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Question

Which of the following pair of mechanisms is correct for the given reactions?
CH3−CH2−CH2−BrCH3CH2OH/CH3CH2O−−−−−−−−−−−−−−−−−→△CH3−CH=CH2
H3C−CH(Br)−CH3CH3CH2OH/H2O−−−−−−−−−−−→△CH3−CH=CH2

A
SN1,E1
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B
SN2,E2
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C
E2,E1
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D
SN1,SN2
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Solution

The correct option is C E2,E1
In first reaction, the reactant is primary halide so it undergoes SN2 or E2 reaction. Smaller and aqueous base will favour SN2 reaction whereas larger and stronger base at high temperature favours E2 reaction. Here, the given conditions are fravourable for E2, thus the mechanism of the reaction is E2.

In second reaction, the reactant is secondary halide which can undergo E1, E2, SN1 and SN2. The alcohols are weak base/nucleophile hence, E1 or SN1 is favoured. Since the condition is at high temperature, the elimination reaction is favoured. Therefore, the mechanism of second reaction is E1.

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