Question Number 81 in ADA Sample 2007 DAT Test: SNS

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melmu001

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I need help with this question. How come this compound is not a meso compound. You have exactly the same substituents on both sides so why its not a meso compound?
 
For meso look for a line of symmetry and chiral centers. Right compound satisfies both but Left compound does not. The compound on the left is not meso because it does not have a plane of symmetry as in the compound on the right. Also since it is a cyclic compound there is no rotation of the CH3 and H so those substituents are stuck in that position. I hope I explained that well enough :/
 
So for a compound to be meso, you need atleast 2 chiral centers (C bound to 4 different groups) and have a line of symmetry (Can be folded on this self)
 
If a meso compound has to have at least 2 chiral centers... you indicated to me that there are 2 chiral centers for both compounds. So I want to know more how they are not meso compounds.
 
The 2nd one is meso because it has a line of symmetry but the first compound doesn't have a line of symmetry. As a result, the compounds are diasteromers or each other. Both are not meso. So C is the right answer
 
If a meso compound has to have at least 2 chiral centers... you indicated to me that there are 2 chiral centers for both compounds. So I want to know more how they are not meso compounds.

This is only half of what he/she said. They said that to be a meso compound it has to have two chiral centers AND have a line of symmetry. A line of symmetry is when you draw an imaginary line down the center and one side looks EXACTLY like the other (keep in mind certain bonds can rotate and may not appear symmetrical at first, although this is not the case here because it's a ring). Compound two meets both criteria. Compound one doesn't.
 
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This is only half of what he/she said. They said that to be a meso compound it has to have two chiral centers AND have a line of symmetry. A line of symmetry is when you draw an imaginary line down the center and one side looks EXACTLY like the other (keep in mind certain bonds can rotate and may not appear symmetrical at first, although this is not the case here because it's a ring). Compound two meets both criteria. Compound one doesn't.
Nicely said!
 
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