Answers

Chapter 8: Nucleophilic Substitution

Qu 1 The first step has to be identifying the reaction involved. This is a nucleophilic substitution of the methyl iodide. This conclusion could by reached by either reflecting on the chemistry of alkyl halides or how bad -NH₂ would be as a leaving group or that the N atom has lone pairs that make it a potential nucleophile.
In both cases the reaction leads to N-methylamines via an S_N² reaction. Methyl systems do not undergo S_N¹ reactions (the carbocation is too high energy).
Knowing what the reaction is, we can then determine that the amine species are the nucleophiles in those reactions. For aniline, resonance of the N lone pair with the adjacent p system decreases the availability of the lone pair and so makes aniline a poorer nucleophile thus resulting in the lower reactivity.

Qu 2
The relative reactivity of the following tertiary alkyl bromides towards hydrolysis in aqueous ethanol solution (favours S_N¹) is due to the requirement for planarity in the carbocation. The 2 ring systems on the right cannot become planar due to steric effects and the cations are not as stable. Hence, they react more slowly. They are prohibited from S_N² reaction by the ring system preventing the "backside" attack.

Qu 3 CH₃CH₂Li > CH₃CH₂S^-> CH₃CH₂O^- > CF₃CH₂O^-> CH₃CO₂^- > (CH3CH2)2O

Relative nucleophilicity depends on electron availablitity. Factors to consider are the charge and the identify of the atom carrying the charge. Remember an organolithium is essentially a carbanion and is very reactive.
S is more nucleophilic than O because of its polarizability. The electronegativity of F causes inductive electron withdrawal, lowering the electron density on O, but not to the same degree as the resonance stabilization in the carboxylate. The neutral O of the ether is last.

Qu 4 We need to use the idea that a good leaving group is the conjugate base of a strong acid, since it infers that the displaced group will be stable:

The groups to be displaced in bold. So the first step is to recognize the acids that we need to consider (in order) : H₃O⁺, H₂O, CH₃CO₂H, HBr and MeOH.
How would we rationalize the pKa's of these or rank their acidity ?
HBr is a very strong acid since the bromine is very capable of accommodating negative charge. Hydronium is also a strong acid with a neutral conjugate base. Carboxylic acids are weak acids, but the resonance stabilization of the carboxylate makes them a stronger acid than an alcohol. Alcohols and water have very similar acidity.

R-Br (HBr = -9) > R-O⁺H₂ (H₃O⁺= -1.7) > RO₂CCH₃ (CH₃CO₂H = 4.8) > ROMe (MeOH = 15.5) = ROH (H₂O = 15.7)

Qu 5

This is just a simple nucleophilic substitution reaction, with a couple of subtleties. First since the Nu = LG, the starting material and the products have equal chemical reactivity towards the iodide, thus, regardless of it being SN1 or SN2 the reaction mixture will always eventually become racemic (50% R, 50%S). since the (S)-product is the enantiomer of the (R)-starting material.
The rate of iodine incorporation tells us how fast the iodide becomes part of the product, the rate of racemization tells us how rapidly the 50:50 mixture is formed......these two things are not quite the same.
If the reaction is S_N¹, then the planar carbocation intermediate will mean that both (R)- and (S)-products will be formed in equal amounts. This means that when two molecules of the (R)-starting material react, one will give (R)-product and the other (statistically) will give (S)-product. The rotations of these will cancel each other out. So this would mean that the rate of iodide incorporation would equal the rate of racemization.
If the reaction is S_N², then inversion occurs, so only the (S)-product will be formed. This means that for every molecule of (R)-starting material that reacts, one molecule of (S)-product is produced. This (S)-enantiomer will also cancel out the rotation of another (R)-starting material. This means that the rate of iodide incorporation would equal the half rate of racemization since each iodide converts an (R) to an (S) that cancels out the rotation of a second (R) molecule.
Therefore the reaction must be occurring via the S_N² pathway.

Qu 6

(substitution with retention of stereochemistry)

Even though we have not yet seen all ot these reactions, given that you have the structure of the reactants and the products, you should be able to draw a mechanism to show the formation of the product based on what you know from earlier chapters.
If this reaction were S_N¹ then we would have loss of stereochemistry due to the carbocation intermediate.
If the reaction were simple S_N², then there should have been inversion....one way to get retention is to have two inversions.... So if the HO^- is not a nucleophile, then it reacts as a base, generating the alkoxide, RO^-, another potential nucleophile. This reacts S_N² fashion intramolecularly to displace the Cl^- with inversion, giving an epoxide.

This epoxide is then attacked by more HO^- in a second S_N² reaction at the less hindered position to give the product with overall retention of configuration.

If this reaction surprises you, since our leaving group is an RO^- (remember that HO^- is a poor leaving group), the extra "fact" helping the process is the relief of ring strain as the 3-membered ring of the epoxide opens.

