Cyclohexane // Hexane (1 Viewer)

[kimmeh]

what is the difference between these two chemicals apart from their name? i mean, their chemical formula is the same..(CH3CH3) or am i wrong ?
is this the same for cyclohexene and hexene?

 

[CM_Tutor]

I'm not really clear on your question, but to answer what I think you are asking:

Hexane is a straight chain alkane, CH₃CH₂CH₂CH₂CH₂CH₃, with molecular formula C₆H₁₄.

Cyclohexane is a cyclic alkane, with the six C's forming a ring, and it has a molecular formula of C₆H₁₂.

Hexene is a straight chain alkene, which has one carbon-to-carbon double bond in its chain. Whilst it has the same molecular formula as cyclohexane, it has a markedly different chemistry. There are a number of possible compounds that could be called hexene. Provided we ignore geometric differences (OK for school, not OK for Uni), they are:

1-hexene: CH₂=CHCH₂CH₂CH₂CH₃

2-hexene: CH₃CH=CHCH₂CH₂CH₃

3-hexene: CH₃CH₂CH=CHCH₂CH₃

Cyclohexene is a cyclic alkene, with its 6 carbons forming a ring, and having a carbon-to-carbon double bond somewhere in the ring. There is only one such compound, so the name is unambiguous (by contrast, the name hexene was ambiguous because it is one of at least 3 possible compounds). The molecular formula of cyclohexene is C₆H₁₀.

 

[Xayma]

Is there any reason why schools would use cyclohexane/cyclohexene over hexane/hexene (unless hexane/hexene is still a gas but any other straight-chained alkane/alkene)?

 

[CM_Tutor]

Yes, there is. Hexane is considerably more toxic than pentane, heptane or cyclohexane. So, for safety reasons, it is good practice to use the pair cyclohexane / cyclohexene, rather than a pair like hexane / 1-hexene. The cyclic option also means that there is only one alkene to consider, rather than having to discuss which of the pentenes or hexenes or heptenes is being used.

 

[kimmeh]

Originally posted by CM_Tutor
Hexane is a straight chain alkane, CH₃CH₂CH₂CH₂CH₂CH₃, with molecular formula C₆H₁₄.

Cyclohexane is a cyclic alkane, with the six C's forming a ring, and it has a molecular formula of C₆H₁₂.

bingo thanks

what about the reaction with bromine water ? would that change the benzene ring thing ? hmm the equations for the hydrocarbons with bromine water would help

 

[Calculon]

Originally posted by kimmeh
bingo thanks

what about the reaction with bromine water ? would that change the benzene ring thing ? hmm the equations for the hydrocarbons with bromine water would help

the double bond of an alkene makes it more reactive, as it undergoes an addition reaction with bromine, rather than the substitution reaction seen in the alkanes reaction, which requires a catalyst such as UV light to occur.

 

[randhi]

cyclohexane with bromine water
C6H12 (aq) + Br2 (aq) - - - > C6H11Br + HBr (with UV)

Cyclohexene

C6H10 (aq) + Br2 (aq) - - - > C6H10Br2

 

[Calculon]

Originally posted by randhi
cyclohexane with bromine water
C6H12 (aq) + Br2 (aq) - - - > C6H11Br + HBr (with UV)

Cyclohexene

C6H10 (aq) + Br2 (aq) - - - > C6H10Br2

The formula for bromine water is BrOH because it reacts with the water to dissolve

 

[kimmeh]

Originally posted by Calculon
The formula for bromine water is BrOH because it reacts with the water to dissolve

we've always done it with Br₂
CM tutor, please verify

 

[Calculon]

Originally posted by kimmeh
we've always done it with Br₂
CM tutor, please verify

Its Br₂ when its dissolved in another solvent

 

[kimmeh]

do i write Br₂ ? or BrOH

 

[Xayma]

Originally posted by kimmeh
bingo thanks

what about the reaction with bromine water ? would that change the benzene ring thing ? hmm the equations for the hydrocarbons with bromine water would help

It isnt a benzene ring thing, but it will still be in a ring the reaction only breaks the double bond. (Benzene has 3 pairs of delocalised electrons which makes it unreactive)

 

[spice girl]

Originally posted by kimmeh
do i write Br₂ ? or BrOH

aqueous bromine is an equilibrium mixture of Br_2(aq) and HOBr (formed when Br₂ reacts with H₂O to form HBr and HOBr).

