Which of the two stereoisomers of 4-t-butylcy | vedclass

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(C) In isomer $A$,the iodine atom is in the axial position. Due to $1,3-$diaxial interactions,the leaving group $(I^-)$ is more sterically crowded and

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$A$ will react faster because the approach of $^{128}I^-$ can occur unhindered.

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(C) In isomer $A$,the iodine atom is in the axial position. Due to $1,3-$diaxial interactions,the leaving group $(I^-)$ is more sterically crowded and experiences repulsion,which facilitates its departure.In isomer $B$,the iodine atom is in the equatorial position,which is less sterically hindered and more stable,making the departure of the leaving group slower.Therefore,the axial isomer $(A)$ undergoes $S_N2$ substitution faster due to the relief of steric strain upon the departure of the leaving group.

Which of the two stereoisomers of $4-t-$butylcyclohexyl iodide $(^{127}I^-)$ will undergo $S_N2$ substitution with $^{128}I^-$ faster,and why?

Similar Questions

For the following reactions,which of the following statements is correct?
$A. \, CH_3CH_2CH_2Br + KOH \rightarrow CH_3CH=CH_2 + KBr + H_2O$
$B. \, (CH_3)_2CHBr + KOH \rightarrow (CH_3)_2CHOH + KBr$
$C. \, C_6H_{10} + Br_2 \rightarrow C_6H_{10}Br_2$

Predict the order of reactivity of the following compounds in $S_N1$ reaction :
$(I)$ $CH_3-CH_2-CH(Br)-CH_3$
$(II)$ $(CH_3)_2CH-CH_2-Br$
$(III)$ $(CH_3)_3CBr$

Among the given halides,which one will give the same product in both $S_{N^1}$ and $S_{N^2}$ reactions?
$(I) \ CH_3-CH(CH_3)-CH_2-CH(Br)-CH_3$
$(II) \ \text{2-methylcyclohexyl chloride}$
$(III) \ \text{Chlorocyclohexane}$
$(IV) \ CH_3-CH(Br)-CH_2-CH_3$

The reaction of $p-$bromobenzyl chloride with $NaCN$ in ethanol gives:

The ascending order of relative rate of solvolysis of the following compounds is:

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