Alkanoic Acids, Esterification, Saponification, Fats, and Oils Practice Questions — JAMB UTME

Question 1Question →

A triglyceride derived from a single saturated alkanoic acid undergoes complete saponification with excess aqueous $\text{NaOH}$ , yielding glycerol and a sodium soap salt. When the isolated soap salt is acidified with excess dilute $\text{HCl}$ , a pure saturated alkanoic acid $X$ is liberated. Neutralization of $10.24\text{ g}$ of acid $X$ requires exactly $40.0\text{ cm}^3$ of a $1.00\text{ mol dm}^{-3}$ $\text{NaOH}$ solution. What is the correct IUPAC name of the alkanoic acid $X$ ? (Atomic masses: $\text{H} = 1$ , $\text{C} = 12$ , $\text{O} = 16$ )

Hexadecanoic acid

Tetradecanoic acid

Pentadecanoic acid

Octadecanoic acid

Show answer & explanation

Answer: Hexadecanoic acid

Answer

Hexadecanoic acid

The acid

X

reacts with

\text{NaOH}

in a 1:1 stoichiometry.

0.0400\text{ mol}

of

\text{NaOH}

neutralizes

0.0400\text{ mol}

of

X

, giving a molar mass of

256\text{ g mol}^{-1}

. Setting the general formula for saturated alkanoic acids

\text{C}_n\text{H}_{2n}\text{O}_2

equal to

256

yields

14n + 32 = 256

, which solves to

n = 16

. The 16-carbon saturated carboxylic acid is Hexadecanoic acid.

Step-by-Step Solution

1

Calculate the number of moles of NaOH used in the neutralization reaction

\text{Moles of NaOH} = \text{Concentration} \times \text{Volume in dm}^3 = 1.00\text{ mol dm}^{-3} \times \frac{40.0}{1000}\text{ dm}^3 = 0.0400\text{ mol}

Neutralization uses volume and molarity to determine mole quantity.

2

Determine the molar mass of the monocarboxylic acid X

Since monocarboxylic acid reacts with NaOH in a 1:1 mole ratio (

\text{R-COOH} + \text{NaOH} \rightarrow \text{R-COONa} + \text{H}_2\text{O}

), moles of

X = 0.0400\text{ mol}

. Thus,

\text{Molar mass of } X = \frac{10.24\text{ g}}{0.0400\text{ mol}} = 256\text{ g mol}^{-1}

.

Molar mass is the mass divided by the amount in moles.

3

Use the general formula for a saturated alkanoic acid to find the number of carbon atoms (n)

The general formula for a saturated monocarboxylic acid is

\text{C}_n\text{H}_{2n}\text{O}_2

. Molar mass

= 12n + 2n + 32 = 14n + 32 = 256 \implies 14n = 224 \implies n = 16

.

Determining

n

gives the total number of carbon atoms in the IUPAC parent chain.

4

Assign the official IUPAC name for a 16-carbon saturated alkanoic acid

\text{C}_{16}\text{H}_{32}\text{O}_2

is named Hexadecanoic acid.

The IUPAC suffix for a 16-carbon alkanoic acid is hexadecanoic acid.

Key Concept

Determination of alkanoic acid stoichiometry from saponification and neutralization data

Estimated Time:3m 0s

Question 2Question →

An organic compound $X$ with the molecular formula $\text{C}_4\text{H}_8\text{O}_2$ does not react with sodium hydrogentrioxocarbonate(IV) solution to liberate gas. Upon refluxing $X$ with dilute sodium hydroxide solution, two organic products, $Y$ and $Z$ , are formed. Acidification of product $Z$ yields ethanoic acid, while mild oxidation of product $Y$ yields an alkanal that gives a positive Tollen's test. Which of the following correctly identifies the IUPAC name of compound $X$ and the chemical nature of its reaction with dilute sodium hydroxide?

Ethyl ethanoate; an irreversible alkaline hydrolysis reaction

Ethyl ethanoate; a reversible esterification equilibrium reaction

Methyl propanoate; an irreversible alkaline hydrolysis reaction

Butanoic acid; a reversible acid-base neutralization reaction

Show answer & explanation

Answer: Ethyl ethanoate; an irreversible alkaline hydrolysis reaction

Answer

The IUPAC name of compound

X

is ethyl ethanoate, and its reaction with dilute sodium hydroxide is an irreversible alkaline hydrolysis reaction.

