Vidaara.orgClass 11 · Chemistry
CodeVID-C11-12-CH-01
Chapter Assignment — Organic Chemistry: Some Basic Principles and Techniques
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- This is a full-length test covering the whole chapter — every topic is included.
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks. Show all working for Sections B, C and D.
Section A — Multiple Choice Questions
6 × 1 = 6 marks
1.
The hybridisation of carbon in ethene ($CH_2=CH_2$) is:
- A.$sp$
- B.$sp^2$
- C.$sp^3$
- D.$sp^3d$
2.
The IUPAC name of $CH_3COCH_3$ is:
- A.propanal
- B.propan-2-one
- C.ethanal
- D.propan-1-ol
3.
Propan-1-ol and propan-2-ol are:
- A.chain isomers
- B.position isomers
- C.functional isomers
- D.metamers
4.
Which species is an electrophile?
- A.$OH^-$
- B.$NH_3$
- C.$NO_2^+$
- D.$CN^-$
5.
The most stable free radical is:
- A.methyl
- B.primary
- C.secondary
- D.tertiary
6.
Nitrogen is confirmed in Lassaigne's test by a:
- A.white precipitate
- B.Prussian-blue colour
- C.violet colour
- D.milky lime-water
Section B — Short Answer (2 marks)
4 × 2 = 8 marks
7.
Define a homologous series and give its two characteristics.
8.
Distinguish a nucleophile from an electrophile with one example each.
9.
Why does butan-2-ol show optical activity while butan-1-ol does not?
10.
State the principle of steam distillation and name one compound it purifies.
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
11.
Arrange the carbocations $CH_3^+$, $(CH_3)_2CH^+$ and $(CH_3)_3C^+$ in order of stability and justify using +I and hyperconjugation.
12.
On combustion, $0.30\ \text{g}$ of a compound gave $0.44\ \text{g}$ $CO_2$ and $0.18\ \text{g}$ $H_2O$. Find $\%C$ and $\%H$.
Section D — Long Answer (5 marks)
2 × 5 = 10 marks
13.
Explain the inductive, resonance and hyperconjugation effects, and use them to explain why $CH_3COO^-$ is more stable than $CH_3O^-$.
14.
Describe, with the percentage formula, the estimation of nitrogen by the Dumas method, and compute $\%N$ if $0.25\ \text{g}$ of a compound gives $40\ \text{mL}$ of $N_2$ at STP (1 mole gas = 22400 mL at STP).
Answer Key
Section A — Multiple Choice Questions
- (B) $sp^2$
- (B) propan-2-one
- (B) position isomers
- (C) $NO_2^+$
- (D) tertiary
- (B) Prussian-blue colour
Section B — Short Answer (2 marks)
- A family of compounds with the same functional group, represented by a general formula, in which successive members differ by a $CH_2$ unit (14 u) and show a gradual change in physical properties with similar chemistry.
- A nucleophile is electron-rich and donates an electron pair (e.g. $OH^-$); an electrophile is electron-deficient and accepts an electron pair (e.g. $NO_2^+$).
- Butan-2-ol has a chiral carbon (C-2) bonded to four different groups ($−H$, $−OH$, $−CH_3$, $−C_2H_5$), so it has non-superimposable mirror images; butan-1-ol has no such carbon.
- A steam-volatile compound immiscible with water distils with steam at a temperature below $100^\circ$C, because the combined vapour pressures equal atmospheric pressure. Example: aniline.
Section C — Short Answer (3 marks)
- $CH_3^+ < (CH_3)_2CH^+ < (CH_3)_3C^+$. More electron-releasing alkyl groups (+I) and more $\alpha$-C−H bonds for hyperconjugation increasingly disperse the positive charge, so the tertiary cation is most stable.
- $\%C=\dfrac{12}{44}\times\dfrac{0.44}{0.30}\times100=40\%$; $\%H=\dfrac{2}{18}\times\dfrac{0.18}{0.30}\times100=6.67\%$.
Section D — Long Answer (5 marks)
- Inductive effect: permanent $\sigma$-bond polarisation by an electronegative atom, transmitted through the chain (+I for alkyl, −I for $−NO_2$/$−Cl$). Resonance (mesomeric): delocalisation of $\pi$/lone-pair electrons over a conjugated system, lowering energy. Hyperconjugation: delocalisation of $\sigma(C−H)$ electrons into an adjacent empty $p$/$\pi$ orbital. In $CH_3COO^-$ the negative charge is delocalised equally over two oxygen atoms by resonance, spreading the charge and stabilising the ion; in $CH_3O^-$ the charge is localised on one oxygen with no resonance, so it is less stable.
- In the Dumas method the compound is heated with CuO in a stream of $CO_2$; nitrogen is set free as $N_2$, collected over KOH (which absorbs $CO_2$) and its volume measured at known $T$ and $P$, corrected to STP. $\%N=\dfrac{28}{22400}\times\dfrac{V_{STP}}{m}\times100$. Substituting $V=40\ \text{mL}$, $m=0.25\ \text{g}$: $\%N=\dfrac{28}{22400}\times\dfrac{40}{0.25}\times100=\dfrac{28}{22400}\times16000=20\%$.
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