A tertiary halide it most likely to react with NaOH by which mechanism? a) S_N 2

Question

A tertiary halide it most likely to react with NaOH by which mechanism? a) S_N 2 b) S_N 1 c) E_2 The mechanism shown here is for an (Ph = phenyl, an aromatic ring) a) E_2 reaction b) S_N 2 reaction c) S_N 1 reaction A primary halide is most likely to react with NaOH by which mechanism? a) E_2 b) S_N 1 c) S_N 2 A tertiary halide is most likely to react with H_2 O by which mechanism? a) E_2 b) S_N 1 c) S_N 2 Which solvent is best for S_N 2 reactions? a) CH_3 OH b) DMSO c) HBr d) H_2 O The leaving group and the H on the adjoining carbon must be anti periplanar to each other in which mechanism? a) S_N 2 b) S_N 1 c) E_2 An S_N 2 reaction goes with: a) inversion b) racemization c) retention d) rotation The first slow step in an S_N 1 reaction is: a) attack by nucleophile/base b) loss of H^+ c) loss of leaving group d) protonation of nucleophile/base Zaitsev's Rule says: a) in the presence of a strong nucleophile substitution is favored b) the H goes to the C that already has the most H's c) the most substituted alkene is favored d) the order of stability of carbocations is 3 degree > 2 degree > 1 degree

Explanation / Answer

Ans 3 b) Sn1

The tertiary halide undergoes unimolecular nucleophilic substitution reaction , which occurs in 2 steps .

In the first step a tertiary carbocation is formed , which is then attacked by the nucleophile to form the substituted product .

The reactivity of the reaction depends upon the stability of carbocation , which is formed stable in the case of tertiary halides , and hence reactive.

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