Compound

X

does not evolve carbon(IV) oxide gas with sodium hydrogentrioxocarbonate(IV), ruling out alkanoic acids and establishing that

X

is an ester. Alkaline hydrolysis of

X

yields sodium ethanoate (

Z

) and ethanol (

Y

), because acidification of

Z

produces ethanoic acid (

2

carbons) and oxidation of ethanol (

Y

) yields ethanal, an alkanal that gives a silver mirror with Tollen's reagent. Therefore,

X

is ethyl ethanoate (

\text{CH}_3\text{COOCH}_2\text{CH}_3

). Base hydrolysis of esters converts the carboxyl moiety into a resonance-stabilized carboxylate ion, preventing the reverse reaction and rendering the hydrolysis irreversible.

Step-by-Step Solution

1

Determine the functional group class of compound

X

Compound

X

is an ester.

Isomers with formula

\text{C}_4\text{H}_8\text{O}_2

can be alkanoic acids or esters. Since

X

does not react with

\text{NaHCO}_3

to evolve

\text{CO}_2

gas, it lacks the free carboxyl acid group (

-\text{COOH}

) and must be an ester.

2

Deduce the structure of the alkanoate (acid) portion of the ester

The alkanoate part is ethanoate (

\text{CH}_3\text{COO}^-

).

Alkaline hydrolysis of ester

X

produces carboxylate salt

Z

. Acidification of

Z

yields ethanoic acid (

\text{CH}_3\text{COOH}

), which contains 2 carbon atoms.

3

Deduce the structure of the alkyl (alcohol) portion of the ester

The alkyl group is ethyl (

-\text{C}_2\text{H}_5

), making

Y

ethanol (

\text{CH}_3\text{CH}_2\text{OH}

).

The total number of carbon atoms in

X

is 4. Subtracting 2 carbons from the acid part leaves 2 carbons for the alcohol

Y

(ethanol). Mild oxidation of ethanol yields ethanal (an alkanal), which reduces Tollen's reagent.

4

Identify the reaction type with dilute

\text{NaOH}

Irreversible alkaline hydrolysis (saponification).

Hydrolysis of an ester with alkali consumes hydroxyl ions (

\text{OH}^-

) to form an unreactive carboxylate anion (

\text{CH}_3\text{COO}^-

), driving the reaction to completion irreversibly.

Key Concept

Chemical differentiation between carboxylic acids and esters, ester hydrolysis kinetics, and structural deduction.

Question 3Question →

Match each chemical reaction or process involving alkanoic acids, esters, fats, or oils on the left with its corresponding chemical outcome or product description on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings match: (1) Acid-catalyzed esterification of propane-1,2,3-triol with hexadecanoic acid to producing tripalmitin fat and water; (2) Alkaline hydrolysis of glyceryl tristearate with aqueous NaOH to yielding propane-1,2,3-triol and sodium octadecanoate soap; (3) Catalytic hydrogenation of glyceryl trioleate to converting an unsaturated liquid triacylglycerol into a saturated solid fat; and (4) Acid-catalyzed reflux of ethyl ethanoate with water to reversibly producing ethanoic acid and ethanol.

The correct pairings logically connect each organic process to its definitive product or reaction characteristic: esterification of hexadecanoic acid with glycerol yields tripalmitin; saponification of glyceryl tristearate using NaOH irreversibly yields glycerol and sodium octadecanoate soap; catalytic hydrogenation saturates double bonds in glyceryl trioleate turning liquid oil into solid fat; and acid hydrolysis of ethyl ethanoate is a reversible equilibrium yielding ethanoic acid and ethanol.

Step-by-Step Solution

1

Analyze the reaction of propane-1,2,3-triol (glycerol) with hexadecanoic acid (palmitic acid).

Identified as esterification forming tripalmitin and water.

Hexadecanoic acid (

C_{15}H_{31}COOH

) esterifies with glycerol to form glyceryl tripalmitin (

C_{51}H_{98}O_6

), a saturated fat.

2

Analyze the alkaline hydrolysis of glyceryl tristearate using aqueous

NaOH

.

Identified as saponification yielding glycerol and sodium octadecanoate.

Base hydrolysis of fats irreversibly converts ester groups into glycerol and carboxylate salts (soap).

3

Analyze the catalytic hydrogenation of glyceryl trioleate.

Identified as hardening of oils.

Addition of

H_2

across the

C=C

double bonds of unsaturated oleic acid residues converts liquid oil into saturated solid fat.

4

Analyze the acid-catalyzed reaction of ethyl ethanoate with water.

