Explain the order of reactivity of $1^o$,$2^o$,and $3^o$ halides towards the $S_N2$ reaction.
(N/A) The order of reactivity for the $S_N2$ reaction is: $CH_3X > 1^o$ halide $> 2^o$ halide $> 3^o$ halide.
In an $S_N2$ reaction,the nucleophile approaches the carbon atom attached to the leaving group. The bulky substituent groups present on this carbon atom create steric hindrance. As the size of these groups increases,the $S_N2$ reaction becomes slower.
Methyl halides react the fastest in $S_N2$ reactions because they only have three small hydrogen atoms attached to the carbon. Tertiary $(3^o)$ alkyl halides are the least reactive in $S_N2$ reactions because the bulky groups provide maximum steric hindrance to the incoming nucleophile.
In an $S_N2$ reaction,the nucleophile approaches the carbon atom attached to the leaving group. The bulky substituent groups present on this carbon atom create steric hindrance. As the size of these groups increases,the $S_N2$ reaction becomes slower.
Methyl halides react the fastest in $S_N2$ reactions because they only have three small hydrogen atoms attached to the carbon. Tertiary $(3^o)$ alkyl halides are the least reactive in $S_N2$ reactions because the bulky groups provide maximum steric hindrance to the incoming nucleophile.
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Similar Questions
The major product of the following reaction is:
For the following reactions,which of the following statements is correct?
$A. \, CH_3CH_2CH_2Br + KOH \rightarrow CH_3CH=CH_2 + KBr + H_2O$
$B. \, (CH_3)_2CHBr + KOH \rightarrow (CH_3)_2CHOH + KBr$
$C. \, C_6H_{10} + Br_2 \rightarrow C_6H_{10}Br_2$
$A. \, CH_3CH_2CH_2Br + KOH \rightarrow CH_3CH=CH_2 + KBr + H_2O$
$B. \, (CH_3)_2CHBr + KOH \rightarrow (CH_3)_2CHOH + KBr$
$C. \, C_6H_{10} + Br_2 \rightarrow C_6H_{10}Br_2$
MediumNEET 2016
View SolutionThe correct reactivity order for $S_N2$ reaction is:
$(i)$ $CH_3COCH_2Cl$
(ii) $CH_2=CH-CH_2Cl$
(iii) $CH_3CH_2CH_2Cl$
$(i)$ $CH_3COCH_2Cl$
(ii) $CH_2=CH-CH_2Cl$
(iii) $CH_3CH_2CH_2Cl$
Difficult
View SolutionAddition of $KI$ accelerates the hydrolysis of primary alkyl halides because
Medium
View SolutionMatch the reactions in List-$I$ with the features of their products in List-$II$ and choose the correct option.
| List-$I$ | List-$II$ |
| :--- | :--- |
| $(P)$ $(-)-1$-Bromo-$2$-ethylpentane $\xrightarrow{aq. NaOH, S_N2}$ | $(1)$ Inversion of configuration |
| $(Q)$ $(-)-2$-Bromopentane $\xrightarrow{aq. NaOH, S_N2}$ | $(2)$ Retention of configuration |
| $(R)$ $(-)-3$-Bromo-$3$-methylhexane $\xrightarrow{aq. NaOH, S_N1}$ | $(3)$ Mixture of enantiomers |
| $(S)$ $(3S, 5S)-5$-bromo-$3$-methylheptane $\xrightarrow{aq. NaOH, S_N1}$ | $(4)$ Mixture of structural isomers |
| | $(5)$ Mixture of diastereomers |
| List-$I$ | List-$II$ |
| :--- | :--- |
| $(P)$ $(-)-1$-Bromo-$2$-ethylpentane $\xrightarrow{aq. NaOH, S_N2}$ | $(1)$ Inversion of configuration |
| $(Q)$ $(-)-2$-Bromopentane $\xrightarrow{aq. NaOH, S_N2}$ | $(2)$ Retention of configuration |
| $(R)$ $(-)-3$-Bromo-$3$-methylhexane $\xrightarrow{aq. NaOH, S_N1}$ | $(3)$ Mixture of enantiomers |
| $(S)$ $(3S, 5S)-5$-bromo-$3$-methylheptane $\xrightarrow{aq. NaOH, S_N1}$ | $(4)$ Mixture of structural isomers |
| | $(5)$ Mixture of diastereomers |
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