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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Halogens
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Mark out the correct comparison of the rates of solvolysis reaction

$\begin{array}{1 1}(a)\;\text{t-BuF in acidic solution >t-BuF in basic solution}\\ (b)\;\text{t-BuCl>2-chloro 2,3,3 trimethyl butane}\\ (c)\;\text{t-BuCl >2-chloro 1,1,1 trifluro 2-methyl propane}\\ (d)\;\text{t-BuCl in 90% }H_2O\text{-10% dioxane>t-BuCl in 90% }D_2O\text{-10% dioxane }\end{array}$

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1 Answer

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(a) t-BuF in acidic solution > t-BuF in basic solution (t-buty fluroide is solvolysed in very acidic solution,because $F^-$ is a poor leaving group but H-bonding with a strong acid encourages its departure,ithis is example of electrophilic catalysis.
(b) t-BuCl > 2-chloro 2,3,3 trimethylbutane (t-Bucl solvolyse more slowly than 2-chloro 2,3,3 trimethyl butane,formation of $Me_3C^+Me-2$) alliviates steric crowding on alpha carbon.This is example of steric accelaration.
(c) t-BuCl > 2 chloro 1,1,1 trifluro 2-methyl propane $F_3C-C^+H-Me$ is destabilized by the strongly electron with drawing $F_3C$ group making the solvolysis slower than the t-BuCl.
(d) t-Bucl in 90% $H_2O$-10% dioxane t-Bucl in 90% $D_2O$-10% dioxane.-D bonds are not as stabilized as -H bonds
Hence (c) is the correct answer.
answered Sep 4, 2013 by sreemathi.v
edited Mar 31, 2014 by sreemathi.v
 
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