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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 SN1 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 SN1 reaction will be. The planar, trigonal carbocation may be attacked equally well from either side by a nucleophile. As a consequence, an SN1 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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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 E1 or SN1 ...
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)
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