Asked by | 29th Feb, 2008, 05:09: AM
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 | 20th Dec, 2017, 04:22: PM
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