Describe separately, tartaric acid and 2-chloro-3-methylpentane, do they obey/disobey the 2^n formula when estimating the possible number of stereoisomers.
Please help/explain, and thank you all for all the help you guys have given me on other organic probs, it really helps me understand it and I appreciate it!!!
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2007-06-19 04:21:34 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
since tartaric acid has 2 chirlal carnbons, it has 4 isomers- RR, RS, SR, and RR. now the catch is, that the molecule as a whole is symmetric (axis of symmetry right in the middle) and therefore u can not distinguish between the RS and SR isomer (draw the RS, turn it upside down, and u get the SR. draw the RR. turn it upside down. u still get the RR and NOT the SS). this form (RS and SR) is called a meso-form. therefore the number of isomers is 3: RR, SS and meso.
in the 2-Cl-3-Me-pentane, the molecule is asymmetric, and u have the normal number of isomers.
2007-06-19 18:11:44 · answer #1 · answered by chem_freak 5 · 0⤊ 0⤋
Tartaric acid is meso, 2-chloro-3-methylpentane does.
2007-06-19 13:07:24 · answer #2 · answered by ag_iitkgp 7 · 0⤊ 0⤋