Question

Each question in this section has four suggested answers of which ONE OR MORE answers will be correct. The feasible reaction (s) among the following is/are (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br}+\mathrm{CH}_{3} \mathrm{ONa} \rightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OCH}_{3}+\mathrm{NaBr}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OK}+\mathrm{CH}_{3} \mathrm{I} \rightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OCH}_{3}+\mathrm{KI}\) (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CONa}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br} \rightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COCH}_{2} \mathrm{CH}_{3}+\mathrm{NaBr}\) (d) \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{CBr}+\mathrm{OH}^{-} \stackrel{\mathrm{CH}_{3} \mathrm{OH}}{\longrightarrow}\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{COH}+\mathrm{Br}^{-}\) (major product)

Short answer

Based on the analysis of the given reactions, the feasible reactions are: (a) C₆H₅Br + CH₃ONa → C₆H₅OCH₃ + NaBr (b) C₆H₅OK + CH₃I → C₆H₅OCH₃ + KI The other reactions (c) and (d) are not highly feasible due to steric hindrance or are more likely to proceed via elimination reactions.

Step-by-step solution

Analyzing Reaction (a)

In this reaction, C\(_{6}\)H\(_{5}\)Br reacts with CH\(_{3}\)ONa to form C\(_{6}\)H\(_{5}\)OCH\(_{3}\) and NaBr. This is a nucleophilic substitution reaction (SN2) of a primary alkyl halide (C\(_{6}\)H\(_{5}\)Br) with a strong nucleophile (CH\(_{3}\)O\(^{-}\)). The reaction is feasible as the nucleophile will backside attack the carbon atom attached to the halide and displace the bromide ion.

Analyzing Reaction (b)

In this reaction, C\(_{6}\)H\(_{5}\)OK reacts with CH\(_{3}\)I to form C\(_{6}\)H\(_{5}\)OCH\(_{3}\) and KI. This is also a nucleophilic substitution (SN2) reaction between a primary alkyl halide (CH\(_{3}\)I) and a strong nucleophile (C\(_{6}\)H\(_{5}\)O\(^{-}\)). This reaction is also feasible as the nucleophile will backside attack the carbon atom attached to the halide and displace the iodide ion.

Analyzing Reaction (c)

In this reaction, (CH\(_{3}\))\(_{3}\)CONa reacts with CH\(_{3}\)CH\(_{2}\)Br to form (CH\(_{3}\))\(_{3}\)COCH\(_{2}\)CH\(_{3}\) and NaBr. This is a nucleophilic substitution reaction (SN2) between a secondary alkyl halide (CH\(_{3}\)CH\(_{2}\)Br) and a weak nucleophile ((CH\(_{3}\))\(_{3}\)CO\(^{-}\)). Due to steric hindrance by the bulky tert-butoxide anion ((CH\(_{3}\))\(_{3}\)CO\(^{-}\)), this reaction is not highly feasible, and it might undergo elimination reaction (E2) as a major pathway instead.

Analyzing Reaction (d)

In this reaction, (CH\(_{3}\)CH\(_{2}\))\(_{3}\)CBr reacts with OH\(^{-}\) in the presence of CH\(_{3}\)OH to form (CH\(_{3}\)CH\(_{2}\))\(_{3}\)COH and Br\(^{-}\). This is an attempted nucleophilic substitution reaction (SN2) between a tertiary alkyl halide and a strong nucleophile (OH\(^{-}\)). However, SN2 reactions with tertiary substrates are unfavorable due to steric hindrance. Instead, due to the basic conditions (presence of OH\(^{-}\)), this reaction is more likely to proceed via elimination (E2) to form an alkene as a major product. In conclusion, the feasible reactions among the given choices are: (a) C\(_{6}\)H\(_{5}\)Br + CH\(_{3}\)ONa → C\(_{6}\)H\(_{5}\)OCH\(_{3}\) + NaBr (b) C\(_{6}\)H\(_{5}\)OK + CH\(_{3}\)I → C\(_{6}\)H\(_{5}\)OCH\(_{3}\) + KI

Key concepts

SN2 Reaction Mechanism

The SN2 reaction mechanism is fundamental for understanding how certain molecules interact during a nucleophilic substitution. SN2 stands for 'Substitution Nucleophilic Bimolecular.' This tells us a few key things about how the reaction works. First, it is a substitution reaction, where one group is replaced by another.

The 'nucleophilic' part refers to the incoming group, which has an affinity for positive centers due to its electron-rich nature. It is attracted to the positive charge of the carbon atom that's bonded to a leaving group, typically a halide. Now, 'bimolecular' indicates that the speed of the reaction depends on the concentration of both the nucleophile and the substrate—the molecule with the leaving group.

In the SN2 mechanism, the nucleophile attacks from the opposite side of the leaving group, leading to what is known as backside attack. This is because the carbon atom is sp3 hybridized, with the group to be substituted located at one apex of a tetrahedral shape. The nucleophile pushes the leaving group out as it bonds, which results in an inversion of the molecule's configuration, often likened to an umbrella turning inside out in a strong wind.

Alkyl Halides Reactivity

When we talk about alkyl halides reactivity, especially in the context of nucleophilic substitution reactions like SN2, the type of alkyl halide matters significantly. Alkyl halides are categorized by the nature of the carbon to which the halide is attached. Those carbons can be primary, secondary, or tertiary.

Primary alkyl halides, where the carbon attached to the halide is connected to only one other carbon, are generally the most reactive in SN2 reactions. This is because there is less steric hindrance, allowing the nucleophile easier access to the carbon atom for the backside attack.

Secondary alkyl halides have a moderate reactivity since the carbon is connected to two other carbons, creating more hindrance. Tertiary alkyl halides are the least reactive in SN2 due to the carbon being connected to three other carbons, which create significant steric hindrance and physically block the nucleophile from attacking the carbon atom efficiently.

Steric Hindrance in Organic Reactions

Steric hindrance plays a crucial role in organic reactions, including both nucleophilic substitution and elimination reactions. It refers to the prevention of chemical reactions by sheer physical obstruction—essentially, the atoms or groups of atoms are so bulky that they get in each other's way.

This physical blocking can prevent the reactants from coming close enough for the reaction to occur, or, as in SN2 reactions, can interfere with the approach of the nucleophile to the substrate. Steric hindrance can not only slow down reactions but also alter the reaction pathway. For example, a reaction that would theoretically proceed via an SN2 mechanism might be diverted to an E2 elimination if the substrate is too hindered for the nucleophile to approach effectively.

This is precisely why in the original exercise, reaction (c) is less feasible and reaction (d) is more likely to undergo an elimination reaction. The bulky groups in both cases significantly decrease the likelihood of the SN2 progression.

Elimination Reactions

Elimination reactions are another pathway for reactants that may not successfully undergo nucleophilic substitution. In these reactions, elements are removed from a molecule, resulting in the formation of a double bond. The two common types of elimination reactions are E1 and E2.

E2 stands for 'Elimination Bimolecular.' This occurs in a single concerted step where the base (nucleophile) removes a hydrogen atom (proton) from a carbon adjacent to the carbon with the leaving group, while the leaving group leaves, resulting in the formation of the double bond (alkene). The E2 mechanism is favored when the substrate is hindered, as seen in the examples from the exercise where more bulky reactants were present.

In the case of the tertiary alkyl halide in reaction (d), the reaction proceeds via E2 because of steric hindrance near the reaction site. This reaction is also regioselective, meaning the more substituted alkene, which is often more stable, will be the major product.