Organic chemistry alkane reactions question?

Question

CH3CH2CH3 + Br2 ---> ?
i need to know how to find three different possible products and how to balance the reaction. above the arrow is the word Light.
i think i should replace the H with Br, but do i do that to all the H or just one.
Thanks for your help!

i really need help balancing

when replace a H with br can i replace it with Br2 or just Br

Answer 1

Lancenigo di Villorba (TV), Italy

Hi dear Bee!
How are you? Bad?
I GIVE YOU "BALANCED REACTIONS" FOR THREE PRODUCTS YOU WROTE!!

BROMINATION
The process is "propane bromination".
You wrote right, a gas mixture of propane and bromine undergoes light action by irradiation with a U.V. lamp.
The U.V. radioactions permits bromine's molecule breakdown, so it generates couples of bromine's atoms (e.g. radicalic) which substain "substitution reaction".
In effect, as you suggest, bromine atom can replace hydrogen atom belonging to propane molecule : so, you obtain alkylic derivatives (e.g. bromo-propanes) and hydrobromic acid (gas form, this is not an aqueous solutions).

BALANCED REACTIONS
As I said, propane can react with bromine, see the followings :

a) CH3CH2CH3 + Br2 ---> CH3CH2CH2Br + HBr
b) CH3CH2CH3 + Br2 ---> CH3CH(Br)CH3 + HBr

where you retrieve 1-bromopropane (n-propyl bromide, see a)) and 2-bromopropane (i-propyl bromide, see b)).
You wait three main products...you wrote right, there is a third compound!! In a succeeding step, as the following :

c) CH3CH2CH2Br + Br2 ---> CH2(Br)CH2CH2Br + HBr

so you retrieve 1,3-dibromopropane.
The latter is a dibromo-substitued compound...why I assume it was the lonely? Why I omit other dibromo-substitued compounds?

BROMO-DERIVATIVES
The alkane's bromination proceeds by means of U.V. irradiation, so it involve radicals intermediate (e.g. very reactive species).
Prediction over product's formation is based on stability of the intermediates (e.g. radicals) involved in chemical mechanism followed in product's formation. Thus, you must evaluate radical's stabilities on all the mechanism followed for create the bromo-compounds.
Criteria for evaluate these stabilities are two, the following :
-) polarity criterion, it is about inductive effect played by eventually substituent groups (e.g. bromine atom exercite strong electron-attractor effect) ;
-) steric criterion, it is about volumic hindering played by some atom having great atomic radius (e.g. bromine atom exercite strong hindering).
Since overwritten criteria, you state that 2-bromopropane cannot proceeds in further substitution, viceversa 1-bromopropane proceeds toward 1,3-dibromopropane. The latter, among the products deriving by 1-bromopropane, is the lonely satisfacting the criteria.

CONCLUSION
When a gas mixture propane-bromine went irradiated by U.V. radioactions, you retrieve four products as 1-bromopropane,
2-bromopropane, 1,3-dibromopropane and hydrobromic acid.
For propane derivatives, it is evaluated relative reactivities.
I think you retrieve different amounts of bromo-derivatives, the greatest in 2-bromopropane's case and smallest in 1,3-dibromopropane's one (e.g. the production of the latter "eats" part of 1-bromopropane).

I hope this helps you.

Answer 2

One at a time.

There will in time be MULTIPLE halogenation on the same molecule, but I would first write down all the MONO-halogenated products first (I see two of those), and then you'll need at least one of the di-halogenated species to get your three products.

That's a poor reaction scheme to be using to actually synthesize a specific halocarbon because the halogenation under these conditions is so random.

But for this question, just pluck off an H and replace it with a Br.

And with all due respect to the others who answered this question, some of you are dead wrong. You can and WILL multihalogenate propane if you keep exposing it to light, a steady supply of Br2, in a closed container. You can and WILL get the di-halogenated, tri-, even tetra-halogenated species if you let it go long enough.

Pick up ANY Organic text please, and look at the section on free-radical halogenation, and you will see that the usual example of methane can be reacted with enough chlorine to make carbon tetrachloride eventually. Though, I say again, this is a lousy choice of synthesis for commercial reasons, it CAN be done.

Answer 3

Any of the hydrogens on the alkane can be replaced by bromine, yielding the bromnated alkane and HBr. But the hydrogens on the ends are the ones most exposed, so the bromine is likely to show up on one of the end carbons. If you push the reaction, you will probably do one end, the other end, and then some in the middle; in principle, you can get a fully brominated hydrocarbon, although I don't know that bromine is small enough to make this work in practice.

Answer 4

Monohalogenated Alkane

Answer 5

CH4 + Cl2 --> CH3Cl + HCl It's a substitution reaction that happens under the presence of light. This happens with all halogens. The reaction is quicker with alkenes, but there the reaction is referred to as addition.

Answer 6

There will be only two products, CH3CH3CH2Br and (CH3)2CHBr. The first is n-propyl bromide or 1-bromopropane. The second is isopropyl bromide or 2-bromopropane. There are no asymmetric carbon atoms in isopropyl bromide, so there is no dl-pair.