NCERT Exemplar Class 11 Chemistry Solutions Chapter 12 Organic Chemistry Some Basic Principles and Technique

Question 2. The IUPAC name for is _______:

(i) 1-hydroxypentane-1,4-dione
(ii) 1,4-dioxopentanol
(iii) 1-carboxybutan-3-one
(iv) 4-oxopentanoic acid
Answer:

The answer is the option (iv) 4-oxopentanoic acid
Explanation: According to the IUPAC naming, in case of presence of more than one different types of functional groups, one functional group will be taken as the main functional group on a priority basis and is mentioned as a suffix, while the other functional group is written as a prefix.

Question 3. The IUPAC name for

(i) 1-Chloro-2-nitro-4-methylbenzene
(ii) 1-Chloro-4-methyl-2-nitrobenzene
(iii) 2-Chloro-1-nitro-5-methylbenzene
(iv) m-Nitro-p-chlorotoluene
Answer:

The answer is the option (ii) 1-Chloro-4-methyl-2-nitrobenzene
Explanation: The substituent of the base compound is considered to be number 1, and then the direction of numbering is chosen such that the next substituent group is getting the lowest number. The substituents appear in the name, arranged in alphabetical order.

Question 4. Electronegativity of carbon atoms depends upon their state of hybridisation. In which of the following compounds, the carbon marked with asterisk is most electronegative?
$(i) CH_3 - CH_{2}-^{\ast }CH_{2} -CH_{3}$
$(ii) CH_{}3 - ^{\ast}CH = CH - CH_{3}$
$(iii) CH_{3} - CH_{2} - C \equiv ^{\ast } CH$
$(iv) CH_{3} - CH_{2} - CH = ^{\ast}CH_{2}$
Answer:

The answer is the option $(iii) CH_{3} - CH_{2} - C \equiv ^{\ast }CH$
Explanation: The electronegativity of the carbon atom depends on its hybridisation state and the percentage of ‘s’ character. It is directly proportional to the ‘s’ character. Here we have sp hybridised carbon having 50 percent s character.

Question 5. In which of the following, functional group isomerism is not possible?
(i) Alcohols
(ii) Aldehydes
(iii) Alkyl halides
(iv) Cyanides
Answer:

The answer is the option (iii) Alkyl halides
Explanation: Isomerism is a phenomenon in which two compounds have the exact same molecular formula, but the different structural formula and such compounds are called isomers, and Functional isomers have same molecular formula but a different functional group. Alcohols are functional isomers of ethers. Aldehydes are functional isomers of ketones. Cyanides are functional isomers of isocyanides. Only alkyl halides do not show functional isomerism.

Question 6. The fragrance of flowers is due to the presence of some steam volatile organic compounds called essential oils. These are generally insoluble in water at room temperature but are miscible with water vapour in vapour phase. A suitable method for the extraction of these oils from the flowers is:
(i) Distillation
(ii) Crystallisation
(iii) Distillation under reduced pressure
(iv) Steam distillation
Answer:

The answer is the option (iv) Steam distillation
Explanation: This is a technique which is applied to separate substances which are steam volatile and are immiscible with water, differently. In case of steam distillation, steam, from a steam generator is passed through a heated flask containing the liquid to be distilled. The mixture of steam and the volatile organic compound is then condensed and collected together. The compound is later separated from water using a separating funnel.

Question 7. During hearing of a court case, the judge suspected that some changes in the documents had been carried out. He asked the forensic department to check the ink used at two different places. According to you which technique can give the best results?
(i) Column chromatography
(ii) Solvent extraction
(iii) Distillation
(iv) Thin layer chromatography
Answer:

The answer is the option (iv) Thin layer chromatography
Explanation: Thin-layer chromatography (TLC) is a method for analysing mixtures by separating the components of the mixture. This method can be used to determine the number of components in the mixture, to identity the compounds and the purity of compounds by observing the appearance of a product or the disappearance of a reactant, whichever is the case.

Question 8. The principle involved in paper chromatography is
(i) Adsorption
(ii) Partition
(iii) Solubility
(iv) Volatility
Answer:

The answer is the option (ii) Partition
Explanation: Partition chromatography is based on the concept of continuous differential partitioning of components of a mixture between stationary and mobile phases.

Question 9. What is the correct order of decreasing stability of the following cations?

(i) II > I > III
(ii) II > III > I
(iii) III > I > II
(iv) I > II > III
Answer:

(i) II > I > III
Explanation: In the case of (I) +vely charged C is attached to two alkyl groups and so +I effect stabilises the carbocation. In the case of (II) +R effect of $OCH_3$ group stabilises the carbocation. In the case of (III), - I effect of $-OCH_3$ gp, destabilises the carbocation; hence, the order of stability will be such: II > I >III.

