Carboxylic Acid and its Derivatives

Working backwards. (E) is a primary alcohol giving a positive iodoform test, so (E) = ethanol, $CH_3CH_2OH$. It is formed from (D) by NaNO$_2$/HCl, so (D) = ethylamine, $CH_3CH_2NH_2$. Hofmann bromamide degradation removes one carbon, so (C) = propanamide, $CH_3CH_2CONH_2$. Then (B) = propanoyl chloride and (A) = propanoic acid.

Reactions

1. $CH_3CH_2COOH + SOCl_2 \longrightarrow CH_3CH_2COCl + SO_2 + HCl$
2. $CH_3CH_2COCl + 2NH_3 \xrightarrow{\Delta} CH_3CH_2CONH_2 + NH_4Cl$
3. $CH_3CH_2CONH_2 + Br_2 + 4KOH \xrightarrow{\Delta} CH_3CH_2NH_2 + K_2CO_3 + 2KBr + 2H_2O$ (Hofmann bromamide reaction)
4. $CH_3CH_2NH_2 + HNO_2 \xrightarrow{NaNO_2/HCl,\ 0-5^\circ C} CH_3CH_2OH + N_2 + H_2O$

Iodoform test: $CH_3CH_2OH + 4I_2 + 6NaOH \longrightarrow CHI_3\downarrow(\text{yellow}) + HCOONa + 5NaI + 5H_2O$.

Decarbonylation is the removal of CO from a molecule. When methanoic acid is heated with conc. $H_2SO_4$, it loses CO:

$HCOOH \xrightarrow{\text{conc. } H_2SO_4,\ \Delta} CO\uparrow + H_2O$

(c) $CH_3COONa$

Acetyl chloride reacts with sodium acetate by nucleophilic acyl substitution to give acetic anhydride:

$CH_3COCl + CH_3COONa \longrightarrow (CH_3CO)_2O + NaCl$

i) Reagents and conditions

• Reaction (1): Malonic acid is heated at about $140^\circ C$ (decarboxylation): $CH_2(COOH)_2 \xrightarrow{\Delta} CH_3COOH + CO_2$
• Reaction (2): Grignard reagent with dry ice. $CH_3MgBr + CO_2 \xrightarrow{\text{dry ether}} CH_3COOMgBr \xrightarrow{H_3O^+} CH_3COOH + Mg(OH)Br$
• Reaction (3): Sodium methoxide with carbon monoxide at $150^\circ C$ and 10 atm, followed by acidification. $CH_3ONa + CO \longrightarrow CH_3COONa \xrightarrow{H^+} CH_3COOH$
• Reaction (4): Dehydration with $P_2O_5$ and heat. $2CH_3COOH \xrightarrow{P_2O_5,\ \Delta} (CH_3CO)_2O + H_2O$

ii) Compound (z)

$LiAlH_4$ reduces the anhydride to two molecules of primary alcohol:

$(CH_3CO)_2O + 8[H] \xrightarrow{LiAlH_4} 2CH_3CH_2OH + H_2O$

(z) = $CH_3CH_2OH$, ethanol.

Decarboxylation is the removal of $CO_2$ from a carboxylic acid. Sodium ethanoate heated with soda lime gives methane:

$CH_3COONa + NaOH \xrightarrow[\Delta]{CaO} CH_4\uparrow + Na_2CO_3$

In the Hell-Volhard-Zelinsky (HVZ) reaction, a carboxylic acid with an $\alpha$-hydrogen is treated with $Cl_2$ or $Br_2$ in the presence of red phosphorus. The $\alpha$-hydrogen is replaced by a halogen.

$CH_3COOH \xrightarrow[\text{red P}]{Cl_2} ClCH_2COOH \xrightarrow{Cl_2/P} Cl_2CHCOOH \xrightarrow{Cl_2/P} Cl_3CCOOH$

Note on the question. A is $C_3H_6O$, and propan-2-ol is formed from it in Reaction 1. Only propanone ($CH_3COCH_3$) is reduced to propan-2-ol. Propanal ($CH_3CH_2CHO$) is the $C_3H_6O$ isomer that actually gives Tollens' and Fehling's tests, but it would give propan-1-ol. The two statements in the question do not agree, so both are shown below. The rest of the sequence follows from propan-2-ol.

i) Compound A and test reaction

A is $C_3H_6O$. The positive Tollens' and Fehling's tests point to the aldehyde propanal, $CH_3CH_2CHO$. The reduction product propan-2-ol requires propanone, $CH_3COCH_3$.

