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A question about the effect of no sunlight on cycloalkanes and cycloalkenes (1 Viewer)

Dragonmaster262

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If any of you have done this practical then can you tell me what will happen if we carry out this investigation in an area of no sunlight. Will it make any difference to the reaction between the hydrocarbons and the bromine water.
 

Fortify

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Cyclohexane requires UV light to react with Bromine water quicker.
 

Sarebs

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Yeah, it must be carried out in the dark because the cyclohexane will slowly react with UV light.

Cyclohexene --> Will go clear with or without UV light
Cyclohexane --> Will go clear slowly with UV light
 

bayan92

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<meta http-equiv="Content-Type" content="text/html; charset=utf-8"><meta name="ProgId" content="Word.Document"><meta name="Generator" content="Microsoft Word 10"><meta name="Originator" content="Microsoft Word 10"><link rel="File-List" href="file:///C:%5CDOCUME%7E1%5CBAYANH%7E1%5CLOCALS%7E1%5CTemp%5Cmsohtml1%5C01%5Cclip_filelist.xml"><link rel="Edit-Time-Data" href="file:///C:%5CDOCUME%7E1%5CBAYANH%7E1%5CLOCALS%7E1%5CTemp%5Cmsohtml1%5C01%5Cclip_editdata.mso"><!--[if !mso]> <style> v\:* {behavior:url(#default#VML);} o\:* {behavior:url(#default#VML);} w\:* {behavior:url(#default#VML);} .shape {behavior:url(#default#VML);} </style> <![endif]--><!--[if gte mso 9]><xml> <w:WordDocument> <w:View>Normal</w:View> <w:Zoom>0</w:Zoom> <w:Compatibility> <w:BreakWrappedTables/> <w:SnapToGridInCell/> <w:WrapTextWithPunct/> <w:UseAsianBreakRules/> </w:Compatibility> <w:BrowserLevel>MicrosoftInternetExplorer4</w:BrowserLevel> </w:WordDocument> </xml><![endif]--><style> <!-- /* Font Definitions */ @font-face {font-family:"Lucida Sans Unicode"; panose-1:2 11 6 2 3 5 4 2 2 4; mso-font-charset:0; mso-generic-font-family:swiss; mso-font-pitch:variable; mso-font-signature:-2147476737 14699 0 0 63 0;} /* Style Definitions */ p.MsoNormal, li.MsoNormal, div.MsoNormal {mso-style-parent:""; margin:0in; margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:12.0pt; font-family:"Times New Roman"; mso-fareast-font-family:"Times New Roman"; mso-ansi-language:EN-AU; mso-fareast-language:EN-AU;} @page Section1 {size:8.5in 11.0in; margin:1.0in 1.25in 1.0in 1.25in; mso-header-margin:.5in; mso-footer-margin:.5in; mso-paper-source:0;} div.Section1 {page:Section1;} --> </style><!--[if gte mso 10]> <style> /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-parent:""; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin:0in; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:10.0pt; font-family:"Times New Roman";} </style> <![endif]--><!--[if gte mso 9]><xml> <o:shapedefaults v:ext="edit" spidmax="1028"/> </xml><![endif]--><!--[if gte mso 9]><xml> <o:shapelayout v:ext="edit"> <o:idmap v:ext="edit" data="1"/> </o:shapelayout></xml><![endif]--> [FONT=&quot]Alkenes are more reactive than alkanes due to the presence of their double bond. Alkenes go through what is called an addition reaction. It is where in an alkene; a double bond is broken and replaced with two new covalent bonds added across double bond where a foreign element/compound attaches itself. <o:p></o:p>[/FONT]
[FONT=&quot] <o:p></o:p>[/FONT]
<!--[if gte vml 1]><v:line id="_x0000_s1026" style='position:absolute;left:0;text-align:left;z-index:1' from="207pt,12.95pt" to="234pt,12.95pt"> <v:stroke endarrow="block"/> </v:line><![endif]--><!--[if !vml]-->[FONT=&quot]e.g. Ethene + Bromine --> 1, 2 – dibromoethane<o:p></o:p>[/FONT]
[FONT=&quot]<o:p> </o:p>[/FONT]​
[FONT=&quot]As mentioned above this is described as an addition reaction because extra elements are being added to the hydrocarbon and breaking the double bond. Alkanes go through a process called a substitution reaction. The substitution needs the presence of UV light to go through.
[/FONT]
[FONT=&quot]
[/FONT]
[FONT=&quot]<o:p></o:p>[/FONT]

[FONT=&quot]<o:p></o:p>[/FONT][FONT=&quot]When for example Ethane and Bromine undergo a substitution reaction the compound HBr is formed. This is due to one of the Br being substituted with an H on the hydrocarbon. Therefore Br<sub>2 </sub>is broken into two, one atom goes and joins the hydrocarbon and the H which the Br replaces attaches on to the free Br to make HBr. <o:p></o:p>[/FONT]
[FONT=&quot]<o:p> </o:p>[/FONT]
 

brenton1987

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Is there any particular reason why bromine water is used in this experiment?
It is coloured.
When it reacts with the alkene it forms a colourless product. The change from coloured to colourless indicates that a reaction has occured.
Whereas if you were going to add water across the double bond, the mixture would start colourless and remain colourless. You cant be certain that anything has happened.
 

