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8.12 Exercise Problems and Solutions

CLASS-XI NCERT CHEMISTRY UNIT-8: Organic Chemistry – Some Basic Principles and Techniques 8.12 Exercise Problems and Solutions

Q1. What are hybridisation states of each carbon atom in the following compounds?
\(
\mathrm{CH}_2=\mathrm{C}=\mathrm{O}, \mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2,\left(\mathrm{CH}_3\right)_2 \mathrm{CO}, \mathrm{CH}_2=\mathrm{CHCN}, \mathrm{C}_6 \mathrm{H}_6
\)

Answer:

Q2. Indicate the \(\sigma\) and \(\pi\) bonds in the following molecules :
\(\mathrm{C}_6 \mathrm{H}_6, \mathrm{C}_6 \mathrm{H}_{12}, \mathrm{CH}_2 \mathrm{Cl}_2, \mathrm{CH}_2=\mathrm{C}=\mathrm{CH}_2, \mathrm{CH}_3 \mathrm{NO}_2, \mathrm{HCONHCH}_3\)

Answer: The \(\sigma\) and \(\pi\) bonds are indicated below :

(i)

\(\mathrm{C}_6 \mathrm{H}_{12}\) can be either :
Straight chain compound
\(
\mathrm{H}_3 \mathrm{C}-\mathrm{H}_2 \mathrm{C}-\mathrm{H}_2 \mathrm{C}-\mathrm{HC}=\mathrm{CH}-\mathrm{CH}_3
\)

(ii)

Q3. Write bond line formulas for : Isopropyl alcohol, 2,3-Dimethylbutanal, Heptan-4-one.

Answer:

Q4. Give the IUPAC names of the following compounds:

Answer: (a) Propylbenzene (b) 3-Methylpentanenitrite (c) 2, 5-Dimethylheptane (d) 3-Bromo-3-chloroheptane (e) 3-Chloropropanal (f) 2, 2-Dichloroethanol

Q5. 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-methylpentane (d) But-3-yn-1-ol or But-4-ol-1-yne.

Answer: The correct IUPAC name out of each set of two compounds is
(a) 2,2-Dimethylpentane
(b) 2,4,7-Trimethyloctane
(c) 2-Chloro-4-methylpentane
(d) But-3-yn-1-ol

Q6. Draw formulas for the first five members of each homologous series beginning with the following compounds. (a) \(\mathrm{H}-\mathrm{COOH}\) (b) \(\mathrm{CH}_3 \mathrm{COCH}_3\) (c) \(\mathrm{H}-\mathrm{CH}=\mathrm{CH}_2\)

Answer:

(a) : \(\mathrm{HCOOH}\)

The given compound is formic acid (a carboxylic acid) and the first five members of the carboxylic acid series are :
(1) \(\mathrm{HCOOH}\)
(2) \(\mathrm{CH}_3 \mathrm{COOH}\)
(3) \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH}\)
(4) \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{COOH}\)
(5) \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{COOH}\)
(b) : \(\mathrm{CH}_3 \mathrm{COCH}_3\)

The given compound, propanone, is a ketone and the first five members of the ketone series are :
(1) \(\mathrm{CH}_3 \mathrm{COCH}_3\)
(2) \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COCH}_3\)
(3) \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{COCH}_3\)
(4) \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{COCH}_3\)
(5) \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{COCH}_3\)
(c) : \(\mathrm{CH}_2=\mathrm{CH}_2\)

Ethene is an alkene and the first five members of the alkene series are :
(1) \(\mathrm{CH}_2=\mathrm{CH}_2\)
(2) \(\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2\)
(3) \(\mathrm{CH}_3 \mathrm{CH}_2-\mathrm{CH}=\mathrm{CH}_2\)
(4) \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2-\mathrm{CH}=\mathrm{CH}_2\)
(5) \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2-\mathrm{CH}=\mathrm{CH}_2\)

Q7. 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

Answer:

Q8. Identify the functional groups in the following compounds

Answer:

Q9. Which of the two: \(\mathrm{O}_2 \mathrm{NCH}_2 \mathrm{CH}_2 \mathrm{O}^{-}\)or \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{O}^{-}\)is expected to be more stable and why?

Answer:

In order to find the more stable one of the two given ions we need to understand that any ion, whether positively charged or negatively charged, is stable if it is somehow able to diminish the charge appearing on it or, is able to delocalise (spread) the charge on more than one atom. Such a delocalisation of charge is possible in I but not in II. Hence, I is more stable. This dispersal/delocalisation of negative charge happens in I due to the presence of the \(-\mathrm{NO}_2\) group which is electron withdrawing in nature. Due to this electron withdrawal, the charge on the oxygen atom disperses (spreads) and the ion becomes more stable.

Q10. Explain why alkyl groups act as electron donors when attached to a \(\pi\) system.

Answer: Alkyl groups attached to a \(\pi\)-bonded system tend to act as electron donors by virtue of hyperconjugation.
Hyperconjugation is the phenomenon wherein \(\sigma\)-electrons of a \(\mathrm{C}-\mathrm{H}\) bond enter into partial conjugation with the attached \(\pi\)-system.
In order to envision this, consider the following \(\pi\)-system :
\(
\mathrm{H}_3 \mathrm{C}-\mathrm{CH}=\mathrm{CH}_2
\)
Here, the \(\pi\)-electron cloud may get polarised as :

As soon as state II is achieved an electron pair from one of the \(\mathrm{C}-\mathrm{H}\) bonds from the neighbouring methyl group moves towards the empty \(p\)-orbital on \(\mathrm{C}^{+}\)and leads to bond formation.

The hydrogen of the \(\mathrm{C}-\mathrm{H}\) bond remains in vicinity and rebonds with the carbon as:

Q11. Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation.
(a) \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}\)
(b) \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NO}_2\)
(c) \(\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCHO}\)
(d) \(\mathrm{C}_6 \mathrm{H}_5-\mathrm{CHO}\)
(e) \(\mathrm{C}_6 \mathrm{H}_5-\stackrel{+}{\mathrm{C}} \mathrm{H}_2\)
(f) \(\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH} \stackrel{+}{\mathrm{C}} \mathrm{H}_2\)

Answer:(a) Resonance structures of phenol are

Phenol is a planar molecule. The \(p\)-orbitals on each carbon atom lie perpendicular to the plane of the molecule, the oxygen atom lying in its \(p\)-orbital is also perpendicular to the molecule. Consequently the p-orbital overlap, between \(\mathrm{O}\) and \(\mathrm{C}\) takes place and the resonance structures arise.

(b) \(-\mathrm{NO}_2\) is an electron withdrawing group. As a result, it polarises the benzene molecule and resonance occurs as shown :

(c) \(\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCHO}\) is an \(\alpha, \beta\)-unsaturated aldehyde where the \(\mathrm{C}=\mathrm{C}\) gets polarised due to the neighbouring electron withdrawing – CHO group.

(d) Oxygen being highly electronegative pulls the \(\pi\)-electron cloud of the \(\mathrm{C}=\mathrm{O}\) towards itself thereby leaving the carboriyl carbon with a positive charge. In order to compensate this electron deficiency, the electron cloud from the ring moves towards it and the resonance structures arise.

(e)

(f)

Q12. What are electrophiles and nucleophiles? Explain with examples.

