Chapter 8: Organic Chemistry – Some Basic Principles and Techniques Application Questions | chemistry Class 11 CBSE

Chapter 8: Organic Chemistry – Some Basic Principles and Techniques Application Questions | chemistry Class 11 CBSE | ByteNotes.in

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 1

Applied Concept

A forensic chemist receives an unknown white solid suspected to be an organic compound. Describe how they would qualitatively detect the presence of nitrogen and sulphur in the sample using Lassaigne's test, including specific observations.

The forensic chemist would fuse the compound with sodium metal in a fusion tube; this converts covalent C-N and C-S bonds to ionic NaCN and Na2S; the fused mass is dissolved in distilled water to obtain sodium fusion extract.
For nitrogen detection: the extract is boiled with FeSO4 solution and acidified with concentrated H2SO4; appearance of Prussian blue colour confirms nitrogen (due to formation of Fe4[Fe(CN)6]3).
For sulphur detection: the extract is acidified with acetic acid and treated with lead acetate; formation of a black precipitate of PbS confirms sulphur; alternatively, addition of sodium nitroprusside gives a violet colour.
If both nitrogen and sulphur are present, blood-red colour (NaSCN + Fe3+) may be observed instead of Prussian blue; excess sodium fusion decomposes the thiocyanate into cyanide and sulphide, allowing individual tests for each.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 2

Applied Concept

In a petroleum refinery, crude oil containing hydrocarbons with boiling points ranging from 30 degrees C to over 400 degrees C is to be separated into useful fractions like LPG, petrol, kerosene, diesel, and lubricating oil. Which distillation technique is used and why?

Fractional distillation is used because crude oil is a mixture of miscible liquids (hydrocarbons) with boiling points that are closely spaced and span a wide range, making simple distillation ineffective for separation.
The crude oil mixture is heated in a furnace and vapours enter a tall fractionating column; components with lower boiling points (more volatile) rise higher in the column before condensing; higher-boiling fractions condense lower in the column.
The fractionating column provides multiple theoretical plates (repeated condensation-vaporisation surfaces), enriching the vapour progressively in the more volatile component and achieving effective separation.
Each fraction (LPG at the top, petrol, kerosene, diesel, and lubricating oil/residue from bottom up) is collected at different heights corresponding to its boiling point range; this is one of the most important industrial applications of fractional distillation.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 3

Applied Concept

A student prepares aspirin in the laboratory and wants to check its purity and identity quickly before proceeding. Describe how TLC would be used for this purpose and what result would indicate a pure product.

The student dissolves a small amount of the product in a suitable solvent and spots it on a TLC plate along with a reference spot of pure aspirin standard, both at the same baseline position.
The TLC plate is developed in an appropriate eluant (mobile phase) in a closed jar; the eluant rises by capillary action, carrying components at different rates depending on their degree of adsorption.
After development, the Rf value of the product spot is calculated (distance moved by spot / distance moved by solvent front) and compared to the Rf of the authentic aspirin standard under identical conditions.
A pure sample would show a single spot with an Rf value matching the standard aspirin; the presence of additional spots or a non-matching Rf indicates impurities or an incorrect product, prompting further purification.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 4

Applied Concept

With reference to the inductive effect, explain why acetic acid (CH3COOH) is a weaker acid than chloroacetic acid (ClCH2COOH), which is in turn weaker than dichloroacetic acid (Cl2CHCOOH).

The acidity of a carboxylic acid depends on the stability of its carboxylate anion; greater stability of the anion means greater acidity.
The chlorine atom exerts a strong electron-withdrawing inductive effect (-I effect), pulling electron density away from the carboxylate group and dispersing the negative charge, thereby stabilising the anion and increasing acidity.
With two chlorine atoms in Cl2CHCOOH, the cumulative electron-withdrawing effect is greater than with one Cl in ClCH2COOH; thus Cl2CHCOOH is more acidic than ClCH2COOH.
Acetic acid (CH3COOH) has a methyl group which donates electrons by the +I effect, slightly destabilising the carboxylate anion and making it the weakest acid among the three; the order of acidity is Cl2CHCOOH > ClCH2COOH > CH3COOH.

