Question

A compound $X$ decolourises $Br_2$ water and reacts slowly with conc. $H_2SO_4$ to give an addition product. $X$ reacts with $HBr$ to form $Y$. $Y$ reacts with $NaOH$ to form $Z$. On oxidation $Z$ gives hexan-$3$-one. $X$, $Y$ and $Z$ in the reactions are

• A. $X = CH_3CH_2CH = CHCH_3$,
  $Y = CH_3CH_2CH(Br)CH(Br)CH_2CH_3$,
  $Z = CH_3CH_2CH_3$
• B. $X = CH_3CH = CHCH_3$,
  $Y = CH_3CH(Br)CH(Br)CH_3$,
  $Z = CH_3CH_2CH_2OH$
• C.
• D. $X = CH_3CH_2CH_2CH = CHCH_3$,
  $Y = CH_3CH_2CH_2CH_2CH_2CH_2Br$,
  $Z = CH_3CH_2CH_2CH_2OH$

Answer 1

Answer: (C)

Correct answer is (c)