Answer: (a) Electrophiles : The word electrophile is derived from electron + philia. ‘Philia’ refers to love/affinity/attraction. Thus, any species which has an attraction or affinity for electrons is termed as an electrophile.
Species that qualify as electrophiles may either be positively charged or without any formal charge, i.e., they are electron deficient species which do not have a complete octet around them. Examples include group 13 halides like \(\mathrm{BCl}_3, \mathrm{BF}_3, \mathrm{AlCl}_3 \text {, etc., or } \stackrel{-}{\mathrm{N}}\mathrm{O}_2, \mathrm{H}^{-}, \mathrm{CH}_3 \stackrel{-}{\mathrm{C}}\mathrm{O} \text {, etc.}\)
(b) Nucleophiles: ‘Nucleo’ refers to any positively charged centre. Thus, species that are attached to or have an affinity for a positively charged centre are termed as nucleophiles. (Nucleo + phile \(=\) Nucleus loving).
Nucleophiles may either be negatively charged \(\text { such as } \mathrm{Cl}^{-}, \mathrm{Br}^{-}, \mathrm{I}^{-}, \mathrm{F}^{-}, \mathrm{CH}_3 \mathrm{COO}^{-}, \mathrm{OH}^{-}, \mathrm{CN}^{-} \text {, etc.}\) etc. or be electron rich species without any negative charge. The best example of such a nucleophile is \(\mathrm{H}_2 \mathrm{O}\) molecule. Here, the oxygen atom has two lone pairs of electrons and therefore acts as a nucleophile.

Q13. Identify the reagents shown in bold in the following equations as nucleophiles or electrophiles:
\(
\text { (a) } \mathrm{CH}_3 \mathrm{COOH}+\mathrm{HO}^{-} \rightarrow \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}_2 \mathrm{O}
\)
\(
\text { (b) } \mathrm{CH}_3 \mathrm{COCH}_3+\stackrel{-}{\mathrm{C}}{\mathbf{N}} \rightarrow\left(\mathrm{CH}_3\right)_2 \mathrm{C}(\mathrm{CN})(\mathrm{OH})
\)
\(
\text { (c) } \mathrm{C}_6 \mathrm{H}_6+\mathrm{CH}_3 \stackrel{+}{\mathrm{C}} \mathrm{O} \rightarrow \mathrm{C}_6 \mathrm{H}_5 \mathrm{COCH}_3
\)

Answer:

The carbonyl carbon is a positive centre, since the oxygen withdraws the carbonyl electron and leaves the carbon atom electron deficient.

Benzene being an electron rich molecule, is most suitable for an electrophilic attack.

Q14. Classify the following reactions in one of the reaction type studied in this unit.
(a) \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br}+\mathrm{HS}^{-} \rightarrow \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{SH}+\mathrm{Br}^{-}\)
(b) \(\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{CH}_2+\mathrm{HCl} \rightarrow\left(\mathrm{CH}_3\right)_2 \mathrm{ClC}-\mathrm{CH}_3\)
(c) \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br}+\mathrm{HO}^{-} \rightarrow \mathrm{CH}_2=\mathrm{CH}_2+\mathrm{H}_2 \mathrm{O}+\mathrm{Br}\)
(d) \(\left(\mathrm{CH}_3\right)_3 \mathrm{C}-\mathrm{CH}_2 \mathrm{OH}+\mathrm{HBr} \rightarrow\left(\mathrm{CH}_3\right)_2 \mathrm{CBrCH}_2 \mathrm{CH}_2 \mathrm{CH}_3+\mathrm{H}_2 \mathrm{O}\)

Answer:

Here, \(\overline{\mathrm{S}} \mathrm{H}\), a nucleophile displaces \(\mathrm{Br}^{-}\)from the bromoalkane. Since, the reaction is brought about by a nucleophile and substitution occurs thereafter, the reaction will be termed as a nucleophilic substitution reaction.

In the given reaction we see that the \(\mathrm{H}^{+}\) and \(\mathrm{Cl}^{-}\)reacted with the alkene and thus, produced a haloalkane. However, no atom has been displaced. Thus, the reaction is an electrophilic addition reaction.

\(
\text { (c) } \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br}+\mathrm{O}^{-}\mathrm{H} \rightarrow \mathrm{CH}_2=\mathrm{CH}_2+\mathrm{H}_2 \mathrm{O}+\mathrm{Br}
\)
Here there is no atom which is displaced or substituted. Although the reaction is brought about by a nucleophile \(\mathrm{O}^{-}\mathrm{H}\) it is not a nucleophilic substitution. An \(\mathrm{HBr}\) molecule has been removed from the reactant and hence, it is an elimination reaction.

(d) \(\left(\mathrm{CH}_3\right)_3 \mathrm{C}-\mathrm{CH}_2-\mathrm{OH}+\mathrm{HBr} \longrightarrow \left(\mathrm{CH}_3\right)_2 \mathrm{CBrCH}_2 \mathrm{CH}_3+\mathrm{H}_2 \mathrm{O}\)
This is an example of a rearrangement followed by nucleophilic substitution. Initially, a \(1^{\circ}\) carbocation is formed which rearranges to produce a more stable \(3^{\circ}\) carbocation. Finally, \(\mathrm{Br}^{-}\)attacks the carbocation and product is formed.

Q15. What is the relationship between the members of following pairs of structures? Are they structural or geometrical isomers or resonance contributors?

Answer: Structural isomerism arises when two compounds have the same molecular formula but differ in structure i.e., arrangement of atoms.

Such structures arise when the arrangement of atoms remains the same but the bonding pattern differs.

Q16. 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.

Answer:

\(
\text { (a) } \mathrm{CH}_3 \mathrm{O}-\mathrm{OCH}_3 \rightarrow \mathrm{CH}_3 \dot{\mathrm{O}}+\dot{\mathrm{O}} \mathrm{C H_3}
\)
This is an example of homolysis which gives rise to free radicals (the word ‘homo’ means same and ‘lysis’ means breakage. The bond breakage which results in species of the same kind e.g., free radicals is called homolysis.). Electron flow :

This is an example of heterolysis where bond breakage results in formation of two different species. The electron flow may be depicted as :

Given reaction is an example of heterolysis where reactive intermediate is a carbocation. Electron flow :

This is heterolysis where intermediate is the arenium (carbocation) ion.

Q17. Explain the terms Inductive and Electromeric effects. Which electron displacement effect explains the following correct orders of acidity of the carboxylic acids?
(a) \(\mathrm{Cl}_3 \mathrm{CCOOH}>\mathrm{Cl}_2 \mathrm{CHCOOH}>\mathrm{ClCH}_2 \mathrm{COOH}\)
(b) \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH}>\left(\mathrm{CH}_3\right)_2 \mathrm{CHCOOH}>\left(\mathrm{CH}_3\right)_3 \mathrm{C} . \mathrm{COOH}\)

Answer: Inductive effect: Polarisation of \(\sigma\)-bond due to the electron donating or withdrawing nature of groups attached is called inductive effect.
e.g. Attachment of \(\mathrm{Cl}\) an electronegative atom to an ethane molecule causes the carbons to carry a \(\delta^{+}\)charge.

This induced polarity due to presence of \(\mathrm{Cl}\) atom is an example of inductive effect.
Electromeric effect : In the presence of an attacking reagent, the \(\pi\)-electron cloud in a multiple bonded compound tends to get polarized. Such an effect is termed as temporary effect and the original condition is restored if the reagent is removed.

The given order of acidity can be explained by \(-I\) effect. \(\mathrm{Cl}\) is an electronegative atom which withdraws electrons from the adjacent carbon. This in turn pulls electrons from the carboxylic group which results in increased acidity. Thus, more the no. of \(\mathrm{Cl}\) atoms, higher is the acidic strength.
(a) -I-effect as shown below:
As the number of halogen atoms decreases, the overall -I- effect decreases and the acid strength decreases accordingly.

(b) +l-effect as shown below:
As the number of alkyl groups increases, the + I-effect increases and the acid strength decreases accordingly.

Q18. Give a brief description of the principles of the following techniques taking an example in each case.
(a) Crystallisation
(b) Distillation
(c) Chromatography

Answer: (a) Crystallisation : It is based on the difference in solubility of the compound and the impurities in a suitable solvent. While at room temperature, the compound is sparingly soluble and crystallizes out of solution but the impurities do not. As a result, they remain in solution and the compound is obtained as a crystal.
The impure compound is dissolved in a solvent and heated. At elevated temperature the compound dissolves as do the impurities. This solution is then gradually cooled. Being less soluble at room temperature it precipitates out in the form of crystals and pure compound is obtained.

