I would appeciate any help with identifying the chiral carbons, if any, in each of the following compounds:

OCH3
2. CH3--CH--CH3

OH O
11. CH3--CH--C--CH3
OH

12. CH3--C--CH3
OH

CH3 O
15.CH3--CH--C--CH3

Br
CH3--C--CH2--CH3
OH

thank you.

5 answers

There is no way anyone can know what you have typed. As I pointed out in an earlier response, the board ignores any space after the first one you make; take a look at your post and you can see there is no way of knowing where the groups are attached that you have typed above and below the line. A a carbon to be chiral, it must have four different groups attached. That's easy enough to count without posting a question. An alternative is to try writing the condensed structure.
Ah, I see what you are talking about.

OCH3 is above the CH
________________________

OH O is above the CH
Also OH is under the CH
________________________

OH is under the C
___________________

CH3 O is above the CH
____________________
Br is above the C
OH is under the C

_____________________________________
any help is appreciated. If it is four groups, then wouldn't all be chiral carbons?
#2 has no chiral center. Why? On the central C, there is an H, an OCH3, and TWO CH3 groups; therefore, it doesn't have four separate groups.

#11. I can't figure out from your description. I think the OH under the ring probably goes on the C and not the CH.

#12.No chiral center for the same reason as #2.

15. Some of the C atoms don't have four bonds. If you meant
CH3CH(OCH3)CH2CH3 (I have added H atoms to the #2 C atom from the right side), that C is chiral. It has a CH3, H, OCH3, and CH2CH3 which makes four different groups attached.

last one is not numbered. It has a chiral center. There is a CH3, Br, OH, and CH2CH3; again four separate groups.
Hey, thanks a lot.

Yes, sorry I did write those incorrectly.Here they are:

11. CH3--CH--C--CH3

OH is above the CH
O is above the C
_________________________________

So 15. does have four separate groups?

15. CH3--CH--C--CH3
CH3 is above the CH
O is above the C
Try writing them in this fashion.

CH3-CH(OCH3)-CH3 [not chiral]

CH3-CH(OH)-C(=O)-CH3 [chiral]

CH3-C(OH)2-CH3 [not chiral]

CH3-CH(CH3)-C(=O)-CH3 [not chiral]

CH3-C(Br)(OH)-CH2-CH3 [chiral]