NCERT Solutions Class 11 Chemistry Chapter 12

Q.1 What are hybridisation states of each carbon atom in the following compounds?

CH₂=C=O, CH₃CH=CH₂, (CH₃)₂CO, CH₂=CHCN, C₆H₆

Ans.

Q.2 Indicate the σ and p bonds in the following molecules:

C₆H₆, C₆H₁₂, CH₂Cl₂, CH₂=C=CH₂, CH₃NO₂, HCONHCH₃

Ans.

C₆H₆ CH₃NO₂

C₆H₁₂ CH₂=C=CH₂

CH₂Cl₂ HCONHCH₃

Q.3 Write bond line formulas for: Isopropyl alcohol, 2,3-Dimethylbutanal, Heptan-4-one.

Ans.

Q.4 Give the IUPAC names of the following compounds:

Ans.

a) Propylbenzene

b) 3-Methylpentanenitrile

c) 2,5-Dimethylheptane

d) 3-Bromo 3-chloroheptane

e) 3-Chloropropanal

f) 2,2-Dichloroethanol

Q.5 Which of the following represents the correct IUPAC name for the compounds concerned? (a) 2,2-Dimethylpentane or 2-Dimethylpentane (b) 2,4,7-Trimethyloctane or 2,5,7-Trimethyloctane (c) 2-Chloro-4-methylpentane or 4-Chloro-2 methyl pentane (d) But-3-yn-1-ol or But-4-ol-1-yne.

Ans.

(a) 2, 2-Dimethylpentane is correct name because two methyl groups are located on the same carbon and its locant is repeated on the same carbon.

(b) 2, 4, 7-Trimethyloctane is the correct one; the locant set 2, 4, 7 is lower than 2, 5, 7.

(c) 2-Chloro-4-methylpentane is the correct one as chlorine atom is placed in the lower locant.

(d) But-3-yn-1-ol is the correct one as the principal functional group ‘ol’ is in lower locant in this nomenclature.

Q.6 Draw formulas for the first five members of each homologous series beginning with the following compounds. (a) H–COOH (b) CH₃COCH₃ (c) H–CH=CH₂

Ans.

Q.7 Give condensed and bond line structural formulas and identify the functional group(s) present, if any, for:

(a) 2, 2, 4-Trimethylpentane

(b) 2-Hydroxy-1, 2, 3-propanetricarboxylic acid

(c) Hexanedial

Ans.

Condensed structure:

a) CH₃C(CH₃)₂CH₂CH(CH₃)CH₃

b) HOOCCH₂C(OH)(COOH)CH₂COOH

c) OHC(CH₂)₄CHO

Bond line formula:

Q.8 Identify the functional groups in the following compounds:

Ans.

Q.9 Will CCl₄ give white precipitate of AgCl on heating it with silver nitrate? Give reason for your answer.

Ans.

O₂NCH₂CH₂O^– is more stable than CH₃CH₂O^– because NO₂ group has –I effect. It tends to disperse the –ve charge on the O– atom.

In CH₃CH₂O^– , CH₃CH₂ group has +I effect, So, it tends to intensify the negative charge on the oxygen atom and hence it destabilises the structure.

Q.10 Explain why alkyl groups act as electron donors when attached to a p system.

Ans.

Alkyl groups can act as electron donors when attached to a p- system due to the hyperconjugation.

Hyperconjugation can be easily understood by the following structures:

Q.11 Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation.

(a) C₆H₅OH (b) C₆H₅NO₂ (c) CH₃CH=CHCHO (d) C₆H₅–CHO (e) C₆H₅CH₂⁺ (f) CH₃CH=CHCH₂⁺.

Ans.

a) Resonance structures of C₆H₅OH:

b) Resonance structures of nitrobenzene:

c) Resonance structures of But-2-en-1-al:

d) Resonance structures of Benzaldehyde:

e) Resonance structures of Benzyl carbocation:

f) Resonance structures of but-2-en-1-yl carbocation:

Q.12 What are electrophiles and nucleophiles? Explain with examples.

Ans.

Electrophiles: Electrophiles are electron deficient species. They may be positively charged or electrically neutral species. E.g: H⁺, NO₂⁺, NO⁺, R⁺ (carbocation) and alkyl free radicals(R^.), carbenes(:CR₂)etc.