(b) Distillation : This method is used to separate either :

Volatile liquids from non-volatile impurities; and
Two liquids with different boiling points. The liquid mixture such as that of chloroform and aniline is taken in a round bottom flask fitted with a condenser.
Upon heating, the vapours of lower boiling liquid are formed first and collected through the condenser. The vapours of the higher boiling liquid are formed later. Thus, the two are separated.

(c) Chromatography:

It is applicable for the separation of virtually all inorganic and organic materials, except very insoluble polymers.
In this technique, the mixture of compounds which needs to be separated is applied onto a stationary phase, which may be a solid or a liquid. Another phase which may be a pure solvent, a mixture of solvents or a gas is allowed to more slowly over the stationary phase.
The components of the mixture which have different solubility in the moving phase, start moving. Since, they have different solubility, they move to different lengths on the stationary phase and become stable there,
Thus, the different components of the mixture are separated.

Q19. Describe the method, which can be used to separate two compounds with different solubilities in a solvent \(\mathrm{S}\).

Answer: Crystallization is the process that may be employed to seperate two compounds with different solubility in a given solvent at room temperature. Upon heating such a solution to a sufficiently high temperature the solubility of the compound which is insoluble at room-temperature, increases and it dissolves. However, when this solution is cooled down to room temperature the lesser soluble or insoluble component precipitates out and is obtained as crystals while its soluble counterpart remains in solution. Thus, the separation is complete.

Q20. What is the difference between distillation, distillation under reduced pressure and steam distillation?

Answer: Distillation is used in case of volatile liquid mixed with non-volatile impurities.
Distillation under reduced pressure: This method is used to purify such liquids which have very high boiling points and which decompose at or below their boiling points.
Steam distillation is used to purify steam volatile liquids associated with water immiscible impurities.

Q21. Discuss the chemistry of Lassaigne’s test.

Answer: The elements nitrogen, sulphur and halogens are tested in an organic compound by Lassaigne’s test. The organic compound ( \(\mathrm{N}, \mathrm{S}\) or halogens) is fused with sodium metal as to convert these elements into ionisable inorganic substances, i.e., nitrogen into sodium cyanide, sulphur into sodium sulphide and halogens into sodium halides.
\(
2 \mathrm{Na}+\mathrm{S} \stackrel{\Delta}{\longrightarrow} \mathrm{Na}_2 \mathrm{~S}
\)
\(
\mathrm{Na}+\mathrm{C}+\mathrm{N} \stackrel{\Delta}{\longrightarrow} \mathrm{NaCN}
\)
\(
\mathrm{Na}+\mathrm{X} \xrightarrow{\Delta} \mathrm{NaX}(X=\mathrm{Cl}, \mathrm{Br}, \mathrm{I})
\)
Once the ions are formed, the inorganic tests can be applied to them and the compound can be analysed.
(i) Test for Nitrogen : The sodium fusion extract is boiled with iron(II) sulphate and then acidified with acid. The formation of Prussian blue colour confirms the presence of nitrogen.
\(
6 \mathrm{CN}^{-}+\mathrm{Fe}^{2+} \longrightarrow\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}
\)
\(
3\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}+4 \mathrm{Fe}^{3+} \longrightarrow \mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3 \cdot x \mathrm{H}_2 \mathrm{O}
\)
\(
\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \text { Prussian blue }
\)

(ii) Test for Sulphur : The sodium fusion extract is acidified with acetic acid and lead acetate is added to it. A black precipitate of lead sulphide indicates the presence of sulphur.
\(
\mathrm{S}^{2-}+\mathrm{Pb}^{2+} \longrightarrow \mathrm{PbS} \downarrow
\)
\(
\quad \quad \quad \quad \quad \quad \quad \text { Black }
\)

(iii) Test for Halogens : The sodium fusion extract is acidified with nitric acid and then treated with silver nitrate. A white precipitate, soluble in ammonium hydroxide shows the presence of chlorine, a yellowish precipitate, sparingly soluble in ammonium hydroxide shows the. presence of bromine and a yellow precipitate, insoluble in ammonium hydroxide shows the presence of iodine.
\(
X^{-}+\mathrm{Ag}^{+} \longrightarrow \mathrm{AgX}
\)
\(X\) represents a halogen \(-\mathrm{Cl}, \mathrm{Br}\) or \(\mathrm{I}\).

(iv) Test for Phosphorus : The compound is heated with an oxidising agent (sodium peroxide). The phosphorus present in the compound is oxidised to phosphate. The solution is boiled with nitric add and then treated with ammonium molybdate. A yellow colouration or precipitate indicates the presence of phosphorus.
\(
\mathrm{Na}_3 \mathrm{PO}_4+3 \mathrm{HNO}_3 \longrightarrow \mathrm{H}_3 \mathrm{PO}_4+3 \mathrm{NaNO}_3
\)
\(
\mathrm{H}_3 \mathrm{PO}_4+12\left(\mathrm{NH}_4\right)_2 \mathrm{MoO}_4+21 \mathrm{HNO}_3 \longrightarrow \left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 \cdot 12 \mathrm{MoO}_3+21 \mathrm{NH}_4 \mathrm{NO}_3+12 \mathrm{H}_2 \mathrm{O}
\)
\(
\quad \quad \text { Ammonium molybdate } \quad \quad \quad \quad \quad \quad \quad \quad \text { Ammonium phosphomolybdate(yellow ppt.) }
\)

Q22. Differentiate between the principle of estimation of nitrogen in an organic compound by (i) Dumas method and (ii) Kjeldahl’s method.

Answer: (i) Dumas method: The organic compound is heated strongly with excess of CuO ‘ (Cupric Oxide) in an atmosphere of \(\mathrm{CO}_2\) when free nitrogen, \(\mathrm{CO}_2\) and \(\mathrm{H}_2 \mathrm{O}\) are obtained.
(ii) Kjeldahl’s method: A known mass of the organic compound is heated strongly with conc. \(\mathrm{H}_2 \mathrm{SO}_4\), a little potassium sulphate and a little mercury (a catalyst). As a result of reaction the nitrogen present in the organic compound is converted to ammonium sulphate.

Q23. Discuss the principle of estimation of halogens, sulphur and phosphorus present in an organic compound.

Answer: Principles of estimation

(1) Estimation of halogens by Carius method: Upon heating an organic compound with fuming \(\mathrm{HNO}_3\) in the presence of \(\mathrm{AgNO}_3\), the halogen present in the compound forms the corresponding silver halide which is collected filtered; washed, dried and weighed. This is the amount of halogen present in the compound.
Let the mass of organic compound taken \(=m \mathrm{~g}\)
Mass of \(\mathrm{AgX}\) formed \(=m_1 \mathrm{~g}\)
\(1 \mathrm{~mol}\) of \(\mathrm{Ag} X\) contains \(1 \mathrm{~mol}\) of \(X\)
Mass of halogen in \(m_1 g\) of \(\mathrm{AgX}\)
\(
=\frac{\text { atomic mass of } X \times m_1 g}{\text { molecular mass of } A g X}
\)
Percentage of halogen
\(
=\frac{\text { atomic mass of } X \times m_1 \times 100}{\text { molecular mass of } \mathrm{AgX} \times m}
\)

(2) Estimation of sulphur : An organic compound containing sulphur is heated with fuming nitric acid. This oxidises the sulphur to sulphuric acid which is precipitated as \(\mathrm{BaSO}_4\) upon reaction with \(\mathrm{Ba}(\mathrm{OH})_2\). The mass of \(\mathrm{BaSO}_4\) tells the % of sulphur present in the compound.
Let the mass of organic compound taken \(=m \mathrm{~g}\) and the mass of barium sulphate formed \(=m_1 g\)
\(1 \mathrm{~mol} {\text {~of }} \mathrm{BaSO}_4=233 \mathrm{~g} \mathrm{BaSO}_4=32 \mathrm{~g}\) sulphur
\(m_1 \mathrm{~g} \mathrm{BaSO}_4\) contains \(\frac{32 \times m_1}{233} \mathrm{~g}\) sulphur
Percentage of sulphur \(=\frac{32 \times m_1 \times 100}{233 \times m}\)