Question 10. Correct IUPAC name for is _____________ .
(i) 2- ethyl-3-methylpentane
(ii) 3,4- dimethyl hexane
(iii) 2-sec-butylbutane
(iv) 2, 3-dimethylbutane
Answer:

(ii) 3,4- dimethyl hexane
Explanation: In the longest continuous chain, both ethyl groups are included, and 3rd and 4th carbon have methyl groups present as the substituent.

Question 11. In which of the following compounds the carbon marked with asterisk is expected to have greatest positive charge?
$(i) ^{\ast}CH_{3} -CH_{2}-Cl$
$(ii) ^{\ast}CH_{3} -CH_{2}-Mg^{+}Cl^{-}$
$(iii) ^{\ast}CH_{3} -CH_{2}-Br$
$(iv) ^{\ast}CH_{3} -CH_{2}-CH_{3}$
Answer:

The answer is the option $(i) ^{\ast}CH_{3} -CH_{2}-Cl$
Explanation: Order of electronegativity is as follows: Cl > Br > C > Mg. The more electronegative group attached to the C will give a more positive charge. Therefore, in (i) case, asterisk C will have the greatest positive charge.

Question 12. Ionic species are stabilised by the dispersal of charge. Which of the following carboxylate ion is the most stable?

Answer:

The answer is the option
Explanation: The dispersal of the negative charge depends on the stabilisation of the carboxylate ion. The negative charge which is shown is dispersed by two factors i.e +R effect of carboxylate ion and -I effect of halogens. In the above cases, +R effect is common, but halogen atoms are different and hence dispersal of the negative charge depends upon halogen atoms, where F is most electronegative.

Question 13. Electrophilic addition reactions proceed in two steps. The first step involves the addition of an electrophile. Name the type of intermediate formed in the first step of the following addition reaction.
$H_{3}C-HC = CH_{2} + H^{+} \rightarrow ?$
$(i) 2^{\circ} Carbanion$
$(ii) 1^{\circ} Carbocation$
$(iii) 2^{\circ} Carbocation$
$(iv) 1^{\circ} Carbanion$
Answer:

The answer is the option $(iii) 2^{\circ} Carbocation$
Explanation: When H⁺ attacks on propene delocalisation of electrons can take place in two possible ways:

Question 14. Covalent bond can undergo fission in two different ways. The correct representation involving a heterolytic fission of $CH_{3}-Br$ is

Answer:

The answer is the option (ii).
Explanation: Br is more electronegative than C, hence heterolytic fission takes place in that case and so electrons displace from carbon to Br. Therefore,$CH_{3}$ gets the positive charge and Br gets the negative charge.

Question 15. The addition of HCl to an alkene proceeds in two steps. The first step is the attack of H⁺ ion to$> C= C<$ portion which can be shown as

(iv) All these are possible
Answer:

The answer is the option (ii) because the double bond is an electron source. The charge flows from higher electron density; thus, the proton is attacked by n electrons of the double bond.

It can be shown as

Question 17. In which of the following representations given below spatial arrangement of group/ atom different from that given in structure ‘A’?


Answer:

The answer is the option (i), (iii) and (iv)
Explanation: The spatial arrangement of groups of atoms can be checked by interchanging and bringing the H below the plane of the paper and then finding out the sequence of the remaining groups in a particular order whether clockwise or anticlockwise where it is started from the atom with the highest atomic number to the atom with the lowest atomic numbers.

Question 18. Electrophiles are electron seeking species. Which of the following groups contain only electrophiles?
$(i)BF_{3}, NH_{3},H_{2}O$
$(ii)AlCl_{3}, SO_{3},NO_{2}^{+}$
$(iii)NO_{2}^{+},CH_{3}^{+},CH_{3}-C^{+}=O$
$(iv)C_{2}H_{5}^{-},C^{\cdot }_{2}H_{5},C_{2}H_{5}^{+}$
Answer:

The answer is the option (ii) and (iii)
Explanation: Electrophiles are the +vely charged species or the electron-deficient species. They are Lewis acids. $AlCl_{3}, SO_{3}$ and are Lewis acids, $(iii)NO_{2}^{+},CH_{3}^{+},CH_{3}-C^{+}=O$ are +vely charged species

Question 19. Which of the following pairs are position isomers?

(i) I and II
(ii) II and III
(iii) II and IV
(iv) III and IV
Answer:

The answer is the option (ii) II and III
Explanation: When two or more compounds have functional groups or substituent atoms attached at different positions on the carbon skeleton, they are termed as position isomers, and this phenomenon is termed as position isomerism.

Question 20. Which of the following pairs are not functional group isomers?