Nucleophiles: Nucleophiles are electron rich species containing at least one lone pair of electrons. They may be negatively charged or electrically neutral species. E.g: H^–(hydride ion), Cl^–, Br^–, OH^–, OR^–, CN^– ,H₂O , NH₃,RNH₂,etc.

Q.13 Identify the reagents shown in bold in the following equations as nucleophiles or electrophiles:

a) CH₃COOH + OH^– → CH₃COO^– + H₂O

b) CH₃COOH₃ + CN^– → (CH₃)₂C(CN)(OH)

c) C₆H₆ + CH₃CO⁺ → C₆H₅COCH₃

Ans.

In reaction (a) hydroxide ion acts as nucleophile and in (b) cyanide ion acts as nucleophile. In reaction (c) the reagent is electrophile.

Q.14 Classify the following reactions in one of the reaction type studied in this unit.

a) CH₃CH₂Br + HS^– → CH₃CH₂SH + Br^–

b) (CH₃)₂C = CH₂ + HCl → (CH₃)₂CCl — CH₃

c) CH₃CH₂Br + HO^– → CH₂ = CH₂ + H₂O + Br^–

c) (CH₃)₃C — CH₂OH + HBr → (CH₃)₂CBrCH₂CH₃ + H₂O

Ans.

a) Nucleophilic substitution reaction

b) Electrophilic addition

c) Bimolecular elimination reaction

d) Nucleophilic substitution reaction with rearrangement

Q.15

What is the relationship between the members of following pairs of structures? Are they structural or geometrical isomers or resonance contributors?

(a)

(b)

(c)

Ans.

a) Structural isomers

b) Geometrical isomers.

c) They are resonance contributors because they differ in position of electrons but not atoms.

Q.16 For the following bond cleavages, use curved-arrows to show the electron flow and classify each as homolysis or heterolysis. Identify reactive intermediate produced as free radical, carbocation and carbanion.

(a) CH₃O — OCH₃ → CH₃O•OCH₃

Ans.

Q.17 Explain the terms Inductive and Electromeric effects. Which electron displacement effect explains the following correct orders of acidity of the carboxylic acids?

(a) Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH

(b) CH₃CH₂COOH > (CH₃)₂CHCOOH > (CH₃)₃C.COOH

Ans.

Inductive effect: It is the displacement of electrons along the chain of carbon atoms caused due to the presence of groups or atoms of different electronegativity at one end of the chain. It is a permanent effect and involves sigma bond. Electronegativity difference causes the shifting of electron density towards more electronegative atom.

Electromeric effect: It is temporary effect and occurs only when a reagent is added to a system and involves transfer of shared pair of pi electrons of the multiple bond to one of the linked atoms.

(a) –I effect explains the order of acidity of the given acids.

As the number of halogen (here it is Cl atom) atoms decreases, the –I effect also decreases and the acid strength decreases accordingly. Due to the –I effect of Cl atom, the electron density over the oxygen atom decreases, it helps in the removal of the proton. Thus more the number of chlorine atoms higher will be the acidity.

(b) +I-effect explains the order of acidity of the given acids.

As the number of alkyl group (alkyl group has +I effect) increases, the +I effect also increases, thus the acid strength decreases. Due to the +I effect of methyl group, the charge over the oxygen atom gets intensified, thus the removal of proton becomes difficult. More the number of methyl groups, lesser will be the acidity.

Q.18 Give a brief description of the principles of the following techniques taking an example in each case. (a)Crystallisation (b) Distillation (c) Chromatography.

Ans.

(a) Crystallisation: Crystallisation is a process of purification by which crystals of pure compound are obtained from impure sample of compound. It is a technique which is used to purify solid compounds. The basic principle used in this technique is the solubility difference of the compound and the impurity present in the suitable solvent. The impure compound is dissolved in a suitable solvent in which it is sparingly soluble at room temperature but soluble in appreciable amount at higher temperature. The solution is then heated and saturated solution of impure compound is prepared. The crystals of the pure compound separate out when the solution is cooled, which are removed/separated by filtration. The remaining solution contains impurity and small amount of the pure compound.