But conq chem is wrong. mechanistically, it is Br₂ that reacts with the double bond.

in short, just write Br₂

 

[cko]

umm.... im confused, what if i wrote "BrOH" in the HSC exam? Would that be marked as wrong??

 

[CM_Tutor]

Kimmeh: As has already been pointed out, there is no 'benzene ring' in cyclohexane, or any other cyclic compound dealt with in the HSC. The reactions of cyclohexane and hexane with bromine have already been provided, and here are the reactions for hexane and 1-hexene:

C₆H₁₄ + Br₂ ---UV Light---> C₆H₁₃Br + HBr

The product here is one of the isomers of bromohexane - ie. one of 1-bromohexane, 2-bromohexane or 3-bromohexane.

C₄H₉-CH=CH₂ + Br₂ ---> C₄H₉CHBr-CH₂Br

The product here is 1,2-dibromohexane.

OK, now to the bromine water issue.

Let's start with the easy bit. Bromine on its own, or bromine in a non-polar solvent (organic chemists like to use carbon tetrachloride, CCl₄) is unequivocally Br₂. It reacts in these circumstances by adding across a double bond in the usual way, ie:

R¹R²C=CR³R⁴ + Br₂ ---> R¹R²CBr-CBrR³R⁴

where R¹, R², R³ and R⁴ are all either hydrogen or carbon chanins.

In water, the situation becomes more complex. As has been implied above, we have a problem as bromine exists in equilibrium in water with hypobromous acid and hydrobromic acid:

Br_{2 (aq)} + H₂O_(l) <---> HOBr_(aq) + HBr_(aq)

That is, there is present the strong acid HBr, the weak acid HOBr, and in addition, unreacted Br₂. Conquering Chemistry maintains that the reactive species in this mixture is the HOBr, and thus that bromine water should be formulated as HOBr. This completely ignores the fact that HBr (or, to be more precise, its ionised form) is a capable of adding across a double bond, and that this equilibrium lies to the left.

Furthermore, it ignores the purpose of the bromine water test. The use of bromine water is as a simply, qualitative, diagnostic test for the presence of unsaturations in the carbon chain - that is, carbon-to-carbon double and triple bonds - and this is done by the colour change that occurs due to the reaction. Since the coloured species in this equilibrium mixture is the Br₂, and not either of the others, in my opinion the critical reaction in this case is still the addition of Br₂, and not the reaction of minor species produced due to the hydrolysis of bromine. Note that the fact that bromine water retains significant orange / brown colour is proof that bromine remains present at equilibrium in substantial quantities.

I might add that I have discussed this matter in the past with a member of the USyd academic staff who is an organic chemist, and he concurs that the important reaction is the one that leads to the colour change, and thus is the reaction of bromine itself.

Now, what about exam questions? Well, if you feel it is necessary to offer an explanation in an exam, then I would note that the equilibrium does exist, but that the colour change is the basis for the use of the test, and thus it is the reaction of the coloured species that should be considered, and not any other reactions.

In conclusion, whilst I would not go as far as to say that the mechanism discussed by Smith is wrong, I would say that it does not focus on the critical issue, and thus I would still write reactions of bromine water using Br₂.

Please post more questions if you are still having troubles with this.

 

[Calculon]

But wouldn't removing BrOH or HBr from solution also remove Br₂ because of the equilibrium being disturbed?

 

[CM_Tutor]

Calculon, you are correct that reacting with HOBr and / or HBr would draw the equilibrium to the right (Le Chatelier's Principle), thereby decreasing the concentration of Br₂. However, it is equally true to say that removing Br₂ by direct reaction will draw the equilibrium to the left.

