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Question

Optically active isomer (A) of $$(C_5H_9Cl)$$ on treatment with one mole of $$H_2$$ gives an optically inactive compound (B) compound (A) will be :           


A
CH3C|CH2ClHCH=CH2
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B
ClC|CH3HCH=CHCH3
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C
CH3C|ClHCH2CH=CH2
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D
CH3CH2C|ClHCH=CH2
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Solution

The correct option is C $$CH_3-CH_2-\underset{Cl}{\underset{|}{C}}H-CH=CH_2$$
As refer to the above image optically active compound gives optically inactive isomers.
1381145_989667_ans_b0d6becf039a4795a27883556d2be6e4.png

Chemistry
NCERT
Standard XI

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