Example: Crystallisation of sugar: A sample of sugar having impurity of common salt. The impure sugar sample is dissolved in ethanol with shaking and heating. The sugar dissolves in the solvent and the salt remains insoluble. The hot solution is filtered out and then concentrated. Then the solution is allowed to cool, the sugar crysytals separate out.

(b) Distillation: This process is used to separate two liquids having difference in boiling point by heating followed by condensation of the vapours. In this process, liquid is converted into vapours by heating, followed by condensation of the vapours by cooling them. The liquids purified by this method, contain non-volatile impurity and are stable at their boiling points.

Example: By this technique, simple organic liquids such as ethanol, benzene, chloroform, acetone, and carbon tetrachloride can be separated from impurities. When the liquid mixture is heated, it gradually starts boiling when the vapour pressure becomes equal to the atmospheric pressure. The more volatile component distills over first, and the less volatile component distills over later. The impurities which are non-volatile are left over the distillation flask.

(c) Chromatography: It is the method of separation in which components of a mixture are separated by differential movement of individual components through a stationary phase under the influence of a mobile phase. In chromatography, the stationary phase can be either solid or tightly bound liquid on solid support, while the mobile phase can be either liquid or gas. Two important types of chromatography which are widely used are column chromatography and thin layer chromatography.

Q.19 Describe the method, which can be used to separate two compounds with different solubilities in a solvent S.

Ans.

The two compounds having different solubility in solvent S can be separated out by fractional crystallization. The saturated solution of the two compounds is heated to prepare saturated solution. Then it is allowed to cool down. The component which is less soluble, crystallizes out first while the more soluble one remains soluble in the solution. The crystals of the first component are separated out from the mother liquor. Then the mother liquor is again concentrated and the hot solution is again cooled down and the crystals of the second component are obtained.

Q.20 What is the difference between distillation, distillation under reduced pressure and steam distillation?

Ans.

Distillation involves the conversion of the liquid into vapour and condensation of the vapour into liquid. On cooling, we get the pure liquid and the non-volatile impurities remain in the flask. The non-volatile impurities are separated out. This method is applicable for the liquids which are quite stable at their boiling points and contain non-volatile impurities.

Distillation under reduced pressure is used for the liquids which tend to decompose on boiling. In this process, the applied pressure acting on the system is less than the atmospheric pressure; the pressure is reduced by using vacuum pump. This kind of method used to purify the liquids which decompose at or below their boiling points. For example, glycerol is purified by this method.

Steam distillation is quite similar to that of distillation under reduced pressure but there is no reduction in the total pressure acting on the solution. This process is used to purify and separate the liquids which are steam volatile.

The vapour pressure of water is quite high than other liquid, so the mixture will boil at a temperature much lower than its normal boiling point, thus decomposition can be avoided. A mixture of water and aniline is separated using this method.

Q.21 Discuss the chemistry of Lassaigne’s test.

Ans.

This test is used for the detection of nitrogen, halogens, phosphorus and sulphur in an organic compound. In this test covalent bonds present in organic compounds are converted into ionic bonds. It is done by fusing organic compound with sodium.

The chemistry involved in it is as follows

The C and N present in the organic compound combine to form sodium cyanide.

\text{Na}\text{+}\text{C}\text{+}\text{N}\stackrel{\text{Δ}}{\to }\text{NaCN}

If S and N both are present in the organic sample, during the fusion they combine to form sodium thiocyanate.

Na + N + S + C → NaSCN

Sod. thiocyanate

Organic compounds containing halogens react with sodium to give NaX.

Q.22 Differentiate between the principles of estimation of nitrogen in an organic compound by (i) Dumas method and (ii) Kjeldahl’s method.

Ans.

i) Dumas method: A known mass of organic compound is heated with excess CuO in CO₂ atmosphere, the N of the organic compound is converted into N₂ gas. By this reaction, the volume of obtained N₂ gas is converted into NTP. Then the volume obtained in N.T.P. further converted into its mass. We will get the percentage of nitrogen by applying the following equation:

\text{\%N =}\left(\text{28/22400}\right)\text{×}\frac{\text{Volume}\text{of}{\text{N}}_{\text{2}}\text{at}\text{N}\text{.T}\text{.P}\text{.}}{\text{Mass}\text{of the substance taken}}\text{×100}

ii) Kjeldahl’s method: A known mass of organic sample is heated with conc. H₂SO₄ in presence of K₂SO₄ and catalyst Hg or CuSO₄ in a long necked flask called Kjeldahl’s flask, N present in the compound is converted into (NH₄)₂SO₄. It is further boiled with excess of NaOH solution to liberate NH₃ gas which is absorbed in a known excess volume of a standard acid such as H₂SO₄ or HCl.

