Question

Predict the major organic product obtained in each of the following reactions. Assume that [O] represents \(\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) in \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (a) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}_{2} \mathrm{OH}+\mathrm{HBr} \longrightarrow\) (b) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}+\mathrm{K} \longrightarrow\) (c) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHOH} \stackrel{[\mathrm{O}]}{\longrightarrow}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{I} \longrightarrow\)

Short answer

The major organic products for these reactions are \((\mathrm{CH}_{3})_{2}CHCH_2Br\), \((\mathrm{CH}_{3})_{3}C + KOH\), \((\mathrm{CH}_{3})_{2}CO\) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2}I + \mathrm{CH}_{3} \mathrm{CH}_{2}OH\).

Step-by-step solution

Prediction of product for reaction (a)

The alcohol \((\mathrm{CH}_{3})_{2}CHCH_2OH\) reacts with \(\mathrm{HBr}\) to form an alkyl bromide. The bromine ion in HBr acts as a nucleophile which attacks the carbon atom of the alcohol group. Therefore, the product will be \((\mathrm{CH}_{3})_{2}CHCH_2Br\).

Prediction of product for reaction (b)

The alcohol \((\mathrm{CH}_{3})_{3}COH\) reacts with \(\mathrm{K}\) to undergo deprotonation due to the basic nature of K. As a result, the hydroxide group (OH) is removed from the molecule, therefore, the product will be \((\mathrm{CH}_{3})_{3}C + KOH\).

Prediction of product for reaction (c)

The alcohol \((\mathrm{CH}_{3})_{2}CHOH\) reacts with the oxidizing agent \([\mathrm{O}]\) (\(\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) in \(\mathrm{H}_{2} \mathrm{SO}_{4}\)). The secondary alcohol is oxidized to a ketone. Therefore, the product will be \((\mathrm{CH}_{3})_{2}CO\).

Prediction of product for reaction (d)

The alcohol \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2}OH\) reacts with the compound \(\mathrm{CH}_{3} \mathrm{CH}_{2}I\). In this case, the iodine ion acts as a nucleophile which attacks the carbon atom of the alcohol group, therefore the product will be \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2}I + \mathrm{CH}_{3} \mathrm{CH}_{2}OH\).

Key concepts

Alkyl Halide Formation

The process of alkyl halide formation involves the replacement of a hydroxyl group (–OH) in an alcohol molecule with a halogen atom (such as chlorine, bromine, or iodine), resulting in the formation of an alkyl halide. This transformation typically occurs through the reaction of an alcohol with a hydrogen halide (like HCl, HBr, or HI).

During the reaction, the halide ion acts as a nucleophile, which means it donates an electron pair to bond with the electrophilic carbon atom bearing the hydroxyl group. For instance, in step 1 of the provided solution, the alcohol (CH_{3})_{2}CHCH_2OH reacts with HBr, leading to the formation of (CH_{3})_{2}CHCH_2Br, an alkyl bromide.

Oxidation of Alcohols

The oxidation of alcohols is a common reaction in organic chemistry where an alcohol is transformed into a ketone, aldehyde, or carboxylic acid by increasing its oxidation state. This reaction typically requires the presence of an oxidizing agent, such as Na_2Cr_2O_7 in H_2SO_4, which is often represented by the symbol [O] in chemical equations.

Primary alcohols are oxidized to aldehydes and further to carboxylic acids, while secondary alcohols are oxidized to ketones. Tertiary alcohols do not normally undergo oxidation due to the lack of a hydrogen atom attached to their alcohol carbon. For example, step 3 shows the oxidation of a secondary alcohol, (CH_3)_{2}CHOH, forming a ketone (CH_3)_{2}CO.

Nucleophilic Substitution Reactions

A nucleophilic substitution reaction is a fundamental mechanism in organic chemistry where a nucleophile displaces a leaving group attached to a carbon atom. This reaction type is often abbreviated as SN1 or SN2, indicating whether the substitution occurs in a one-step or two-step process.

Nucleophiles are species that are attracted to positively charged or electron-deficient centers, often carrying a negative charge or a lone pair of electrons. In the context of the exercise, we can see nucleophilic substitution in both alkylation and oxidation reactions. For instance, in step 4, the iodide ion from CH_{3}CH_{2}I acts as a nucleophile, attacking the electrophilic carbon of the alcohol group to form CH_{3}CH_{2}CH_{2}I.