KCC's Quizzes AQQ247 about Chiral Molecules

With  a single carbon atom, the criteria often used is to see it one of the 4 companions of the C atom is duplicated. 
If the three companions are different, there is a "left" version and a "right" version of the molecule  (like the Cartesian x-y-z axis can be left hand or right hand version).
So, 1, 3, and 4 have and a left and a right version, while molecule 2 has a single version.

With two carbon atoms, things are a little bit more complex 
Molecule 6 is the easiest one (with 2 carbons) since the carbons do have different companions than the other:   
            C of the left:   OH, COH , Cl, and the second carbon GROUP  
            C  of the right: OH,  CH3, CH3  and the first carbon GROUP
So, for each carbon, each carbon has 4 different companions, and the previous criteria is applicable.

BUT it 3 companions of the two carbons are mutually the same (as for molecule 5 and 7) that becomes more complex since the "carbon GROUP" companion may or may not carry the decision.

If we look straight from a companion to one of the C atoms, the 3 other companions can be at 12 o'clock, 4 and 8 while, for the second C atom, its three other companions can be at the same matching position, OR, and often more stable, be at  2 o'clock, 6 and 10. So they will be spatially different. But if we take OFF (don't consider) that possibility, since we can turn the molecule around the single bound between the two C, then:

molecule 5 has the two carbons in a mirror like but nothing forbids the OH of one C to be in opposition of the COOH  or of the CH3.
(And indeed, the single bound C-C  is said to be stiffer, in "torsion", than say, a double bound as in ethylene HH C =  C HH  )
So, here, molecule 5  MAY be chiral... or not, given the possible stiff arrangements (and the "attraction/repulsion" of the companions facing each other, or at 60 degree phase of facing each other)
Molecule 7 is easier that molecule 5 since while each carbon has the same three other companions, they are NOT in the same order:  if one is  in the order  a-b-c, or b-c-a, or c-a-b (considering the possible rotation along the single bound between the two C atoms), the other has the order a-c-b, or c-b-a, or b-a-c  and so, has definitively a "left" and a "right" version.

So.  has a POSSIBLE left and right version:   1, 3, 4, 5*, 6 and 7
        are (possibly) not:  2 and 5*

Rectification: a double or triple bound is stiffer than a single bound, generally.

See the comment put earlier

See the comment put earlier