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Chapter 5: Problem 59

Observe the reaction titled "Bimolecular nucleophilic substitution." Notice that a bromide is being displaced from a carbon atom, in this case \(\mathrm{C}(2)\) of 2-bromopropane. We can imagine a closely related reaction in which displacement takes place in bromocyclopropane. All we have (mentally) done is to connect the two end carbons of 2-bromopropane with a carbon-carbon bond. Would the reaction with bromocyclopropane be faster or slower (it can't be the same!) as the reaction with 2-bromopropane? Hint: Think about the transition state for this reaction-what is the hybridization of carbon in the transition state?

Short Answer

Expert verified
The reaction with bromocyclopropane would be slower.

Step by step solution

01

Understand the Reactions

Identify the type of reaction we are discussing, which is Bimolecular Nucleophilic Substitution (SN2). The bromide (a leaving group) is displaced by a nucleophile in both 2-bromopropane and bromocyclopropane.
02

Examine Hybridization in Transition State

Consider the transition state of the carbon where the displacement is happening. In an SN2 reaction, the carbon is typically sp2 hybridized during the transition state.
03

Analyze Transition State in 2-bromopropane

For 2-bromopropane, during the transition state, the carbon changes hybridization from sp3 to sp2, which is relatively stable.
04

Analyze Transition State in Bromocyclopropane

In bromocyclopropane, the three-membered ring structure makes the carbons highly strained. During the transition state, the ring would become even more strained due to the bond angles required for sp2 hybridization.
05

Compare Reaction Rates

Since the transition state for bromocyclopropane is more strained and less stable, the activation energy required for this reaction is higher, making the reaction slower compared to 2-bromopropane.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

SN2 Reaction
An SN2 reaction, otherwise known as a Bimolecular Nucleophilic Substitution, involves a nucleophile displacing a leaving group on a carbon atom.
The '2' signifies that this reaction is bimolecular, meaning that two molecules are involved in the rate-determining step.
The nucleophile attacks the carbon from the opposite side of the leaving group.
This backside attack causes an inversion of the carbon's configuration, known as the Walden inversion.
Because of this, SN2 reactions are particularly sensitive to the steric hindrance around the carbon.
Less crowding around the carbon leads to a faster reaction.
Transition State
In an SN2 reaction, the transition state is the high-energy state during the process where old bonds are partially broken, and new bonds are partially formed.
The transition state for a carbon going through an SN2 reaction is planar and sp2 hybridized.
This state is less stable due to its partial bond formation and breakage, making it a key factor in determining the reaction rate.
Ease of the nucleophile's approach and the stability of this transition state can substantially impact the reaction speed.
The more stable the transition state, the faster the reaction will proceed.
Hybridization
Hybridization refers to the mixing of atomic orbitals in an atom to generate new hybrid orbitals that form covalent bonds.
In the context of an SN2 reaction, during the transition state, the central carbon is sp2 hybridized.
Initially, the carbon is sp3 hybridized with four sigma bonds.
When the nucleophile attacks, the carbon briefly adopts a planar geometry, signifying sp2 hybridization.
This intermediate state is critical to understanding the energy dynamics of the reaction.
Reaction Rate
The reaction rate of an SN2 reaction is directly influenced by several factors including the nature of the leaving group, the nucleophile, and steric hindrance around the carbon.
For the reactions discussed, comparing 2-bromopropane and bromocyclopropane, the overall reaction rate will differ due to strain in the transition state.
Bromocyclopropane has a highly strained transition state due to its three-membered ring, resulting in a slower reaction rate.
In contrast, 2-bromopropane's transition state is relatively stable, hence the reaction proceeds faster.
Strain Energy
Strain energy in molecules arises from deviations in bond angles or bond lengths from their ideal values.
Bromocyclopropane has significant ring strain due to its three-membered ring, which forces the carbon atoms into bond angles much smaller than the ideal tetrahedral angle of 109.5 degrees.
During an SN2 reaction, this strain increases as the carbon must adopt a planar transition state with approximately 120-degree bond angles.
This additional strain requires more energy, thus increasing the activation energy and slowing down the reaction rate for bromocyclopropane compared to 2-bromopropane.

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Most popular questions from this chapter

Draw the possible chair forms of cis- and trans-1-isopropyl-3-methylcyclohexane. Are the two forms identical, enantiomeric, or diastereomeric? In each case, indicate which chair form will be more stable and explain why. Is either of these molecules chiral?

Write structures for the following compounds: (a) 2,2 -dibromo- 3,3 -dichloro-5,5-difluoro6,6 -diiodobicyclo[2.2.1]heptane (b) bicyclo[1.1.1]pentan-1,3-diol (c) hexamethylbicyclo \([2.2 .0]\) hexa- 2,5 -diene

Draw a bicyclic molecule that has a three-carbon bridge. Use cis-decalin as your starting point.

Draw the planar representations of the three isomers of tetramethylcyclohexane that have all four of the methyl groups cis on the ring. Now draw each of the molecules in a chair conformation. Draw the ring-flipped chair for each of your molecules. Circle the more stable chair conformation for each of the three molecules.

Let's say you are cleaning out your low-temperature freezer and you discover a bottle with a label that has been partially eaten by low-temperature, label- eating bugs. Fortunately, enough of the label remains for you to see that the bottle contains either \((R, R)-1,2\)-dimethylcyclohexane or \((R, S)-1,2\)-dimethylcyclohexane. To resolve (pun intended) the identity of the substance, you remove the bottle from the freezer, and the next day you take an optical rotation measurement, thinking you can match it to the known value for one of the two possible structures. Much to your surprise you find that the optical rotation value is 0 . There is no optical rotation. Draw \((R, R)-1,2\)-dimethylcyclohexane and \((R, S)-1,2-\) dimethylcyclohexane. Which stereoisomer is in the bottle?
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