Question

The order of rate constant for cleavage of \(\mathrm{X}_{3} \mathrm{C}-\mathrm{C}\) Bond during reaction between \(\mathrm{CX}\) CC(=O)c1ccccc1 and \(\stackrel{\ominus}{\mathrm{O}} \mathrm{H}\) is (where X is halogen) (A) \(\mathrm{F}>\mathrm{Cl}>\mathrm{Br}>\mathrm{I}\) (B) \(\mathrm{Cl}>\mathrm{F}>\mathrm{Br}>\mathrm{I}\) (C) \(\mathrm{I}>\mathrm{Br}>\mathrm{Cl}>\mathrm{F}\) (D) \(\mathrm{I}>\mathrm{Br}>\mathrm{F}>\mathrm{Cl}\)

Short answer

The correct order of rate constant for cleavage of the \(\mathrm{X}_{3}\mathrm{C}-\mathrm{C}\) bond is (C) \(\mathrm{I}>\mathrm{Br}>\mathrm{Cl}>\mathrm{F}\). This is due to the decreasing bond strength and electronegativity, and increasing leaving group ability as we move down the halogen group from fluorine to iodine.

Step-by-step solution

Identify the reaction type and the effect of halogen on the rate constant

In this exercise, we are considering the cleaving of the \(\mathrm{X}_{3}\mathrm{C}-\mathrm{C}\) bond during a nucleophilic substitution reaction between a halogen-containing compound (which has the structure \(\mathrm{CX}\)) and a hydroxide ion (\( OH^{-} \)). The rate of the reaction depends on the ease of breaking the \(\mathrm{X}_{3}\mathrm{C}-\mathrm{C}\) bond, and the subsequent formation of a negative ion. The easier it is to break the bond, the higher the rate constant for the reaction will be.

Examine the effect of the halogen size on bond strength

The strength of the \(\mathrm{X}_{3}\mathrm{C}-\mathrm{C}\) bond depends on the size of the halogen atom. The bond strength usually decreases with increasing atomic size because the larger the halogen atom, the weaker the bond between it and the neighboring carbon atom. As a halogen atom gets bigger, it is also less electronegative, making it more willing to give up electrons and thus more reactive.

Consider the effect of leaving group ability on the rate constant

The leaving group ability is an important factor influencing the rate constant for this reaction. A good leaving group is one that, once cleaved from the initial molecule, is stable as a separate negative ion. This stability typically correlates inversely with the basicity of the leaving group.

Determine the order of the rate constants based on halogen size and leaving group ability

From Step 2, we know that the bond strength decreases as we move from fluorine (\(\mathrm{F}\)) to iodine (\(\mathrm{I}\)). From Step 3, we know that less basic leaving groups, which lead to more stable negative ions, are the better leaving groups. Comparing these factors, we can determine that as we move down the halogen group, the bond strength and the electronegativity decrease. Hence, the leaving group ability increases, which in turn would increase the rate constant for the reaction. Therefore, the order of rate constant for cleavage of the \(\mathrm{X}_{3}\mathrm{C}-\mathrm{C}\) bond is: (C) \(\mathrm{I}>\mathrm{Br}>\mathrm{Cl}>\mathrm{F}\)

Key concepts

Halogen Bond Cleavage

In the context of a nucleophilic substitution reaction, halogen bond cleavage involves breaking the bond between a halogen atom and a carbon atom. The concept primarily focuses on how easily these bonds can break based on the size and electronegativity of the halogen involved. As you progress down the periodic table from fluorine to iodine, halogen atoms increase in size. This size increase results in a weaker bond with the carbon atom because larger atoms cannot bond as tightly with smaller carbon atoms.

• Fluorine has the strongest bond due to its small size and high electronegativity.
• Iodine has the weakest bond because of its large atomic size.

This means iodine bonds break more easily compared to other halogens, thus playing a critical role in increasing the rate of the reaction. Breaking these bonds is a key aspect because it allows the substitution reaction to take place efficiently.

Leaving Group Ability

The ability of a group to leave a molecule in a substitution reaction is known as its leaving group ability. A good leaving group is typically stable when it exists on its own as a separate ion. The halogen atoms' stability as ions determines their ability to be good leaving groups.
The general rule is that larger halogens make better leaving groups:

• Larger halogens like iodine form more stable ions, making them better leaving groups.
• Smaller halogens like fluorine are less effective as leaving groups since they are less stable as separate ions.

Overall, the leaving group's effectiveness increases from fluorine to iodine, making iodine the best leaving group among the halogens in these reactions.

Rate Constant Order

The rate constant order in a nucleophilic substitution reaction tells us how fast the reaction proceeds based on the halogen involved. The cleavage of a halogen bond and the ability of the group to leave both influence this rate. The rate constant increases as the ease of bond breakage increases and the leaving group becomes more competent.
For our exercise, this is demonstrated in the following order:

• Iodine (\(\mathrm{I}\)): Most efficient bond cleavage and leaving group.
• Bromine (\(\mathrm{Br}\)): Slightly less efficient than iodine.
• Chlorine (\(\mathrm{Cl}\)): Less effective than bromine and iodine.
• Fluorine (\(\mathrm{F}\)): Least effective due to strong bonds.

Hence, the order of rate constants is (C) \(\mathrm{I} > \mathrm{Br} > \mathrm{Cl} > \mathrm{F}\), which aligns with how bond strength and leaving group ability decrease with increasing atomic size.

Electronegativity and Reactivity

Electronegativity refers to an atom's ability to attract electrons. In the case of halogens, fluorine is the most electronegative, but that doesn't necessarily mean it's the most reactive in nucleophilic substitution reactions.
With decreasing electronegativity, from fluorine to iodine:

• Fluorine's strong electronegativity creates strong bonds, making it less reactive in these reactions.
• Iodine, with lower electronegativity, creates weaker bonds, leading to higher reactivity and an efficient cleavage process.

This trend showcases that while fluorine is extremely electronegative, it isn't the most reactive when it comes to these reactions, as its high electronegativity makes the bonds tougher to break. Hence, iodine is more reactive and better suited for these changes due to its balance between size and reactivity.