Delta, Δ
Badges: 7
Rep:
?
#1
Report Thread starter 7 years ago
#1
Hello Guys! I have 2 questions.
1. Say if I had CClFBrI (chloroflourobromoiodomethane), I know this is an optical isomer as it is non superimposeable on its mirror image. But since you can rotate single bonds, can't you rearrange the bonds to a position where the mirror image be superimposable?

2. If I had say HIC=O (iodomethanal), dose this compound have an optical isomer? as all the groups around the carbon different. If not, why so?

Thanks Guys! =)
0
reply
Ari Ben Canaan
Badges: 16
Rep:
?
#2
Report 7 years ago
#2
(Original post by Delta, Δ)
Hello Guys! I have 2 questions.
1. Say if I had CClFBrI (chloroflourobromoiodomethane), I know this is an optical isomer as it is non superimposeable on its mirror image. But since you can rotate single bonds, can't you rearrange the bonds to a position where the mirror image be superimposable?

2. If I had say HIC=O (iodomethanal), dose this compound have an optical isomer? as all the groups around the carbon different. If not, why so?

Thanks Guys! =)
So lets start with a slightly different compound to iodomethanal. Lets imagine we had DCHO so that's the same as methanal but one hydrogen has been replaced by a Deuterium atom.

So, what do you know about the geometry of the deuterated molecule ? What is the hybridisation of the Carbon ?

You should realise that this is an sp2 Carbon and hence the molecule is trigonal planar. There is no orientation of the substituents that would render a mirror image that is non-superimposable... Basically, no matter how you place the D and H atoms you will always have the same molecule if you simply flip it 180 degrees about the C=O bond.

Also, for your purposes at A Level you need to have 4 completely different groups for chirality when this compound only possesses 3.

Lastly, we come to your first point. There is no easy way to explain this and I would strongly suggest making a model by attaching some coloured pens to a central ball or obtaining a model.

You should see that there is no way to orientate those bonds to obtain the same molecule.
1
reply
Delta, Δ
Badges: 7
Rep:
?
#3
Report Thread starter 7 years ago
#3
(Original post by Ari Ben Canaan)
So lets start with a slightly different compound to iodomethanal. Lets imagine we had DCHO so that's the same as methanal but one hydrogen has been replaced by a Deuterium atom.

So, what do you know about the geometry of the deuterated molecule ? What is the hybridisation of the Carbon ?

You should realise that this is an sp2 Carbon and hence the molecule is trigonal planar. There is no orientation of the substituents that would render a mirror image that is non-superimposable... Basically, no matter how you place the D and H atoms you will always have the same molecule if you simply flip it 180 degrees about the C=O bond.

Also, for your purposes at A Level you need to have 4 completely different groups for chirality when this compound only possesses 3.

Lastly, we come to your first point. There is no easy way to explain this and I would strongly suggest making a model by attaching some coloured pens to a central ball or obtaining a model.

You should see that there is no way to orientate those bonds to obtain the same molecule.
Thanks for the reply!
So you cannot rotate the bonds in the 3d plane of the molecule, but you can in the 2d plane?
0
reply
DashStrike
Badges: 0
Rep:
?
#4
Report 7 years ago
#4
You can rotate bonds in all planes, however think of it like trying to superimpose your right hand with your left hand. You can turn your left hand upside down and superimpose, but it wouldn't be a NORMAL left hand, it would be a FLIPPED left hand. The same with molecules. You can rotate the bonds, but in the end you end still have 2 different isomers, each with a different effect on plane polarised light.

As for the HIC=O, no, it does not have optical isomers, but the Carbon has a potential to become chiral when undergoing addition reaction and become one. As Mr Canaan rightly pointed out, the molecule is trigonal planar, and thus the carbon is susceptible to attacks BOTH from ABOVE AND BELOW, in a 50/50 chance, creating a racemic mixture (a mixture of 2 optical isomers in EQUAL AMOUNTS which has no effect on plane polarised light due to vectorial addition of the effect on it of both isomers). It might be worth having a look at chemguide : http://www.chemguide.co.uk/basicorg/...m/optical.html
Attached files
1
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Should 'Mental health support' be included on league tables?

Yes (220)
74.83%
No (74)
25.17%

Watched Threads

View All