Identified as reversible ester hydrolysis.

Acid hydrolysis of esters is reversible, establishing an equilibrium mixture of the parent alkanoic acid (ethanoic acid) and alkanol (ethanol).

Key Concept

Reactivity and Interconversion of Alkanoic Acids, Esters, Fats, and Oils

Question 4Question →

A student subjects separate samples of methyl propanoate to two different laboratory reactions:

Reaction 1: Refluxing with dilute $\text{HCl}(aq)$
Reaction 2: Refluxing with aqueous $\text{NaOH}(aq)$

Which of the following correctly compares the reversibility of these reactions and the nature of the organic products formed?

Reaction 1 is reversible yielding propanoic acid and methanol, whereas Reaction 2 is irreversible yielding sodium propanoate and methanol.

Both Reaction 1 and Reaction 2 are fully reversible equilibrium processes yielding propanoic acid and methanol.

Reaction 1 is irreversible yielding sodium propanoate, whereas Reaction 2 is reversible yielding propanoic acid.

Both Reaction 1 and Reaction 2 are irreversible single-step neutralizations yielding sodium propanoate and methanol.

Show answer & explanation

Answer: Reaction 1 is reversible yielding propanoic acid and methanol, whereas Reaction 2 is irreversible yielding sodium propanoate and methanol.

Answer

Reaction 1 is reversible yielding propanoic acid and methanol, whereas Reaction 2 is irreversible yielding sodium propanoate and methanol.

The statement specifying that Reaction 1 is reversible yielding propanoic acid and methanol, while Reaction 2 is irreversible yielding sodium propanoate and methanol, is correct. Acid-catalyzed hydrolysis of an ester is an equilibrium process that forms the parent alkanoic acid and alkanol. In contrast, alkaline hydrolysis (saponification) uses hydroxide ions to deprotonate the acid as it forms, producing an unreactive alkanoate salt (sodium propanoate) which prevents the reverse reaction, making the process quantitative and irreversible.

Step-by-Step Solution

1

Analyze Reaction 1 (Acid Hydrolysis)

Methyl propanoate reacts with water in the presence of

\text{H}^+

catalyst to form propanoic acid (

\text{C}_2\text{H}_5\text{COOH}

) and methanol (

\text{CH}_3\text{OH}

).

Acid hydrolysis of an ester is the reverse of esterification and reaches dynamic equilibrium (it is reversible).

2

Analyze Reaction 2 (Alkaline Hydrolysis / Saponification)

Methyl propanoate reacts with

\text{OH}^-

ions to form the propanoate anion (

\text{C}_2\text{H}_5\text{COO}^-

) as sodium propanoate salt and methanol (

\text{CH}_3\text{OH}

).

The base reacts un-reversibly with the carboxylic acid product to form a carboxylate salt, pulling the equilibrium completely to the right and making the overall saponification irreversible.

3

Compare both reactions

Reaction 1 is reversible (producing acid + alcohol), while Reaction 2 is irreversible (producing salt + alcohol).

Differentiates reversible acid-catalyzed ester hydrolysis from irreversible base-promoted ester hydrolysis.

Key Concept

Difference between acid hydrolysis (reversible) and alkaline hydrolysis/saponification (irreversible) of esters.

Question 5Question →

Match each chemical process involving alkanoic acids, esters, or fats on the left with its correct chemical description on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Esterification pairs with 'Reversible reaction between an alkanoic acid and an alkanol forming an ester and water'. Saponification pairs with 'Alkaline hydrolysis of fats yielding soap and glycerol'. Hydrogenation of oils pairs with 'Addition of hydrogen gas across C=C double bonds to convert liquid oils into solid fats'.

Esterification is defined as the reversible condensation between an alkanoic acid and an alkanol. Saponification is the base-catalyzed hydrolysis of esters/fats to produce soap and glycerol. Hydrogenation reduces unsaturation in vegetable oils by adding hydrogen across carbon-carbon double bonds.

Step-by-Step Solution

1

Analyze Esterification

Esterification combines an alkanoic acid and an alkanol in a reversible equilibrium reaction to form an ester and water.

This matches the second description.

2

Analyze Saponification

Saponification breaks down triacylglycerols (fats/oils) using sodium or potassium hydroxide, yielding glycerol and salts of fatty acids (soap).

This matches the first description.

3

Analyze Hydrogenation of oils

Hydrogenation saturates the double bonds in unsaturated vegetable oils using a nickel catalyst to turn liquid oils into solid margarine.

