From the chapter Organic Chemistry: Some Basic Principles and Techniques, some of the important questions are covered in this question paper. The questions are covers from the book back and the previous year questions.

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Questions:

Generic 2

8 x 2 = 16

Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound?

What is meant by delocalization?

Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens.

Draw the possible resonance structure for \(C{ H }_{ 3 }-\overset { \bullet \bullet }{ \underset { \bullet \bullet }{ O } } -\overset { + }{ C } { H }_{ 2 }\) and predict which of the structures is more stable. Give reason for your answer.

How many \(\sigma\) and \(\pi\) bonds are present in each of the following molecules?
CH₃CH₂C \(\equiv\) N

Name a suitable technique of separation of the components from a mixture of calcium sulphate and camphor.

Name the type organic reaction involved in the following reaction.
CH₃CH₂I + KOH(aq) \(\longrightarrow \) CH₃ CH₂OH + KI

What are primary and secondary suffixes as applied to IUPAC nomenclature?

Generic 3

5 x 3 = 15

A sample of 0.50 g of an organic compound was treated according to Kjeldahl’s method. The ammonia evolved was absorbed in 50 ml of 0.5 M H₂SO₄. The residual acid required 60 mL of 0.5 M solution of NaOH for neutralisation. Find the percentage composition of nitrogen in the compound.

10.

During estimation of nitrogen present in an organic compound by Kjeldahl’s method, the ammonia evolved from 0.5 g of the compound in Kjeldahl’s estimation of nitrogen, neutralized 10 mL of 1 M H₂SO₄ .Find out the percentage of nitrogen in the compound.

11.

In Carius method of estimation of halogen, 0.15 g of an organic compound gave 0.12 g of AgBr. Find out the percentage of bromine in the compound.

12.

Consider structure I to VII and answer the following questions (i) and (ii).
I. \({ CH }_{ 3 }-{ CH }_{ 2 }-{ CH }_{ 2 }-{ CH }_{ 2 }-{ OH }\)
II. \({ CH }_{ 3 }-{ CH }_{ 2 }-{ CH }-{ CH }_{ 3 }\\ \quad \quad \quad \quad \quad \quad \quad |\\ \quad \quad \quad \quad \quad \quad \quad { OH }\)
III. \({ \quad \quad \quad \quad CH }_{ 3 }\\ \quad \quad \quad \quad |\\ { { CH }_{ 3 } }-{ C }-{ CH }_{ 3 }\\ \quad \quad \quad \quad |\\ \quad \quad \quad \quad { OH }\)
IV. \({ CH }_{ 3 }-{ CH }-{ CH_{ 2 } }-{ OH }\\ \quad \quad \quad \quad |\\ \quad \quad \quad \quad { CH }_{ 3 }\)
V. \({ CH }_{ 2 }-{ CH }_{ 2 }-{ O }-{ CH }_{ 2 }-{ CH }_{ 3 }\)
VI. \({ CH }_{ 2 }-{ O }-{ CH }_{ 2 }-{ CH }_{ 2 }-{ CH }_{ 3 }\)
VII. \({ CH }_{ 3 }-{ O }-{ CH }-{ CH }_{ 3 }\\ \quad \quad \quad \quad \quad \quad |\\ \quad \quad \quad \quad \quad \quad { CH }_{ 3 }\)
Identify the pairs of compound which are functional group isomers.

13.

Write structural formula for all the isomeric amines with molecular formula C₄H₁₁ N.

Generic 5

1 x 5 = 5

14.

In Dumas’ method for estimation of nitrogen, 0.3g of an organic compound gave 50mL of nitrogen collected at 300K temperature and 715mm pressure. Calculate the percentage composition of nitrogen in the compound. (Aqueous tension at 300K=15 mm)

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Answers Key:

Generic 2

8 x 2 = 16

CO₂ is slightly acidic in nature therefore, it reacts with the strong base KOH to form K₂CO₃ and from the weight of the CO₂ obtained, the percentage of carbon in the organic compound is calculated.
2 KOH + CO₂ \(\longrightarrow \) K₂CO₃ + H₂O
The increase in the weight of U-tube containing KOH gives the weight of CO2 produced and from the weight of the CO2 obtained, the percentage of carbon in the organic compound is calculated as
% of carbon = \(\frac { 12 }{ 44 } \times \frac { Weight \ of \ CO_{ 2 } \ formed }{ Weight \ of \ substance \ taken } \times 100\)

Delocalisation means that pairs of bonding electrons extend over three or more atoms and belong to the whole molecule. Delocalisation \(\pi \) orbital are much larger than the localized \(\pi \) orbital and are therefore more stable.

Nitrogen, sulphur and halogens are present in an organic compound in covalent form so their detection is not easy.By fusing with Na metal,these elements are converted into NaCN or Na₂S or NaX i.e in an ionic form.In ionic form, these can be easily detected by ionic reactions.

The given carbocation has two resonance structures. Structures (II) is more stable because both the carbon atoms and the oxygen atom have an octet of electrons.

