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word problem??

Asked by 29th February 2008, 5:09 AM
Answered by Expert
Answer:

an optically active compound 'A' C4H9Br, on reduction, gives optically inactive compound 'B'. 'A' on reaction with KOH(alc.) gives 'C' which on reaction with HBr gives 'A'  back. identify A,B and C and write the chemical reactions involved?

A  is CH3CH2CHBrCH3 will be optically active with the chiral carbon atom.

B is CH3CH2CH2CH3

C isCH3CH=CHCH3 elimination will be by satzeffs rule.

 

Answered by Expert 20th December 2017, 4:22 PM
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