This matches the third description.

Key Concept

Reactions and industrial processes of alkanoic acids, esters, fats, and oils

Question 6Question →

During the esterification reaction between ethanoic acid ( $\text{CH}_3\text{COOH}$ ) and ethanol ( $\text{C}_2\text{H}_5\text{OH}$ ), concentrated tetraoxosulfate(VI) acid ( $\text{H}_2\text{SO}_4$ ) is added to the reaction mixture. Which of the following statements correctly accounts for the dual function of concentrated $\text{H}_2\text{SO}_4$ in maximizing the equilibrium yield of ethyl ethanoate?

It functions as a catalyst to increase the reaction rate and as a dehydrating agent to absorb water, shifting the equilibrium position to the right.

It acts as a strong inorganic acid reactant that undergoes irreversible ionic neutralization with ethanol to produce ethyl ethanoate complete to completion.

It serves solely as a catalyst, which selectively increases the rate of the forward reaction and permanently increases the equilibrium constant (

K_c

).

It acts primarily as an oxidizing agent that converts ethanol into ethanoic acid before esterification begins.

Show answer & explanation

Answer: It functions as a catalyst to increase the reaction rate and as a dehydrating agent to absorb water, shifting the equilibrium position to the right.

Answer

Concentrated tetraoxosulfate(VI) acid acts as both a catalyst to speed up the reaction rate and a dehydrating agent to remove water, shifting the reversible equilibrium toward the formation of ethyl ethanoate.

The correct response accurately identifies both functions of concentrated tetraoxosulfate(VI) acid in esterification: it acts as a catalyst to increase the rate at which equilibrium is reached and as a strong dehydrating agent that removes water from the system, driving the reversible equilibrium to the right to maximize ester production according to Le Chatelier's principle.

Step-by-Step Solution

1

Analyze the chemical equation for esterification

\text{CH}_3\text{COOH}(l) + \text{C}_2\text{H}_5\text{OH}(l) \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5(l) + \text{H}_2\text{O}(l)

Esterification is a reversible organic reaction between an alkanoic acid and an alkanol to form an ester and water.

2

Identify the catalytic role of concentrated

\text{H}_2\text{SO}_4

Provides

\text{H}^+

ions to protonate the carbonyl oxygen, accelerating both forward and backward rates.

Lowering activation energy allows the system to reach equilibrium faster.

3

Identify the dehydrating role and apply Le Chatelier's principle

Concentrated

\text{H}_2\text{SO}_4

absorbs water (

\text{H}_2\text{O}

), decreasing product concentration.

Removing a product from a reversible system shifts the equilibrium position to the right, increasing the yield of the ester.

Key Concept

Reversibility of Esterification and Dual Role of Concentrated H2SO4

Question 7Question →

Which of the following chemical changes occurs during the industrial conversion of a liquid vegetable oil into solid margarine via catalytic hydrogenation?

The ester linkages are hydrolyzed to yield glycerol and free fatty acids

The degree of unsaturation decreases as carbon-carbon double bonds are converted to single bonds

The alkanoic acid chains undergo dehydration to produce gaseous alkynes

The triacylglycerol molecules are oxidized to form short-chain dicarboxylic acids

Show answer & explanation

Answer: The degree of unsaturation decreases as carbon-carbon double bonds are converted to single bonds

Answer

The degree of unsaturation decreases as carbon-carbon double bonds are converted to single bonds.

Vegetable oils are naturally liquid at room temperature due to their high degree of unsaturation (presence of multiple

C=C

double bonds in the fatty acid tails). During industrial hydrogenation, hydrogen gas (

H_2

) is reacted with the oil using a nickel catalyst at elevated temperatures. This addition reaction converts unsaturated

C=C

bonds into saturated

C-C

single bonds, thereby decreasing the degree of unsaturation, increasing the packing efficiency of the molecules, and raising the melting point to produce solid margarine.

Step-by-Step Solution

1

Identify the chemical structure of vegetable oils

Vegetable oils are liquid triacylglycerols (esters of glycerol and long-chain fatty acids) containing a high proportion of unsaturated carbon-carbon double bonds (

C=C

).

Liquid fats have lower melting points due to kinks created by double bonds in the hydrocarbon chains.

2

Determine the reaction mechanism of catalytic hydrogenation

Addition of hydrogen gas (

H_2

) in the presence of a finely divided nickel catalyst converts

C=C

double bonds into

C-C

single bonds.

