Question 1

Question 2

Molecule A: The electronegative Cl atoms are electron-withdrawing and thus the carbonyl carbon atom is extremely electrophilic.  As such in water hydration is most likely to occur.

Molecule B: The carbonyl carbon atom is not extremely stable despite the electron-donating inductive effect because of steric-instability caused by ring strain on the cyclopropane ring.  As such water hydration is more likely to occur compared to molecule C.

Molecule C: The electron-donating inductive effect from the neighbouring groups makes the carbon of the carbonyl group less electrophilic and thus hydration is least likely to occur here.

Molecule D: The three carbonyl groups cause the carbon atoms of the carbonyl groups to be electrophilic due to the electron-withdrawing inductive effects of oxygen.  As such, water hydration is likely to occur in molecule D.

Question 3

Question 1

Question 2

Question 3

Question 4

Question 5

Question 1

Question 2

Question 3

Question 1

Question 1: Synthesis of the tranquilizer via electrophilic additon

Question 2

Question 2: Markovnikov and anti-Markovnikov reagents

Question 2bi – Syn addition of diol groups

Question 2bii: anti-addition of diol groups

Question 3

Question 4

Question 4a: From an activating amino group to a deactivating ammonium group.

As such, both groups direct the substituent towards the ortho positions on the benzene ring with respect to the methyl group to form the below product:

Question 4b: Activation of the benzene ring via resonance.

The resonance of the lone pair from the oxygen strongly activates the benzene ring and thus the methyl group is ortho and para directing with respect to the C-O ether bond to form the product below:

Question 1

a) This step undergoes an Sn2 nucleophilic substitution.  In a medium of NaOH(aq), 2-ethoxyphenol is a good nucleophile but a weak base.  This reaction causes an inversion of configuration.  The process is as seen in the pictures below:

Figure 1ai: The substitution/addition of 2-ethoxyphenol via an Sn2 mechanism.

Figure 1aii: Regeneration of OH- base

b) The process undergoes an Sn1 nucleophilic substitution.  This is because the -OH group is a poor nucleophile as the ROH group is a stronger base than even NaOH.  As such, Sn2 nucleophilic substitution does not occur.  The Sn1 nucleophilic substitution is favoured as -OMs is a very good leaving group attached to a primary carbon.

Though normally Sn1 nucleophilic additions are racemic in nature, the stereochemistry of the -OH group that is added to the carbocation in the second step ensures that only one stereoisomeric product is formed.  The process is as seen in the pictures below:

Figure 1bi: The formation of the carbocation

Figure 1bii: The addition of the nucleophile and the subsequent formation of water by reaction with the OH- base.

c) Under neutral or acidic conditions, it is possible to catalyze the nucleophilic substitution of an epoxide group using a weak base such as ammonia or R-NH2.  This is done via the Sn1 nucleophilic substitution mechanism and the formation of a carbocation.  Due to the formation of a carbocation, in this case the products formed will be enantiomers of each other and roughly equal in proportion to form a racemic mixture.  The mechanism is as shown below:

Figure 1ci: Formation of the carbocation in neutral/acid solutions via the presence of acid catalysts

Figure 1cIi: Sn1 nucleophilic substitution mechanism and the regeneration of the acid catalyst

d) Since the substrate carbon is a primary carbon, the reaction must be an Sn2 nucleophilic substitution.  However, since -OH is a poor nucleophile/weak base, we must first use sodium to convert it to an alkoxide ion, which is a good nucleophile.  Following which, the reaction undergoes an Sn2 nucleophilic reaction.  This reaction causes an inversion of configuration.  The process would be as in the picture below:

Figure 1di: The formation of the alkoxide ion

Figure 1dii: The Sn2 nucleophilic substitution process.

Question 2

Under basic conditions, Hydrogen Sulphide occurs in the form of the hydrosulfide ion.  As such, this hydrosulfide ion is a good nucleophile but a weak base.  Thus the mechanism for substitution here will be Sn2 nucleophilic substitution.  As the R-SH group is still considered to be a weak acid, it reacts with the strong OH- base to give rise to a negatively charged ion that is a strong nucleophile, which will again cause an Sn2 nucleophilic substitution.

There is only one possible product formed as it has a line of internal symmetry and thus the mirror images are in fact identical to each other.  The reaction is as shown below:

Figure 2a: The first Sn2 nucleophilic substitution reaction

Figure 2b: An intramolecular Sn2 substitution caused by the strong nucleophile generated from the presence of an OH- base

Question 3

NaOEt is a good nucleophile and a strong base.  Since it is non-bulky, it greatly favours the E2 mechanism as the -OAc leaving group is bonded to a secondary carbon atom.

By converting the structure of the initial product of the following reaction, we have the following Newman projection.

Figure 3a: The Newman projection of the initial reactant

As such, it can be seen that it is possible for this molecule to under an elimination reaction.  The following is the major product of the following reaction, along with the mechanism of the reaction.

Figure 3b: The E2 mechanism of the reaction, and the final major product formed.