Test reaction (Tollens', for the aldehyde):

$CH_3CH_2CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \longrightarrow CH_3CH_2COO^- + 2Ag\downarrow + 4NH_3 + 2H_2O$

Silver mirror formed.

ii) Reactions 1 and 2

• Reaction 1 (reduction): $CH_3COCH_3 + 2[H] \xrightarrow{H_2/Ni \text{ or } LiAlH_4} CH_3CH(OH)CH_3$
• Reaction 2 (esterification): $CH_3CH(OH)CH_3 + CH_3COOH \xrightarrow{\text{conc. } H_2SO_4,\ \Delta} CH_3COOCH(CH_3)_2 + H_2O$

B = isopropyl ethanoate (propan-2-yl ethanoate), $C_5H_{10}O_2$.

iii) Structure of D

D = propan-2-ol:

   CH3 - CH - CH3
         |
         OH

iv) Identification of C and D

$CH_3COOCH(CH_3)_2 + H_2O \xrightarrow{H^+} CH_3COOH + CH_3CH(OH)CH_3$

C = ethanoic acid, D = propan-2-ol.

v) B with sodium ethoxide

Two molecules of the ester condense. One molecule supplies an $\alpha$-carbanion, which attacks the carbonyl of the other, and an alcohol is lost. This gives a $\beta$-keto ester:

$2CH_3COOCH(CH_3)_2 \xrightarrow{C_2H_5ONa} CH_3COCH_2COOCH(CH_3)_2 + (CH_3)_2CHOH$

This is the Claisen condensation. Isopropyl 3-oxobutanoate is formed, the analogue of ethyl acetoacetate.

Any two of the following.

1. From a Grignard reagent: $CH_3MgBr + CO_2 \xrightarrow{\text{dry ether}} CH_3COOMgBr \xrightarrow{H_3O^+} CH_3COOH$
2. By oxidation of ethanol: $CH_3CH_2OH + 2[O] \xrightarrow{K_2Cr_2O_7/H_2SO_4} CH_3COOH + H_2O$
3. By hydrolysis of methyl cyanide: $CH_3CN + 2H_2O \xrightarrow{H^+,\ \Delta} CH_3COOH + NH_3$

i) With $P_2O_5$. It is a dehydrating agent. Two molecules of ethanoic acid lose one molecule of water to form ethanoic anhydride:

$2CH_3COOH \xrightarrow{P_2O_5,\ \Delta} (CH_3CO)_2O + H_2O$

ii) With $NH_3$. First, ammonium ethanoate (a salt) forms. On heating, it loses water and gives ethanamide:

$CH_3COOH + NH_3 \longrightarrow CH_3COONH_4 \xrightarrow{\Delta} CH_3CONH_2 + H_2O$

(b) Ethanoic anhydride

Z is $C_3H_4O_4$, malonic acid $CH_2(COOH)_2$. On heating, it loses $CO_2$ to give Y = ethanoic acid. Two molecules of ethanoic acid are dehydrated by $P_2O_5$ to give X = ethanoic anhydride, $(CH_3CO)_2O$.

$CH_2(COOH)_2 \xrightarrow{\Delta} CH_3COOH + CO_2, \qquad 2CH_3COOH \xrightarrow{P_2O_5} (CH_3CO)_2O + H_2O$

(d) $CH_3COOH$

(Options (a) and (c) are printed identically in the question.) The $-I$ effect of Cl atoms stabilises the carboxylate ion and increases acidity. $CCl_3COOH$ has the strongest effect, then $ClCH_2COOH$. In $CH_3COOH$ the methyl group has a $+I$ effect, which destabilises the anion, so it is the weakest acid.

(i) From ethanamide: hydrolysis

Boiling with dilute acid or alkali hydrolyses the amide:

$CH_3CONH_2 + H_2O \xrightarrow{H^+,\ \Delta} CH_3COOH + NH_4^+$

or, with nitrous acid, $CH_3CONH_2 + HNO_2 \longrightarrow CH_3COOH + N_2 + H_2O$.

(ii) From sodium methoxide: carbonylation

Sodium methoxide is heated with carbon monoxide at $150^\circ C$ and about 10 atm. Sodium ethanoate forms and is acidified:

$CH_3ONa + CO \xrightarrow{150^\circ C,\ 10\text{ atm}} CH_3COONa \xrightarrow{H^+} CH_3COOH$

(i) Methane. Convert the acid to its sodium salt, then heat with soda lime (decarboxylation).