Adding hydrogen saturates the hydrocarbon chains, causing them to pack more tightly and raising their melting point to solidify the oil into margarine.

Key Concept

Hardening of Oils and Hydrogenation of Unsaturated Lipids

Estimated Time:1m 0s

Question 8Question →

Arrange the following sequential steps involved in the laboratory preparation and isolation of soap (saponification) from vegetable oil in the correct chronological order from first to last:

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order of steps for the preparation and isolation of soap is: first, refluxing the vegetable oil with concentrated sodium hydroxide (saponification); second, adding concentrated sodium chloride solution (salting out); third, filtering the mixture to separate the solid soap curd; and fourth, washing the solid soap with cold distilled water to remove impurities.

Saponification begins by refluxing vegetable oil (a triglyceride ester) with concentrated sodium hydroxide, hydrolyzing the ester links to yield sodium alkanoates (soap) and glycerol. Next, concentrated sodium chloride solution is added (salting out) to precipitate the soap by decreasing its solubility. The mixture is then filtered to isolate the solid soap curd from the liquid filtrate, and finally, the solid residue is washed with cold distilled water to remove excess sodium hydroxide.

Step-by-Step Solution

1

Reflux vegetable oil with aqueous sodium hydroxide solution.

Complete alkaline hydrolysis of the triester (triglyceride) into glycerol and sodium alkanoates (soap).

Base-catalyzed ester hydrolysis is required to break down the vegetable oil into fatty acid sodium salts.

2

Add concentrated sodium chloride solution (brine) to the reaction mixture.

Precipitation of sodium alkanoate (soap) as a floating curd.

Increasing the concentration of sodium ions forces the soap salt out of solution due to the common-ion effect and reduced solubility.

3

Filter the mixture using a funnel and filter paper.

Crude solid soap is retained on the filter paper while glycerol and brine pass through as filtrate.

Filtration separates the insoluble soap precipitate from the soluble glycerol byproduct.

4

Rinse the collected soap residue with cold water.

Purified soap free from unreacted sodium hydroxide and salt.

Cold water dissolves remaining surface impurities without dissolving significant amounts of the solid soap.

Key Concept

Saponification process and salting out of soap

Estimated Time:1m 30s

Question 9Question →

Arrange the following organic compounds, each having a relative molecular mass of approximately $58-60\text{ g/mol}$ , in order of INCREASING boiling point (from lowest boiling point to highest boiling point).

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence in order of increasing boiling point is Butane (

\text{C}_4\text{H}_{10}

), followed by Methyl formate (

\text{HCOOCH}_3

), then Propan-1-ol (

\text{C}_3\text{H}_7\text{OH}

), and finally Ethanoic acid (

\text{CH}_3\text{COOH}

).

Boiling points depend on the relative strength of intermolecular forces when molecular masses are comparable (~60 g/mol). Butane is non-polar and exhibits only weak dispersion forces (lowest boiling point). Methyl formate is polar and exhibits dipole-dipole attractions. Propan-1-ol forms strong hydrogen bonds via its hydroxyl group. Ethanoic acid forms even stronger hydrogen bonds and stable cyclic dimers, giving it the highest boiling point.

Step-by-Step Solution

1

Identify the primary type of intermolecular force present in each compound of similar molar mass (

\approx 58-60\text{ g/mol}

).

Butane has London dispersion forces; Methyl formate has dipole-dipole forces; Propan-1-ol has hydrogen bonding; Ethanoic acid has extensive hydrogen bonding and dimer formation.

Boiling point increases as the strength of intermolecular forces holding the liquid molecules together increases.

2

Compare the compounds without hydrogen bonding capabilities (Butane vs. Methyl formate).

Butane is non-polar and exhibits only weak dispersion forces. Methyl formate has a polar carbonyl group (

\text{C=O}

) causing dipole-dipole attractions, making its boiling point higher than butane.

Permanent dipole-dipole attractions are stronger than instantaneous dispersion forces for molecules of similar size.

3

Compare the hydrogen-bonded compounds (Propan-1-ol vs. Ethanoic acid).

Propan-1-ol forms intermolecular hydrogen bonds through its single hydroxyl group. Ethanoic acid forms stronger hydrogen bonds using both its carbonyl oxygen and hydroxyl hydrogen to form stable cyclic dimers.

Dimerization in alkanoic acids effectively doubles the molecular interaction area, requiring significantly more thermal energy to break apart during vaporization.