(i) II and III
(ii) II and IV
(iii) I and IV
(iv) I and II
Answer:

The answer is the option (i) and (iii)
Explanation: Isomerism is a phenomenon in which two compounds have the exact same molecular formula, but the different structural formula and such compounds are called isomers, and Functional isomers have same molecular formula but a different functional group.

Question 21. Nucleophile is a species that should have
(i) a pair of electrons to donate
(ii) positive charge
(iii) negative charge
(iv) electron deficient species
Answer:

The answer is the option (i) and (iii)
Explanation: Nucleophiles are -vely charged or electron rich (lone pair of electrons) species.

Question 22. Hyperconjugation involves delocalisation of ______.
(i) electrons of carbon-hydrogen $\sigma$ bond of an alkyl group directly attached to an atom of unsaturated system.
(ii) electrons of carbon-hydrogen $\sigma$ bond of alkyl group directly attached to the positively charged carbon atom.
(iii) $\pi$-electrons of carbon-carbon bond
(iv) lone pair of electrons
Answer:

The answer is the option (i) and (ii)
Explanation: Hyperconjugation means the delocalisation of electrons of the C-H bond of an alkyl group directly attached to an atom of unsaturated system or to an atom with an unshared p orbital. Electrons of C-H bond of the alkyl group enter into the partial conjugation with the attached unsaturated system or with the unshared p orbital. Thus, Hyperconjugation is a permanent effect.

Question 24. Identify the pairs of compounds which are functional group isomers.

Answer:

Isomerism is a phenomenon in which two compounds have the exact same molecular formula, but the different structural formula and such compounds are called isomers, and Functional isomers have same molecular formula but a different functional group. Hence, I and V, I and VI, I and VII; II and V, II and VI, II and VII; III and V, III and VI; III and VII; IV and V, IV and VI, IV and VI are functional group isomers.

Question 25. Identify the pairs of compounds that represents position isomerism.

Answer:

When two or more compounds have functional groups or substituent atoms attached at different positions on the carbon skeleton, they are termed as position isomers, and this phenomenon is termed as position isomerism. In the given structures, I and II, III and IV, and VI and VII are position isomers.

Question 26. Identify the pairs of compounds that represents chain isomerism.

Answer:

compounds I and III, I and IV, II and III and II and IV represent chain isomerism.

Question 27. For testing halogens in an organic compound with $AgNO_{3}$ solution, sodium extract (Lassaigne’s test) is acidified with dilute $HNO_{3}$. What will happen if a student acidifies the extract with dilute $H_{2}SO_{4}$ in place of dilute $HNO_{3}$ ?
Answer:

Elements like nitrogen, halogens, sulphur and phosphorous present in organic compounds converted into ions by fusing with sodium metal which is then plucked in distilled water getting sodium extract. On adding diluted $H_{2}SO_{4}$ in place of diluted HNO₃ for testing halogens by $AgNO_{3}$, a white precipitate of $Ag_2SO_{4}$ is formed. This will lead to the wrong result of chloride. Hence, only $HNO_{3}$ is used instead of diluted $H_{2}SO_{4}$.

Question 28. What is the hybridisation of each carbon in $H_{2}C = C = CH_{2}$.
Answer:

All three carbon atoms are linked to each other by double bonds only. Carbon at 1 and 3 are $sp^{2}$ hybridised because it has $3\sigma$ and $1\pi$ bonds, whereas carbon at 2 has 2 $\sigma$ bonds and $2\pi$ bonds, so it is sp hybridised.

Question 29. Explain, how is the electronegativity of carbon atoms related to their state of hybridisation in an organic compound?
Answer:

Electronegativity of carbon is directly proportional to the ‘s’ character. If C is $sp^{3}$hybridised then V character is 25%, $sp^{2}$ hybridised then V character is 33% and if sp hybridised then V character is 50%. Hence, sp hybridised carbon has strong hybridisation, s electrons are more strongly attracted by nucleus than p-electrons thus electronegativity of carbon increases with increase in ‘s’ character.

Question 30. Show the polarisation of carbon-magnesium bond in the following structure.
$CH_{3}-CH_{2}-CH_{2}-CH_{2}-Mg-X$
Answer:

Carbon is more electronegative than magnesium; hence Mg has a partial positive charge, and C has a partial negative charge because the bonded pair of electrons are attracted towards carbon.
$CH_{3}-CH_{2}-CH_{2} -CH_{2}^{-\delta} -Mg^{+\delta}- X$

Question 31. 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:

These are Metamers. When two or more compounds have differnet arrangement of atoms attached at different positions on the carbon skeleton, they are termed as metamers, and this phenomenon is termed as Metamarism.

Question 32. Which of the following selected chains is correct to name the given compound according to IUPAC system.

