Orgo 2: ochem week 6 part 3

Exam Information and Reminders

Orgo 1 Review: Substitution and Elimination

• Importance: Need to remember substitution and elimination concepts for the upcoming test.

• Tertiary Alkyl Halides: Preferred $S<em>N1$ over $S</em>N2$

  • Reason 1: Steric hindrance prevents $S_N2$ backside attack.

  • Reason 2: More stable carbocation intermediate (a key feature of $S_N1$).

• Stereochemistry in $S_N1$: Not a concern because the $sp^3$ carbon loses its stereochemistry when it becomes an $sp^2$ carbocation, leading to racemization (or an equal mix of possible stereoisomers).

Transforming Alcohols: Bad Leaving Groups to Good Leaving Groups

• Problem: Alcohols ($-OH$ groups) are terrible leaving groups.

• Solution: Convert alcohols into good leaving groups using specific reagents.

Using $SOCl<em>2$ (Thionyl Chloride) and $PBr</em>3$ (Phosphorus Tribromide)

• Purpose: These reagents replace an alcohol ($-OH$) with a halogen ($-Cl$ or $-Br$).

• Applicability: Primarily used with primary and secondary alcohols.

• Mechanism (for $SOCl_2$ as an example, note this is the instructor's preferred mechanism variant):

  1. The oxygen of the alcohol attacks the sulfur of $SOCl_2$.

  2. The $S=O$ pi bond breaks, placing electrons on the oxygen.

  3. The pi bond reforms, kicking off a chlorine atom ($-Cl^-$).

  4. An intermediate is formed where the alcohol oxygen is now bonded to sulfur, and the original carbon-oxygen bond is still intact. A hydrogen is still on the oxygen.

  5. A base (typically pyridine, not explicitly shown but implied for proton removal) deprotonates the oxygen, and the halogen (the $-Cl^-$ that was kicked off) performs a backside attack on the carbon that was originally bonded to the alcohol oxygen.

  6. This backside attack forces the newly formed oxygen-sulfur-chlorine group to leave.

  7. Result: An alkyl halide is formed, along with byproducts ($SO_2$ and $HCl$).

• $S<em>N2$ Characteristics: This sequence of reactions is an $S</em>N2$ mechanism.

  • Stereochemistry: Due to the backside attack of the halide, there is an inversion of stereochemistry at the reactive carbon center (if it's chiral).

• Significance: An alcohol (terrible leaving group) is converted into an alkyl halide (good leaving group), which then allows for many other reactions such as further $S_N2$ reactions, elimination reactions, or the formation of Grignard reagents.

Practice Problem: Grignard Synthesis (Challenge Problem)

• Step 1: Alcohol to Alkyl Halide: React a primary alcohol with $PBr_3$. The $-OH$ is replaced by $-Br$.

  • Since it's a primary alcohol and does not have an asymmetric center, stereochemistry is not a concern, but if it were an asymmetric center, $S_N2$ inversion would occur.

• Step 2: Grignard Formation and Reaction (Intramolecular):

  1. Form Grignard Reagent: Insert magnesium (Mg) into the C-Br bond (e.g., using $Mg, Et_2O$). It is crucial to remember to