Dragonmaster262

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<link rel="File-List" href="file:///C:%5CDOCUME%7E1%5CBAYANH%7E1%5CLOCALS%7E1%5CTemp%5Cmsohtml1%5C01%5Cclip_filelist.xml"><link rel="Edit-Time-Data" href="file:///C:%5CDOCUME%7E1%5CBAYANH%7E1%5CLOCALS%7E1%5CTemp%5Cmsohtml1%5C01%5Cclip_editdata.mso"><!--[if !mso]> <style> v\:* {behavior:url(#default#VML);} o\:* {behavior:url(#default#VML);} w\:* {behavior:url(#default#VML);} .shape {behavior:url(#default#VML);} </style> <![endif]--><!--[if gte mso 9]><xml> <w:WordDocument> <w:View>Normal</w:View> <w:Zoom>0</w:Zoom> <w:Compatibility> <w:BreakWrappedTables/> <w:SnapToGridInCell/> <w:WrapTextWithPunct/> <w:UseAsianBreakRules/> </w:Compatibility> <w:BrowserLevel>MicrosoftInternetExplorer4</w:BrowserLevel> </w:WordDocument> </xml><![endif]--><style> <!-- /* Font Definitions */ @font-face {font-family:"Lucida Sans Unicode"; panose-1:2 11 6 2 3 5 4 2 2 4; mso-font-charset:0; mso-generic-font-family:swiss; mso-font-pitch:variable; mso-font-signature:-2147476737 14699 0 0 63 0;} /* Style Definitions */ p.MsoNormal, li.MsoNormal, div.MsoNormal {mso-style-parent:""; margin:0in; margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:12.0pt; font-family:"Times New Roman"; mso-fareast-font-family:"Times New Roman"; mso-ansi-language:EN-AU; mso-fareast-language:EN-AU;} @page Section1 {size:8.5in 11.0in; margin:1.0in 1.25in 1.0in 1.25in; mso-header-margin:.5in; mso-footer-margin:.5in; mso-paper-source:0;} div.Section1 {page:Section1;} --> </style><!--[if gte mso 10]> <style> /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-parent:""; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin:0in; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:10.0pt; font-family:"Times New Roman";} </style> <![endif]--><!--[if gte mso 9]><xml> <o:shapedefaults v:ext="edit" spidmax="1028"/> </xml><![endif]--><!--[if gte mso 9]><xml> <o:shapelayout v:ext="edit"> <o:idmap v:ext="edit" data="1"/> </o:shapelayout></xml><![endif]--> [FONT=&quot]Alkenes are more reactive than alkanes due to the presence of their double bond. Alkenes go through what is called an addition reaction. It is where in an alkene; a double bond is broken and replaced with two new covalent bonds added across double bond where a foreign element/compound attaches itself. <o>:p></o>:p>[/FONT]
[FONT=&quot] <o>:p></o>:p>[/FONT]
<!--[if gte vml 1]><v:line id="_x0000_s1026" style='position:absolute;left:0;text-align:left;z-index:1' from="207pt,12.95pt" to="234pt,12.95pt"> <v:stroke endarrow="block"/> </v:line><![endif]--><!--[if !vml]-->[FONT=&quot]e.g. Ethene + Bromine --> 1, 2 – dibromoethane<o>:p></o>:p>[/FONT]
[FONT=&quot]<o>:p> </o>:p>[/FONT]​
[FONT=&quot]As mentioned above this is described as an addition reaction because extra elements are being added to the hydrocarbon and breaking the double bond. Alkanes go through a process called a substitution reaction. The substitution needs the presence of UV light to go through.
[/FONT]
[FONT=&quot]
[/FONT]
[FONT=&quot]<o>:p></o>:p>[/FONT]

[FONT=&quot]<o>:p></o>:p>[/FONT][FONT=&quot]When for example Ethane and Bromine undergo a substitution reaction the compound HBr is formed. This is due to one of the Br being substituted with an H on the hydrocarbon. Therefore Br<sub>2 </sub>is broken into two, one atom goes and joins the hydrocarbon and the H which the Br replaces attaches on to the free Br to make HBr. <o>:p></o>:p>[/FONT]
[FONT=&quot]<o>:p> </o>:p>[/FONT]
Copy pasted directly from Ahmad's Notes.
 

annabackwards

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Hey can someone complete and balance this equation:

C6H12 +Br2 + H20----->?
Water doesn't react in the equation, it's just the solvent.

C6H12 +Br2-----> Br2C6H10 + H2 --> All aqeous except for H2 which is a gas

If you draw it, the carbons are in a hexagon shape and the above reaction is just an addition reaction, where the 2 Br atoms takes the place of 2 H atoms.

Note that although cyclohexane would react with the bromine with UV light, it takes a very long time so you don't need to do the experiment in a dark room - cyclohexene would react very quickly so you just need to note which one decolourises the bromine water quickly.

