NEETChemistryHydrocarbons
An alkyl halide C 5 H 11 Br (A) reacts with ethanolic KOH to give an alkene ‘B’, which reacts with Br 2 to give a compound ‘C’, which on dehydrobromination gives an alkyne ‘D’. On treatment with sodium metal in liquid ammonia one mole of ‘D’ gives one mole of the sodium salt of ‘D’ and half a mole of hydrogen gas. Complete hydrogenation of ‘D’ yields a straight chain alkane. Identify C and D.
Options
- AC is CH 3 —CH 2 —CH 2 —CH(Br)—CH 2 Br and D is CH 3 —CH 2 —CH 2 ≡CH
- BC is CH 3 —CH 2 —CH 2 —CH(Br)—CH 2 Br and D is CH 3 —CH 2 —CH 2 —CH 2 ≡CH
- CC is CH 3 —CH(Br)—CH 2 Br and D is CH 3 —CH 2 —CH 2 —CH 2 ≡CH
- DC is CH 3 —CH(Br)—CH 2 Br and D is CH 3 —CH 2 —CH 2 ≡CH
Correct answer
B. C is CH 3 —CH 2 —CH 2 —CH(Br)—CH 2 Br and D is CH 3 —CH 2 —CH 2 —CH 2 ≡CH
Step-by-step solution
Correct Option is : (B) C is CH 3 —CH 2 —CH 2 —CH(Br)—CH 2 Br and D is CH 3 —CH 2 —CH 2 —CH 2 ≡CH