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Organic : 3 Questions
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 Message 1 of 2 in Discussion 
From: MSN NicknameAlbert1145  (Original Message)Sent: 4/29/2008 2:53 AM
Hi Steve,
Which, if any, of the following compounds can exist in optically active form? (Select ALL correct answers)
A. 2,3-dichlorohexane
B. 2,4-dichlorohexane
C. 2,2-dichlorohexane

Select the active nitrating species in aromatic nitration reactions involving a mixture of concentrated nitric and concentrated sulfuric acids:
A. NO2+
B. HO-NO2
C. NO2-
D. [H2O-NO2]+
E. HO-SO2-ONO2



In the structure below, which atom possesses exactly TWO lone pairs (or unshared pairs) of electrons.
CH3-CH-OH=CH-CH-Br-CH2-NH2

I really have no idea, could you help me
Thank you




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 Message 2 of 2 in Discussion 
From: MSN Nickname·Steve·Sent: 4/29/2008 5:05 AM
1. If a carbon has four different groups attached to it, it is a chirality or stereogenic center with R or S configurations.  In 2.3-dichlorohexane, the second carbon has four groups attached:  1) CH3, 2) Cl, 3) H, and 4) the rest of the molecule.  Thus carbon 2 is a stereogenic carbon.  Same with carbon 3.  In 2,4-dichlorohexane, you have the same situation.  In 2,2-dichlorohexane, no carbon has four different groups attached, so it is not optically active.
 
If you have an even number of stereogenic centers in a molecule, it is possible for it to still be optically inactive, if it possesses a plane of symmetry.  Such a molecules is classified as being meso.  But that is not the case with 2,3- and 2,4-dichlorohexane.
 
2.  NO2+, the nitronium ion, is the electrophile in aromatic nitrations using nitric acid and sulfuric acid.
 
3.  CH3-CH-OH=CH-CH-Br-CH2-NH2  There seems to be an extra H, but if the oxygen atom has two bonds attached to it, it will have two lone pairs on it.  Just as in the H2O molecule.
 
 
Steve