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Application Question

Easy difficulty • Concept in a practical situation

Easy

Question 5

Applied Concept

A chemist needs to separate a mixture of camphor (sublimable) and common salt (non-sublimable). Which purification technique should be used and why is it specifically suitable for this mixture?

Sublimation is the appropriate technique; camphor sublimes (converts directly from solid to vapour without passing through the liquid state) when gently heated.
The mixture is placed in a china dish covered with an inverted funnel plugged with cotton; on gentle heating, camphor sublimes and the vapours condense on the cooler funnel walls as pure camphor crystals.
Common salt (NaCl) is non-sublimable and remains behind in the china dish; the separation is clean and efficient with no solvent needed.
This technique is ideal specifically because the two components differ fundamentally in their ability to sublime; it would not work if both components sublimed or neither sublimed, making it a highly selective purification method.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 6

Applied Concept

In a reaction, bromomethane (CH3Br) reacts with hydroxide ion (OH-). With reference to the chapter concepts of nucleophile, electrophile, and heterolytic cleavage, explain the mechanism of this reaction.

In CH3Br, the C-Br bond is polar because Br is more electronegative than C; the carbon atom carries a partial positive charge (electrophilic centre) due to the electron-withdrawing inductive effect of Br.
Hydroxide ion (OH-) carries a lone pair of electrons and a negative charge; it is a nucleophile (electron pair donor) that attacks the electrophilic carbon of CH3Br.
The curved arrow from the lone pair of OH- points to the electrophilic carbon; simultaneously the C-Br bond breaks heterolytically with both electrons going to Br, forming Br- (the leaving group).
This is a nucleophilic substitution reaction (specifically SN2); the products are CH3OH (methanol) and Br-; the reaction proceeds via ionic (heterolytic or polar) mechanism as both carbocation/carbanion or direct displacement pathways are possible.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 7

Applied Concept

A student incorrectly names the compound CH3CH2CH(CH3)CH2CH3 as '3-methylpentane'. Using IUPAC rules, identify the error and give the correct name.

The student chose a five-carbon parent chain and placed the methyl substituent at C-3; however, the correct IUPAC name requires identifying the longest continuous carbon chain.
The longest chain in CH3CH2CH(CH3)CH2CH3 contains six carbon atoms (hexane), not five; the methyl group is a branch on this six-carbon chain.
Numbering from either end, the methyl branch is at C-3; since the same locant results from either direction, the name is 3-methylhexane.
The error was selecting an incorrect (shorter) parent chain; IUPAC rule 1 states that the longest continuous carbon chain must be chosen as the parent chain, regardless of how the formula is written.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 8

Applied Concept

With reference to the concept of resonance, explain why the carboxylate ion (CH3COO-) is more stable than expected from a single Lewis structure, and how this relates to the higher acidity of acetic acid compared to ethanol.

A single Lewis structure of CH3COO- would show one C-O single bond and one C=O double bond; this predicts two different C-O bond lengths, but experimentally both C-O bonds are equivalent (bond length intermediate between single and double).
The actual structure is a resonance hybrid of two equivalent canonical structures in which the negative charge is delocalized over both oxygen atoms; this charge delocalization provides extra stability called resonance stabilisation energy.
In ethanol (CH3CH2OH), deprotonation gives ethoxide ion (CH3CH2O-) in which the negative charge is localised on a single oxygen atom with no delocalization; it is less stable than the delocalized carboxylate ion.
Since the carboxylate ion is more stable (lower energy) than the ethoxide ion, acetic acid loses its proton more readily (has a lower pKa) than ethanol; resonance stabilisation of the conjugate base is the key factor making carboxylic acids much stronger acids than alcohols.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 9

Applied Concept

A compound has the molecular formula C4H10O and gives a positive test for halogens in Lassaigne's test with different silver halide precipitates. Can this be correct? Also, how many structural isomers are possible for C4H10O as an ether?

