post by KaylaB

No.

You have the benzene electron pair attacking a bromine atom with no charge.

I have a benzene electron pair attacking a (partially) positive bromine atom ...

Reply 9

KaylaB

Original post by charco

No.

You have the benzene electron pair attacking a bromine atom with no charge.

I have a benzene electron pair attacking a (partially) positive bromine atom ...

Ah okay I understand now, thanks for explaining that :smile:

Reply 10

Kallisto

Entertainment Forum Helper

Original post by maths_4_life

Thank you!! Ohhhh I forgot to say this is for halogenation of benzene. Does that make any of your answers different? Sorry about that.

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No, it makes no difference in my view. Benzene has an aromatic structure, so electrophilic substitution would work for it. Electrophilic substitution in my given example is a reaction for aromatics.