Orgo 2: ochem week 6 part 3

Why do tertiary alkyl halides prefer $S<em>N1$ over $S</em>N2$?

Tertiary alkyl halides prefer $S<em>N1$ over $S</em>N2$ due to steric hindrance preventing $S<em>N2$ backside attack and the formation of a more stable carbocation intermediate, which is a key feature of $S</em>N1$.

What is the stereochemical outcome in an $S*N1$ reaction?

In an $S*N1$ reaction, stereochemistry is not a concern because the $sp^3$ carbon loses its stereochemistry when it becomes an $sp^2$ carbocation, leading to racemization (an equal mix of possible stereoisomers).

What is the problem with alcohols ($-OH$ groups) in reactions?

Alcohols ($-OH$ groups) are terrible leaving groups.

How can alcohols be converted into good leaving groups?

Alcohols can be converted into good leaving groups using specific reagents like $SOCl<em>2$ (Thionyl Chloride) or $PBr</em>3$ (Phosphorus Tribromide).

What is the purpose of using $SOCl<em>2$ or $PBr</em>3$ with alcohols?

These reagents replace the alcohol group ($-OH$) with a halogen ($-Cl$ or $-Br$), converting a bad leaving group into a good one.

What types of alcohols are primarily used with $SOCl<em>2$ and $PBr</em>3$?

These reagents are primarily used with primary and secondary alcohols.

What are the byproducts of converting an alcohol to an alkyl halide using $SOCl*2$?

The byproducts are $SO_2$ and $HCl$.

What is the stereochemical outcome when converting a chiral alcohol into an alkyl halide using $SOCl<em>2$ (via the instructor's preferred $S</em>N2$-like mechanism)?

Due to the backside attack of the halide, there is an inversion of stereochemistry at the reactive carbon center.

Why is it significant to convert an alcohol into an alkyl halide?

Converting an alcohol (terrible leaving group) into an alkyl halide (good leaving group) allows for many other reactions, such as further $S*N2$ reactions, elimination reactions, or the formation of Grignard reagents.

Step 1 of Grignard Synthesis (from alcohol):

React a primary alcohol with $PBr_3$ to replace the $-OH$ with $-Br$, forming an alkyl halide. If an asymmetric center were involved, $S*N2$ inversion would occur.

Step 2 of Grignard Synthesis (formation of Grignard Reagent):

Insert magnesium (Mg) into the C-Br bond of the alkyl halide (e.g., using $Mg, Et_2O$) to form the Grignard reagent.