$CH_3COOH + NaOH \longrightarrow CH_3COONa + H_2O$ $CH_3COONa + NaOH \xrightarrow[\Delta]{CaO} CH_4 + Na_2CO_3$

(ii) Methyl ethanoate. Fischer esterification with methanol in the presence of conc. $H_2SO_4$.

$CH_3COOH + CH_3OH \xrightarrow[\Delta]{\text{conc. } H_2SO_4} CH_3COOCH_3 + H_2O$

Identification

Reactions

1. A to B (hydrolysis): $(CH_3CO)_2O + H_2O \longrightarrow 2CH_3COOH$
2. B to C: $CH_3COOH + SOCl_2 \longrightarrow CH_3COCl + SO_2 + HCl$ (or with $PCl_5$)
3. C to D (Rosenmund reduction): $CH_3COCl + H_2 \xrightarrow{Pd/BaSO_4} CH_3CHO + HCl$
4. D to E (aldol condensation): $2CH_3CHO \xrightarrow{\text{dil. NaOH}} CH_3CH(OH)CH_2CHO$

Conversion of D (ethanal) into propanone

• Ethanal reacts with methylmagnesium bromide, then hydrolysis gives propan-2-ol: $CH_3CHO + CH_3MgBr \xrightarrow{\text{dry ether}} CH_3CH(OMgBr)CH_3 \xrightarrow{H_3O^+} CH_3CH(OH)CH_3$
• Oxidation of propan-2-ol gives propanone: $CH_3CH(OH)CH_3 \xrightarrow{K_2Cr_2O_7/H^+} CH_3COCH_3$

Identification. C is an aldehyde giving Tollens' test, so C = ethanal. Hence A = ethanoic acid, B = ethanoyl chloride, D = ethanol and E = chloroethane.

Reactions

1. $CH_3COOH + SOCl_2 \longrightarrow CH_3COCl + SO_2 + HCl$ (or $PCl_5$)
2. $CH_3COCl + H_2 \xrightarrow{Pd/BaSO_4} CH_3CHO + HCl$ (Rosenmund reduction)
3. $CH_3CHO + 2[H] \xrightarrow{NaBH_4 \text{ or } H_2/Ni} CH_3CH_2OH$
4. $CH_3CH_2OH + SOCl_2 \xrightarrow{\text{pyridine}} CH_3CH_2Cl + SO_2 + HCl$ (or $PCl_5$, or HCl/anhydrous $ZnCl_2$)

Methanoic acid, $H-COOH$, has both a carboxyl group and an aldehyde group (H-C=O attached to OH).

As an acid. It turns blue litmus red and neutralises bases:

$HCOOH + NaOH \longrightarrow HCOONa + H_2O$

It also liberates $CO_2$ from carbonates: $2HCOOH + Na_2CO_3 \longrightarrow 2HCOONa + H_2O + CO_2$.

As an aldehyde. It is a reducing agent. With Tollens' reagent it gives a silver mirror:

$HCOOH + 2[Ag(NH_3)_2]^+ + 2OH^- \longrightarrow CO_2 + 2Ag\downarrow + 4NH_3 + 2H_2O$

It also reduces Fehling's solution to red $Cu_2O$ and decolourises acidified $KMnO_4$.

Methanoic acid is converted to methanol, then to methyl cyanide, which is hydrolysed. This adds one carbon.

1. Reduction: $HCOOH + 4[H] \xrightarrow{LiAlH_4} CH_3OH + H_2O$
2. $CH_3OH + PCl_5 \longrightarrow CH_3Cl + POCl_3 + HCl$ (or $CH_3OH + HI \longrightarrow CH_3I + H_2O$)
3. $CH_3Cl + KCN \xrightarrow{\text{alc.}} CH_3CN + KCl$
4. Hydrolysis: $CH_3CN + 2H_2O \xrightarrow{H^+,\ \Delta} CH_3COOH + NH_3$

Preparation of ethanoic acid from $CH_3MgBr$. Pass dry $CO_2$ (or add dry ice) to the Grignard reagent in dry ether, then hydrolyse:

$CH_3MgBr + CO_2 \xrightarrow{\text{dry ether}} CH_3COOMgBr \xrightarrow{H_3O^+} CH_3COOH + Mg(OH)Br$

