A tertiary halide is most likely to react with NaOH by which mechanism? a) SN_2

Question

A tertiary halide is most likely to react with NaOH by which mechanism? a) SN_2 B) SN1 c) E2 The mechanism shown here is for and (ph = phenyl, an aromatic ring) a0 E_1 reaction c) Sm1 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_1O 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 pentanal 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 over elimination 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

Sn1 reaction is unimolecular nucleophilic substitution reaction which occurs in 2 steps. A tertiary carbocation is formed in 1st step , which is then attacked by the nucleophile and forms product.

Tertiary halide leads to formation of stable carbocation in step 1 , so it undergoes Sn1 reaction .

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