The volume of acid unused (unreacted) is determined by titration against standard alkali solution. The %N can be calculated by the following equation:

\text{\%N =}\frac{\text{1}\text{.4}\text{×}\text{Molarity of the acid × basicity of the acid × Vol}\text{. of the acid used}}{\text{Mass}\text{of the substance taken}}

Q.23 Discuss the principle of estimation of halogens, sulphur and phosphorus present in an organic compound.

Ans.

The principle of estimation of halogens: A known mass of organic sample is heated with fuming nitric acid and few drops of silver nitrate in a sealed hard glass tube (Carius tube). The C and H present in the sample are converted into the CO₂ and H₂O respectively while the silver salt is converted into silver halide. The precipitates of silver halide formed are filtered out, washed, dried and weighed. Knowing the mass of the substances taken and the mass of the precipitates formed, the % of halogen can be calculated by using the following formula:

\text{\%}\text{of halogen in compound =}\frac{\text{Atomic mass of X}}{\text{108 + At}\text{. mass of X}}\text{×}\text{m}\text{×}\left(\text{100}\right)

X = Halogen present, m=mass of the silver halide,

w = mass of the substance taken.

The principle of estimation of sulphur: A known mass of organic sample is heated with fuming nitric acid or sodium peroxide in Carius glass tube. The C and H present in the sample are converted into the CO₂ and H₂O respectively while S present is oxidized to sulphuric acid, which then reacts with excess of BaCl₂ solution and precipitates as BaSO₄.

\begin{array}{l}\text{C + 2O}\left(\text{from}{\text{HNO}}_{\text{3}}\right)\stackrel{\text{Δ}}{\to }{\text{CO}}_{\text{2}}\text{2H+O}\left(\text{from}{\text{HNO}}_{\text{3}}\right)\stackrel{\text{Δ}}{\to }{\text{H}}_{\text{2}}\text{O}\\ \text{S}\text{+}{\text{H}}_{\text{2}}\text{O}\text{+}\text{3O}\left(\text{from}{\text{HNO}}_{\text{3}}\right)\stackrel{\text{Δ}}{\to }{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{+}{\text{BaCl}}_{\text{2}}\stackrel{\text{Δ}}{\to }\underset{\text{White}\text{ppt}\text{.}}{{\text{BaSO}}_{\text{4}}\text{+}}\text{2HCl}\end{array}

The ppt. of BaSO₄ is collected, washed, dried and weighed. The % of S can be calculated by using the following formula:

\text{\% of S =}\frac{\text{32}}{\text{233}}\text{×}\frac{\text{Mass}\text{of}{\text{BaSO}}_{\text{4}}\text{formed}}{\text{Mass of substance taken}}\text{×100}

The principle of estimation of phosphorus: A known mass of organic sample is heated with fuming nitric acid in a Carius tube; C and H are oxidized to CO₂ and H₂O respectively, while P present in the compound is oxidized to phosphoric acid, later it reacts with magnesia mixture (MgCl₂ + NH₄Cl + NH₃). A ppt. of magnesium ammonium phosphate is formed which further ignited to obtain magnesium pyrophosphate.