The premise is incorrect: C4H10O does not contain any halogen; Lassaigne's test for halogens would give no precipitate with AgNO3 since no halogen atoms are present in the molecule.
C4H10O as an ether (R-O-R) would show the molecular formula by having the oxygen atom between two alkyl groups; the total alkyl carbons must add to 4.
The ether isomers are: methoxyproPane (CH3OC3H7), ethoxyethane (C2H5OC2H5), and methoxy-2-methylpropane (CH3OCH(CH3)2 or (CH3)2CHOCH3).
The phenomenon where C4H10O gives two different ethers (methoxypropane and ethoxyethane) with different alkyl chains on either side of the -O- group is called metamerism, a type of structural isomerism.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 10

Applied Concept

An organic compound on combustion with CuO gives 0.44 g of CO2 and 0.18 g of H2O from 0.30 g of the compound. Calculate the percentage of C and H and determine whether oxygen is also present.

Percentage of carbon: (12/44) x (0.44/0.30) x 100 = (0.3/0.30) x 100 = (12 x 0.44 x 100) / (44 x 0.30) = 40%.
Percentage of hydrogen: (2/18) x (0.18/0.30) x 100 = (2 x 0.18 x 100) / (18 x 0.30) = 6.67%.
Sum of percentages of C and H = 40 + 6.67 = 46.67%; since this is less than 100%, the remaining 53.33% is likely oxygen (assuming only C, H, O are present).
Presence of oxygen is confirmed by difference from 100%; this method of estimating oxygen by subtracting the percentages of all other elements from 100 is the standard approach in quantitative organic analysis.

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Chapter 8: Organic Chemistry – Some Basic Principles and Techniques | Application Questions

A forensic chemist receives an unknown white solid suspected to be an organic compound. Describe how they would qualitatively detect the presence of nitrogen and sulphur in the sample using Lassaigne's test, including specific observations.

In a petroleum refinery, crude oil containing hydrocarbons with boiling points ranging from 30 degrees C to over 400 degrees C is to be separated into useful fractions like LPG, petrol, kerosene, diesel, and lubricating oil. Which distillation technique is used and why?

A student prepares aspirin in the laboratory and wants to check its purity and identity quickly before proceeding. Describe how TLC would be used for this purpose and what result would indicate a pure product.

With reference to the inductive effect, explain why acetic acid (CH3COOH) is a weaker acid than chloroacetic acid (ClCH2COOH), which is in turn weaker than dichloroacetic acid (Cl2CHCOOH).

A chemist needs to separate a mixture of camphor (sublimable) and common salt (non-sublimable). Which purification technique should be used and why is it specifically suitable for this mixture?

In a reaction, bromomethane (CH3Br) reacts with hydroxide ion (OH-). With reference to the chapter concepts of nucleophile, electrophile, and heterolytic cleavage, explain the mechanism of this reaction.

A student incorrectly names the compound CH3CH2CH(CH3)CH2CH3 as '3-methylpentane'. Using IUPAC rules, identify the error and give the correct name.

With reference to the concept of resonance, explain why the carboxylate ion (CH3COO-) is more stable than expected from a single Lewis structure, and how this relates to the higher acidity of acetic acid compared to ethanol.

A compound has the molecular formula C4H10O and gives a positive test for halogens in Lassaigne's test with different silver halide precipitates. Can this be correct? Also, how many structural isomers are possible for C4H10O as an ether?

An organic compound on combustion with CuO gives 0.44 g of CO2 and 0.18 g of H2O from 0.30 g of the compound. Calculate the percentage of C and H and determine whether oxygen is also present.

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