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Question

A compound X decolourises Br2 water and reacts slowly with conc. H2SO4 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 :
929195_40a6471d4bef4278b9e3a6a9be793019.png

A
X=CH3CH2CH=CHCH3,

Y=CH3CH2CH(Br)CH(Br)CH2CH3,

Z=CH3CH2CH3
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B
X=CH3CH=CHCH3,

Y=CH3CH(Br)CH(Br)CH3

Z=CH3CH2CH2OH
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C
X=CH3CH2CH=CHCH2CH3,

Y=CH3CH2CH|BrCH2CH2CH3,

Z=CH3CH2CH2CH|OHCH2CH3
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D
X=CH3CH2CH2CH=CHCH3,

Y=CH3CH2CH2CH2CH2CH2Br,

Z=CH3CH2CH2CH2OH
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Solution

The correct option is C X=CH3CH2CH=CHCH2CH3,

Y=CH3CH2CH|BrCH2CH2CH3,

Z=CH3CH2CH2CH|OHCH2CH3

  • Br2 and H2SO4 will give the addition product,
  • The nucleophilic part of NaOH, (OH) will substitute the Br
  • Oxidising agent will oxidise the secondary alcohol to ketone.


Option C is correct.

1015200_929195_ans_3bbfbfd9f9134f8f8abd4dbb5e70d7e8.png

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