4

Arrange the compounds in order of increasing boiling point.

Butane < Methyl formate < Propan-1-ol < Ethanoic acid.

The progression of intermolecular force strength directly dictates the trend in boiling points.

Key Concept

Intermolecular Forces and Boiling Point Trends in Alkanoic Acids, Esters, Alkanols, and Alkanes

Question 10Question →

An organic compound $X$ with the molecular formula $\text{C}_4\text{H}_8\text{O}_2$ reacts with aqueous sodium hydroxide upon heating to yield ethanol and a salt $Y$ . What is the IUPAC name of compound $Y$ ?

Sodium ethanoate

Ethanoic acid

Sodium propanoate

Sodium methanoate

Show answer & explanation

Answer: Sodium ethanoate

Answer

Sodium ethanoate

Alkaline hydrolysis of the four-carbon ester ethyl ethanoate (

\text{CH}_3\text{COOCH}_2\text{CH}_3

) with sodium hydroxide breaks the ester linkage to produce ethanol (

\text{C}_2\text{H}_5\text{OH}

) and the sodium salt of ethanoic acid, which is sodium ethanoate (

\text{CH}_3\text{COONa}

).

Step-by-Step Solution

1

Identify the structure of the ester from its molecular formula and hydrolysis alcohol product.

The ester has molecular formula

\text{C}_4\text{H}_8\text{O}_2

and produces ethanol (

\text{C}_2\text{H}_5\text{OH}

). The alkyl group attached to oxygen contains 2 carbons, leaving 2 carbons for the acyl group. Thus, the ester is ethyl ethanoate (

\text{CH}_3\text{COOCH}_2\text{CH}_3

).

Esters have the general formula

\text{RCOOR}'

, where

\text{R}'\text{OH}

is the alcohol component.

2

Determine the products of alkaline hydrolysis.

Heating ethyl ethanoate with aqueous sodium hydroxide (

\text{NaOH}

) cleaves the ester link to yield ethanol and sodium ethanoate (

\text{CH}_3\text{COONa}

).

Alkaline hydrolysis (saponification) yields the alcohol and the alkali metal salt of the alkanoic acid.

Key Concept

Alkaline Hydrolysis (Saponification) of Esters

Question 11Question →

An organic compound $P$ with the molecular formula $\text{C}_4\text{H}_8\text{O}_2$ reacts with aqueous sodium hydroxide upon heating to produce a sodium salt $Q$ and an alkanol $R$ . Complete oxidation of alkanol $R$ with acidified potassium dichromate(VI) yields an alkanoic acid identical to the acid produced when salt $Q$ is acidified with dilute hydrochloric acid. What is the IUPAC name of compound $P$ ?

Methyl propanoate

Ethyl ethanoate

Propyl methanoate

Butanoic acid

Show answer & explanation

Answer: Ethyl ethanoate

Answer

Ethyl ethanoate

Ethyl ethanoate has the molecular formula

\text{C}_4\text{H}_8\text{O}_2

. Alkaline hydrolysis of ethyl ethanoate with sodium hydroxide yields sodium ethanoate (salt

Q

) and ethanol (alkanol

R

). Acidification of sodium ethanoate yields ethanoic acid (2 carbons). Complete oxidation of ethanol with acidified potassium dichromate(VI) also yields ethanoic acid (2 carbons). Because both pathways yield the exact same acid (ethanoic acid), ethyl ethanoate satisfies all conditions.

Step-by-Step Solution

1

Determine the functional group of compound P.

Compound P (

\text{C}_4\text{H}_8\text{O}_2

) reacts with

\text{NaOH}

to yield a salt and an alkanol, identifying

P

as an ester with general formula

\text{R}^1\text{COOR}^2

.

Alkaline hydrolysis (saponification) of esters produces a carboxylate salt and an alkanol.

2

Analyze the carbon distribution from the reaction products.

Acidifying salt

Q

(

\text{R}^1\text{COONa}

) gives alkanoic acid

\text{R}^1\text{COOH}

(containing

n_1 + 1

carbon atoms). Oxidation of primary alkanol

R

(

\text{R}^2\text{OH}

) gives alkanoic acid

\text{R}'\text{COOH}

(containing

n_2

carbon atoms).

Primary alkanols undergo complete oxidation with acidified

\text{K}_2\text{Cr}_2\text{O}_7

to produce alkanoic acids with the same number of carbon atoms as the alkanol.

3

Equate the carbon counts of the two acids formed.

