# Preparation of Alcohols from Grignard Reagent

We were given this question in the class:

Pentan-2-ol

I got 2-Methyl Butanol. Why doesn't the methyl attack the 2nd carbon atom from the oxygen atom and form 2-Methyl Butanol instead of attacking the 1st carbon atom from oxygen atom forming Pentan-2-ol. Is it just because of easy access or is there any other reason for this happening?

• Why do you think the grignard should attack the second C? That carbon is not next to an oxygen so it's not electrophilic, and there is no leaving group. There is no way for the grignard reagent to attack the second C Jun 1 at 8:11
• The ring strain inherent in a [4] ring makes it more reactive Jun 1 at 11:21

TL;DR The Grignard reagent reacts via the SN2 mechanism and attacks the less sterically hindered carbon.

Now, this reaction is similar to the opening of an epoxide ring by a strong nucleophile, wherein the less sterically hindered position is favored for attack by the nucleophile.

According to SN2 type Reactions of Epoxides from Organic Chemistry, 5th edition by Francis Carey,

Scenario 2:
The $$\ce{Nu}$$ attacks the least hindered end of the epoxide at $$180^\circ$$ to the $$\ce{C-O}$$ bond that breaks.

Experimental results show that scenario 2 is observed by these stronger nucleophiles such as Grignard reagents, $$\ce{RMgBr}$$.

Therefore a similar reaction is seen here leading to the formation of 2-pentanol.

As @Shoubhik R Maiti said, Grignard reagent won't attack the second $$\ce{C}$$ as, being farther away from the electronegative atom $$\ce{O}$$, it has less partial polarity.

Now, Grignard being a strong nucleophile, would attack the carbon with highest partial positive charge, hence we are left with a $$3°$$ carbon and a $$2°$$ carbon adjacent to $$\ce{O}$$ where attack is possible. For the $$3°$$ carbon, however, due to +I effect of $$2$$ $$\ce{Me}$$ group, partial positive is less compared to $$2°$$ carbon.

Also, $$2°$$ carbon is less sterically hindered. Hence attack is most preferred there.