Question

30. Late papers not accepted. Work Independently. 1. Predict the most likely reaction mechanism (SN1, SN2, E1, E2) and draw all products. CH₂ Na O=CH₂ Hic Br 2 Draw complete mechanism with arrows for the SNI/E1 mechanism. Check for carbocation rearrangements CH3 + H2O 3. Which is the major product for the following reaction? Y CH + NaOH CH HO HO H₃ CH₂ H₃ HOCH CH₂ H₃ CHE CH₂ CH₂ CH3 CH

Answer 1

Reaction mechanism:

1)    In the first reaction, sodium methoxide (strong base) used. It will act as a base, not a nucleophile (so no S_N1 reaction). So it will abstract the beta-hydrogen near to bromine and gives alkene (elimination product). E1 mechanism not possible because bromide leaving create primary carbocation. If it is tertiary carbocation means E1 will happen.

Mechanism: E2

2) In this reaction water molecule act as a nucleophile. Water is a weak nucleophile so this reaction won't give S_N2 mechanism. For S_N2 mechanism we need strong nucleophile. The starting material not suitable for following the S_N1 mechanism also because after chlorine leaving it will creat secondary carbocation (less stable than tertiary carbocation).

So this reaction follows the carbocation rearrangement mechanism. After chlorine leaving secondary carbocation will form. Then methyl migration will happen after that hydride shift happened which will create stable tertiary carbocation. Finally, the formed carbocation attacked by weak nucleophile water which will create a tertiary-alcoholic compound.

Mechanism: carbocation rearrangement

3) In this reaction, NaOH was used. NaOH is strong base So it will abstract the beta-hydrogen near to chlorine and gives alkene (elimination product). E1 mechanism not possible because bromide leaving create secondary carbocation. If it is tertiary carbocation means E1 will happen.

Mechanism: E2

4) The alkene is Z; Chiral carbon is R (Plese go through the below image)

5) In this reaction tertiary-butoxide used as a base. tertiary-butoxide is a bulky as well as a strong base so it will abstract the hydrogen from the less hindered side and gives a less substituted alkene.

Na OCH3 CH3 BY E2 H H CH3 CH2 .CH3 2) carbocation Yearargarort CH3 H- CH3 O H tertiary-Altohol more stab le tertiany Carbocation 3) Na CH3 E 2 CH 3 Alkenz CH3 н CH3 ff,z)-6-brome-2,2- dimothyhept H Bro 3-ona R'

5)

bulky base Н E2 Н "CHз Нас Less substituted alkene