Since both processes yield the identical acid, the acid must have 2 carbon atoms (ethanoic acid). Thus,

n_1 + 1 = 2 \implies n_1 = 1

(methyl group

\text{CH}_3-

) and

n_2 = 2

(ethyl group

-\text{C}_2\text{H}_5

).

The total number of carbon atoms in ester

P

is 4 (

1 + 1 + 2 = 4

). Dividing 4 total carbons equally between the acyl and alkoxy portions yields ethanoic acid derivative and ethanol derivative.

4

Deduce the structure and IUPAC name of ester P.

Ester

P

is

\text{CH}_3\text{COOCH}_2\text{CH}_3

, which has the IUPAC name ethyl ethanoate.

The IUPAC name of an ester consists of the alkyl group attached to the oxygen followed by the alkanoate chain.

Key Concept

Ester Saponification and Oxidation of Alkanols

Question 12Question →

Propanoic acid is heated under reflux with ethanol in the presence of concentrated tetraoxosulfate(VI) acid to produce a sweet-smelling liquid ester. This ester is separated and subsequently boiled under reflux with aqueous potassium hydroxide until saponification is complete. Which pair of organic products is isolated from the saponification mixture?

Potassium propanoate and ethanol

Potassium ethanoate and propan-1-ol

Propanoic acid and ethanol

Ethanoic acid and propan-1-ol

Show answer & explanation

Answer: Potassium propanoate and ethanol

Answer

The isolated products of the alkaline hydrolysis are potassium propanoate and ethanol.

In the initial esterification reaction, propanoic acid and ethanol react in the presence of concentrated tetraoxosulfate(VI) acid catalyst to form ethyl propanoate (

\text{C}_2\text{H}_5\text{COOCH}_2\text{CH}_3

). When ethyl propanoate is subsequently hydrolyzed under basic conditions with aqueous potassium hydroxide (saponification), the carbonyl-oxygen ester linkage is irreversibly cleaved. This forms the potassium salt of the carboxylic acid (potassium propanoate) and regenerates the alcohol (ethanol).

Step-by-Step Solution

1

Identify the ester formed in the esterification step

Propanoic acid (

\text{C}_2\text{H}_5\text{COOH}

) + Ethanol (

\text{C}_2\text{H}_5\text{OH}

)

\xrightarrow{\text{conc. H}_2\text{SO}_4}

Ethyl propanoate (

\text{C}_2\text{H}_5\text{COOCH}_2\text{CH}_3

) + Water (

\text{H}_2\text{O}

)

Esterification combines the alkanoic acid acyl group (

\text{C}_2\text{H}_5\text{CO-}

) with the alkoxy group (

\text{-OCH}_2\text{CH}_3

) of the alkanol.

2

Determine the saponification reaction of ethyl propanoate with potassium hydroxide

Ethyl propanoate (

\text{C}_2\text{H}_5\text{COOCH}_2\text{CH}_3

) +

\text{KOH(aq)} \rightarrow

Potassium propanoate (

\text{C}_2\text{H}_5\text{COOK}

) + Ethanol (

\text{C}_2\text{H}_5\text{OH}

)

Alkaline hydrolysis cleaves the ester bond to produce the potassium salt of the alkanoic acid and the free alkanol.

Key Concept

Alkaline Hydrolysis (Saponification) of Esters

Estimated Time:2m 0s

Question 13Question →

Match each chemical reaction or process involving alkanoic acid derivatives on the left with its corresponding principal product on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The reaction of ethanoic acid and ethanol (esterification) pairs with Ethyl ethanoate and water. The alkaline hydrolysis of oil (saponification) pairs with Sodium salt of fatty acid (soap) and glycerol. The addition of hydrogen to liquid oil (catalytic hydrogenation) pairs with Solid saturated fat (margarine).

Esterification of ethanoic acid and ethanol yields ethyl ethanoate and water. Saponification of vegetable oils with aqueous sodium hydroxide yields fatty acid sodium salts (soap) and glycerol. Catalytic hydrogenation of unsaturated vegetable oils yields solid saturated fats (margarine).

Step-by-Step Solution

1

Identify the products of esterification

Ethanoic acid reacts with ethanol in the presence of concentrated tetraoxosulfate(VI) acid to produce ethyl ethanoate and water.

The hydroxyl group of the acid combines with hydrogen from the alkanol to form water, linking the remaining fragments into an ester.

