Question

($sec$-Butyl bromide) undergoes alkaline hydrolysis by:

• A. $S_{N}1$
• B. $S_{N}2$
• C. both (A) and (B)
• D. none

Answer 1

The correct option is A $S_{N}1$

Consider following factors while determining the type of mechanism (between $S{N}_1$ and $S_{N}2$):

• The $S_{N}2$ reaction is concerted. That is, the $S_{N}2$ occurs in one step, and both the nucleophile and substrate are involved in the rate determining step. Therefore the rate is dependent onboth the concentration of substrate and that of the nucleophile.
• The $S_{N}1$ reaction proceeds stepwise. The leaving group first leaves, whereupon a carbocation forms that is attacked by the nucleophile.

The Big Barrier – this is the most important thing to understand about each reaction. What’s the one key factor that can prevent this reaction from occurring?

• In the $S_{N}2$ reaction, the big barrier is steric hindrance. Since the $S_{N}2$ proceeds through a backside attack, the reaction will only proceed if the empty orbital is accessible. The more groups that are present around the vicinity of the leaving group, the slower the reaction will be. That’s why the rate of reaction proceeds from primary (fastest) > secondary >> tertiary (slowest)
• In the $S_{N}1$ reaction, the big barrier is carbocation stability. Since the first step of the $S_{N}1$ reaction is loss of a leaving group to give a carbocation, the rate of the reaction will be proportional to the stability of the carbocation. Carbocation stability increases with increasing substitution of the carbon (tertiary > secondary >> primary) as well as with resonance.

The dependence of rate upon the substrate

• For the $S_{N}2$, since steric hindrance increases as we go from primary to secondary to tertiary, the rate of reaction proceeds from primary (fastest) > secondary >> tertiary (slowest).
• For the $S_{N}1$, since carbocation stability increases as we go from primary to secondary to tertiary, the rate of reaction for the $S_{N}1$ goes from primary (slowest) << secondary < tertiary (fastest)

Remember that $S_{N}1$ and $S_{N}2$ reactions only occur for alkyl halides (and related compounds like tosylates and mesylates). If the leaving group is directly attached to an alkene or alkyne, $S_{N}1$ or $S_{N}2$ will not occur.

The Nucleophile

• The $S_{N}2$ tends to proceed with strong nucleophiles; by this, generally means negatively charged nucleophiles such as $(C{H}_{3}O{)}^{-},C{N}^{-},R{S}^{-},(N_3{)}^{-},(HO{)}^{-}$, and others.
• The $S_{N}1$ tends to proceed with weak nucleophiles – generally neutral compounds such as solvents like $C{H}_{3}OH,H_{2}O,C{H}_{3}C{H}_{2}OH$ and so on.

The Solvent

• The $S_{N}2$ reaction is favored by polar aprotic solvents – these are solvents such as acetone, DMSO, acetonitrile, or DMF that are polar enough to dissolve the substrate and nucleophile but do not participate in hydrogen bonding with the nucleophile.
• The $S_{N}1$ reaction tends to proceed in polar protic solvents such as water, alcohols, and carboxylic acids. These also tend to be the nucleophiles for these reactions as well.

Stereochemistry

• Since the $S_{N}2$ proceeds through a backside attack, if a stereocenter is present the $S_{N}2$ reaction will give inversion of stereochemistry.
• By contrast, if the $S_{N}1$ leads to the formation of a stereocenter, there will be a mixture ofretention and inversion since the nucleophile can attack from either face of the flat carbocation.

So does the story about the hobo on the bench make sense now?

• In the $S_{N}2$, the nucleophile (you) forms a bond to the substrate (bench) at the same time the leaving group (hobo) leaves.
• In the $S_{N}1$, the leaving group (hobo) leaves the substrate (bench), and then the nucleophile (you) forms a bond.