Question

Compound \(\mathrm{X}\) has the molecular formula \(\mathrm{C}_{4} \mathrm{H}_{8}\). Compound \(\mathrm{Y}\) is obtained when hydrogen bromide is added to \(\mathrm{X}\). Compound Y reacts with AgOH to form a tertiary alcohol. Identify all structures.

Short answer

Compound X is 2-methylpropene with the structure \(\mathrm{CH_3C(CH_3)=CH_2}\). When it reacts with hydrogen bromide (HBr), it forms compound Y, which is 2-bromo-2-methylpropane with the structure \(\mathrm{CH_3C(CH_3)CH_2Br}\). Compound Y then reacts with AgOH to form a tertiary alcohol, which is 2-methyl-2-propanol with the structure \(\mathrm{CH_3C(CH_3)CH_2OH}\).

Step-by-step solution

Determine the possible structures of compound X

From the molecular formula C4H8, we can determine that compound X is an alkene because the general formula for alkenes is CnH2n. Hence, compound X would have one double bond. There are four possible isomers for this formula: 1-butene, 2-butene (cis and trans), and 2-methylpropene.

Identify the structures of compound X that can react with HBr

We now look at the four possible isomers of compound X and consider how they can react with HBr. HBr will react with the double bond of the alkene in an electrophilic addition reaction, following the Markovnikov rule. Let's examine each possible isomer: 1. 1-butene: With HBr addition, 2-bromobutane can be formed. 2. cis-2-butene: With HBr addition, 2-bromobutane would be formed. 3. trans-2-butene: With HBr addition, 2-bromobutane would be formed. 4. 2-methylpropene: With HBr addition, 2-bromo-2-methylpropane would be formed. At this stage, we are not yet able to exclude any of the isomers, so in the next step, we will use the further reaction with AgOH to find the correct structure.

Determine which structures of compound Y can react with AgOH

Now, we need to identify which of the possible structures of compound Y can react with AgOH to form a tertiary alcohol. AgOH would replace the bromide in compound Y with a hydroxyl group through a nucleophilic substitution reaction. Let's consider the potential reactions for each isomer: 1. 2-bromobutane (derived from 1-butene, cis-2-butene, and trans-2-butene): reacts with AgOH to form 2-butanol, which is a secondary alcohol. 2. 2-bromo-2-methylpropane (derived from 2-methylpropene): reacts with AgOH to form 2-methyl-2-propanol, which is a tertiary alcohol. Since we are looking for a compound Y that can form a tertiary alcohol, we can exclude 2-bromobutane, and thus, compound X must be 2-methylpropene. In conclusion, compound X is 2-methylpropene, and compound Y is 2-bromo-2-methylpropane.

Key concepts

Markovnikov's Rule

Markovnikov's rule is an essential guide in organic chemistry that helps predict the outcome of electrophilic addition reactions involving unsymmetrical alkenes and electrophiles. It states that during the addition of a protic acid (such as HBr) to an unsymmetrical alkene, the hydrogen atom (H) attaches itself to the carbon with more hydrogen substituents, and the other part (Bromine, in this case) bonds with the carbon that has fewer hydrogen atoms.

This rule aligns with the stability of carbocations; a more substituted carbocation is more stable and favored in the reaction mechanism. For instance, when hydrogen bromide (HBr) is added to an alkene like 2-methylpropene, the hydrogen attaches to the carbon that is already part of two other carbon bonds (thereby following Markovnikov's rule), resulting in the formation of 2-bromo-2-methylpropane, a more substituted alkyl bromide. This understanding is crucial for students tackling synthesis problems involving alkenes, as it allows them to accurately determine the major product of such additions.

Electrophilic Addition Reaction

The electrophilic addition reaction is a typical reaction of alkenes, where an electrophile and a nucleophile add to the carbon atoms of a double bond. In the case of our exercise, the alkene present in compound X reacts with hydrogen bromide (HBr). The double bond is electron-rich, making it attractive to electrophiles, such as H⁺ from HBr. Once the H⁺ adds to one of the carbon atoms, a carbocation is formed.

The stability of this positively charged intermediate directly impacts the reaction's outcome. For instance, in the addition of HBr to 2-methylpropene, the resulting tertiary carbocation is more stable due to the inductive effect from nearby alkyl groups and hyperconjugation. It's important to emphasize that students should visualize the reaction mechanism, seeing the electronegative bromine atom eventually attacking the positively charged intermediate to form the final addition product. Electrophilic addition is not limited to reactions with HBr but can involve a wide array of electrophiles and alkenes.

Nucleophilic Substitution Reaction

The nucleophilic substitution reaction is another fundamental type of reaction in organic chemistry. In our exercise, this reaction is illustrated when compound Y reacts with AgOH. The nucleophilic substitution involves a nucleophile (OH^- in this case) displacing a leaving group (the bromide ion from the bromoalkane) in a single step or two-step process. The silver ion, Ag⁺, helps facilitate this reaction by first forming a precipitate with bromide, effectively removing the leaving group and allowing the nucleophile to attack.

For the compound in the exercise, 2-bromo-2-methylpropane, the reaction with AgOH leads to the formation of a tertiary alcohol, 2-methyl-2-propanol, showcasing a typical nucleophilic substitution by an OH group. It's valuable for students to recognize that the nature of the substrate, the leaving group, and the nucleophile will significantly affect the reaction rate and pathway (S_N1 or S_N2 mechanism). Understanding these elements helps students to predict the configurations and types of alcohol that result from such reactions.