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Question Video: Identifying the Ideal Halogenoalkane (Alkyl Halide) for Producing Alcohol Molecules through Alkaline Hydrolysis Reactions Chemistry

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 [D] Alkyl fluoride

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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.

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