\begin{array}{l}\text{C}\text{+}\text{2O}\left(\text{from}{\text{HNO}}_{\text{3}}\right)\stackrel{\text{Δ}}{\to }{\text{CO}}_{\text{2}}\text{2H}\text{+}\text{O}\left(\text{from}{\text{HNO}}_{\text{3}}\right)\stackrel{\text{Δ}}{\to }{\text{H}}_{\text{2}}\text{O}\\ \text{2P}\text{+}\text{5O}\left(\text{from}{\text{HNO}}_{\text{3}}\right)\stackrel{\text{Δ}}{\to }{\text{P}}_{\text{2}}{\text{O}}_{\text{5}}{\text{P}}_{\text{2}}{\text{O}}_{\text{5}}\text{+}{\text{3H}}_{\text{2}}\text{O}\stackrel{\text{Δ}}{\to }\underset{\text{Phosphoric}\text{acid}}{{\text{2H}}_{\text{3}}{\text{PO}}_{\text{4}}}\end{array} {\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\stackrel{\text{Magnesia}\text{mixture}}{\to }\underset{\begin{array}{l}\text{Magnesium}\text{ammonium}\\ \text{phosphate}\end{array}}{{\text{MgNH}}_{\text{4}}{\text{PO}}_{\text{4}}} {\text{2MgNH}}_{\text{4}}{\text{PO}}_{\text{4}}\stackrel{\text{Δ}}{\to }\underset{\begin{array}{l}\text{Magnesium}\\ \text{pyrophosphate}\end{array}}{{\text{Mg}}_{\text{2}}{\text{P}}_{\text{2}}{\text{O}}_{\text{7}}\text{+}}{\text{2NH}}_{\text{3}}{\text{+ H}}_{\text{2}}\text{O}

The yellow ppt. is collected, washed and weighed. The % of P can be calculated by knowing the mass of the substance taken and the mass of the ppt., by using the following formula:

\text{\%P =}\frac{\text{62}}{\text{222}}\text{×}\frac{\text{Mass}\text{of}{\text{Mg}}_{\text{2}}{\text{P}}_{\text{2}}{\text{O}}_{\text{7}}\text{formed}}{\text{Mass}\text{of}\text{substance}\text{taken}}\text{×100}

Q.24 Explain the principle of paper chromatography.

Ans.

In paper chromatography, chromatographic paper (water trapped in it), acts as the stationary phase. A spot of the solution of the mixture is made on the paper strip which is then suspended in a suitable solvent, which acts as the mobile phase. By capillary action, this solvent rises up the chromatographic paper. The solvent flows over the spot, the components are selectively retained on the paper (because they have different partition coefficient in these two phases). The different components present in the spot travel with the mobile phase to different heights. The paper so obtained is known as a chromatogram.

Q.25 Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens?

Ans.

When we carry out the Lassaigne’s test for the detection of halogens, the sodium extract is first boiled with dilute nitric acid. The acid is added to decompose NaCN to HCN and Na₂S to H₂S and to expel these gases. So, if any nitrogen and sulphur are present in the form of NaCN and Na₂S, they are removed. Otherwise, these will react with AgNO₃ and will interfere with the test. The chemical equations involved in the reaction are represented as:

NaCN + HNO₃ → NaNO₃ + HCN ↑

Na₂S + 2HNO₃ → 2NaNO₃ + H₂S ↑

Q.26 Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens.

Ans.

Nitrogen, sulphur and halogens are covalently bonded in organic compounds. For their detection, they have to be first converted to ionic form. This is done by fusing the organic compound with sodium metal. This test is called “Lassaigne’s test”. The chemical reactions involved in this test are:

\text{Na}\text{+}\text{C}\text{+}\text{N}\stackrel{\text{Δ}}{\to }\text{NaCN}

Na + N + S + C → NaSCN

2Na + S → Na₂S

Na + X → NaX

X = Halogens

Q.27 Name a suitable technique of separation of the components from a mixture of calcium sulphate and camphor.

Ans.

The process which is used to separate a mixture of camphor and calcium sulphate is known as sublimation. In this process, the sublimable compound changes from solid to vapour state without passing through the liquid state. Camphor is sublimable while calcium sulphate is non-sublimable. Therefore, on heating a mixture containing camphor and calcium sulphate, camphor will form vapours leaving behind calcium sulphate.

Q.28

Explain, why an organic liquid vapourises at a temperature below its boiling point in its steam distillation?

Ans.

In steam distillation, the organic liquid starts to boil when the sum of vapour pressure due to the organic liquid (p₁) and the vapour pressure due to water (p₂) becomes equal to atmospheric pressure (p), i.e., p=p₁+p₂.Since p₁< p₂ (as water has high specific heat and boiling point than other liquids) organic liquid will vapourise at a lower temperature than its boiling point.

Q.29 Will CCl₄ give white precipitate of AgCl on heating it with silver nitrate? Give reason for your answer.

Ans.

CCl₄ will not give white precipitate of AgCl on heating it with AgNO₃ because the chlorine atoms are covalently bonded to carbon atom in CCl₄. To get the precipitate Cl atom should be bonded by ionic bond.

