word problem??

Asked by  | 29th Feb, 2008, 05:09: AM

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.


C isCH3CH=CHCH3 elimination will be by satzeffs rule.


Answered by  | 20th Dec, 2017, 04:22: PM