# Balanced Reactions and Mechanisms

1. Jan 10, 2005

### MusicMonkey

Write an equation for the reaction with an excess of H2. Give all possible isomeric products.

I see the most obvious isomer adding a total of 6 H's (due to the equation) and getting rid of the double bonds but I don't understand how to get the other 3 isomers. I know that there are a total of 4 answers.

Balanced equations and Mechanisms
Create a balanced equation and write out the mechanism of the following reactions:
1) 1-octene, tested with Br2 in CH2Cl2

I got CH3(CH2)5CH=CH2 + Br2 ---> CH3(CH2)5CH2Br-CH2Br
Is this correct?
Would the mechanism just be a halogenation reaction?

2) n-heptane tested with Br2 in CH2Cl2, illuminating the Br2 solution.

I got C7H16 + Br* -----> C7H15Br + H*
* - electron
Is this correct?
Would the mechanism just be a free radical bromination and would it only be 2 steps resulting in the product of the balanced equation without the H*?

3) 1-octene tested with KMnO4

I got C8H16 + KMnO4 + H20 ----> C2H6O2 + C2H2O2 + MnO2
I do not feel that this is correct?
Is the mechanism just the 1-octene reacted with KMnO4/H2) to produce CH2OH-CH2OH resulting in further oxidation and ending up with CH2=O + CH2=O + MnO2?

4) Napthalene tested with AlCl3

I got C10H8 + CHCl3 ----> (C6H5)3CH + 3HCl
Is this correct? If so, how do I balance it?
I do not understand the mechanism for this reaction.

2. Jan 11, 2005

### Gokul43201

Staff Emeritus
What reaction ? You haven't stated what is reacting with the hydrogen.

Actually, it should be, CH3(CH2)5CHBr-CH2Br, but I'm sure that was just an omission of carelessness. The reaction is certainly a halogenation, but I suspect you might be required to draw the mechanism, which is a symmetric electrophilic addition across the double bond.

The mechanism is a free-radical bromination. But it's a little more involved that what you've shown.

Essentially,
Br2 ----hv----> Br* + Br*
Br* + CH3(CH2)5CH3 -----> HBr + CH3(CH2)5CH2*
CH3(CH2)5CH2* + Br* ------> CH3(CH2)5CH2Br

Now, there are other possibilities as well, which occur at lower probability (depending on how much Br is available, etc). For instance :

CH3(CH2)5CH2* + CH3(CH2)5CH2* -----> CH3(CH2)12CH3

KMnO4 is an oxidizing agent. It will oxidize 1-octene to 1,2 - octanediol. Most likely, the reaction won't stop here, unless you use very cold dilute KMnO4. The alcohol will be further oxidized to a dicarboxylic acid (through an aldehyde/ketone), unless it is a tertiary alcohol, which in this case, it is not. I suggest you go over the section of your text dealing with oxidation of alkenes). There are probably good discussions online as well.

I think you are talking about the alkylation of naphthalene by electrophilic aromatic substitution, with AlCl3 catalyst. But I could be wrong...it's been years since I've done any organic. Perhaps you should wait for someone else to come along, but until then, I would recommend you read up electrophilic aromatic substitution, as well. The mechanism may involve the resonance stabilization of a carbocation, but I'm not certain.

Also, it would be useful for others to know what level you are at, and what course this is for. The approach to helping a high schooler will need be very different from that for a chemistry major in college.

3. Jan 11, 2005

### MusicMonkey

Thank you very much for your suggestions and your help. I figured out majority of the problems and I also handed in my assignment after reading a little more on the certain reactions and reagents involved in these examples and in general. Thank you for your help.

4. Jan 11, 2005

### Gokul43201

Staff Emeritus
I just noticed that #1 and #2 involve dichloromethane. I don't know what its role is.

5. Jan 13, 2005

### chem_tr

About question 1, I can say yes; the mechanism involves a partially positive and partially negative bromine atoms, with partially negatively charged =CH2 and partially positively charged -CH. This causes the negatively charged bromine to attack the CH and =CH2 to attack the positively charged bromine. So, you are right about the final product.

Illumination causes hydrocarbons to form radicalic species as you showed. However, Br-Br bond is cleaved homolytically, producing two radicals, and the hydrogen radical formed from the reaction immediately combines with the other bromine radical to form HBr. This is my reasoning, but your assumption is not very wrong.

I am not sure about this, but if I remember correctly, this test is done with cold and basic KMnO4. This will produce a diol from the alkene, just like the dibromination product. However, the diol will be a vicinal and syn (same-face-looking) diol, due to the proper approaching of permanganate. Permanganate will approach and give its two oxygen atoms to the double bond do oxidize it, and reduced itself to form manganese dioxide.

This is not correct in my opinion. Aromatic compounds like benzene or naphthalene is tested with dry aluminum chloride. These are heated together, and a colored appearance is expected if the compound is really aromatic. However, if chloroform is really present, your assumption is okay; this will probably form trinaphthylmethane (C10H7)3CH), but the steric hindrance of naphthyl group is too high, so dinaphthylmethane must also be considered. The mechanism involves carbon-aluminum bonding, and [AlCl4]- complex is formed. Therefore, negatively charged naphthyl ring will attack to the dechlorinated chloroform cation, producing a C-C bond between these two. The reaction will proceed for one more time (maybe two) to give dinaphthyl derivative.

By the way, dichloromethane, chloroform, or carbon tetrachloride for the first bromination reactions are just solvents to dissolve bromine effectively. You all know that "similar dissolves similar".

6. Jan 13, 2005

### Gokul43201

Staff Emeritus
Phew ! Glad you're back ... I really should stay away from organic.