Reactions of ethanoic acid

i) With $NH_3$: $CH_3COOH + NH_3 \longrightarrow CH_3COONH_4 \xrightarrow{\Delta} CH_3CONH_2 + H_2O$ (ethanamide)

ii) With $SOCl_2$: $CH_3COOH + SOCl_2 \longrightarrow CH_3COCl + SO_2 + HCl$ (ethanoyl chloride)

iii) With $P_2O_5$: $2CH_3COOH \xrightarrow{P_2O_5,\ \Delta} (CH_3CO)_2O + H_2O$ (ethanoic anhydride)

iv) With $LiAlH_4$: $CH_3COOH + 4[H] \xrightarrow{LiAlH_4,\ \text{dry ether}} CH_3CH_2OH + H_2O$ (ethanol)

Working from the hint. Clemmensen reduction of benzaldehyde gives toluene, so C = toluene. Toluene is formed from benzene (B) by Friedel-Crafts alkylation, and benzene comes from decarboxylation of sodium benzoate (A). Oxidation of toluene gives D = benzaldehyde.

The printed "$CHCl_3/AlCl_3$" must be read as $CH_3Cl/AlCl_3$, since $CHCl_3$ would not give toluene.

Reactions

1. $C_6H_5COONa + NaOH \xrightarrow[\Delta]{CaO} C_6H_6 + Na_2CO_3$
2. $C_6H_6 + CH_3Cl \xrightarrow{\text{anhyd. } AlCl_3} C_6H_5CH_3 + HCl$
3. $C_6H_5CH_3 + [O] \xrightarrow{\text{mild oxidant}} C_6H_5CHO + H_2O$

Check: $C_6H_5CHO + 4[H] \xrightarrow{Zn-Hg/HCl} C_6H_5CH_3 + H_2O$, which confirms C.

The carbon chain must be shortened by one carbon, which is done by the Hofmann bromamide reaction and then oxidation:

$$CH_3COOH \xrightarrow{NH_3,\ \Delta} CH_3CONH_2 \xrightarrow{Br_2/KOH} CH_3NH_2 \xrightarrow{HNO_2} CH_3OH \xrightarrow{[O]} HCHO \xrightarrow{[O]} HCOOH$$

The oxidation of methanol uses acidified $K_2Cr_2O_7$ (an excess of oxidant takes it through methanal to methanoic acid).

Ascending order of acidic strength:

$CH_3CH_2COOH < C_6H_5COOH < ClCH_2CH_2COOH$

(pKa about 4.87, 4.20 and 4.0 respectively.)

Justification

• $CH_3CH_2COOH$: the ethyl group has a $+I$ effect. It pushes electrons towards the carboxylate and destabilises $CH_3CH_2COO^-$, so this is the weakest acid.
• $C_6H_5COOH$: the phenyl carbon is $sp^2$ and is more electronegative than an alkyl carbon. It has a $-I$ effect and withdraws electrons. This stabilises $C_6H_5COO^-$, so benzoic acid is stronger than propanoic acid.
• $ClCH_2CH_2COOH$: chlorine is strongly electronegative and has a $-I$ effect. It withdraws electrons and disperses the negative charge on the carboxylate. This gives the most stable anion and the strongest acid, even though Cl is two carbons away from the COOH group.

A more stable conjugate base means a stronger acid.

$C_2H_4O_2$ has two functional isomers. A is ethanoic acid, $CH_3COOH$ (turns blue litmus red). B is methyl methanoate, $HCOOCH_3$ (fruity smell, an ester).

a) Formation of A using Grignard's reagent

$CH_3MgBr + CO_2 \xrightarrow{\text{dry ether}} CH_3COOMgBr \xrightarrow{H_3O^+} CH_3COOH + Mg(OH)Br$

b) Conversion of A into B

1. $CH_3COOH + NH_3 \longrightarrow CH_3COONH_4 \xrightarrow{\Delta} CH_3CONH_2 + H_2O$
2. $CH_3CONH_2 + Br_2 + 4KOH \longrightarrow CH_3NH_2 + K_2CO_3 + 2KBr + 2H_2O$ (Hofmann)
3. $CH_3NH_2 + HNO_2 \longrightarrow CH_3OH + N_2 + H_2O$
4. Part of the methanol is oxidised: $CH_3OH \xrightarrow{K_2Cr_2O_7/H^+} HCOOH$
5. $HCOOH + CH_3OH \xrightarrow{\text{conc. } H_2SO_4} HCOOCH_3 + H_2O$

c) Test to distinguish A from methanoic acid

Use Tollens' reagent. Methanoic acid gives a silver mirror; ethanoic acid does not:

$HCOOH + 2[Ag(NH_3)_2]^+ + 2OH^- \longrightarrow CO_2 + 2Ag\downarrow + 4NH_3 + 2H_2O$

(Fehling's solution can also be used: methanoic acid gives red $Cu_2O$.)

d) Decreasing order of acidic strength

$FCH_2COOH > ClCH_2COOH > C_6H_5COOH > CH_3COOH$

Reason: Electron-withdrawing groups ($-I$ effect) stabilise the carboxylate anion and increase acidity. F is more electronegative than Cl, so its $-I$ effect is stronger. The phenyl group is weakly electron-withdrawing, so benzoic acid is stronger than ethanoic acid. In $CH_3COOH$ the methyl group has a $+I$ effect, which destabilises the anion, so it is the weakest.

e) Part (e) refers to reactions that are not printed in the paper, so it cannot be answered from the given text. From the data, A = $CH_3COOH$ (ethanoic acid) and B = $HCOOCH_3$ (methyl methanoate).

(c) C > B > A

Ethanoic acid forms very strong hydrogen bonds and exists as a hydrogen-bonded dimer, so it has the highest boiling point (118 °C). Ethanol has intermolecular hydrogen bonding but weaker than the acid dimer (78 °C). Ethanal has no O–H, so it has only dipole–dipole forces and boils lowest (21 °C).

(a) $P_2O_5$

$P_2O_5$ is a strong dehydrating agent. It removes one molecule of water from two molecules of the acid:

$$2CH_3COOH \xrightarrow{P_2O_5,\ \Delta} (CH_3CO)_2O + H_2O$$

$CaO$ only neutralises the acid, while $LiAlH_4$ and $Na/C_2H_5OH$ are reducing agents.

A carboxylic acid reacts with an alcohol in the presence of a few drops of conc. $H_2SO_4$ (catalyst and dehydrating agent) to form an ester and water:

$$CH_3COOH + C_2H_5OH \underset{}{\overset{conc.\ H_2SO_4,\ \Delta}{\rightleftharpoons}} CH_3COOC_2H_5 + H_2O$$

The products are ethyl ethanoate (ethyl acetate) and water.

The five compounds can be linked in the following order:

propanoic acid → propanamide → ethanamine → ethanol → chloroethane

1. Propanoic acid to propanamide. Treat with $NH_3$ to form ammonium propanoate, then heat it:

$$CH_3CH_2COOH + NH_3 \rightarrow CH_3CH_2COONH_4 \xrightarrow{\Delta} CH_3CH_2CONH_2 + H_2O$$

2. Propanamide to ethanamine (Hofmann bromamide reaction). The amide loses one carbon:

$$CH_3CH_2CONH_2 + Br_2 + 4KOH \rightarrow CH_3CH_2NH_2 + K_2CO_3 + 2KBr + 2H_2O$$

3. Ethanamine to ethanol. Treat with nitrous acid ($NaNO_2 + HCl$, 0–5 °C):

$$CH_3CH_2NH_2 + HNO_2 \rightarrow CH_3CH_2OH + N_2 + H_2O$$

4. Ethanol to chloroethane. Use $SOCl_2$ (or $PCl_5$, or HCl/anhyd. $ZnCl_2$):

$$CH_3CH_2OH + SOCl_2 \rightarrow CH_3CH_2Cl + SO_2 + HCl$$

Closing the cycle. Chloroethane returns to propanoic acid by heating with KCN (giving propanenitrile) and then hydrolysing with dilute acid:

$$CH_3CH_2Cl \xrightarrow{KCN} CH_3CH_2CN \xrightarrow{H_3O^+,\ \Delta} CH_3CH_2COOH$$

i. From $CH_3MgBr$. Pass dry $CO_2$ (or add solid $CO_2$) into the Grignard reagent, then hydrolyse with dilute acid:

$$CH_3MgBr + CO_2 \rightarrow CH_3COOMgBr \xrightarrow{H_3O^+} CH_3COOH + Mg(OH)Br$$

ii. From $CH_3CN$. Hydrolyse by boiling with dilute HCl (or aqueous NaOH, followed by acidification):

$$CH_3CN + 2H_2O \xrightarrow{H^+,\ \Delta} CH_3COOH + NH_4^+$$

iii. From $CH_3ONa$. Heat with carbon monoxide under pressure (about 10 atm, 150 °C) to get sodium ethanoate, then acidify:

$$CH_3ONa + CO \xrightarrow{\Delta,\ \text{pressure}} CH_3COONa \xrightarrow{HCl} CH_3COOH + NaCl$$