MC test: Nucleophilic substitution

Multiple choice test on 20.1(1) Nucleophilic substitution

Use the following 'quiz' to test your knowledge and understanding of this sub-topic. You will need access to a periodic table (Section 6 of the IB data booklet).

If you get an answer wrong, read through the explanation carefully to learn from your mistakes.

What does a 'curly arrow' represent when used in organic reaction mechanisms?

Curly arrows are used to show the movement of pairs of electrons in organic reaction mechanisms. They start at the origin of a pair of electrons and the 'head of the arrow' shows where they end up.

 

Which alcohol can be formed directly from a halogenoalkane that has predominantly reacted with warm aqueous sodium hydroxide via the SN1 mechanism?

Tertiary halogenoalkanes react with hydroxide ions by the SN1 mechanism to form tertiary alcohols. The only tertiary alcohol in the list is 2-methylpropan-2-ol.

 

Which factors explain why tertiary halogenoalkanes tend to react via the SN1 mechanism rather than via the SN2 mechanism?

I. Steric hindrance - there is less room around the central carbon atom in tertiary halogenoalkanes to form a transition state involving five entities bonded to the carbon atom.

II. Homolytic fission - the C−halogen bond in tertiary halogenoalkanes undergoes homolytic fission more readily than in primary halogenoalkanes.

III. Positive inductive effect - the R− groups tend to 'push' electrons towards the central carbon atom which helps to stabilise the carbocation when it is formed.

The first step (the rate determining step) in the SN1 mechanism is the heterolytic breaking of the C−halogen bond.

 

Which is the main reason why iodooalkanes are more readily substituted by nucleophiles than chloroalkanes?

The C−I bond enthalpy (228 kJ mol−1) is considerably less than the C−Cl bond enthalpy (324 kJ mol−1) so the activation energy will be lower for iodoalkanes.

 

Which can be used as an aprotic polar solvent?

Protic solvents contain an hydrogen atom bonded directly to an oxygen or nitrogen atom. Aprotic solvents are polar but do not have an hydrogen atom bonded directly to an oxygen or nitrogen atom. Hexane is a non-polar solvent. 

 

Which row gives all the correct information concerning the reaction between 1-bromopropane and warm aqueous sodium hydroxide?

Row Mechanism Rate equation
1 SN1 rate = k[C3H7Br][OH−]
2 SN1 rate = k[C3H7Br]
3 SN2 rate = k[C3H7Br][OH−]
4 SN2 rate = k[C3H7Br]

It is a primary halogenoalkane so the mechanism is SN2 and the rate depends upon the concentration of both the halogenoalkane and the hydroxide ion as both are involved in the rate determining step.

 

Which is the transition state in the reaction between bromoethane and warm dilute sodium hydroxide solution?

The transition state has five entities around the central carbon atom and, as it is formed by the OH− beginning to bond with the carbon atom before the C−Br bond has broken, it will have a delocalized negative charge.

 

Which statements provides evidence that some secondary halogenoalkanes can undergo nucleophilic substitution by an SN2 mechanism?

I. The organic product rotates plane-polarized light in the opposite direction to the organic reactant.

II. The experimental rate equation shows that the reaction is first order with respect to hydroxide ions.

III. The reaction works best using a protic solvent rather than an aprotic solvent.

When the two R− groups are different, the secondary halogenoalkane will be optically active as it contains an asymmetric carbon atom. If  one of the enantiomers is reacted, the OH− 'attacks' the carbon atom on the opposite side to the C−Br bond so the product is inverted and will rotate plane-polarized light in the opposite direction. Protic solvents favour the SN1 mechanism not the SN2 mechanism.

 

Which statements are correct about  the rate determining step in the SN1 mechanism?

I. The molecularity of the reaction is 1.

II. It is affected by the nature of the nucleophile.

III. It involves the formation of a carbocation.

The rate determining step for SN1 is R3C−X → R3C+ + X− (where R is an alkyl group and X is a halogen). The rate equation is rate = k[R3C−X] and the step is unimolecular. Since the concentration of the nucleophile is not involved in the rate equation the nature and concentration of the nucleophile does not affect it.

 

Which factors explain why bromobenzene does not undergo nucleophilic substitution with warm dilute sodium hydroxide solution.

I. Too much energy is required to overcome the delocalization energy of the aromatic ring.

II. The nucleophile will be repelled by the electrons in the delocalized π bond above and below the benzene ring.

III. The delocalization can extend to two more p electrons from a non-bonding orbital on the bromine atom making the C−Br bond stronger and so harder to break.

Since the reaction involves substitution, not addition, the delocalization energy of the aromatic ring does not have to be overcome.

 

Total Score:
All materials on this website are for the exclusive use of teachers and students at subscribing schools for the period of their subscription. Any unauthorised copying or posting of materials on other websites is an infringement of our copyright and could result in your account being blocked and legal action being taken against you.