plz tell me dat when will tetiary butyl halide undergo s1i,e1 and e2 reaction. by giving some examples........

Dear student,

SN1 reaction

In an S_N1 reaction the key step is the loss of the leaving group to form the intermediate carbocation. This step is the slow, rate determining step of the reaction. The carbocation is then attacked by a nucleophile in a fast second step to form the product. The more stable the carbocation is, the easier it is to form, and the faster the S_N1 reaction will be. The planar, trigonal carbocation may be attacked equally well from either side by a nucleophile. As a consequence, an S_N1 reaction leads to a racemization, in which both retention and inversion of configuration at a chiral center occur to the same extent. Optically active tertiary haloalkanes produce a mixture of two enantiomers (mirror image isomers).

   

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Askiitians Expert

Sagar Singh

B.Tech, IIT Delhi

sagarsingh24.iitd@gmail.com

tertiary halide will never undergo SN² reaction due to staric  hinderence ...it mainly undergoes substitution by SN¹ & elimination by both E₁,E₂ ....

if solvent is highly basic then there are more chances that reaction will be E₂ but in presence of weak base the reaction may be E_{1 or SN}1 ...

eg1) tertiary halide + C2H5OH or H2O (solvent) => ether or alkene      (by E1 or Sn1 & carbocation is formed in both)

 Cl-C(CH3)3 + C2H5OH => CH3CH2-0-C(CH3)3 (ETHER)  +  CH2=C-(CH3)2  (alkene)

eg2)tertiary halide + C₂H₅O^- (solvent) => only alkene            (by E₂ ,due to presence of strong base)

Cl-(CH₃)₃ + C₂H₅O^- => CH₂=C-(CH₃)₂      (ALKENE)