(3) Estimation of phosphorus : A known mass of an organic compound is heated with fuming nitric acid whereupon phosphorus present in the compound is oxidised to phosphoric acid. It is precipitated as ammonium phosphomolybdate, \(\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 \cdot 12 \mathrm{MoO}_3\), by adding ammonia and ammonium molybdate. Alternatively, phosphoric acid may be precipitated as \(\mathrm{MgNH}_4 \mathrm{PO}_4\) by adding magnesia mixture which on ignition yields \(\mathrm{Mg}_2 \mathrm{P}_2 \mathrm{O}_7\).
Let the mass of organic compound taken \(=m \mathrm{~g}\) and mass of ammonium phosphomolybdate\(=m_1 \mathrm{~g}\)
\(
\begin{aligned}
& \quad \begin{array}{c}
\mathrm{Mg}_2 \mathrm{P}_2 \mathrm{O}_7 \equiv 2 \mathrm{P} \\
222 \mathrm{~g} \quad \quad 62 \mathrm{~g}
\end{array} \\
& \text { Percentage of phosphorus }=\frac{62 \times m_1 \times 100}{222 \times m}
\end{aligned}
\)

Q24. Explain the principle of paper chromatography.

Answer: The underlying principle of paper chromatography is that of partition chromatography which is based on continuous differential partitioning of components of mixture between stationary and mobile phases. In paper chromatography, the paper used has water trapped in it which acts as the stationary phase while a suitable solvent or a mixture of solvents is used as a mobile phase. A strip of chromatography paper spotted at the base with the solution of the mixture is suspended in a suitable solvent or a mixture of solvents. As the mobile phase moves over the paper, it carries the mixture with it. Since the different components have different solubility, they travel to different extents on the paper and become stationary at different lengths on the paper and are thus, separated.

Q25. Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens?

Answer: Halogens are detected by \(\mathrm{AgNO}_3\) due to the formation of \(\mathrm{AgX}\) ppt. But, if the Lassaigne’s extract contains \(\mathrm{S}^{2-}\) or \(\mathrm{CN}^{-}\)ions due to the compound, then they may interfere in the test for halogens.
\(\mathrm{CN}^{-}\)and \(\mathrm{S}^{2-}\) form \(\mathrm{AgCN}\) and \(\mathrm{Ag}_2 \mathrm{~S}\) with \(\mathrm{AgNO}_3\) and may produce erroneous results. Therefore, the Lassaigne’s extract is treated with \(\mathrm{HNO}_3\) to decompose \(\mathrm{HCN}\) and \(\mathrm{H}_2 \mathrm{~S}\) (vapours) so as to prevent interference.

Nitric acid is added to sodium extract so as to decompose
\(
\begin{aligned}
& \mathrm{NaCN}+\mathrm{HNO}_3 \longrightarrow \mathrm{NaNO}_3+\mathrm{HCN} \\
& \mathrm{Na}_2 \mathrm{~S}+2 \mathrm{HNO}_3 \rightarrow 2 \mathrm{NaNO}_3+\mathrm{H}_2 \mathrm{~S}
\end{aligned}
\)

Q26. Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens.

Answer:Lassaigne’s test is used for the detection of extra elements such as N, S and X by applying the inorganic tests of analysis to these. Since, in organic compounds, the elements are present in covalent form and inorganic tests can be applied only to ions, therefore these extra elements are first converted into their inorganic (ionic) forms by fusing with sodium metal.

Organic compound is fused with sodium metal so as to convert organic compounds into \(\mathrm{NaCN}, \mathrm{Na}_2 \mathrm{~S}, \mathrm{NaX}\) and \(\mathrm{Na}_3 \mathrm{PO}_4\). Since these are ionic compounds and become more reactive and thus can be easily tested by suitable reagents.

Q27. Name a suitable technique of separation of the components from a mixture of calcium sulphate and camphor.

Answer: Camphor is sublimable compound while \(\mathrm{CaSO}_4\) being ionic is not. Therefore, the two can be separated by the method of sublimation. If a mixture of the two is heated in a China dish covered with a porous paper and an inverted funnel over it, we will find the crystals of camphor forming on the inside walls of the inverted funnel. Thus, pure \(\mathrm{CaSO}_4\) will be left in the China dish.

Q28. Explain, why an organic liquid vaporises at a temperature below its boiling point in its steam distillation?

Answer: We know that any liquid boils when its vapour pressure is equal to the atmospheric pressure. There are certain liquids such as aniline which need very high temperature in order to start boiling. It is quite likely that at such elevated temperatures the molecules may just disintegrate. Therefore, to prevent this, steam distillation is employed. Here, the mixture of organic liquids containing the high boiling liquid say, aniline is mixed with water and heated. On doing so, at a temperature close to but less than \(100^{\circ} \mathrm{C}\) (b.p. of water) the vapour pressure of water equals the atmospheric pressure and it boils. Since, in the mixture, aniline is present in conjugation with water it vapourises and moves out of the mixture.
The mixture of water and aniline is separated using a separating funnel. Steam distillation is used extensively in perfumery to separate essential oils.

Q29. Will \(\mathrm{CCl}_4\) give white precipitate of \(\mathrm{AgCl}\) on heating it with silver nitrate? Give reason for your answer.

Answer: \(\mathrm{AgNO}_3\) solution is ionic in nature. It contains \(\mathrm{Ag}^{+}\)ions which when react with \(\mathrm{Cl}\) ions produce a white ppt. of \(\mathrm{AgCl}\). In \(\mathrm{CCl}_4\) the \(\mathrm{Cl}\) atoms are covalent in nature. They are not present as ions. Therefore, when \(\mathrm{AgNO}_3\) is added it, it does not produce a white ppt. of \(\mathrm{AgCl}\).

Q30. Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound?

Answer: \(\mathrm{KOH}\) reacts with \(\mathrm{CO}_2\) to produce \(\mathrm{K}_2 \mathrm{CO} 3\) which is a solid. The \(\mathrm{K}_2 \mathrm{CO}_3\) formed may be weighed and estimated to know the carbon content of the organic compound.
\(
2 \mathrm{KOH}+\mathrm{CO}_2 \longrightarrow \mathrm{K}_2 \mathrm{CO}_3+\mathrm{H}_2 \mathrm{O}
\)

Q31. Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test?

Answer: Sulphuric acid cannot be used for acidification of sodium extract because it would oxidize the sulphur to sulphur dioxide which would not give the black ppt. of \(\mathrm{PbS}\), which is otherwise obtained upon reaction with lead acetate.
\(
\underset{\text { lead acetate }}{\mathrm{Pb}\left(\mathrm{OCOCH}_3\right)_2}+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{PbSO}_4 \downarrow+2 \mathrm{CH}_3 \mathrm{COOH}
\)

Q32. An organic compound contains \(69 \%\) carbon and \(4.8 \%\) hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when \(0.20 \mathrm{~g}\) of this substance is subjected to complete combustion.

