Question

4. Propose a mechanism for the following reaction, as in show all the arrows and all intermediates (if any) that would eventually provide you the product from the alkyl iodide to the resulting alkene (a) is the product above the Zaitsev product? Why? (yes/no) (b) How would you change the reaction conditions to specifically get the alkene product drawn below instead: Base Solvent Temperature (c) How would you change the reaction conditions to specifically get the following substitution product drawn below instead: сн. CHE OCH Base Solvent Temperature

Answer 1

Mechanism

methyl shift 0 F cao 2° Carbocation 3º Carbocation + -60- CH3OH Major product 3° Carbocation CH3OH is act as a weak base to abstract the suitable hydrogen

It’s a dehydrohalogenation of alkyl halides reaction to form alkenes. This is the E₁ elimination reaction.

The reaction proceeds through the carbocation intermediate formation, the tertiary carbocation formed by the methyl shift from secondary carbocation.

Eventually, this tertiary carbocation gives the final major product through the abstraction of hydrogen through methanol as the methanol act here as a weak base.

(a)

Yes, the above product is Zaitsev’s product, because according to Zaitsev’s rule in the elimination reactions, the most substituted alkene product will be the most stable, and therefore this will be the most favoured too.

Here three products are possible in above reaction. It can be seen from the bellow mechanism that highly substituted product is the given in the above reaction itself as this alkene (Product) contains 4 substitutions in it therefore it is the most stable and favoured product.

i.e.

MeOH 100 °C Major product Minor product Minor product MeOH Di substituted alkene 3° Carbocation MeOH + 1 Tri substituted alkene 3° Carbocation 151 = Substitution Tetra substituted alkene 3 Carbocation | --- MeOH

(C)                                                           

When a strong base like tertiary butoxide (t-BuOK) is used with alcohol below 100 °C temperature then the Major product criteria will be different than the previous reaction. Tertiary butoxide going to abstract the hydrogen which is more accessible and therefore the product is different.

i.e.

Alcohol k t-BuOK

(d)

When alkoxides (CH₃ONa, CH₃OK etc.) are reacted with alkyl halides then this type of products are formed.

i.e.

CHONa 01