Is this a possible structure of C6H12O2 which is optically active?

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irfy
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#1
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Need to find a possible structure of C6H12O2 which is optically active

Marksheme says this:

C (CH3) (CH3) (CH3) (COOCH3)

or if its easier to read: (CH3) C (COOCH3)


But I got this:


C (H) (CH3) (CH2-CH2-CH3) (COOH)

and I cant see why its wrong?
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shengoc
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(Original post by irfy)
Need to find a possible structure of C6H12O2 which is optically active

Marksheme says this:

C (CH3) (CH3) (CH3) (COOCH3)

or if its easier to read: (CH3) C (COOCH3)


But I got this:


C (H) (CH3) (CH2-CH2-CH3) (COOH)

and I cant see why its wrong?
The mark scheme is not optical isomer - they dont have four different groups on a central carbon. Your answer is a correct one as well, first of all you can count the atoms and tell that they are isomers, secondly, you can see that there are four different substituents on your central carbon, try draw a 3D molecular structure and draw its mirror image, and you can realise that they are non superimposable, so they are enantiomers.

One thing to note, C6H12O6 is a sugar molecule(monosaccharides), there are more than one chiral centre, but if they specifically say they want a sugar, then maybe it has to be a specific isomer, but if not, your answer is correct!
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Eklipz89
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(Original post by irfy)
Need to find a possible structure of C6H12O2 which is optically active

Marksheme says this:

C (CH3) (CH3) (CH3) (COOCH3)

or if its easier to read: (CH3) C (COOCH3)


But I got this:


C (H) (CH3) (CH2-CH2-CH3) (COOH)

and I cant see why its wrong?
C (CH3) (CH3) (CH3) (COOCH3)

I dont see how that be optically active - doesnt the central C need 4 different groups attached to it??


Yours seem right

Damn i havent done this topic in ages =/
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Kaeroll
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Their suggested isomer would be optically inactive. Yours is a better suggestion. Might want to double check the wording of the question, it seems odd for the mark scheme to be so obviously wrong.

Off topic: Eklipz89, long signature is long!
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MrCynical
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The mark scheme compound is very clearly not optically active (simple check - are there four different groups around a carbon?). Check the question - that's a rather blatant mistake to make it into a mark scheme. Your suggested compound is optically active.
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irfy
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hawwww ****, i did copied out the wrong compound. Ill write it all again:



C (CH2CH3) (CH3) (H) (COOCH)



But I got this:


C (H) (CH3) (CH2-CH2-CH3) (COOH)

and I cant see why its wrong?
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shengoc
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(Original post by irfy)
hawwww ****, i did copied out the wrong compound. Ill write it all again:



C (CH2CH3) (CH3) (H) (COOCH)



But I got this:


C (H) (CH3) (CH2-CH2-CH3) (COOH)

and I cant see why its wrong?
both compounds are optically active, but the one you have got is not an isomer of the first one, count the number of hydrogen on each one, i think you got two extra hydrogen for the bottom structure.
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Kaeroll
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That's probably because s/he's forgotten to add two more protons to the methyl ester in the mark scheme's compound. Remember kids, always proof-read your posts...

Both are optically active and there's nothing wrong with them. Did they specify they want an ester?
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bingeonshows
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(Original post by irfy)
Need to find a possible structure of C6H12O2 which is optically active

Marksheme says this:

C (CH3) (CH3) (CH3) (COOCH3)

or if its easier to read: (CH3) C (COOCH3)


But I got this:


C (H) (CH3) (CH2-CH2-CH3) (COOH)

and I cant see why its wrong?
its not an ester
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charco
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(Original post by bingeonshows)
its not an ester
If you had held off for another 6 months you could have answered as a tenth birthday post ...

:clap2::congrats:
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