\({ \sigma }_{ C-C }=2, \ { { \sigma }_{ C-H }=5, \ }{ { \sigma }_{ C-N } }=1,\ { { \pi }_{ C-N } }=2\)

A mixture of CaSO₄ and camphor can be separated by sublimation. Because camphor is sublimate but CaSO₄ is not therefore, sublimation of the mixture gives camphor on the sides of funnel while CaSO₄ is left in the China dish.

This is an example of nucleophilic substitution reaction because the nucleophile I^- is replaced by the nucleophile OH^-.

The primary suffix indicates whether the carbon chain is saturated or unsaturated while the secondary suffix indicates the functional group present in the molecule.

Generic 3

5 x 3 = 15

Volume of acid taken = 50 mL of 0.5 M H₂SO₄
= 25 mL of 1.0 M H₂SO₄
Volume of alkali used for neutralisation of excess acid
= 60 mL of 0.5 M NaOH
= 30 mL of 1.0 M NaOH
H₂SO₄ + 2NaOH\(\longrightarrow \)Na₂SO₄ + 2H₂O
1 mole of H₂SO₄ = 2 moles of NaOH
Hence 30 mL of 1.0 M NaOH
= 15 mL of 1.0 M H₂SO₄
\(\therefore \) Volume of acid used by ammonia = 25 - 15 = 10 mL
% of nitrogen = 1.4 x N₁x Vol. of acid used/w
(where N₁ = Normality of acid and w = mass of the organic compound taken)
% of nitrogen = (1.4 x 2 x 10/0.5) = 56.0.

10.

1 M of 10 mL H₂SO₄=1M of 20 mL NH₃
1000 mL of 1M ammonia contains 14 g nitrogen 20 mL of 1M ammonia contains \(=\frac{14 \times 12}{1000} g\) nitrogen
Percentage of nitrogen \(=\frac{14 \times 20 \times 100}{1000 \times 0.5} = 56.0\%\)

11.

Molar mass of AgBr = 108 + 80 = 188 g mol^-1
188 g AgBr contains 80 g bromine
0.12 g AgBr contains \(= \frac{80 \times 0.12}{188}\) g bromine
Percentage of bromine \(=\frac { 80\times 0.12\times 100 }{ 188\times 0.15 } =34.04\%\)

12.

I and V, I and VI, I and VII; II and V, II and VI, II and VII; III and V, II and VI; III and VI; III and VII; IV and V; IV and VI and IV and VII are functional group of isomers.

13.

(i) \({ CH }_{ 3 }-{ CH }_{ 2 }-{ CH }_{ 2 }-{ CH }_{ 2 }-{ NH }_{ 2 }\)
(ii)
\({ CH }_{ 3 }-{ CH }_{ 2 }-{ CH }-{ CH }_{ 3 }\\ \quad \quad \quad \quad \quad \quad \quad |\\ \quad \quad \quad \quad \quad \quad \quad { NH }_{ 2 }\)
(iii)
\({ CH }_{ 3 }-{ CH }-{ CH_{ 2 } }-{ NH }_{ 2 }\\ \quad \quad \quad \quad |\\ \quad \quad \quad \quad { CH }_{ 3 }\)
(iv)
\({ \quad \quad \quad \quad CH }_{ 3 }\\ \quad \quad \quad \quad |\\ { { CH }_{ 3 } }-{ C }-{ NH }_{ 2 }\\ \quad \quad \quad \quad |\\ \quad \quad \quad \quad { CH }_{ 3 }\)
(v)
\({ CH }_{ 3 }-{ CH }_{ 2 }-{ CH }_{ 2 }-{ NH }-{ CH }_{ 3 }\)
(vi)
\({ CH }_{ 3 }-{ CH }-{ NH }-{ CH }_{ 3 }\\ \quad \quad \quad \quad |\\ \quad \quad \quad \quad { CH }_{ 3 }\)
(vii)
\({ CH }_{ 3 }-{ CH }_{ 2 }-{ NH }-{ CH }_{ 2 }-{ CH }_{ 3 }\)
(viii)
\({ CH }_{ 3 }-{ CH }_{ 2 }-{ N }-{ CH }_{ 3 }\\ \quad \quad \quad \quad \quad \quad \quad |\\ \quad \quad \quad \quad \quad \quad \quad { CH }_{ 3 }\)

Generic 5

1 x 5 = 5

14.

Volume of nitrogen collected at 300K and 715mm pressure is 50 mL Actual pressure = 715-15 =700 mm
Volume of nitrogen at STP = \(\frac { 273\times 700\times 50 }{ 300\times 760 } =41.9 \ mL\)
22,400 mL of N₂ at STP weighs = 28 g
41.9mL of nitrogen weighs = \(\frac { 28\times 41.9 }{ 22400 } \)g
Percentage of nitrogen =\(\frac { 28\times 41.9\times 100 }{ 22400\times 0.3 } =17.46%\)%