Answer:Step I. Calculation of mass of \(\mathrm{CO}_2\) produced

Mass of compound \(=0.20 \mathrm{~g}\)
Percentage of carbon \(=69 \%\)
\(
\begin{aligned}
\text { Percentage of carbon } & =\frac{12}{44} \times \frac{\text { Mass of carbon dioxide formed }}{\text { Mass of compound }} \times 100 \\
69 & =\frac{12}{44} \times \frac{\text { Mass of carbon dioxide formed }}{(0.20 \mathrm{~g})} \times 100 \\
\therefore \text { Mass of } \mathrm{CO}_2 \text { formed } & =\frac{69 \times 44 \times(0.20 \mathrm{~g})}{12 \times 100}=0.506 \mathrm{~g}
\end{aligned}
\)
Step II. Calculation of mass of \(\mathrm{H}_2 \mathrm{O}\) produced

Mass of compound \(=0.20 \mathrm{~g}\)

Percentage of hydrogen \(=4.8 \%\)
Percentage of hydrogen \(=\frac{2}{18} \times \frac{\text { Mass of water formed }}{\text { Mass of compound }} \times 100\)
\(4.8=\frac{2}{18} \times \frac{\text { Mass of water formed }}{(0.20 \mathrm{~g})} \times 100\)
\(\therefore\) Mass of \(\mathrm{H}_2 \mathrm{O}\) formed \(=\frac{4.8 \times 18 \times(0.20 \mathrm{~g})}{2 \times 100}=\mathbf{0 . 0 8 6 4} \mathrm{g}\)

Q33. A sample of \(0.50 \mathrm{~g}\) of an organic compound was treated according to Kjeldahl’s method. The ammonia evolved was absorbed in \(50 \mathrm{ml}\) of \(0.5 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4\). The residual acid required \(60 \mathrm{~mL}\) of \(0.5 \mathrm{M}\) solution of \(\mathrm{NaOH}\) for neutralisation. Find the percentage composition of nitrogen in the compound.

Answer: Given:
Mass of compound taken \(=0.50 \mathrm{~g}\)
Vol. of \(\mathrm{H}_2 \mathrm{SO}_4=50 \mathrm{~mL}\)
Molarity of \(\mathrm{H}_2 \mathrm{SO}_4=0.5 \mathrm{M}\)
Vol. of \(\mathrm{NaOH}\) required \(=60 \mathrm{~mL}\)
Molarity of \(\mathrm{NaOH}\) required \(=0.5 \mathrm{M}\)
Method adopted : Kjeldahl’s method
Formula used :
\(
\% \text { of } \mathrm{N}=\frac{1.4 \times M \times 2\left[V-\frac{V_1}{2}\right]}{m}=\frac{1.4 \times \text { Volume of acid used } \times \text { Normality of acid used }}{\text { Mass of the compound }}
\)
\(
\begin{aligned}
& \left.M=0.5 \mathrm{M} \text { (Molarity of } \mathrm{H}_2 \mathrm{SO}_4\right) \\
& V=50 \mathrm{~mL} \\
& V_1=60 \mathrm{~mL} \\
& m=0.5 \mathrm{~g} \text { (mass of compound) }
\end{aligned}
\)
By substituting the values in the formula, we get,
\(\%\) of \(\mathrm{N}=\frac{1.4 \times 0.5 \times 2(50-60 / 2)}{0.5}=56\)
\(\therefore \quad \%\) of \(\mathrm{N}\) in the given compound \(=56 \%\)

Alternate:

Step I. Calculation of volume of unused acid
Volume of \(\mathrm{NaOH}\) solution required \(=60 \mathrm{~cm}^3\)
Normality” of \(\mathrm{NaOH}\) solution \(=1 / 2 \mathrm{~N}\)
Normality of \(\mathrm{H}_2 \mathrm{SO}_4\) solution \(=1 / \mathrm{N}\)
Volume of unused acid can be calculated by applying normality equation
\(
\underbrace{N_1 V_1}_{\text {Acid }}=\underbrace{N_1 V_1}_{\text {Base }}
\)
\(
1 \times \mathrm{V}=\frac{1}{2} \times 60=30 \mathrm{~cm}^3
\)

Step II. Calculation of volume of acid used
Volume of acid added \(=50 \mathrm{~cm}^3\)
Volume of unused acid \(=30 \mathrm{~cm}^3\)
Volume of acid used \(=(50-30)=20 \mathrm{~cm}^3\)

Step III. Calculation of percentage of nitrogen
Mass of compound \(=0.50 \mathrm{~g}\)
Volume of acid used \(=20 \mathrm{~cm}^3\)
Normality of acid used \(=1 \mathrm{~N}\)
\(
\begin{aligned}
\text { Percentage of } \mathrm{N} & =\frac{1.4 \times \text { Volume of acid used } \times \text { Normality of acid used }}{\text { Mass of the compound }} \\
& =\frac{1.4 \times 20 \times 1}{0.50}=56 \%
\end{aligned}
\)

Q34. \(0.3780 \mathrm{~g}\) of an organic chloro compound gave \(0.5740 \mathrm{~g}\) of silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound.

Answer: Given :
Mass of organic compound \(=0.3780 \mathrm{~g}\)
Mass of silver chloride \(=0.5740 \mathrm{~g}\)
\(1 \mathrm{~mol}\) of \(\mathrm{AgCl}=143.5 \mathrm{~g}\)
\(143.5 \mathrm{~g}\) of \(\mathrm{AgCl}\) contains \(35.5 \mathrm{~g}\) of Cl
\(\therefore \quad 0.5740 \mathrm{~g}\) of \(\mathrm{AgCl}\) contains
\(
\frac{35.5 \times 0.574}{143.5}=0.142 \mathrm{~g}
\)
\(0.142 \mathrm{~g}\) of \(\mathrm{Cl}\) is present in \(0.3780 \mathrm{~g}\) of the compound
\(\Rightarrow \quad \%\) of \(\mathrm{Cl}\) in compound
\(
=\frac{0.142}{0.378} \times 100=37.56 \%
\)

Q35. In the estimation of sulphur by Carius method, \(0.468 \mathrm{~g}\) of an organic sulphur compound afforded \(0.668 \mathrm{~g}\) of barium sulphate. Find out the percentage of sulphur in the given compound.

Answer: Given:
Mass of organic sulphur compound \(=0.468 \mathrm{~g}\)
Mass of \(\mathrm{BaSO}_4=0.668 \mathrm{~g}\)
\(1 \mathrm{~mol~} {\text {of }} \mathrm{BaSO}_4=233 \mathrm{~g}\)
\(233 \mathrm{~g}\) of \(\mathrm{BaSO}_4\) contains \(32 \mathrm{~g}\) of S
\(\therefore \quad 0.668 \mathrm{~g}\) of \(\mathrm{BaSO}_4\) contains
\(
\frac{0.668 \times 32}{233}=0.09174 \mathrm{~g}
\)
\(0.09174 \mathrm{~g} \mathrm{~S}\) is present in \(0.468 \mathrm{~g}\) of the compound.
\(\Rightarrow \quad \%\) of \(\mathrm{S}\) in compound
\(
=\frac{0.09174}{0.468} \times 100=19.6 \%
\)

Q36. In the organic compound \(\mathrm{CH}_2=\mathrm{CH}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{C} \equiv \mathrm{CH}\), the pair of hydridised orbitals involved in the formation of: \(\mathrm{C}_2-\mathrm{C}_3\) bond is:
(a) \(s p-s p^2\)
(b) \(s p-s p^3\)
(c) \(s p^2-s p^3\)
(d) \(s p^3-s p^3\)

Answer: (c) The carbon atom in the given compound may be numbered as :
\(
\mathrm{H}_2 \stackrel{1}{\mathrm{C}}=\stackrel{2}{\mathrm{C}} \mathrm{H}-\stackrel{3}{\mathrm{C}} \mathrm{H}_2-\stackrel{4}{\mathrm{C}} \mathrm{H}_2-\stackrel{5}{\mathrm{C}} \equiv \stackrel{6}{\mathrm{C}} \mathrm{H}
\)
Thus, \(\mathrm{C}_2-\mathrm{C}_3\) are bonded by \(s p^2-s p^3\) hybrid orbitals.

Q37. In the Lassaigne’s test for nitrogen in an organic compound, the Prussian blue colour is obtained due to the formation of:
(a) \(\mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]\)
(b) \(\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3\)
(c) \(\mathrm{Fe}_2\left[\mathrm{Fe}(\mathrm{CN})_6\right]\)
(d) \(\mathrm{Fe}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]_4\)

Answer: The Prussian blue colour in the Lassaigne’s test is due to the formation of \(\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3\).

