(b) Suggest the structural formula of the final organic product formed when reacted with ethanal, $$ \mathrm{CH}_{3} \mathrm{CHO} $$, in a similar two-step process. MgBr is [1] (c) The Grignard reagent, $$ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{MgBr} $$, can be readily converted into a carboxylic acid by using carbon dioxide. Suggest a two-step reaction sequence for this conversion, giving structural formulae for the compounds involved. $$ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{MgBr} \cdot \mathrm{CO}_{2} \xrightarrow{\text { step } 1} \square \xrightarrow{\text { stop II }} \square \square \mathrm{Mg}(\mathrm{OH}) \mathrm{Br} $$ N10/1/4 12 In 1834 the French chemist Eugene Peligot published his investigations into the action of heat on the Group II salts of carboxylic acids. He discovered that the salts decomposed to give the metal carbonate and a carbonyl compound. This calcium ethanoate produces propanone. $$ \left(\mathrm{CH}_{3} \mathrm{CO}_{2}\right)_{2} \mathrm{Ca} \longrightarrow \mathrm{CaCO}_{3}+\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO} $$ When this reaction is carried out by heating a $$ 1: 1 $$ molar mixture of calcium ethanoate and calcium methanoate, $$ \left(\mathrm{HCO}_{2}\right)_{2} \mathrm{Ca} $$, a mixture of three carbonyl compounds is formed, all having $$ \mathrm{M}

Question

(b) Suggest the structural formula of the final organic product formed when reacted with ethanal, $$ \mathrm{CH}_{3} \mathrm{CHO} $$, in a similar two-step process. MgBr is [1] (c) The Grignard reagent, $$ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{MgBr} $$, can be readily converted into a carboxylic acid by using carbon dioxide. Suggest a two-step reaction sequence for this conversion, giving structural formulae for the compounds involved. $$ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{MgBr} \cdot \mathrm{CO}_{2} \xrightarrow{\text { step } 1} \square \xrightarrow{\text { stop II }} \square \square \mathrm{Mg}(\mathrm{OH}) \mathrm{Br} $$ N10/1/4 12 In 1834 the French chemist Eugene Peligot published his investigations into the action of heat on the Group II salts of carboxylic acids. He discovered that the salts decomposed to give the metal carbonate and a carbonyl compound. This calcium ethanoate produces propanone. $$ \left(\mathrm{CH}_{3} \mathrm{CO}_{2}\right)_{2} \mathrm{Ca} \longrightarrow \mathrm{CaCO}_{3}+\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO} $$ When this reaction is carried out by heating a $$ 1: 1 $$ molar mixture of calcium ethanoate and calcium methanoate, $$ \left(\mathrm{HCO}_{2}\right)_{2} \mathrm{Ca} $$, a mixture of three carbonyl compounds is formed, all having $$ \mathrm{M}<60 $$. (i) Suggest the structures of the three carbonyl compounds, and the ratio in which they might be produced. (ii) Suggest two chemical tests that could be used to distinguish the three carbonyl compounds from each other. You should state what you would observe for each compound in each test. [5] N07/II/

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### Part (b)

We need to suggest the structural formula of the final organic product formed when the Grignard reagent $$ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{MgBr} $$ reacts with ethanal $$ \mathrm{CH}_{3} \mathrm{CHO} $$.

1. **Reaction of Grignard Reagent with Ethanal:**
   - The Grignard reagent acts as a nucleophile and attacks the carbonyl carbon of ethanal.
   - This forms a tetrahedral intermediate.

2. **Protonation:**
   - Upon workup (usually with water or dilute acid), the intermediate is protonated to form the final alcohol.

The final product is a secondary alcohol, specifically 2-pentanol.

**Structural Formula:**
$$
\text{Final Product: } \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C(OH)(CH}_{3}) \mathrm{CH}_{2} \mathrm{CH}_{3}
$$

### Part (c)

We need to suggest a two-step reaction sequence for converting the Grignard reagent $$ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{MgBr} $$ into a carboxylic acid using carbon dioxide.

1. **Step 1: Reaction with Carbon Dioxide**
   - The Grignard reagent reacts with carbon dioxide to form a carboxylate intermediate.
   - The structure of the intermediate is $$ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COO}^{-} $$.

2. **Step 2: Protonation**
   - Upon acid workup, the carboxylate ion is protonated to form the carboxylic acid.

**Reaction Sequence:**
$$
\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{MgBr} + \mathrm{CO}_{2} \xrightarrow{\text{step 1}} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COO}^{-} \xrightarrow{\text{step 2}} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH} + \mathrm{Mg(OH)Br}
$$

### Part (i)

We need to suggest the structures of the three carbonyl compounds formed from the reaction of calcium ethanoate and calcium methanoate.

1. **Calcium Ethanoate Decomposition:**
   - Produces propanone $$ \mathrm{(CH}_{3})_{2}\mathrm{CO} $$.

2. **Calcium Methanoate Decomposition:**
   - Produces formaldehyde $$ \mathrm{HCHO} $$ and methanol $$ \mathrm{CH}_{3}\mathrm{OH} $$.

**Three Carbonyl Compounds:**
1. Propanone $$ \mathrm{(CH}_{3})_{2}\mathrm{CO} $$
2. Formaldehyde $$ \mathrm{HCHO} $$
3. Acetaldehyde $$ \mathrm{CH}_{3}\mathrm{CHO} $$

**Ratio of Production:**
- The ratio of these compounds can vary, but a possible ratio could be 1:1:1.

### Part (ii)

We need to suggest two chemical tests to distinguish the three carbonyl compounds.

1. **Test 1: 2,4-Dinitrophenylhydrazine (DNPH) Test**
   - **Observation:**
     - All three compounds will react to form yellow/orange precipitates of hydrazones.
     - The melting points of the hydrazones can be used to distinguish them.

2. **Test 2: Tollens' Test**
   - **Observation:**
     - Propanone (ketone) will not react, while formaldehyde and acetaldehyde (aldehydes) will reduce Tollens' reagent to form a silver mirror.
     - Thus, a positive result (silver mirror) indicates the presence of aldehydes, while a negative result indicates a ketone.

In summary, the three carbonyl compounds are propanone, formaldehyde, and acetaldehyde, and they can be distinguished using DNPH and Tollens' tests.
**Part (b):** The final product is 2-pentanol. **Part (c):** 1. Grignard reagent reacts with CO₂ to form a carboxylate intermediate. 2. Protonation converts the carboxylate to carboxylic acid. **Part (i):** Three carbonyl compounds formed are propanone, formaldehyde, and acetaldehyde. **Part (ii):** 1. **2,4-Dinitrophenylhydrazine (DNPH) Test:** All three compounds form yellow/orange hydrazones, but their melting points can distinguish them. 2. **Tollens' Test:** Formaldehyde and acetaldehyde produce a silver mirror, while propanone does not.

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