A hydrocarbon 'X' on reaction with Br2 in presence of CCl4 decolourises bromine water. Also 'X' reacts with KMnO4 in the presence of an acid to give two moles of the same carboxylic acid. The structure of 'X' is:
A
CH3CH=CHCH2CH2CH3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
CH3CH2CH=CHCH2CH3
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
CH3CH2CH2−CH=CHCH3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
CH2=CHCH2CH2CH3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is BCH3CH2CH=CHCH2CH3 Since it decolorizes bromine water, so X must be an alkene. SO (c) is incorrect. The reaction of the process: CH3CH2CH=CHCH2CH3+Br2→CH3CH2CHBr−CHBrCH2CH3
Since it gives two moles of same carboxylic acid on reaction with oxidizing agent potassium permanganate hence it must be symmetrical alkene. Here the only alkene in option B is symmetrical alkene. In presence of acidic KMnO4 alkene is oxidised to form acid. CH3CH2CH=CHCH2CH3+KMnO4/H+−−−−−−−−→K2Cr2O72CH3CH2COOH So the structure of X is CH3CH2CH=CHCH2CH3 Thus option B is correct.