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Haloalkanes And Haloarenes

Question
CBSEENCH12011323

Consider the following reaction,


ethanol space rightwards arrow with PBr subscript 3 on top space straight X space rightwards arrow with alc. space KOH on top space straight Y space rightwards arrow from left parenthesis ii right parenthesis space straight H subscript 2 straight O comma space heat to left parenthesis straight i right parenthesis space straight H subscript 2 SO subscript 4 of space straight Z semicolon
the product Z, is

  • CH2 = CH2

  • CH3CH2-O-CH2-CH3

  • CH3-CH2-O-SO3H

  • CH3CH2OH

Solution

D.

CH3CH2OH

i) PBr3 is a halogenating agent, ie, converts -OH group into -Br
ii) Alc. KOH is a dehydrohalogenation agent.
iii) H2SO4 and H2O converts an olefin into alcohol

straight C subscript 2 straight H subscript 5 OH space rightwards arrow with PBr subscript 3 on top space straight C subscript 2 straight H subscript 5 Br space space rightwards arrow from negative KBr comma space minus straight H subscript 2 straight O to Alc. KOH of space CH subscript 2 space equals space CH subscript 2
rightwards arrow with straight H subscript 2 SO subscript 4 on top space CH subscript 3 minus CH subscript 2 OSO subscript 3 straight H space rightwards arrow from negative straight H subscript 2 straight O to straight H subscript 2 straight O divided by increment of CH subscript 3 CH subscript 2 OH