Edit: Equation should be C6H12 + Br2 --> C6H12Br2 with all states being aqueous
 
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brenton1987

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Water doesn't react in the equation, it's just the solvent.

C6H12 +Br2-----> Br2C6H10 + H2 --> All aqeous except for H2 which is a gas

If you draw it, the carbons are in a hexagon shape and the above reaction is just an addition reaction, where the 2 Br atoms takes the place of 2 H atoms.

Note that although cyclohexane would react with the bromine with UV light, it takes a very long time so you don't need to do the experiment in a dark room - cyclohexene would react very quickly so you just need to note which one decolourises the bromine water quickly.
Youve drawn a substitution reaction. Your product is still an alkene.

Bromine + alkene is an addition reaction to form an alkane.

C6H12 + Br2 --> C6H12Br2

Although technically the reaction is:
Br2 + H2O --> HBr + HOBr
and both C6H12Br2 and C6H12BrOH are formed.
 

clintmyster

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Youve drawn a substitution reaction. Your product is still an alkene.

Bromine + alkene is an addition reaction to form an alkane.

C6H12 + Br2 --> C6H12Br2

Yeah brenton is right. Btw this whole UV stuff tbh is EXTRA knowledge. You actually don't want to talk about UV at all because then your not controlling your variables. The only thing you want to change is the functional group so alkanes/alkenes. The other thing is the aim of the experiment is to distinguish between the reactivities of the two so by exposing to UV light, you won't be able to do this.
 

annabackwards

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Youve drawn a substitution reaction. Your product is still an alkene.

Bromine + alkene is an addition reaction to form an alkane.

C6H12 + Br2 --> C6H12Br2

Although technically the reaction is:
Br2 + H2O --> HBr + HOBr
and both C6H12Br2 and C6H12BrOH are formed.
I stand corrected; it's been less than a week and i'm forgetting details... at least i remembered that it was an addition reaction XD

Yeah, double bond breaks open and Br2 adds across :)
 

iMAN2

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Is there any particular reason why bromine water is used in this experiment?
As 1, 2-dibromohexane (aq) is clear and so we can see a colour change.

This point isn't as important as knowing why hexane+hexene, cyclohexane+cyclohexene were used:

The hexenes have double bonds (so can react with Br2)
Liquids at room temp
Readily available at school
Cheaper than other hydrocarbons
 

iMAN2

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Hey can someone complete and balance this equation:

C6H12 +Br2 + H20----->?
In your title you mention CYLCOhexane and CYCLOhexene.

Thus the equations are as such:

Cyclohexane:
C6H12 (l) + Br2 (aq) ----> C6H12 (l) + Br2 (aq) (No Reaction)

Cylcohexene:
C6H10 (l) + Br2 (aq) ----> C6H10Br2 (aq)

The product is 1, 2-dibromocyclohexane (aq)

For hexane and hexene (an addition reaction):
Hexane:
C6H14 (l) + Br2 (aq) ----> C6H14 (l) + Br2 (aq) (No Reaction)

Hexene:
A critical mistake many people make: Remember, to put 1-hexene or 2-hexene etc (depending on which one you used, we used 1-hexene). As there is a double bond we have to say where in the chain it is located.

C6H12 (l) + Br2 (aq) --> C6H12Br2 (aq)

Although technically the reaction is:
Br2 + H2O --> HBr + HOBr
and both C6H12Br2 and C6H12BrOH are formed.
UV Light and its effects
If any of you have done this practical then can you tell me what will happen if we carry out this investigation in an area of no sunlight. Will it make any difference to the reaction between the hydrocarbons and the bromine water.
UV Light is controlled as the hexanes would undergo a substitution reaction due to the uv light acting as a catalyst, thus invalidating the experiment. Thus, 'NO UV light' is a controlled variable - using aluminium foil to cover and then removing slightly to look at colour change or keeping beakers in shade.

Hexane (with UV as catalyst):

C6H14 (l) + Br2 (aq) ----> C6H13Br (aq) + HBr (g)

Product names: 1-Bromohexane (aq) and Hydrogen Bromide (g)

Cylcohexane (with UV as catalyst):

C6H12 (l) + Br2 (aq) ----> C6H11Br (aq) + HBr (g)

Product names: 1-Bromocylcohexane (aq) and Hydrogen Bromide (g)
 
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iMAN2

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Bromine water can be written as Br2 (aqeuous) - meaning dissolved in water. That's all you need, trust me and the many past papers that i've gone through.
So it's Br2 (l) + H20 (l) --> Br2 (aq)

If you really want to go into the depth then use this:
Br2 (l) + H2O (l) --> HBr (aq) + HOBr (aq)
 
Last edited:

brenton1987

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So it's Br2 (g) + H20 (l) --> Br2 (aq)

If you really want to go into the depth then use this:
Br2 (g) + H2O (l) --> HBr + HOBr
Br2(l) + H2O(l) --> Br2(aq)
Br2(l) + H2O(l) --> HBr(aq) + HOBr(aq)
 

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