[Rutherford]

1.Why is the name 2-methyl-cyclohexanal inappropriate? Doesn't it refer to the following molecule http://imageshack.us/photo/my-images/580/cyclo.gif/
2.Is 2-methyl-buthan-1-ol chiral and/or optical active? I would say chiral, but not optical active. True?

[Dan]

1.Why is the name 2-methyl-cyclohexanal inappropriate? Doesn't it refer to the following molecule http://imageshack.us/photo/my-images/580/cyclo.gif/

Cyclohexanal is not the IUPAC name, it should be cyclohexanecarbaldehyde

2.Is 2-methyl-buthan-1-ol chiral and/or optical active? I would say chiral, but not optical active. True?

It is indeed chiral, can you explain why you think it is not optically active?

[Rutherford]

It has no symmetry=chiral, but there isn't a single C atom that has 4 different groups attached, so it shouldn't be optical active, or should it?

[discodermolide]

1.Why is the name 2-methyl-cyclohexanal inappropriate? Doesn't it refer to the following molecule http://imageshack.us/photo/my-images/580/cyclo.gif/
2.Is 2-methyl-buthan-1-ol chiral and/or optical active? I would say chiral, but not optical active. True?

2-methyl-cyclohexanal is ok as far as I know

2-methyl-butan-1-ol is chiral and as written not optically active as you imply the racemate.
If had written  2-(S)-methyl-butan-1-ol you imply one of the enantiomers

[Rutherford]

I am a little confused, chirality arises when:
-there is a C atom that has 4 different groups attached
-the molecule is not symmetric
The second statement should apply for chirality, so the molecule is chiral only if it is asymmetric, it doesn't depend on C atoms that have 4 different groups attached (meaning that the first statement isn't always true). Right?

[sjb]

It has no symmetry=chiral, but there isn't a single C atom that has 4 different groups attached, so it shouldn't be optical active, or should it?

Let's take each of the carbons in turn. There are 2 primary carbons (methyl groups), so these are not asymmetric. There are also 2 methylenes (>CH₂) so these are not asymmetric. However, the carbon that has only one hydrogen attached to it is asymmetric, it is also attached to a methyl group, an ethyl group and a CH₂OH group, right?

So although the gross mixture may not be optically active, each individual molecule is.

[Rutherford]

Yes, now I know why I was confused, I was only looking at the nearest groups to that C atom, so there were 2 CH₂ groups. Seems that chirality after all requires a C atom attached to 4 different groups. Thanks for helping me.