A chemical reaction is one in which old bonds are broken and new bonds are made. During the course of these changes a variety of intermediates are formed before a starting material is converted to final products. Formation of these intermediates depend on several factors like bond energies, kinetics of the reactions etc. In which of the following choices, the correct order is not indicated. (a) \(\mathrm{F}^{-}<\mathrm{Cl}^{-}<\mathrm{Br}^{-}<\mathrm{I}^{-}(\) the order of base strength \()\) (b) \(\mathrm{ClCH}_{2} \underset{\oplus}{\mathrm{C}}\left(\mathrm{CH}_{3}\right)_{2}<\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}<\left(\mathrm{CH}_{3}\right)_{2} \mathrm{COCH}_{3}\) (stability of carbocation) (c) \(\mathrm{O}_{2} \mathrm{~N}-\mathrm{CH}_{2}>\mathrm{CH}_{3}>\mathrm{CH}_{3} \mathrm{CH}_{2}\) (stability of carbanion) (d) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2}<\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCH}_{3}<\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{C}\left(\mathrm{CH}_{3}\right)_{2}\) (stability of alkenes)

Step by step solution

01

Analyze the order of base strength in option (a)

For option (a), we have the order of base strength as: \(\mathrm{F}^{-}<\mathrm{Cl}^{-}<\mathrm{Br}^{-}<\mathrm{I}^{-}\). As we move down the group in the periodic table, the size of the anion increases and its electron cloud becomes more diffused. This makes the larger anions better bases as they can accommodate the extra electron more easily. So, the order given in option (a) is correct.

02

Analyze the stability of carbocations in option (b)

For option (b), we have the order of carbocation stability as: \(\mathrm{ClCH}_{2} \underset{\oplus}{\mathrm{C}}\left(\mathrm{CH}_{3}\right)_{2}<\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}<\left(\mathrm{CH}_{3}\right)_{2} \mathrm{COCH}_{3}\). We have the following carbocations: Chloroalkyl carbocation, tertiary carbocation, and alkyl carbonyl carbocation. The stability of carbocations can be determined based on the hyperconjugation and inductive effects. Tertiary carbocations are more stable than secondary carbocations, which are more stable than primary carbocations. Chloroalkyl carbocations are usually less stable than alkyl carbocations because of the electron-withdrawing inductive effect of the chlorine atom. Alkyl carbonyl carbocation can be stabilized by the resonance effect with the oxygen atom. So, the order given in option (b) is correct.

03

Analyze the stability of carbanions in option (c)

For option (c), we have the order of carbanion stability as: \(\mathrm{O}_{2} \mathrm{~N}-\mathrm{CH}_{2}>\mathrm{CH}_{3}>\mathrm{CH}_{3} \mathrm{CH}_{2}\). The stability of carbanions is determined by the ability of the negative charge to be delocalized and distributed. Alkyl carbanions are not stabilized by resonance, so they are typically less stable. The first carbanion in the order has nitrogen atoms bearing oxygen atoms, which are highly electronegative atoms. These electronegative atoms can delocalize the negative charge and stabilize the carbanion. The order given in option (c) is incorrect since the alkyl carbanions should have lower stability compared to the one with electronegative atoms.

04

Analyze the stability of alkenes in option (d)

For option (d), we have the order of alkene stability as: \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2}<\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCH}_{3}<\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{C}\left(\mathrm{CH}_{3}\right)_{2}\). The stability of alkenes is determined by the number of substituents attached to the double bond. The more substituted the double bond, the more stable the alkene. This order is based on the hyperconjugative stabilization effect by the alkyl groups. The order given in option (d) is correct, as it showcases the increasing stability of alkenes with increasing substitution. The incorrect order is present in option (c).

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