Q.30 Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound?

Ans.

CO₂ is acidic gas and potassium hydroxide is a strong base. Hence, carbon dioxide reacts with potassium hydroxide to form potassium carbonate and water. The reaction is as follows:

2KOH + CO₂ → K₂CO₃ + H₂O

Q.31 Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test?

Ans.

During the test of sulphur, the sodium extract is acidified with acetic acid because lead acetate is soluble and does not interfere with the test. Instead of acetic acid, if H₂SO₄ is used, lead acetate will react with it to form white precipitate of lead sulphate which will interfere with the test.

(CH₃COO)₂Pb +H₂SO₄ → PbSO₄ + 2CH₃COOH

Q.32 An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this substance is subjected to complete combustion.

Ans.

We know that,

\text{\%C =}\frac{\text{12}}{\text{44}}\text{×}\frac{\text{Mass}\text{of}{\text{CO}}_{\text{2}}\text{formed}}{\text{Mass}\text{of}\text{substance}\text{taken}}\text{× 100}

Thus, substituting the values of %C and mass of the substance taken, we will get the mass of carbon dioxide formed.

\text{69 =}\frac{\text{12}}{\text{44}}\text{×}\frac{\text{Mass}\text{of}{\text{CO}}_{\text{2}}\text{formed}}{\text{0}\text{.2g}}\text{× 100}

Or,

{\text{Mass of CO}}_{\text{2}}\text{formed}\text{=}\frac{\text{69 × 44 × 0}\text{.2}}{\text{12 × 100}}\text{= 0}\text{.506g}

Similarly,

\text{\%H}\text{=}\frac{2}{18}\times \frac{\text{Mass}\text{of}{\text{H}}_{\text{2}}\text{O}\text{}\text{formed}}{\text{Mass}\text{of}\text{substance}\text{}\text{taken}}\text{× 100}

Substituting the values in the above equation, we have

\text{4}\text{.8}\text{=}\frac{\text{2}}{\text{18}}\text{×}\frac{\text{Mass}\text{of}{\text{H}}_{\text{2}}\text{O}\text{}\text{formed}}{\text{0}\text{.2}}\text{× 100} {\text{Mass of H}}_{\text{2}}\text{O}\text{formed}\text{=}\frac{\text{4}\text{.8}\text{×}\text{18}\text{×}\text{0}\text{.2}}{\text{2 ×}\text{100}}\text{= 0}\text{.0864g}\text{.}

Q.33 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.

Ans.

Total mass of organic compound taken= 0.50 g

60 mL of 0.5 M solution of NaOH was required by residual acid for neutralisation.

Now 1 mole of H₂SO₄ neutralises 2 moles of NaOH.

Thus,

60 mL of 0.5 M NaOH solution =

(60/30) mL of 0.5 M H₂SO₄ solution º

= 20 mL of 0.5 M H₂SO₄ solution

Acid consumed in absorption of evolved ammonia is

(50–30) mL = 20 mL

1 mole of H₂SO₄ neutralizes 2 moles of NH_3.

20 mL of 0.5 M H₂SO₄ neutralises 40 mL of 0.5 M NH₃

Since, 1000 mL of 1 M NH₃ contains 14 g of nitrogen

Therefore, 40 mL of 0.5 M NH₃ will contain=

\frac{\text{14 × 40 × 0}\text{.5}}{\text{1000}}\text{= 0}\text{.28g}\text{of}\text{N}

Therefore, percentage of nitrogen in 0.50 g of organic compound=

\frac{\text{0}\text{.28 × 100}}{\text{0}\text{.5}}\text{= 56\%}

Q.34 0.3780 g of an organic chloro compound gave 0.5740 g of silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound.

Ans.

