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 :
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=CH3CH2−CH|Br−CH2CH2CH3,
Z=CH3CH2CH2−CH|OH−CH2CH3
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D
X=CH3CH2CH2CH=CHCH3,
Y=CH3CH2CH2CH2CH2CH2Br,
Z=CH3CH2CH2CH2OH
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Solution
The correct option is CX=CH3CH2CH=CHCH2CH3,
Y=CH3CH2−CH|Br−CH2CH2CH3,
Z=CH3CH2CH2−CH|OH−CH2CH3
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.