Chapter 11, Problem 11.35P

Interpretation Introduction

(a)

Interpretation:

The ${\text{S}}_{\text{N}}\text{2}$ mechanism using curved arrow notation in which ${\text{OH}}^{-}$ acts as a nucleophile is to be stated.

Concept introduction:

The curved-arrow notation is used to show the transfer of electrons from one atom to another. The curved arrow has two barbs (head and tail) which represent the direction of electron flow.

The ${\text{S}}_{\text{N}}\text{2}$ reaction is a nucleophilic substitution reaction in which the addition of nucleophile and removal of leaving group takes place simultaneously.

Answer to Problem 11.35P

The ${\text{S}}_{\text{N}}\text{2}$ reaction of $\text{sodium-2-bromopropanoate}$ with ${\text{OH}}^{-}$ using curved arrow notation is shown below.

Explanation of Solution

The given reaction is shown below.

Figure 1

The ${\text{S}}_{\text{N}}\text{2}$ reaction of $\text{sodium-2-bromopropanoate}$ with ${\text{OH}}^{-}$ using curved arrow notation is shown below.

Figure 2

Figure 2 shows that in the first step ${\text{OH}}^{-}$ acts as a nucleophile which attacks at the $s{p}^3$ hybridized carbon atom. The $\text{C}-\text{OH}$ bond forms and $\text{C}-\text{Br}$ bond breaks simultaneously. This leads to the removal of the leaving group $\left(\text{Br}\right)$. The product formed using curved arrow notation is shown in Figure 2.

Conclusion

The ${\text{S}}_{\text{N}}\text{2}$ reaction of $\text{sodium-2-bromopropanoate}$ with ${\text{OH}}^{-}$ using curved arrow notation is shown in Figure 2.

Interpretation Introduction

(b)

Interpretation:

The curved arrow notation for an intramolecular mechanism is to be stated.

Concept introduction:

The curved-arrow notation is used to show the transfer of electrons from one atom to another. The curved arrow has two barbs (head and tail) which represent the direction of electron flow.

The ${\text{S}}_{\text{N}}\text{2}$ reaction is a nucleophilic substitution reaction in which the addition of nucleophile and removal of leaving group takes place simultaneously.

Answer to Problem 11.35P

The curved arrow notation for the intramolecular mechanism is shown below.

Explanation of Solution

The curved arrow notation for the intramolecular mechanism is shown below.

Figure 3

Figure 3 shows the intramolecular displacement of the bromide ion by an oxygen atom. Due to which a three-membered ring is formed as an unstable intermediate. As a result, the ring-opening of three-membered takes place due to the attack of the nucleophile $\left({\text{OH}}^{-}\right)$. The product formed is shown in Figure 3.

Conclusion

The curved arrow notation for the intramolecular mechanism is shown in Figure 3.

Interpretation Introduction

(c)

Interpretation:

The proximity effect for the intramolecular reaction is to be stated.

Concept introduction:

The ${\text{S}}_{\text{N}}\text{2}$ reaction is a nucleophilic substitution reaction in which the addition of nucleophile and removal of leaving group takes place simultaneously. These reactions depend upon the nucleophilicity and concentration of the nucleophile.

Answer to Problem 11.35P

The proximity effect for the intramolecular reaction is $0.1875M$.

Explanation of Solution

It is given that the first-order rate constant $\left(k_1\right)$ is $1.2\times {10}^{-4}{\text{s}}^{-1}$ and second-order rate constant $\left(k_2\right)$ is $6.4\times {10}^{-4}{M}^{-1}{\text{s}}^{-1}$.

The formula to determine proximity is shown below.

$\text{Proximity}\text{effect}=\frac{k_1}{k_2}$ ...(1)

Where,

• $k_1$ is the first-order rate constant for intramolecular reaction.
• $k_2$ is the second-order rate constant for intermolecular reaction.

Substitute the value of $k_1$ as $1.2\times {10}^{-4}{\text{s}}^{-1}$ and the value of $k_2$ as $6.4\times {10}^{-4}{M}^{-1}{\text{s}}^{-1}$ in equation (1).

$\begin{array}{rll}\text{Proximity}\text{effect}& =& \frac{k_1}{k_2}\\ & =& \frac{1.2\times {10}^{-4}{\text{s}}^{-1}}{6.4\times {10}^{-4}{M}^{-1}{\text{s}}^{-1}}\\ & =& 0.1875M\end{array}$

Therefore, the proximity for the reaction is $0.1875M$.

Conclusion

The proximity effect for the reaction is $0.1875M$.

Interpretation Introduction

(d)

Interpretation:

The concentration of $\text{NaOH}$ at which reaction proceed by two mechanisms at the same rate is to be stated.

Concept introduction:

The ratio of rate constants of intramolecular reaction over intermolecular reaction is known as proximity effect. It is also known as effective molarity. The unit used to represent the proximity effect is $M$.

Answer to Problem 11.35P

The concentration of $\text{NaOH}$ at which reaction proceed by two mechanisms at the same rate is $0.1875M$.

Explanation of Solution

The proximity effect is also termed as effective molarity. The value of the proximity effect for the reaction is $0.1875M$. Therefore, effective molarity is $0.1875M$. This shows that the concentration of $\text{NaOH}$ at which reaction proceeds by two mechanisms at the same rate is $0.1875M$.

Conclusion

The concentration of $\text{NaOH}$ at which reaction proceed by two mechanisms at the same rate is found to be $0.1875M$.

Interpretation Introduction

(e)

Interpretation:

The predominant mechanism in $\text{1}M\text{NaOH}$ is to be stated.

Concept introduction:

The replacement or substitution of one functional group with another different functional group in any chemical reaction is termed as substitution reaction. In nucleophilic substitution reaction, nucleophile takes the position of leaving group by attacking on the electron deficient carbon atom.

Answer to Problem 11.35P

The nucleophilic displacement reaction, ${\text{S}}_{\text{N}}\text{2}$ is predominant in $\text{1}M\text{NaOH}$.

Explanation of Solution

At $\text{1}M\text{NaOH}$, the nucleophilic displacement reaction, ${\text{S}}_{\text{N}}\text{2}$ is predominant. The ${\text{S}}_{\text{N}}\text{2}$ reaction is a nucleophilic substitution reaction in which the addition of nucleophile and removal of leaving group takes place simultaneously.

Therefore, as the nucleophilic addition increases, the rate of reaction also increases.

Conclusion

The nucleophilic displacement reaction, ${\text{S}}_{\text{N}}\text{2}$ is predominant in .

Interpretation Introduction

(f)

Interpretation:

An explanation for the unstable intermediate derived in part (b) than the other epoxides is to be stated.

Concept introduction:

The cyclic ether containing a ring of atoms, carbon, oxygen and carbon is known as an epoxide. The strained cyclic ring is found to be highly reactive in nature. In presence of strong base, it opens from the less hindered site. In presence of strong acid, it opens from more hindered site.

Answer to Problem 11.35P

The intermediate formed in part (b) is unstable due to steric hindrance.

Explanation of Solution

The intermediate formed in part (b) is shown below.

Figure 4

The intermediate formed in part (b) is a three-membered ring. Due to steric hindrance, the three-membered rings are unstable in nature. The structure shown in Figure 4 is three-membered ring and it also contains a double bond. As a result, this ring is highly unstable as compared to other epoxides.

Therefore, ring opening takes place by the attack of incoming nucleophile.

Conclusion

The intermediate formed in part (b) is unstable due to a double bond in three-membered ring.