Q.

 In the following sequence of reactions 

CH3CH2CH2Bralc. KOH(A)HBr(B)aq. KOH(C)

The product (C) is 

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a

 Propan-2-ol 

b

Propyne

c

 Propene 

d

 Propan-l-ol 

answer is A.

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Detailed Solution

When alkyl halide reacts with alcoholic KOH, then alkene forms with the removal of HBr.

CH3CH2CH2Bralc. KOHCH3CH=CH2A+HBr

Addition of HBr in alkene follows Markovnikov's rule, where Br attached to the carbon atom that has a fewer number of hydrogens.

CH3-CH=CH2+HBrCH3-CH(Br)-CH3B

In the presence of aqueous KOH, alkyl halide forms corresponding alcohol

CH- 3 CH(Br)-CH3+aq.KOHCH3-CH(OH)-CH3C (propan-2-ol)

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