Which of these interpretations is correct? It depends on which reaction is more likely, the reactions with HOBr
and / or HBr, or the reaction with Br₂.

How would we determine this? Well, it can be determined, just not by HSC methods. Here is an explanation (you'll have to trust me on the calculations, but anyone doing Chemistry at Uni is welcome to check me, as they're only first year level).

We have a reaction Br_{2 (aq)} + H₂O_(l) <---> HOBr_(aq) + H⁺_(aq) + Br^-_(aq)
This is a redox reaction, as we have oxidation of Br₂ to HOBr and reduction of Br₂ to Br^-. We can calculate that
E = -0.50 V under standard conditions. This means (applying the Nernst equation) that the equillibrium constant for the reaction at room temperature is 3.50 x 10^-9 mol²L^-2, and from this it is possible to calculate that approximately 1 in every 660 bromine molecules has undergone reaction, and the remaining 659 bromine molecules remains as Br₂. It is thus vastly more likely that the alkene molecules will encounter a bromine molecule and react than it will encounter a molecule of HOBr or an H⁺ ion.

I conclude that the major reaction is the direct reaction of Br₂ with the alkene, and it is this reaction, rather than the shifting of the equilibrium, that is primarily responsible for the colour change.

 

[spice girl]

well, the thing with 660 molecules of Br2 to every molecule of HOBr (OBr- anion) is that the argument will work if u assume that both species are equally likely to react with the double bond.

but in terms of mechanisms, the reason why the conq chem is dodge is that HOBr is actually H+ and OBr- ions. this means if it reacts u get -H attached to one carbon, and -OBr attached to the other. which isn't the case in reality. also, another reason is that in things like cyclohexene, u find that the product -Br and -OH is 'trans' to each other (i.e. on opposite sides of the ring). This can only be explained if Br2 reacted (forming an intermediate where the bromine is joined to both carbons, blocking the way of water attack on the same side). I guess u can argue that the Br can come from OBr-, but the OBr- anion is negatively charged, so an electron-dense double bond is less likely to attack it.

 

[Calculon]

Thankyou, my curiousity is satiated.

 

[CM_Tutor]

Originally posted by spice girl
but in terms of mechanisms, the reason why the conq chem is dodge is that HOBr is actually H+ and OBr- ions. this means if it reacts u get -H attached to one carbon, and -OBr attached to the other. which isn't the case in reality. also, another reason is that in things like cyclohexene, u find that the product -Br and -OH is 'trans' to each other (i.e. on opposite sides of the ring). This can only be explained if Br2 reacted (forming an intermediate where the bromine is joined to both carbons, blocking the way of water attack on the same side). I guess u can argue that the Br can come from OBr-, but the OBr- anion is negatively charged, so an electron-dense double bond is less likely to attack it.

Sorry, but this isn't correct. HOBr is a weak acid, with a pK_a of 8.62, and this means that less than 1 in every 20000 HOBr molecules have actually undergone ionisation. Thus, hypobromous acid is overwhelming present as unionised HOBr molecules, and not as ionised H⁺ and OBr^- ions. Furthermore, the reaction of hypohalous acids, HOX, with alkenes to give the haloalcohol is a well known organic reaction, ie:

R¹R²C=CR³R⁴ + HOX ---> R¹R²CX-C(OH)R³R⁴

where R¹, R², R³ and R⁴ are all either hydrogen or carbon chanins.

This reaction starts with an attack by the electrons in the double bond onto the electrophilic halogen atom (noting that the halogen has as oxidation state of +I in HOX), producing a cyclic halonium ion of the same type as is formed when reacting with X₂. The hydroxide ion then attacks from the reverse side, giving trans addition of HOX, just like occurs for X₂.

I would agree with you that there was a problem if HOBr was substantially ionised, but it is not, and so the 'mechanism' that Smith is using is chemically reasonable, it is just not the primary mechanism responsible for the observed phenomenon.