Snowie9
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Size:  370.8 KBCan someone please explain the double bond part? I need help ):
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TheTennisOne
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(Original post by Snowie9)
Name:  IMG_5150.jpg
Views: 109
Size:  370.8 KBCan someone please explain the double bond part? I need help ):

Well you know the Cyclohexene part has a formula C6H9 which is bonded to the long aliphatic chain which hence has formula C14H19O,

Then you can just draw out in rough working 4 double bonds in the straight chain with an aldehyde on the end and see if it has the right number of hydrogens
Then if not - try putting 5 double bonds in your aliphatic chain which will yield the correct result.

(It does say between carbon 11 and 12 but you dont actually have to draw it out like that as it makes things complicated just draw out the double bonds)
so just do

C6H9-CH=C=C=C=C-C8H16-CHO
and
C6H9-CH=C=C=C=C=C-C7H14-CHO

add up the total numbers of hydrogens and compare to the start is quite quick to do so should be a relatively easy mark and u got the right answer there sooo good stuff!
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_NMcC_
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(Original post by Snowie9)
Name:  IMG_5150.jpg
Views: 109
Size:  370.8 KBCan someone please explain the double bond part? I need help ):
All they've told you is that it's a aliphatic side chain. That is, a straight chain containing only carbon and hydrogen. Aliphatic chains can be saturated or unsaturated.

The answer is A.

B and D are both trans, that eliminates both of them.

Here is a quick method for determining the number of double bonds and rings (without having to mess about with structures),called the degree of unsaturation

DOS = nCarbons - nHydrogens/2 - nHalogens/2 + nNitrogens/2 + 1

So 20 Carbons - 14 - 0 + 0 + 1 = 7

So how do we make 7? (Which actually means there's 6 double bonds).

Bear with me.

You know that there is a cyclohexene ring in the structure.

DOS for rings = nDouble bonds in ring + 1. (This is important for later)**

In this case with cyclohexene, it's 1 double bond + 1 = DOS of 2

We also know there is an aldehyde in the structure, DOS for C=O = 1 so 2 + 1 = 3

So to make a degree of unsaturation of 7, we must have a further 4 double bonds, as each double bond has a DOS of 1. So 3 + 4 = 7

The question actually asks for number of C=C double bonds, so we ignore the aldehyde C=O.

So if we take 7-2 **(ignoring the extra DOS for the ring and the aldehyde double bond), we must have 5 carbon to carbon double bonds in the structure. Therefore the answer is A.


You can also do it via the trial and error method but it takes longer.

**You can prove the DOS for rings = nDouble bonds + 1. Say you took cyclohexene on its own.

C6H10

nCarbons - nHydrogens/2 +1 (note: there's no halogens or Nitrogens)

that gives:

6-5 + 1 = 2

Or with 2 double bonds

6-4+1 = 3

or with Benzene = 4


Always just use the formula first to get the degree of unsaturation, then look for any other clues as to if there are any special groups like Rings or Aldehydes in the structure. You would need the clues otherwise 7 could mean any number of structures!

******Note the edit to ignore the C=O since it only asks for number of C=C bonds*********
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TheTennisOne
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(Original post by _NMcC_)
All they've told you is that it's a aliphatic side chain. That is, a straight chain containing only carbon and hydrogen. Aliphatic chains can be saturated or unsaturated.

The answer is C.

B and D are both trans, that eliminates both of them.

Here is a quick method for determining the number of double bonds and rings (without having to mess about with structures),called the degree of unsaturation

DOS = nCarbons - nHydrogens/2 - nHalogens/2 + nNitrogens/2 + 1

So 20 Carbons - 14 - 0 + 0 + 1 = 7

So how do we make 7? (Which actually means there's 6 double bonds).

Bear with me.

You know that there is a cyclohexene ring in the structure.

DOS for rings = nDouble bonds in ring + 1. (This is important for later)**

In this case with cyclohexene, it's 1 double bond + 1 = DOS of 2

We also know there is an aldehyde in the structure, DOS for C=O = 1 so 2 + 1 = 3

So to make a degree of unsaturation of 7, we must have a further 4 double bonds, as each double bond has a DOS of 1. So 3 + 4 = 7

So if we take 7-1 **(ignoring the extra DOS for the ring), we must have 6 double bonds in the structure. Therefore the answer is C.


You can also do it via the trial and error method but it takes longer.

**You can prove the DOS for rings = nDouble bonds + 1. Say you took cyclohexene on its own.

C6H10

nCarbons - nHydrogens/2 +1 (note: there's no halogens or Nitrogens)

that gives:

6-5 + 1 = 2

Or with 2 double bonds

6-4+1 = 3

or with Benzene = 4
But you've included the C=O of the aldehyde? Is it not only asking for C=C double bonds?
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Snowie9
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#5
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(Original post by TheTennisOne)
Well you know the Cyclohexene part has a formula C6H9 which is bonded to the long aliphatic chain which hence has formula C14H19O,

Then you can just draw out in rough working 4 double bonds in the straight chain with an aldehyde on the end and see if it has the right number of hydrogens
Then if not - try putting 5 double bonds in your aliphatic chain which will yield the correct result.

(It does say between carbon 11 and 12 but you dont actually have to draw it out like that as it makes things complicated just draw out the double bonds)
so just do

C6H9-CH=C=C=C=C-C8H16-CHO
and
C6H9-CH=C=C=C=C=C-C7H14-CHO

add up the total numbers of hydrogens and compare to the start is quite quick to do so should be a relatively easy mark and u got the right answer there sooo good stuff!
How do you get the formula of C14H19O?
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TheTennisOne
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(Original post by Snowie9)
How do you get the formula of C14H19O?
Original Formula C20H28O take away the formula of the Cyclohexene ring C6H9 which gives C14H19O
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_NMcC_
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(Original post by TheTennisOne)
But you've included the C=O of the aldehyde? Is it not only asking for C=C double bonds?
Ah you're right. So it's actually A. The DOS method still works, just take 2 off from 7. to ignore the aldehyde C=O.
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TheTennisOne
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(Original post by _NMcC_)
Ah you're right. So it's actually A. The DOS method still works, just take 2 off from 7. to ignore the aldehyde C=O.
Yep Im taking notes from your method as it is completely sound XD Very nice lol
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_NMcC_
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(Original post by TheTennisOne)
Yep Im taking notes from your method as it is completely sound XD Very nice lol
I've never seen it taught at A level for some reason, It should be. It's a 1st year undergrad thing that is primarily used in Spectroscopy and other Analysis problems. The maths behind it is very simple though!
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