The relative reactivity of the ethyl halides C2H5X toward E2 elimination by NaOEt is

C2H5I > C2H5Br > C2H5Cl > C2H5F

C2H5Br > C2H5I > C2H5Cl > C2H5F

C2H5Cl > C2H5Br > C2H5I > C2H5F

C2H5F > C2H5Cl > C2H5Br > C2H5I

C2H5I > C2H5Br > C2H5Cl > C2H5F

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JEE Main levelSingle correct optionDifficulty 2 of 3o4101

The relative reactivity of the ethyl halides $C X_{2} H X_{5} X$ toward E2 elimination by $NaOEt$ is

A $C X_{2} H X_{5} Br > C X_{2} H X_{5} I > C X_{2} H X_{5} Cl > C X_{2} H X_{5} F$
B $C X_{2} H X_{5} Cl > C X_{2} H X_{5} Br > C X_{2} H X_{5} I > C X_{2} H X_{5} F$
C $C X_{2} H X_{5} F > C X_{2} H X_{5} Cl > C X_{2} H X_{5} Br > C X_{2} H X_{5} I$
D $C X_{2} H X_{5} I > C X_{2} H X_{5} Br > C X_{2} H X_{5} Cl > C X_{2} H X_{5} F$

Answer: D

The C-X bond is broken in the rate-determining E2 step, so the rate depends on the leaving group. Iodide is the best leaving group and fluoride the poorest, giving the order

I > Br > Cl > F

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The relative reactivity of the ethyl halides CX2​HX5​X toward E2 elimination by NaOEt is

Answer: D

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The relative reactivity of the ethyl halides $C X_{2} H X_{5} X$ toward E2 elimination by $NaOEt$ is