Q38. Which of the following carbocation is most stable?
(a) \(\left(\mathrm{CH}_3\right)_3 \mathrm{C} \cdot {\stackrel{+}{\mathrm{C}}} \mathrm{H}_2\)
(b) \(\left(\mathrm{CH}_3\right)_3 \stackrel{+}{\mathrm{C}}\)
(c) \(\mathrm{CH}_3 \mathrm{CH}_2 \stackrel{+}{\mathrm{C}} \mathrm{H}_2\)
(d) \(\mathrm{CH}_3 \stackrel{+}{\mathrm{C}}\mathrm{H} \mathrm{CH}_2 \mathrm{CH}_3\)

Answer:(b) \(\left(\mathrm{CH}_3\right)_3 \stackrel{+}{\mathrm{C}}\) or tert-butyl cation is the most stable of all the given cations. The stability is due to the presence of three
methyl groups attached to the carbocation centre. The hydrogens in the methyl group, by hyperconjugation, diminish the electron deficiency (+ve charge) on the carbocation and hence, stabilize it.

In tert-butyl cation the no. of such \(\mathrm{H}\) atoms which are capable of hyperconjugation is maximum. Thus, it is most stable. In short, greater the no. of \(\alpha-\mathrm{H}\) more stable will be the carbocation.

Q39. The best and latest technique for isolation, purification and separation of organic compounds is:
(a) Crystallisation
(b) Distillation
(c) Sublimation
(d) Chromatography

Answer: (d) Chromatography

Q40. The reaction:
\(
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{I}+\mathrm{KOH}(\mathrm{aq}) \rightarrow \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}+\mathrm{KI}
\)
is classified as :
(a) electrophilic substitution
(b) nucleophilic substitution
(c) elimination
(d) addition

Answer: (b) \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{I}+\mathrm{KOH}(\mathrm{aq}) \rightarrow \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}+\mathrm{KI}\)

In the given reaction, the \(\mathrm{I}^{-}\)from the alkyl iodide is replaced by the \(\mathrm{OH}^{-}\)ion. Thus, it is substitution reaction. The substitution is brought about by the \(\mathrm{OH}^{-}\)ion which is a nucleophile.
\(\because\) The reaction is a nucleophilic substitution reaction.

Exemplar

Consider structures I to VII and answer the questions Q41-Q44.

Q41. Which of the above compounds form pairs of metamers?

Answer: \(\mathrm{V}\) and \(\mathrm{VI}, \mathrm{VI}\) and \(\mathrm{VII}\) and \(\mathrm{V}\) and \(\mathrm{VII}\) form pairs of metamers because they have different alkyl chains on either side of ethereal oxygen (-O-).

Q42. Identify the pairs of compounds which are functional group isomers.

Answer: Two or more compounds having the same molecular formula but different functional groups are called functional isomers.
Compounds I to IV, i.e., alcohols, and V to VII, i.e., ethers, are functional group isomers with molecular formula \(\mathrm{C}_4 \mathrm{H}_{10} \mathrm{O}\) and different functional groups ( \(-\mathrm{OH}\) in I to IV and – \(\mathrm{O}\) – in \(\mathrm{V}\) to \(\mathrm{VII}\) ).
Hence, I and V, I and VI, I and VII; II and V, II and VI, II and VII; \(\mathrm{III}\) and \(\mathrm{V}, \mathrm{III}\) and \(\mathrm{VI} ; \mathrm{III}\) and \(\mathrm{VII} ; \mathrm{IV}\) and \(\mathrm{V}\), IV and \(\mathrm{VI}, \mathrm{IV}\) and \(\mathrm{VII}\) are functional group isomers.

Q43. Identify the pairs of compounds that represents position isomerism.

Answer: I and II, III and IV, VI and VII are position isomers due to different positions of – \(\mathrm{OH}\) group and –\(\mathrm{O}\)-group.
When two or more compounds differ in position of substituent atom or functional group on the carbon skeleton, they are position isomers. In the given structures, I and II; III and IV, and VI and VII are position isomers.

Q44. Identify the pairs of compounds that represents chain isomerism.

Answer: I and III, I and IV, II and III, II and IV

Q45. For testing halogens in an organic compound with \(\mathrm{AgNO}_3\) solution, sodium extract (Lassaigne’s test) is acidified with dilute \(\mathrm{HNO}_3\). What will happen if a student acidifies the extract with dilute \(\mathrm{H}_2 \mathrm{SO}_4\) in place of dilute \(\mathrm{HNO}_3\)?

Answer: On adding dilute \(\mathrm{H}_2 \mathrm{SO}_4\) for testing halogens in an organic compound with \(\mathrm{AgNO}_3\), white precipitate of \(\mathrm{Ag}_2 \mathrm{SO}_4\) is formed. This will interfere with the test of chlorine and this \(\mathrm{Ag}_2 \mathrm{SO}_4\) may be mistaken for white precipitate of chlorine as \(\mathrm{AgCl}\). Hence, dilute \(\mathrm{HNO}_3\) should be used instead of dilute \(\mathrm{H}_2 \mathrm{SO}_4\).

Q46. What is the hybridisation of each carbon in \(\mathrm{H}_2 \mathrm{C}=\mathrm{C}=\mathrm{CH}_2\).

Answer: In \(\mathrm{H}_2 \stackrel{(1)}{\mathrm{C}}=\stackrel{(2)}{\mathrm{C}}=\stackrel{(3)}{\mathrm{C}} \mathrm{H}_2\), carbon (1) and (3) are \(s p^2\) hybridised and carbon (2) is \(s p\) hybridized.

In allene, carbon atoms 1 and 3 are \(\mathrm{sp}^2\)-hybridised as each one of them is joined as a double bond. And, carbon atom 2 is sp-hybridised as it has two double bonds at each of its side. Therefore, the two \(\pi\)-bonds are perpendicular to each other, in allene as shown below.

\(\mathrm{H}_{\mathrm{a}}\) and \(\mathrm{H}_{\mathrm{b}}\) lie in the plane of paper while \(\mathrm{H}_{\mathrm{c}}\) and \(\mathrm{H}_{\mathrm{d}}\) lie in a plane perpendicular to the plane of the paper. Hence, the allene mole as a whole is non-planar.

Q47. Explain, how is the electronegativity of carbon atoms related to their state of hybridisation in an organic compound?

Answer: Electronegativity increases with increasing s-character. This is because s-electrons are more strongly attracted by the nucleus than p-electrons.
\({sp}^3-25 \%\) s-character, \(75 \%\) P-character
\(s p^2-33 \%\) s-character, \(67 \%\) P-character
\(s p-50 \%\) s-character, \(50\%\) P-character
Hence, the order of electronegativity is \(\mathrm{sp}^3<\mathrm{sp}^2<\mathrm{sp}\)

Q48. Show the polarisation of carbon-magnesium bond in the following structure.
\(
\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{Mg}-\mathrm{X}
\)

Answer: Carbon (2.5) is more electronegative than magnesium (1.2) therefore, Mg acquires a partial positive charge attached to it acquires a partial negative charge.
\(
\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\stackrel{-\delta}{\mathrm{C}} \mathrm{H}_2-\stackrel{+\delta}{\mathrm{Mg}}-\mathrm{X}
\)

Since electronegativity of carbon is more than magnesium it will behave as \(\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\overline{\ddot{\mathrm{C}}} \mathrm{H}_2 \stackrel{+}{\mathrm{M}}\mathrm{gBr}\)

Q49. Compounds with same molecular formula but differing in their structures are said to be structural isomers. What type of structural isomerism is shown by

Answer: The two isomers which differ in the position of the functional group on the carbon skeletoon are called position isomers and this phenomenon as position isomerism.
Thus, \((A)\) and \((B)\) may be regarded as position isomers and further they cannot be regarded as metamers since metamers are those isomers which have different number of carbon atoms on either side of the functional group.
But here, the number of carbon atoms on either side of sulphur atom (functional group) is the same i.e., 1 and 3.

