Q

# Help me answer: - Organic Compounds containing Halogens - JEE Main

Which of the following is the correct order of decreasing $S_{N}2$  reactivity?

• Option 1)

$R_2CHX> R_3C\, X> RCH_2X$

( $X$ is a halogen)

• Option 2)

$RCHX> R_3CX> R_2CHX$

( $X$ is a halogen)

• Option 3)

$RCH_2X> R_2CHX> R_3CX$

( $X$ is a halogen)

• Option 4)

$R_3CX> R_2CH X> RCH_2X$

( $X$ is a halogen)

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As we learnt ,

Reactivity Order of alkyl halides -

$1^{\circ}Alkyl\: Halide> 2^{\circ}Alkyl\: Halide> 3^{\circ}Alkyl \: Halide$

- wherein

$1^{\circ}Alkyl\: Halide> 2^{\circ}Alkyl\: Halide> 3^{\circ}Alkyl \: Halide$

Inhibition by steric hindrance SN2 reactions are particularly sensitive to steric factors, since they are greatly retarded by steric hindrance (crowding) at the site of reaction. In general, the order of reactivity of alkyl halides in SN2 reactions is: methyl  > 1> 2o> 3o

Lesser the steric hindrance at c faster the SN2 reaction rate .

So the correct order is :

$RCH_2X> R_2CHX> R_3CX$

Option 1)

$R_2CHX> R_3C\, X> RCH_2X$

( $X$ is a halogen)

Option 2)

$RCHX> R_3CX> R_2CHX$

( $X$ is a halogen)

Option 3)

$RCH_2X> R_2CHX> R_3CX$

( $X$ is a halogen)

Option 4)

$R_3CX> R_2CH X> RCH_2X$

( $X$ is a halogen)

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