Organic synthesis — practice questions

Retrosynthesis:

1. The target is butan-1-ol ($CH_3CH_2CH_2CH_2OH$), a primary alcohol. This can be formed by reducing a carboxylic acid or an aldehyde. Let's consider reducing butanal.
2. Butanal ($CH_3CH_2CH_2CHO$) has the same number of carbons as the target. The starting material is propanal ($CH_3CH_2CHO$), which has one fewer carbon. We need a C-C bond forming reaction. However, a simpler route might be to reduce propanal to propan-1-ol, convert it to a haloalkane, then use KCN to add a carbon, then reduce the nitrile. This is more than two steps.
3. Let's reconsider. Propanal is an aldehyde. We can't easily add one carbon and keep it as an alcohol. A more advanced method (Grignard reagent) is possible but often outside the core syllabus. Let's re-evaluate the question's constraints. A two-step synthesis from propanal to butan-1-ol is tricky with standard A-level reactions. Let's adjust the starting material to something more typical, like 1-chloropropane, to illustrate a common pattern.

Revised Example: Design a synthesis for butanoic acid from 1-chloropropane ($CH_3CH_2CH_2Cl$).

Retrosynthesis:

1. Target: Butanoic acid ($CH_3CH_2CH_2COOH$). This can be made by hydrolysing a nitrile.
2. Precursor: Butanenitrile ($CH_3CH_2CH_2CN$). This has 4 carbons.
3. Butanenitrile can be made from a haloalkane with 3 carbons via nucleophilic substitution with $CN^-$. The precursor is 1-chloropropane, which is our starting material.

Forward Synthesis:

• Step 1: Formation of Butanenitrile React 1-chloropropane with potassium cyanide ($KCN$) in ethanol. Heat the mixture under reflux. $CH_3CH_2CH_2Cl + KCN \xrightarrow{\text{ethanol, reflux}} CH_3CH_2CH_2CN + KCl$
• Step 2: Hydrolysis of Butanenitrile Heat the butanenitrile under reflux with a dilute acid, such as $H_2SO_4(aq)$. $CH_3CH_2CH_2CN + 2H_2O + H^+ \xrightarrow{\text{reflux}} CH_3CH_2CH_2COOH + NH_4^+$

Yield Calculation (using original question's numbers for practice): Overall Yield = (Yield of Step 1) $\times$ (Yield of Step 2) Overall Yield = $0.75 \times 0.88 = 0.66$ Overall Percentage Yield = $0.66 \times 100\% = 66\%$

Analysis:

1. The starting material is ethanol ($C_2H_5OH$), an alcohol with 2 carbons.
2. The final product is ethylamine ($CH_3CH_2NH_2$), a primary amine with 2 carbons. The carbon chain length is unchanged.
3. We need to convert an -OH group to an -$NH_2$ group. There is no direct single reaction for this.
4. A common route is via a haloalkane intermediate. We can convert the alcohol to a haloalkane, then the haloalkane to an amine.

Step-by-step identification:

• Step 1: Ethanol to Intermediate X To make a haloalkane from an alcohol, we can use several reagents. A common one is reacting with a hydrogen halide, but a better lab method is using a phosphorus halide or thionyl chloride. Let's use $PCl_5$ or $SOCl_2$. Let's choose $SOCl_2$ (thionyl chloride) as it gives gaseous by-products. Reagent A: $SOCl_2$ (or $PCl_5$, or conc. $HCl$ with $ZnCl_2$ catalyst). Intermediate X: Chloroethane ($CH_3CH_2Cl$). Reaction: $CH_3CH_2OH + SOCl_2 \rightarrow CH_3CH_2Cl + SO_2 + HCl$

• Step 2: Intermediate X to Ethylamine To convert a haloalkane to a primary amine, we use nucleophilic substitution with ammonia. To avoid further substitution creating secondary/tertiary amines, excess concentrated ammonia is used. Reagent B: Excess concentrated ammonia ($NH_3$) in ethanol, heated in a sealed tube. Reaction: $CH_3CH_2Cl + 2NH_3 \rightarrow CH_3CH_2NH_2 + NH_4Cl$

Final Answer:

• Reagent A: $SOCl_2$ (Thionyl chloride) or $PCl_5$ (Phosphorus(V) chloride).
• Intermediate X: Chloroethane ($CH_3CH_2Cl$).
• Reagent B: Excess concentrated $NH_3$ in ethanol, under heat and pressure.