Q50. Which of the following selected chains is correct to name the given compound according to IUPAC system.

Answer: Among the following given compounds, according to IUPAC, the longest carbon chain having maximum number of functional gap is being selected.

Thus, carbon-chain containing 4-carbon atoms and which also includes both functional groups will be selected (Selected chain should have maximum number of functional groups). While other three \(\mathrm{C}\)-chains are incorrect since none of them contains both the functional groups.

Q51. In DNA and RNA, nitrogen atom is present in the ring system. Can Kjeldahl method be used for the estimation of nitrogen present in these? Give reasons.

Answer: No. DNA and RNA have nitrogen in the heterocyclic rings which cannot be removed as ammonia.
Nitrogen present in rings, azo and nitro groups cannot be converted into ammonium sulphate \(\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4\). That is why Kieldahl method cannot be used for the estimation of nitrogen present in DNA and RNA.

Q52. If a liquid compound decomposes at its boiling point, which method (s) can you choose for its purification. It is known that the compound is stable at low pressure, steam volatile and insoluble in water.

Answer: If a compound decomposes at its boiling point but is steam volatile, water-insoluble and stable at low pressure, steam distillation can be used for its purification. This technique is applied to separate substances which are steam volatile and immiscible with water.

Note : Answer the questions 53 to 56 on the basis of information given below:
“Stability of carbocations depends upon the electron releasing inductive effect of groups adjacent to positively charged carbon atom involvement of neighbouring groups in hyperconjugation and resonance.”

Q53. Draw the possible resonance structures for and predict which of the structures is more stable. Give reason for your answer.

Answer: The given carbocation has two resonance structures

Structure (II) is more stable because both the carbon atoms and the oxygen atom have an octet of electrons.

Q54. Which of the following ions is more stable? Use resonance to explain your answer.

Answer: Carbocation (A) is more stable than carbocation (B). Carbocation (A) is more planar and hence is stabilised by resonance while carbocation (B) is non-planar and hence it does not undergo resonance. Further, double bond is more stable within the ring in comparison to outside the ring.

Structure A is more stable due to resonance. (See resonance structure ‘A’ and ‘B’). No resonance is possible in structure B.

Q55. The structure of triphenylmethyl cation is given below. This is very stable and some of its salts can be stored for months. Explain the cause of high stability of this cation.

Answer: Triphenylmethyl cation is a tertiary carbocation which can show nine possible canonical structures and hence is very stable.

Since, there are three benzene rings, hence, there are, in all, nine resonance structures. Thus, triphenylmethyl cation is highly stable due to these nine resonance structures.

Q56. Write structures of various carbocations that can be obtained from 2-methylbutane. Arrange these carbocations in order of increasing stability.

Answer: Four possible carbocations are

Therefore, the overall stability of these four carbocations increases in the order.
\(
\text { I }<\text { IV }<\text { II }<\text { III }
\)

Q57. Three students, Manish, Ramesh and Rajni were determining the extra elements present in an organic compound given by their teacher. They prepared the Lassaigne’s extract (L.E.) independently by the fusion of the compound with sodium metal. Then they added solid \(\mathrm{FeSO}_4\) and dilute sulphuric acid to a part of Lassaigne’s extract. Manish and Rajni obtained prussian blue colour but Ramesh got red colour. Ramesh repeated the test with the same Lassaigne’s extract, but again got red colour only. They were surprised and went to their teacher and told him about their observation. Teacher asked them to think over the reason for this. Can you help them by giving the reason for this observation. Also, write the chemical equations to explain the formation of compounds of different colours.

Answer:

In Lassaigne’s test \(\mathrm{SCN}^{-}\)ions are formed due to the presence of sulphur and nitrogen both. These give red colour with \(\mathrm{Fe}^{3+}\) ions. This happens when fusion is not carried out in the excess of sodium. With excess of sodium the thiocyanate ion, if formed, is decomposed as follows:
\(
\mathrm{NaSCN}+2 \mathrm{Na} \longrightarrow \mathrm{NaCN}+\mathrm{Na}_2 \mathrm{~S}
\)

Q58. Name the compounds whose line formulae are given below :

Answer:

3-ehtyl-4-methylhept-5-en-2-one ( \(\mathrm{C}\)-atoms of the longest possible chain are numbered in such a way that the functional group, \(>\mathrm{C}=\mathrm{O}\), gets the lowest possible locant)
3-nitrocyclohex-1-en (C-atoms of the ring are numbered in such a manner that double bond gets the lowest possible locant followed by the – \(\mathrm{NO}_2\) group)

Q59. Write structural formulae for compounds named as-
(a) 1-Bromoheptane
(b) 5-Bromoheptanoic acid

Answer:

Q60. Draw the resonance structures of the following compounds;

Answer:

Q61. Identify the most stable species in the following set of ions giving reasons :
(i) \(\stackrel{+}{\mathrm{C}} \mathrm{H}_3, \stackrel{+}{\mathrm{C}} \mathrm{H}_2 \mathrm{Br}, \stackrel{+}{\mathrm{C}} \mathrm{HBr}_2, \stackrel{+}{\mathrm{C}} \mathrm{Br}_3\)
(ii) \(\stackrel{\ominus}{\mathrm{C}} \mathrm{H}_3, \stackrel{\ominus}{\mathrm{C}} \mathrm{H}_2 \mathrm{Cl}, \stackrel{\ominus}{\mathrm{C}} \mathrm{HCl}_2, \stackrel{\ominus}{\mathrm{C}} \mathrm{Cl}_3\)

Answer: (a) \(\stackrel{+}{\mathrm{C}} \mathrm{H}_3\) is the most stable. The replacement of \(\mathrm{H}\) by \(\mathrm{Br}\) increases positive charge on carbon atom because \(\mathrm{Br}\) is more electronegative than \(\mathrm{H}\) and consequently the species becomes less stable.
(b) \(\stackrel{\ominus}{\mathrm{C}} \mathrm{Cl}_3\) is the most stable because \(\mathrm{Cl}\) is more electronegative than hydrogen. On replacing hydrogen by chlorine, negative charge on \(\mathrm{C}\) is reduced and the species becomes stable.

Q62. Give three points of differences between inductive effect and resonance effect.

Answer:

Inductive effect	Resonance effect
Inductive effect involves -electrons displacement and occurs only in saturated compounds. Use \(\sigma\)-electrons.	It involves -electrons or lone pair of electrons and occurs only in unsaturated and conjugated system. Use \(\pi\)-electrons or lone pair of electrons.
Inductive effect can move upto 3-carbon atom	It is applicable all along the length of conjugated system.
In inductive effect, there is a slight displacement of electrons and thus only partial positive and negative charges appear.	In resonance effect, there is complete transfer of electrons, and thus full positive and negative charges appear.

Q63. Which of the following compounds will not exist as resonance hybrid. Give reason for your answer:
(i) \(\mathrm{CH}_3 \mathrm{OH}\)
(ii) \(\mathrm{R}-\mathrm{CONH}_2\)
(iii) \(\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCH}_2 \mathrm{NH}_2\)

Answer:

\(\mathrm{CH}_3 \mathrm{OH}\); Any possible contributing structure will have charge separation and incomplete octet of electrons on atoms. So the structure will be unstable due to high energy. e.g., \(\stackrel{\oplus}{\mathrm{C}} \mathrm{H}_3 \stackrel{\ominus}{\mathrm{O}} \mathrm{H}\).

(a) \(\mathrm{CH} 3 \mathrm{OH}\) does not contain-electrons, hence, it cannot exist as resonance hybrid.
(b) Due to the presence of -electrons in \(\mathrm{C}=\mathrm{O}\) bond and lone pair of electrons on \(\mathrm{N}\), amide can be represented by the following resonating structures.