The mass of the substance (chloro compound) taken = 0.3780 g

Mass of AgCl formed = 0.5740 g

Now,

1 mole of AgCl = 1 g atom of Cl

or (108+35.5) = 143.5 g of AgCl = 35.5 g of Cl

Applying the formula

\text{\% of Cl}\text{=}\frac{\text{35}\text{.5}}{\text{143}\text{.5}}\text{×}\frac{\text{Mass}\text{of}\text{AgCl}\text{formed}}{\text{Mass}\text{of}\text{substance}\text{taken}}\text{×}\text{100} \text{\% of Cl}\text{=}\frac{\text{35}\text{.5}}{\text{143}\text{.5}}\text{×}\frac{\text{0}\text{.5740}}{\text{0}\text{.3780}}\text{×}\text{100 = 37}\text{.566\%}

Thus, the % of Cl present in the organic compound is 37.566%

Q.35 In the estimation of sulphur by Carius method, 0.468 g of an organic sulphur compound afforded 0.668 g of barium sulphate. Find out the percentage of sulphur in the given compound.

Ans.

The mass of the substance taken= 0.468g

Mass of BaSO₄ formed= 0.668g

1 mole of BaSO₄ contain 1 mole of S

Or, [137 + 32+ 4 x 16) = 233 g] of BaSO₄ contain 32 g of S

Applying the formula of % of S, we get

\text{\% of S}\text{=}\frac{\text{32}}{\text{233}}\text{×}\frac{\text{Mass of}{\text{BaSO}}_{\text{4}}\text{formed}}{\text{Mass}\text{of}\text{substance}\text{taken}}\text{×}\text{100} \text{\% of S}\text{=}\frac{\text{32}}{\text{233}}\text{×}\frac{\text{0}\text{.668}}{\text{0}\text{.468}}\text{×}\text{100 = 19}\text{.60\%}

Thus, the % of S present in the organic compound is 19.60%

Q.36 In the organic compound CH₂=CH–CH₂–CH₂–C≡CH, the pair of hybridized orbitals involved in the formation of C₂– C₃ bond is:

(a) sp – sp² (b) sp – sp³ (c) sp²– sp³ (d) sp³– sp³

Ans.

Both the double bond and triple bond are present at equivalent position, and the double bond is given preference over triple bond while numbering the carbon atoms in main carbon chain.

{\text{H}}_{\text{2}}\underset{{\text{sp}}^{\text{2}}}{\overset{\text{1}}{\text{C}}}\text{=}\underset{{\text{sp}}^{\text{2}}}{\overset{\text{2}}{\text{CH}}}\text{—}\underset{{\text{sp}}^{\text{3}}}{\overset{\text{3}}{{\text{CH}}_{\text{2}}}}\text{—}\underset{{\text{sp}}^{\text{3}}}{\overset{\text{4}}{{\text{CH}}_{\text{2}}}}\text{—}\underset{\text{sp}}{\overset{\text{5}}{\text{C}}}\equiv \underset{\text{sp}}{\overset{\text{6}}{\text{CH}}}

Thus, C₂-C₃ bond is formed by the overlap of sp²-sp³ orbitals.
Option (c) is the correct answer.

Q.37 In the Lassaigne’s test for nitrogen in an organic compound, the Prussian blue colour is obtained due to the formation of: (a) Na₄[Fe(CN)₆] (b) Fe₄[Fe(CN)₆]₃ (c) Fe₂[Fe(CN)₆] (d) Fe₃[Fe(CN)₆]_4.

Ans.

The Prussian blue colour is obtained due to the formation of Fe₄[Fe(CN)₆]₃.

Thus option (b) is the right answer.

Q.38 Which of the following carbocation is most stable?

Ans.

The stability of the carbocations follows the order: 3°>2°>1°. (CH₃)₃C⁺ is a tertiary carbocation. A tertiary carbocation is the most stable carbocation due to the + I effect of three methyl groups. An increased + I effect by the three methyl groups stabilises the positive charge on the carbocation. Thus, option (b) is the correct answer.

Q.39 The best and latest technique for isolation, purification and separation of organic compounds is:

(a) Crystallisation (b) Distillation (c) Sublimation (d) Chromatography

Ans.

Chromatography is the best and latest technique. Thus, option (d) is the correct answer.

Q.40 The reaction:

CH₃CH₂I + KOH(aq) → CH₃CH₂OH + KI

is classified as (a) electrophilic substitution (b) nucleophilic substitution (c) elimination (d) addition

Ans.

It is an example of nucleophilic substitution reaction. The hydroxyl group (OH^–) of KOH acts as a nucleophile and substitutes iodide ion (I^–) from CH₃CH₂I and forms ethanol. Thus, option (b) is a correct answer.