selectivity of hydrogens

[inaccensa]

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I'm a little confused about this concept. So EK says for 3>2>1 we follow 1:4:5. This implies that to form a primary product, the substrate will require 5X as many collisions than a tertiary product? I don't understand this since the example they gave us 2-methyl butane has 9 primary, 2 secondary and 1 tertiary hydrogen. This applies for all types of reactions and is not limited to alkanes, right? Since bromine is more selective, should we ignore this rule?

 

[inaccensa]

anyone? plz

 

[loveoforganic]

Your question is very unclear.

 

[DJtiesto]

If you are trying to see the distribution of products, multiply each type of hydrogen on the reactant by the 1:4:5.

For example:

Molecule: H3CCH2CH2 (Propane)

Number of Primary Hydrogens: 6 * 1 = 6

Number of Secondary Hydrogens: 2 * 4= 8

Thus the likelihood of the product being chlorinated on the secondary carbon is 8/14.

This is only for chlorination. Bromination is much slower and more selective so it will always select for the most stable product (in this case distribution will be almost all secondary bromination).

 

[inaccensa]

If you are trying to see the distribution of products, multiply each type of hydrogen on the reactant by the 1:4:5.

For example:

Molecule: H3CCH2CH2 (Propane)

Number of Primary Hydrogens: 6 * 1 = 6

Number of Secondary Hydrogens: 2 * 4= 8

Thus the likelihood of the product being chlorinated on the secondary carbon is 8/14.

This is only for chlorination. Bromination is much slower and more selective so it will always select for the most stable product (in this case distribution will be almost all secondary bromination).

Thanks. EK confused the hell out of me. What do they mean by the probabilty of collisions?

 

[dragon529]

How are there 6 primary hydrogens?

 

[inaccensa]

How are there 6 primary hydrogens?

he meant CH3CH2CH3

 

[zmatrix]

I'm a little confused about this concept. So EK says for 3>2>1 we follow 1:4:5. This implies that to form a primary product, the substrate will require 5X as many collisions than a tertiary product? I don't understand this since the example they gave us 2-methyl butane has 9 primary, 2 secondary and 1 tertiary hydrogen. This applies for all types of reactions and is not limited to alkanes, right? Since bromine is more selective, should we ignore this rule?

You should go back and reread that section in depth.

1:4:5 is selectivity for radical halogenation of plain alkanes by chlorine ONLY
Also, the 1:4:5 ordering is for primary:secondary:tertiary

Bromine selectivity is around 1:82:1640

Fluorine is close to 1:1:1 IIRC

To solve these types of problems do the following:
Selectivity = 1(primary):4(secondary):5(tertiary)
2-methyl butane = 9 primary H, 2 secondary H, 1 tertiary H

(9 x 1) + (4 x 2) + (5 x 1) = 9(prim) + 8(sec) + 5(tert) = 22(total)

Therefore I would expect:
9/22 = 41% Primary chlorinated product
8/22 = 36% Secondary chlorinated product
5/22 = 23% Tertiary chlorinated product

 

[loveoforganic]

It's difficult to radical halogenate fluorine due to the large change in enthalpy - go boom!

 

[inaccensa]

thanks everyone🙂 thats makes a lot of sense👍

 

[Tokspor]

Why is bromine the most selective radical and not fluorine?

 

[inaccensa]

Why is bromine the most selective radical and not fluorine?

You need to think about the stability of the radicals

For electronegativity trends F>Cl>Br> I
For Size F<Cl<Br<I
For radical stability F.<Cl.<Br.<I.
For reactivity F.>Cl.>Br.>I.

Chlorine is less stable and more reactive than bromine radical.If you had propane and you abstract the hydrogen from the 2 carbon, you'd form a secondary radical. The bromine will prefer a secondary because its more stable. Chlorine is more reactive and less choosy. i don't think Iodine is very reactive and like someone else said that flourine will give you a big boom.