Qu 7 The question is based on the radical halogenation of a hydrocarbon H followed by an analysis of the S_N¹ or S_N² reactivity of the chloroalkanes. The possible isomeric structures for H are:

The fact that 4 monochlorinated products were obtained tells us that the original hydrocarbon must have contained 4 types of aliphatic H.
Note that an sp² C-H bond is not usually halogenated because it is a stronger bond and the radical formed is not very stable.
Anyway, you should also recall that vinyl halides (i.e.. sp² C-Cl) do NOT undergo S_N¹ or S_N² reactions.
Therefore, the only isomer of H that would fit is methylcyclopropane, which leads to the chloroalkanes shown:

If you check % yields you will find that A = B > C > D
The AgNO₃/ EtOH / H₂O reaction is S_N¹ and tells us the relative stability of the carbocations, this helps decide which structure is which. C will be fastest since it gives a tertiary carbocation, A and B give the same secondary carbocation and D will the slowest as it is only primary.
The NaI / acetone reaction is S_N² and tells us about the steric hindrance at the Cl center. D is fastest as it is primary, C is the slowest as it is tertiary.
A is faster than B because the methyl group in B will hinder the approach of the nucleophile at 180^o to the leaving group during the inversion.
A = cis-1-chloro-2-methylcyclopropane, or (1S,2R)-1-chloro-2-methylcyclopropane (as drawn)
B = trans-1-chloro-2-methylcyclopropane, or (1R,2R)-1-chloro-2-methylcyclopropane (as drawn)
C = 1-chloro-1-methylcyclopropane
D = (chloromethyl)cyclopropane
If bromination were used, the major product would be 1-bromo-1-methylcyclopropane because radical bromination reactions are much more selective and can be predicted based on the most stable radical which would the tertiary radical.
The calculated % yield = 83 % = (1x1640)/[(1x1640)+(4x82)+(3x1)]

Qu 8

The mechanism is a substitution reaction of an alcohol with HCl to give a rearranged chloroalkane (that is the new substituent is not attached to the same carbon skeleton as the starting material). The fact that the carbon skeleton rearranges indicates that a carbocation was formed, and therefore, that the process is S_N¹ rather than S_N².
The key parts are protonating the -OH to make a good leaving group (-OH will not act as a leaving group to give hydroxide), loss of the leaving group (H₂O) to give the secondary carbocation, rapid rearrangement to the more stable tertiary carbocation and attack by the chloride nucleophile. An S_N² reaction would have given the un-rearranged product, 3-chloro-2,2-dimethylbutane.

Qu 1	The first step has to be identifying the reaction involved. This is a nucleophilic substitution of the methyl iodide. This conclusion could by reached by either reflecting on the chemistry of alkyl halides or how bad -NH₂ would be as a leaving group or that the N atom has lone pairs that make it a potential nucleophile. In both cases the reaction leads to N-methylamines via an S_N² reaction. Methyl systems do not undergo S_N¹ reactions (the carbocation is too high energy). Knowing what the reaction is, we can then determine that the amine species are the nucleophiles in those reactions. For aniline, resonance of the N lone pair with the adjacent p system decreases the availability of the lone pair and so makes aniline a poorer nucleophile thus resulting in the lower reactivity.


Qu 2	The relative reactivity of the following tertiary alkyl bromides towards hydrolysis in aqueous ethanol solution (favours S_N¹) is due to the requirement for planarity in the carbocation. The 2 ring systems on the right cannot become planar due to steric effects and the cations are not as stable. Hence, they react more slowly. They are prohibited from S_N² reaction by the ring system preventing the "backside" attack.


Qu 3	CH₃CH₂Li > CH₃CH₂S^-> CH₃CH₂O^- > CF₃CH₂O^-> CH₃CO₂^- > (CH3CH2)2O
	Relative nucleophilicity depends on electron availablitity. Factors to consider are the charge and the identify of the atom carrying the charge. Remember an organolithium is essentially a carbanion and is very reactive. S is more nucleophilic than O because of its polarizability. The electronegativity of F causes inductive electron withdrawal, lowering the electron density on O, but not to the same degree as the resonance stabilization in the carboxylate. The neutral O of the ether is last.

Qu 4	We need to use the idea that a good leaving group is the conjugate base of a strong acid, since it infers that the displaced group will be stable:
	The groups to be displaced in bold. So the first step is to recognize the acids that we need to consider (in order) : H₃O⁺, H₂O, CH₃CO₂H, HBr and MeOH. How would we rationalize the pKa's of these or rank their acidity ? HBr is a very strong acid since the bromine is very capable of accommodating negative charge. Hydronium is also a strong acid with a neutral conjugate base. Carboxylic acids are weak acids, but the resonance stabilization of the carboxylate makes them a stronger acid than an alcohol. Alcohols and water have very similar acidity. R-Br (HBr = -9) > R-O⁺H₂ (H₃O⁺= -1.7) > RO₂CCH₃ (CH₃CO₂H = 4.8) > ROMe (MeOH = 15.5) = ROH (H₂O = 15.7)

