Problem 36 - Basic Principles of Organic Chemistry

Question

Nitriles, RCN, can be prepared by \(S_{\mathrm{N}} 2\) displacement of alkyl derivatives, \(\mathrm{RX}\), by using sodium or potassium cyanide: $$ \mathrm{RX}+\mathrm{NaCN} \rightarrow \mathrm{RCN}+\mathrm{NaX} $$ a. Which of the following solvents would be most suitable for this reaction: water, 2 -propanone, ethanol, benzene, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~S}=\mathrm{O}\), or pentane? Give reasons for your choice. b. Which of the six isomeric monobromoderivatives of 1 -methylcyclohexane would you expect to react most rapidly with sodium cyanide? Why? c. If you wished to make 2 -phenylethanenitrile, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CN}\), which of the following phenylmethyl compounds, \(\mathrm{RCH}_{2} \mathrm{X}\), would you select to convert to the nitrile? \(\mathrm{X}=-\mathrm{F},-\mathrm{OH},-\mathrm{OCOCH}_{3},-\mathrm{H},-\mathrm{NH}_{2},-\mathrm{O}_{3} \mathrm{SCH}_{3}\), \(-\mathrm{SO}_{3}-\mathrm{CH}_{3} .\) Why?

Step-by-Step Solution

Answer

a. DMSO is the best solvent due to its polar aprotic nature. b. 1-Bromo-1-methylcyclohexane reacts fastest. c. Use compound with \(\mathrm{X}=-\mathrm{O}_{3}\mathrm{SCH}_{3}\).

Step 1: Choosing the Suitable Solvent

In an \(S_{N}2\) reaction, polar aprotic solvents are favored because they can stabilize the charged transition state without strongly interacting with the nucleophile. Among the given solvents, 2-propanone (acetone) and dimethyl sulfoxide \((\mathrm{DMSO})\) \((\left(\mathrm{CH}_{3}\right)_{2}\mathrm{S}=\mathrm{O})\), are polar aprotic solvents. DMSO is known for being highly effective in an \(S_{N}2\) reaction due to its ability to dissolve both polar and non-polar substances and not solvate the nucleophile strongly, making it the most suitable solvent for this reaction.

Step 2: Reactivity of Monobromoderivatives

In an \(S_{N}2\) mechanism, primary halides react faster than secondary halides due to less steric hindrance and easier access to the electrophilic carbon. The six isomeric monobromoderivatives of 1-methylcyclohexane consist of bromine substituents at different positions. The one with the bromine on the least hindered position, specifically the primary bromide (like bromomethylcyclohexane), will react the fastest with sodium cyanide. Thus, the 1-bromo-1-methylcyclohexane, being the primary position in the methyl group, is the most reactive.

Step 3: Choosing the Best Phenylmethyl Halide

For the synthesis of 2-phenylethanenitrile \(\left(\mathrm{C}_{6}\mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CN}\right)\), the phenylmethyl compound with the best leaving group should be selected. In an \(S_{N}2\) reaction, the leaving group ability follows the order: \(-\mathrm{O}_{3}\mathrm{SCH}_{3} > -\mathrm{OCOCH}_{3} > -\mathrm{F} > -\mathrm{OH} > -\mathrm{NH}_{2} > -\mathrm{H}\). Therefore, \(\mathrm{CH}_{3}\mathrm{SO}_{3} \) (mesylate) is the best leaving group, making \(\mathrm{R}=\mathrm{CH}_{3}\mathrm{SO}_{3}\) the best choice for conversion into the nitrile.

Key Concepts

SN2 Reaction Mechanism

The SN2 reaction mechanism is a type of nucleophilic substitution where the nucleophile attacks the electrophilic carbon atom from the opposite side of the leaving group. This process is called a "bimolecular nucleophilic substitution" because two molecules are involved in the rate-determining step: the substrate and the nucleophile. The term "bimolecular" indicates that the reaction rate depends on both the concentration of the nucleophile and the substrate.

• Direct Attack: The nucleophile performs a backside attack on the carbon atom bonded to the leaving group.
• Transition State: As the nucleophile approaches, a transition state is formed where the bond between the carbon and the incoming nucleophile and the bond between carbon and the leaving group coexist.
• Inversion of Configuration: After the nucleophile binds, the molecule's configuration is inverted, resembling an umbrella turned inside out.

This mechanism is most efficient with primary substrates because they have less steric hindrance, allowing the nucleophile to easily approach and attack the electrophilic carbon.

Polar Aprotic Solvents

Polar aprotic solvents play a crucial role in SN2 reactions by stabilizing the reaction environment without solvating the nucleophile. Aromatic or aliphatic solvents, including acetone and dimethyl sulfoxide (DMSO), are common examples.

• No Hydrogen Bonding: Polar aprotic solvents lack hydrogen atoms bound to highly electronegative atoms, such as oxygen or nitrogen, which prevents them from forming hydrogen bonds with the nucleophile. This absence allows nucleophiles to be more "free" and reactive.
• Stabilization of Transition State: These solvents stabilize the charged transition states, facilitating the reaction.

DMSO is particularly effective because it has both high polarity and the ability to dissolve various substances, enhancing the reactivity of nucleophiles in SN2 reactions.

Steric Hindrance in Reactivity

Steric hindrance is a key factor influencing the reactivity of substrates in SN2 reactions. It occurs when bulky groups around the electrophilic carbon atom impede the approach of the nucleophile.

• Effect of Size: Larger substituents near the reactive site on a molecule prevent the nucleophile from effectively colliding and reacting with the substrate. As a result, SN2 reactions are faster with less hindered, or primary, carbons.
• Primary vs. Secondary vs. Tertiary: Primary carbons, having fewer large groups, are more reactive in SN2 reactions, while tertiary carbons, hobbled by large, obstructive groups, are unreactive under SN2 conditions.

Understanding steric hindrance helps in predicting the reactivity order of different isomeric compounds during nucleophilic substitution processes.