(c) \(\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCH}_2 \mathrm{NH}_2\) : Since lone pair of electrons on nitrogen atom is not conjugated with the -electrons, therefore, resonance is not possible.

Q64. Why does \(\mathrm{SO}_3\) act as an electrophile?

Answer: Three highly electronegative oxygen atoms are attached to sulphur atom. This makes sulphur atom electron deficient. Due to resonance, sulphur also acquires positive charge. Both these factors make \(\mathrm{SO}_3\) an electrophile.

Explanation:

Q65. Resonance structures of propenal are given below. Which of these resonating structures is more stable? Give reason for your answer.

Answer: \(\mathrm{CH}_2=\mathrm{CH}-\mathrm{CH}=\mathrm{O}\) is more stable than \(\stackrel{+}{\mathrm{C}} \mathrm{H}_2-\mathrm{CH}=\mathrm{CH}-\mathrm{O}^{-}\)since the octet of \(\mathrm{C}\) and \(\mathrm{O}\) is complete and no separation of opposite charges is there. \(\text { I }>\text { II }\)

Q66. By mistake, an alcohol (boiling point \(97^{\circ} \mathrm{C}\) ) was mixed with a hydrocarbon (boiling point \(68^{\circ} \mathrm{C}\) ). Suggest a suitable method to separate the two compounds. Explain the reason for your choice.

Answer: Simple distillation can be used because the two compounds have a difference of more than \(20^{\circ}\) in their boiling points and therefore, both the liquids can be distilled without any decomposition.

Q67. Which of the two structures (A) and (B) given below is more stabilised by resonance? Explain.
\(\mathrm{CH}_3 \mathrm{COOH}\) and \(\mathrm{CH}_3 \mathrm{CO}\stackrel{\ominus}{\mathrm{O}}\)
\(\text {(A)} \quad \quad \quad \quad \quad \quad \text {(B)}\)

Answer: Resonating structures are as follows:.

(B) is more stable because it does not involve charge separation.

Q68. What is meant by hybridisation? Compound \(\mathrm{CH}_2=\mathrm{C}=\mathrm{CH}_2\) contains \(s p\) or \(s p^2\) hybridised carbon atoms. Will it be a planar molecule?

Answer: No. It is not a planar molecule. Central carbon atom is \(s p\) hybridised and its two unhybridised \(p\)-orbitals are perpendicular to each other. The \(p\)-orbitals in one plane overlap with one of the \(p\)-orbital of left terminal carbon atom and the \(p\)-orbital in other plane overlaps with \(p\)-orbital of right side terminal carbon atom. This fixes the position of two terminal carbon atoms and the hydrogen atoms attached to them in planes perpendicular to each other. Due to this the pair of hydrogen atoms attached to terminal carbon atoms are present in different planes.

Hybridisation is mixing of atomic orbitals to form new hybrid orbitals. The new orbital have the same total electron capacity as the old ones. The properties and energies of the new hybridised orbitals are an average of the unhybridised orbitals.
\(
\underset{3 \sigma}{\stackrel{1}{\mathrm{C}} \mathrm{H}_2}=\underset{2 \sigma}{\stackrel{2}{\mathrm{C}}}=\underset{3 \sigma}{\stackrel{3}{\mathrm{C}} \mathrm{H}_2}
\)
Hybridisation can be found by counting number of \(\sigma\) – bonds around the carbon atom.
\(3 \sigma=\mathrm{sp}^2\) – hybridisation
\(2 \sigma=\mathrm{sp}-\) hybridisation
In allene, carbon atoms 1 and 3 are \(\mathrm{sp}^2\)-hybridised as each one of them is joined by a double bond. And, carbon atom 2 is \(sp\)-hybridised as it has two double bonds at each of its side. Therefore, the two \(\pi\)-bonds are perpendicular to each other, in allene, as shown below.

Q69. Benzoic acid is a organic compound. Its crude sample can be purified by crystallisation from hot water. What characteristic differences in the properties of benzoic acid and the impurity make this process of purification suitable?

Answer: Benzoic acid can be purified by hot water because of following characteristics
(i) Benzoic acid is more soluble in hot water and less soluble in cold water.
(ii) Impurities present in benzoic acid are either insoluble in water or more soluble in water to such an extent that they remain in solution as the mother liquor upon crystallisation.

Q70. Two liquids (A) and (B) can be separated by the method of fractional distillation. The boiling point of liquid (A) is less than boiling point of liquid (B). Which of the liquids do you expect to come out first in the distillate? Explain.

Answer: If the difference in boiling points of two liquids is not much, fractional distillation is used to separate them. In this technique, fractionating column is fitted over the mouth of the round bottom flask.
When vapours of a liquid mixture are passed through a fractionating column, the vapours of the low boiling of liquid (A) will move up while those of the high boiling liquid will condense and fall back into the flask. Therefore, liquid (A) with low boiling point will distill first.

Q71. You have a mixture of three liquids A, B and C. There is a large difference in the boiling points of \(\mathrm{A}\) and rest of the two liquids i.e., \(\mathrm{B}\) and \(\mathrm{C}\). Boiling point of liquids \(\mathrm{B}\) and \(\mathrm{C}\) are quite close. Liquid \(\mathrm{A}\) boils at a higher temperature than \(\mathrm{B}\) and \(\mathrm{C}\) and boiling point of \(\mathrm{B}\) is lower than \(\mathrm{C}\). How will you separate the components of the mixture. Draw a diagram showing set up of the apparatus for the process.

Answer: Liquids which have different boiling points vaporize at different temperatures. The vapours are cooled and then liquids so formed are collected separately. Liquid A can be separated from B and C due to the large difference in boiling points. Liquid \(\mathrm{B}\) and Liquid \(\mathrm{C}\) have boiling points very close to each other and cannot be separated by simple distillation and are separated by fractional distillation. Liquid B distilled first since the order of boiling points of \(A, B\) and \(C\) is as follows: \(B<C<A\).

Q72. Draw a diagram of bubble plate type fractionating column. When do we require such type of a column for separating two liquids. Explain the principle involved in the separation of components of a mixture of liquids by using fractionating column. What industrial applications does this process have?

Answer: Whenever the difference in the boiling points of two liquids is not much than simple distillation technique cannot be used to separate them. The vapours of such liquids are formed within the same temperature range and are condensed simultaneously and hence the technique of fractional distillation is used in such cases. In this technique, vapours of a liquid mixture are passed through a fractionating column before condensation. The fractionating column is fitted over the mouth of the round-bottomed flask. The vapours of the liquid with higher boiling point condense before the vapours of the liquid with lower boiling point. The vapours rising up in the fractionating column become richer in more volatile components and by the time the vapours reach to the top of the fractionating column, these become rich in more volatile components.

Q73. A liquid with high boiling point decomposes on simple distillation but it can be steam distilled for its purification. Explain how is it possible?

Answer: Steam distillation is a special type of separation technique for temperature-sensitive materials such as natural organic compounds. Few organic compounds tend to decompose at higher temperatures and normal distillation does not fit for this purpose. So, water is added to the apparatus and the temperature of the compounds is depressed, evaporating them at lower temperature. Once the distillation is accomplished, the vapours are condensed and there occurs, the separation of the constituents at ease.

In steam distillation, the distillating mixture consists of steam and the vapour of organic substance. In steam distillation, the liquid boils when the sum of the vapour pressure of the organic substance \(\left(p_1\right)\) and that of steam \(\left(p_2\right)\) becomes equal to the atmospheric pressure \((p)\) at the temperature of distillation.
\(
\mathrm{p}=\mathrm{p}_1+\mathrm{p}_2 \text { or } \mathrm{p}_1=\mathrm{p}-\mathrm{p}_2
\)

Since, the vapour pressure of the organic substance is lower than p, it vaporises below its normal boiling point without decomposition e.g., aniline which normally boils at \(457 \mathrm{~K}\) can be distilled at \(371.5 \mathrm{~K}\) by this process.

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