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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 :
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 $$ X= CH_3CH_2CH=CHCH_2CH_3,$$

$$Y=CH_3CH_2-\underset { \underset { \displaystyle Br }{ | } }{ CH } -CH_ 2CH_2CH_ 3,$$

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

  • $$Br_2$$ and $$H_2SO_4$$ 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

Chemistry

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