Video Transcript
Alcohols are prepared in the
laboratory by the hydrolysis of alkyl halides in a strong alkali like potassium
hydroxide, KOH. Which of the following alkyl
halides is the best choice for this reaction? (A) Alkyl bromide, (B) alkyl
iodide, (C) alkyl chloride, or (D) alkyl fluoride.
To help us visualize the
description provided in the question, let’s write out a generic chemical reaction
for the preparation of alcohols through the hydrolysis of alkyl halides. Alkyl halides, also known as
halogenoalkanes, are defined as compounds that contain a halogen atom covalently
bonded to an alkyl group. They are shown as RX in the
chemical reaction, where R represents an alkyl group and X represents the halogen
Cl, Br, or I. When primary, secondary, or
tertiary alkyl halides are heated in the presence of a strong base, like an aqueous
solution of either sodium hydroxide or potassium hydroxide, alcohols are produced
through what is known as a substitution reaction.
The definition of a substitution
reaction is a chemical reaction where part of a molecule is removed and is replaced
by something else. In this chemical reaction, the
carbon-halogen bond is broken so that the halogen is removed and replaced with the
OH group from the strong alkali used in the reaction. We will notice the question is
asking which alkyl halide is the best choice for this reaction. Another way to think of it is to
determine which alkyl halide facilitates the progress of this reaction the best or
which alkyl halide increases the rate of the reactants turning into products. Since the carbon-halogen bond
within the alkyl halide needs to be broken in order for the reaction to form the
desired alcohol, we will be focusing on the strength of the carbon-halogen bonds
within the alkyl halides mentioned in the answer choices.
We will notice the alkyl halide in
answer choice (D) lists fluorine as the halogen, indicated by fluoride in the name
provided. But fluorine has not been included
in the list of potential halogens for this reaction. This is because alkyl fluoride
molecules tend to react very slowly and ineffectively with hydroxide ions because
the carbon-fluorine bond is very strong and as a result is very difficult to break,
which means we can eliminate answer choice (D). Since the breaking of the
carbon-halogen bond is essential to the reaction proceeding, it makes sense that
alkyl halides tend to react with hydroxide ions more rapidly if the carbon-halogen
bond is relatively weak. Carbon-halogen bonds tend to be
weaker when the halogen atoms are larger and the carbon-halogen bonds are
longer.
To compare the size of the three
halogens in our remaining potential answer choices, we can use their placement on
the periodic table. Going from fluorine down to iodine,
atomic radius increases and therefore so does atomic size. As we can see, iodine is the
largest atom and will therefore demonstrate the weakest and longest carbon-halogen
bond of the alkyl halide mentioned in the answer choices. As a result, alkyl iodide molecules
react more quickly with hydroxide ions than alkyl bromide or alkyl chloride
molecules. So we can eliminate answer choices
(A) and (C), with the correct answer being answer choice (B).
Therefore, which of the following
alkyl halides is the best choice for this reaction? The answer is option (B) alkyl
iodide.