Question

A compound x decolourises $Br_{2}$water and reacts slowly with conc. $H_{2}SO_{4}$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. $x = CH_{3}CH_{2}CH = CHCH_{3},$ $Y = CH_{3}CH_{2}CH(Br)CH_{2}CH_{3}$
• B. $X = CH_{3}CH = CHCH_{3}$ $Y = CH_{3}CH(Br)CH(Br)CH_{3}$ $Z = CH_{3}CH_{2}CH_{2}OH$
• C. $X = CH_{3}CH_{2}CH = CHCH_{2}CH_{3}$ $Y = CH_{3}CH_{2} - \underset{Br}{\underset{|}{CH}} - CH_{2}CH_{3}$ $Z = CH_{3}CH_{2}CH_{2} - \underset{OH}{\underset{|}{CH}} - CH_{2}CH_{3}$
• D. $x = CH_{3}CH_{2}CH_{2}CH = CHCH_{3}$ $Y = CH_{3}$

Answer 1

Answer: (C)

$X = CH_{3}CH_{2}CH = CHCH_{2}CH_{3}$

$Y = CH_{3}CH_{2} - \underset{Br}{\underset{|}{CH}} - CH_{2}CH_{3}$

$Z = CH_{3}CH_{2}CH_{2} - \underset{OH}{\underset{|}{CH}} - CH_{2}CH_{3}$

Answer 2

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