Question

Which compound of each of the following sets would you expect to give the higher yield of substitution product under conditions for bimolecular reaction? (a) ethyl bromide or \(\beta\) -phenylethy 1 bromide; (b) \(\alpha\) -phenylethy 1 bromide or \(\beta\) -phenylethyl bromide; (c) isobutyl bromide or \(\mathrm{n}\) -butyl bromide; (d) isobuty1 bromide or tert-buty1 bromide.

Short answer

(a) Ethyl bromide: lesser steric hindrance than β-phenylethyl bromide. (b) β-phenylethyl bromide: lesser steric hindrance than α-phenylethyl bromide. (c) n-butyl bromide: lesser steric hindrance than isobutyl bromide. (d) Isobutyl bromide: lesser steric hindrance than tert-butyl bromide.

Step-by-step solution

(a) Compare ethyl bromide and β-phenylethyl bromide

To determine which compound gives a higher yield of substitution product in an S_N2 reaction, we must consider the steric hindrance around the carbon atom attached to the bromine. Ethyl bromide has a smaller steric hindrance compared to β-phenylethyl bromide due to the phenyl group attached to the β-carbon. Therefore, ethyl bromide is expected to give a higher yield of substitution product in an S_N2 reaction.

(b) Compare α-phenylethyl bromide and β-phenylethyl bromide

In α-phenylethyl bromide, the phenyl group is directly attached to the carbon bearing the bromine atom, causing greater steric hindrance. On the other hand, in β-phenylethyl bromide, the phenyl group is attached to the neighboring (β) carbon atom, leading to lesser steric hindrance compared to α-phenylethyl bromide. Thus, we can expect that β-phenylethyl bromide will give a higher yield of substitution product under S_N2 conditions.

(c) Compare isobutyl bromide and n-butyl bromide

Both isobutyl bromide and n-butyl bromide contain four carbon atoms, but differ in their structural arrangement. The n-butyl bromide has a linear chain structure, while isobutyl bromide has a branched structure. Linear chains have lesser steric hindrance; therefore, n-butyl bromide would give a higher yield of substitution product under S_N2 conditions compared to isobutyl bromide.

(d) Compare isobutyl bromide and tert-butyl bromide

Isobutyl bromide and tert-butyl bromide differ in the degree of substitution at the carbon atom attached to the bromine. The carbon in isobutyl bromide is secondary (two alkyl groups attached), while the carbon in tert-butyl bromide is tertiary (three alkyl groups attached). Tertiary carbons have greater steric hindrance and are less likely to undergo S_N2 reactions. Therefore, we can expect that isobutyl bromide will give a higher yield of substitution product under S_N2 conditions compared to tert-butyl bromide.

Key concepts

Steric Hindrance

When discussing reactions, like the S_N2 mechanism, steric hindrance plays a crucial role.
S_N2 reactions are bimolecular and occur in a single step where the nucleophile attacks the electrophilic carbon, displacing the leaving group. Steric hindrance refers to how the size and orientation of groups around this carbon can block or hinder the approach of the nucleophile.

For example:

• Ethyl bromide has less steric hindrance compared to β-phenylethyl bromide because there are fewer bulky groups near the reactant site. Thus, it fosters a more efficient nucleophilic attack.
• Similarly, n-butyl bromide, with its linear structure, provides less hindrance than isobutyl bromide which has a branched configuration.

Hence, compounds with less steric hindrance typically yield more substitution products in S_N2 reactions.

Substitution Product

In an S_N2 reaction, the substitution product is the new compound formed when the nucleophile replaces the leaving group, usually a halide like bromide.
These reactions favor primary and some secondary carbon structures due to their lower steric hindrance. Thus, primary bromides tend to form the substitution product more readily.

• For instance, in comparing ethyl bromide and β-phenylethyl bromide: Ethyl bromide, with lesser interference, forms a substitution product more efficiently.
• In comparing α-phenylethyl bromide to β-phenylethyl bromide: The latter forms a substitution product more effectively, owning to reduced steric hindrance.

Understanding how the substitution product gets formed assists in predicting reaction outcomes.

Carbon Structure

The carbon structure of a compound determines its suitability for S_N2 reactions.
Primary carbons, which have only one alkyl group attached, are the most conducive for these reactions.
Here's why:

• N-butyl bromide has a primary carbon structure and therefore provides an easier path for nucleophiles to attack, allowing a swift S_N2 reaction.
• Isobutyl bromide contains a secondary carbon, which leads to moderate steric hindrance when compared to the linear, primary structure of n-butyl bromide.

On the other hand, tertiary carbons, such as in tert-butyl bromide, have three alkyl groups. This increased substitution restricts a nucleophile's approach and can hinder S_N2 reactions.

Bromide Compounds

Bromide compounds play a significant role in S_N2 reactions due to their good leaving group nature.
The bromide ion, upon being displaced, stabilizes and exits the reaction, paving the way for nucleophilic substitution.

• In comparisons like isobutyl bromide versus tert-butyl bromide, it's evident that despite bromide being present in both, the carbon structure affects its viability in S_N2 reactions.
• Bromide compounds with primary or secondary carbons, where there's less steric hindrance, generally result in higher yields of substitution products.

Therefore, while bromide is the common factor, the overall ability to undergo a substitution reaction in S_N2 is more dependent on factors like steric hindrance and carbon structure than the halogen itself.