2

Identify the products of saponification

Triglycerides react with boiling aqueous sodium hydroxide to yield sodium alkanoates (soap) and propane-1,2,3-triol (glycerol).

Alkaline cleavage of ester linkages in fats yields carboxylate salts and frees the triol backbone.

3

Identify the products of catalytic hydrogenation

Unsaturated fatty acid chains in liquid vegetable oils undergo addition of hydrogen to form saturated chains, hardening the oil into solid fat.

Reducing double bonds increases the melting point, transforming liquid oils into margarine.

Key Concept

Reactions and Industrial Products of Alkanoic Acids, Esters, Fats, and Oils

Question 14Question →

Which of the following statements correctly distinguishes the esterification reaction between ethanoic acid and ethanol from the neutralization reaction between ethanoic acid and aqueous sodium hydroxide?

Esterification is a slow, reversible covalent reaction catalyzed by an acid, whereas neutralization is a rapid, irreversible ionic reaction.

Esterification is a rapid, irreversible ionic reaction, whereas neutralization is a slow reversible reaction that requires refluxing.

Esterification yields a alkanoic acid salt and water, whereas neutralization yields an ester and glycerol.

Concentrated tetraoxosulfate(VI) acid acts as an oxidizing agent in esterification, while sodium hydroxide acts as a dehydrating catalyst in neutralization.

Show answer & explanation

Answer: Esterification is a slow, reversible covalent reaction catalyzed by an acid, whereas neutralization is a rapid, irreversible ionic reaction.

Answer

Esterification is a slow, reversible covalent reaction catalyzed by an acid, whereas neutralization is a rapid, irreversible ionic reaction.

The statement identifying esterification as a slow, reversible covalent reaction catalyzed by an acid and neutralization as a rapid, irreversible ionic reaction is correct because organic esterification involves covalent bond rearrangement reaching dynamic equilibrium, whereas neutralization is an ionic combination forming water completely.

Step-by-Step Solution

1

Analyze the nature of esterification

Esterification takes place between an alkanoic acid and an alkanol in the presence of concentrated

\text{H}_2\text{SO}_4

. It is a reversible, molecular (covalent) reaction that proceeds slowly to reach equilibrium.

Covalent bond cleavage and formation require activation energy and proceed reversibly.

2

Analyze the nature of neutralization

Neutralization takes place between ethanoic acid and a strong base (

\text{NaOH}

), forming sodium ethanoate salt and water completely.

The reaction involves free ions in solution (

\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}

), making it rapid and virtually quantitative (irreversible).

3

Compare the key characteristics

Esterification is slow, reversible, and acid-catalyzed; neutralization is rapid, irreversible, and ionic.

This highlights the fundamental difference between organic ester formation and acid-base salt formation.

Key Concept

Reversibility and Kinetics of Esterification vs Neutralization

Estimated Time:1m 0s

Question 15Question →

A vegetable oil containing glyceryl tristearate is boiled with aqueous sodium hydroxide during soap manufacturing. Which of the following sets of products is formed from this saponification reaction?

Propane-1,2,3-triol and sodium stearate

Stearic acid and sodium hydroxide

Propane-1,2,3-triol and octadecan-1-ol

Octadecane and sodium trioxocarbonate(IV)

Show answer & explanation

Answer: Propane-1,2,3-triol and sodium stearate

Answer

Propane-1,2,3-triol and sodium stearate

Saponification is the alkaline hydrolysis of fats and oils (triesters of glycerol). When glyceryl tristearate is heated with aqueous sodium hydroxide, the ester bonds are irreversibly broken to yield propane-1,2,3-triol (glycerol) and sodium stearate, which is a soap.

Step-by-Step Solution

1

Identify the functional group and reactants

Glyceryl tristearate is a triacylglycerol (fat/ester) reacting with a strong alkali (

\text{NaOH}

).

Understanding the nature of the reactants helps determine the type of chemical process taking place.

2

Determine the reaction mechanism (saponification)

Alkaline hydrolysis cleaves the three ester ester bonds in the triglyceride.

Base-catalyzed hydrolysis of esters is irreversible and yields the alcohol component and the carboxylate salt.

3

Identify the resulting products

The alcohol component formed is propane-1,2,3-triol (glycerol) and the salt formed is sodium stearate (soap).

The glycerol backbone is released as propane-1,2,3-triol while the long-chain fatty acid chains form sodium carboxylate salts.

Key Concept

Saponification of Fats and Oils

Alkanoic Acids, Esterification, Saponification, Fats, and Oils

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