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Chapter 13

Fundamental Principles of Organic Chemistry

NEB Class 11 past exam questions

Past questions and answers

6 questions set from this chapter, 1 of them more than once; 1 is most repeated (set, or a close variant set, in 3 or more exams). Most repeated first.

  • Most repeated · 4 of 4 exams
  • Asked 4 times
  • 2074 (2017) · Regular, 112 'A' · 5 marks
  • 2074 (2018) · Upgrading (Supplementary) · 5 marks
  • 2075 (2018) · 112 'B' · 5 marks
  • 2076 (2019) · 112 'B' · 5 marks

How would you detect the presence of nitrogen in an organic compound by Lassaigne test method?

Answer

Principle: nitrogen in the organic compound is converted by fusion with sodium into ionic sodium cyanide, which is then detected as Prussian blue.

Procedure

  1. Preparation of sodium extract: a small piece of dry sodium metal is heated in a fusion tube until it melts, and a little of the organic compound is added. The tube is heated strongly to red hot and plunged into distilled water in a china dish. It is ground, boiled and filtered. The filtrate is the Lassaigne's (sodium fusion) extract.

Na+C+N→ΔNaCN\mathrm{Na + C + N \xrightarrow{\Delta} NaCN}

  1. Test: to a portion of the extract add freshly prepared FeSO4_4 solution and boil. Sodium hexacyanoferrate(II) forms:

6NaCN+FeSO4→Na4[Fe(CN)6]+Na2SO4\mathrm{6NaCN + FeSO_4 \rightarrow Na_4[Fe(CN)_6] + Na_2SO_4}

  1. Add a few drops of FeCl3_3 solution (or warm and acidify with dilute H2_2SO4_4/HCl, which dissolves the iron(II) hydroxide). Hexacyanoferrate(II) reacts with Fe3+^{3+} to give

3Na4[Fe(CN)6]+4FeCl3→Fe4[Fe(CN)6]3+12NaCl\mathrm{3Na_4[Fe(CN)_6] + 4FeCl_3 \rightarrow Fe_4[Fe(CN)_6]_3 + 12NaCl}

Result: a Prussian blue colour or precipitate confirms nitrogen. (If sulphur is also present, a blood-red colour of Fe(SCN)2+^{2+} may form instead.)

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2075 (2018) · 112 'B' · 1+1 marks

Define the terms. (i) Inductive effect (ii) Electrophile

Answer

(i) Inductive effect: the permanent displacement of the shared electrons of a σ bond along a carbon chain, caused by an electronegative atom or group attached to it. Example: in Cl ⁣→ ⁣CH2 ⁣− ⁣CH3\mathrm{Cl\!\to\!CH_2\!-\!CH_3}, Cl pulls the electrons toward itself.

(ii) Electrophile: an electron-deficient species that can accept an electron pair from another species to form a covalent bond, e.g. H+\mathrm{H^+}, NO2+\mathrm{NO_2^+}, AlCl3\mathrm{AlCl_3}.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2076 (2019) · 112 'B' · 1+1 marks

Define inductive effect and write its one use.

Answer

Inductive effect: the permanent polarisation (displacement) of σ\sigma-electrons along a saturated carbon chain, caused by the presence of an electron-withdrawing (−I-I) or electron-donating (+I+I) atom or group.

Use: it explains the relative acid strengths of carboxylic acids. For example, ClCH2COOH\mathrm{ClCH_2COOH} is a stronger acid than CH3COOH\mathrm{CH_3COOH}, because the −I-I effect of chlorine stabilises the carboxylate ion. It likewise explains the basic strength of amines and the stability of carbocations.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • NEB Class 11 Chemistry Model Question · 1 mark

One of the examples of homologous series is given below. CH3_3OH CH3_3CH2_2OH X CH3_3CH2_2CH2_2CH2_2OH i. Define homologous series.

Answer

A homologous series is a series of organic compounds having the same functional group, in which each successive member differs from the previous one by a –CH₂– unit (i.e., by a molecular mass of 14 amu), and which can be represented by a common general formula. All members show similar chemical properties and a gradual change in physical properties.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • NEB Class 11 Chemistry Model Question · 2 marks

Find the mass difference between successive member of above homologous series and calculate the molecular mass of X.

Answer

Mass difference. Successive members differ by one −CH2−-\text{CH}_2- unit:

Mass difference=12+2(1)=14 u\text{Mass difference} = 12 + 2(1) = 14\ \text{u}

Molecular mass of X. The series is CH3_3OH, CH3_3CH2_2OH, X, CH3_3CH2_2CH2_2CH2_2OH, so X is the third member, propan-1-ol, CH3CH2CH2OH\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} (C3H8O\text{C}_3\text{H}_8\text{O}).

Mr=3(12)+8(1)+16=60 uM_r = 3(12) + 8(1) + 16 = 60\ \text{u}

Check: ethanol is 46 u, so 46+14=6046 + 14 = 60 u.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • NEB Class 11 Chemistry Model Question · 2 marks

What is the reason behind the highest boiling point but least solubility of the fourth member in the given series?

Answer

The "fourth member" of the homologous series of alcohols is butan-1-ol (C₄H₉OH).

Highest boiling point: Boiling point increases down a homologous series because molecular mass and molecular size increase, which strengthens the van der Waals (London dispersion) forces between molecules. Among the first four alcohols (methanol → butanol), butanol has the largest molecular mass and longest carbon chain, so it has the strongest intermolecular forces and hence the highest boiling point.

Least solubility in water: Solubility in water is due to the polar –OH group, which forms hydrogen bonds with water. As the hydrocarbon (alkyl) chain lengthens, the non-polar part grows and dominates over the small polar –OH group. The alkyl part is hydrophobic and does not interact with water; it actually disrupts the water–water hydrogen bonding. So in butanol the non-polar character outweighs the polar character, making it the least soluble of the four.

In short: The same factor — the increasing size of the non-polar alkyl group — raises the boiling point (stronger van der Waals forces) but lowers water solubility (weaker interaction with water relative to the hydrophobic chain).

Answer checked by Claude Sonnet · Oct 21, 2026

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Questions from NEB Class 11 Chemistry Model Question (New Course, Code 3011) (NEB/CDC model paper, Grade 11, 2077/2078), Practice questions written for this site (Written from the official syllabus and specification grid where no past papers cover a chapter), NEB Grade XI 2074 (2017) Chemistry 112 'A' (old course) (Old HSEB/NEB Grade XI regular board exam, 2074 (2017). Old-course paper; only questions fitting the current outline taken.), NEB Grade XI 2074 (2018) Chemistry 112 Upgrading (Supplementary) (Old-course NEB Grade XI upgrading/supplementary exam, 2074 (2018).), NEB Grade XI 2075 (2018) Chemistry 112 'B' (old course) (Old-course NEB Grade XI board exam, 2075 (2018), set B.) and NEB Grade XI 2076 (2019) Chemistry 112 'B' (old course) (Old-course NEB Grade XI board exam, 2076 (2019), set B.). Answers are written for this site; check them against your class notes.

Chapter titles and hours from the official syllabus ↗