Qu 5	This is just a simple nucleophilic substitution reaction, with a couple of subtleties. First since the Nu = LG, the starting material and the products have equal chemical reactivity towards the iodide, thus, regardless of it being SN1 or SN2 the reaction mixture will always eventually become racemic (50% R, 50%S). since the (S)-product is the enantiomer of the (R)-starting material. The rate of iodine incorporation tells us how fast the iodide becomes part of the product, the rate of racemization tells us how rapidly the 50:50 mixture is formed......these two things are not quite the same. If the reaction is S_N¹, then the planar carbocation intermediate will mean that both (R)- and (S)-products will be formed in equal amounts. This means that when two molecules of the (R)-starting material react, one will give (R)-product and the other (statistically) will give (S)-product. The rotations of these will cancel each other out. So this would mean that the rate of iodide incorporation would equal the rate of racemization. If the reaction is S_N², then inversion occurs, so only the (S)-product will be formed. This means that for every molecule of (R)-starting material that reacts, one molecule of (S)-product is produced. This (S)-enantiomer will also cancel out the rotation of another (R)-starting material. This means that the rate of iodide incorporation would equal the half rate of racemization since each iodide converts an (R) to an (S) that cancels out the rotation of a second (R) molecule. Therefore the reaction must be occurring via the S_N² pathway.


Qu 6
	(substitution with retention of stereochemistry) Even though we have not yet seen all ot these reactions, given that you have the structure of the reactants and the products, you should be able to draw a mechanism to show the formation of the product based on what you know from earlier chapters. If this reaction were S_N¹ then we would have loss of stereochemistry due to the carbocation intermediate. If the reaction were simple S_N², then there should have been inversion....one way to get retention is to have two inversions.... So if the HO^- is not a nucleophile, then it reacts as a base, generating the alkoxide, RO^-, another potential nucleophile. This reacts S_N² fashion intramolecularly to displace the Cl^- with inversion, giving an epoxide. This epoxide is then attacked by more HO^- in a second S_N² reaction at the less hindered position to give the product with overall retention of configuration. If this reaction surprises you, since our leaving group is an RO^- (remember that HO^- is a poor leaving group), the extra "fact" helping the process is the relief of ring strain as the 3-membered ring of the epoxide opens.

Qu 7	The question is based on the radical halogenation of a hydrocarbon H followed by an analysis of the S_N¹ or S_N² reactivity of the chloroalkanes. The possible isomeric structures for H are:
	The fact that 4 monochlorinated products were obtained tells us that the original hydrocarbon must have contained 4 types of aliphatic H. Note that an sp² C-H bond is not usually halogenated because it is a stronger bond and the radical formed is not very stable. Anyway, you should also recall that vinyl halides (i.e.. sp² C-Cl) do NOT undergo S_N¹ or S_N² reactions. Therefore, the only isomer of H that would fit is methylcyclopropane, which leads to the chloroalkanes shown: If you check % yields you will find that A = B > C > D The AgNO₃/ EtOH / H₂O reaction is S_N¹ and tells us the relative stability of the carbocations, this helps decide which structure is which. C will be fastest since it gives a tertiary carbocation, A and B give the same secondary carbocation and D will the slowest as it is only primary. The NaI / acetone reaction is S_N² and tells us about the steric hindrance at the Cl center. D is fastest as it is primary, C is the slowest as it is tertiary. A is faster than B because the methyl group in B will hinder the approach of the nucleophile at 180^o to the leaving group during the inversion. A = cis-1-chloro-2-methylcyclopropane, or (1S,2R)-1-chloro-2-methylcyclopropane (as drawn) B = trans-1-chloro-2-methylcyclopropane, or (1R,2R)-1-chloro-2-methylcyclopropane (as drawn) C = 1-chloro-1-methylcyclopropane D = (chloromethyl)cyclopropane If bromination were used, the major product would be 1-bromo-1-methylcyclopropane because radical bromination reactions are much more selective and can be predicted based on the most stable radical which would the tertiary radical. The calculated % yield = 83 % = (1x1640)/[(1x1640)+(4x82)+(3x1)]

Qu 8
	The mechanism is a substitution reaction of an alcohol with HCl to give a rearranged chloroalkane (that is the new substituent is not attached to the same carbon skeleton as the starting material). The fact that the carbon skeleton rearranges indicates that a carbocation was formed, and therefore, that the process is S_N¹ rather than S_N². The key parts are protonating the -OH to make a good leaving group (-OH will not act as a leaving group to give hydroxide), loss of the leaving group (H₂O) to give the secondary carbocation, rapid rearrangement to the more stable tertiary carbocation and attack by the chloride nucleophile. An S_N² reaction would have given the un-rearranged product, 3-chloro-2,2-dimethylbutane.