Fascinating Chemistry!!
Let's take a common compound Cyclopropane .
Carry out the following reactions on it in sequence ..
- Add B r X 2 in dark.
- Add aq. K O H (excess)
- Then add H B r ( 1 equivalent ).
- Add alc. K O H
- Now add Dilute H X 2 S O X 4 .
- Add periodate and consider the product with more number of Carbons.
- Add L i A l H X 4
- Now add H X + at 170°C.
- Add B r X 2 .
- Add N a N H X 2 .
- Add N i ( C N ) X 2 .
Molecular Mass of the product is A .
Aromaticity of the product is B .
Values assigned ⎩ ⎪ ⎨ ⎪ ⎧ Non-aromatic = 2 Aromatic = 4 Anti-aromatic = 6
Enter your answer as A × B .
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The answer is 208.
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2 solutions
Bro it should be N i ( C N ) X 2 not N i ( C N ) X 4
But wont there be free radical substitution with br2?
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No, halogens give free radical substitution in presence of sunlight or some other radical generatong source only.
Isn't there a chance for carbocation rearrangement in the reaction?
C H X 2 ( O H ) − C H X 2 − C H X 2 ( O H ) with H B r
The product formed in this case would be C H X 3 − C H ( B r ) − C H X 2 ( O H ) instead of C H X 2 ( B r ) − C H X 2 − C H X 2 ( O H ) .
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No chance , as at that position C+ is unstable Due to -I effect of O
after adding Br2 to ethene ,the Br should be anti position so NaNH2 should not eliminate??
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what are you saying! the molecule is symmetric , there is nothing like anti in it!
After adding LiAlH4, you added H+ you must also mention heating as ethene is formed at 170°C.
Forgot , done !
Here is the solution