Answer:

IUPAC says that the selected longest carbon chain must have maximum functional groups present in the compound. Hence 4 carbon chain is correct to name the given compound according to IUPAC system

Question 33. 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:

In the case of DNA and RNA, nitrogen is present in the heterocyclic base and in the ring not as a substituent. Therefore, nitrogen present in the ring cannot be converted into$(NH_{4} )_{2}SO_{4}$. Hence, it cannot be estimated by the Kjeldahl method.

Question 34. 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:

Steam distillation can be used for its purification for a compound when the compound decomposes at its boiling point but is steam volatile and insoluble in water and is also stable at low pressures

Question 35. Draw the possible resonance structures for

and predict which of the structures is more stable. Give a reason for your answer.
Answer:

C has a more stable structure than A because. This is because in C, the octet of all the atoms is complete, but, in A, C-atom with a positive charge does not have 8 electrons in the valence shell.

Question 36. Which of the following ions is more stable? Use resonance to explain your answer.

Answer:

Structure A is more stable. Carbocation A is more Planar, and $\pi$ electrons from the ring shift to side group and are stabilised by resonance. Moreover, the double bond is more stable within the ring as compared to the side chain.

Question 37. 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 very stable because the +ve charge of methyl carbon is delocalised in three phenyl rings. In each phenyl ring, +ve charge is developed on 2 ortho position and para position, i.e. three resonating structures. Thus, the total resonating structures given by triphenylmethyl cation are nine. Hence, it is very stable.

Question 38. Write structures of various carbocation’s that can be obtained from 2- methylbutane. Arrange these carbocation’s in order of increasing stability.
Answer:

The stability order according to increasing stability is (III) > (II) > (IV) > (I)
because (I) and (IV) are primary, (II) is secondary carbocation, and (III) is tertiary carbocation.

Question 39. 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 $FeSO_{4}$ 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 colors.
Answer:

If an organic compound contains N and S both, then on fusion with Na metal, the compound gives NaSCN or NaCN and $Na_{2} S$ depending on what is the quantity of Na metal. If Na metal is less, then only NaSCN is formed. In that case L.E on treating with $FeSO_{4}$ and $H_{2}SO_{4}$ gives red colour due to the formation of ferric thiocyanide $Fe (SCN)_{3}$ . In case of NaCN, L.E on treating with $FeSO_{4}$ and $H_{2}SO_{4}$ gives a total Prussian blue colour.
Chemical reactions:
$NaCNS+2Na\rightarrow NaCN+Na_{2}S$
$3NaCNS+FeCl_{3}\rightarrow Fe(CNS)_{3}+NaCl$
Ferric sulphocynanide
(Blood red in colour)

Question 40. Name the compounds whose line formulae are given below:

Answer:

(i) 3-Ethyl-4-methylheptan-5-en-2-one (because the longest chain of carbon atoms is selected in such a way that the functional group > C = O gets lowest possible locant number)
(ii) 3-Nitrocyclohex-1-ene. (Carbon atoms of the ring are numbered in such a way that double bonded carbon gets the lowest number followed by the nitro group –$NO_{2}$)

Question 41. Write structural formulae for compounds named as- (a) 1-Bromoheptane (b) 5-Bromoheptanoic acid
Answer:

(a) $CH_{3}-CH_{2}-CH_{2}-CH_{2}-CH_{2}-CH_{2}-CH_{2}-Br$
7 6 5 4 3 2 1
(b)

Question 42. Draw the resonance structures of the following compounds:
(i)

$(ii)CH_{2}=CH-CH=CH_{2}$
(iii)

Answer:

Question 43. Identify the most stable species in the following set of ions giving reasons:

Answer:

(i) $CH_{3}^{+}$ will be more stable. This is because bromine atom destabilises the positive charge on a carbon atom. Bromine atom has a lone pair of electrons and is from the electron-withdrawing group.
(ii) $^-CCl_{3}$ will be most stable. This is because chlorine is more electron-withdrawing atom. The negative charge on carbon will be stabilised by the chlorine atom. As the number of chlorine atoms that are attached to carbocation increase, its stability also enhances.

Question 44. Give three points of differences between inductive effect and resonance effect.
Answer:

Question 45. Which of the following compounds will not exist as resonance hybrid. Give reason for your answer:
$(i) CH_{3} OH$
$(ii) R-CONH_{2}$
$(iii) CH_{3}CH = CHCH_{2}NH_{2}$
Answer:

$(i) CH_{3} OH$: It does not show resonance because there are no electrons.
$(ii) R-CONH_{2}$ : It shows resonance due to the presence of lone pair of electrons on N atom and electrons on C = O bond. Hence it can be represented by three resonating structures.
$(iii) CH_{3}CH = CHCH_{2}NH_{2}$ : Lone pair of electrons on N atom, is not conjugated with the $\pi$ electrons of double bond, so no resonance.

Question 46. Why does $SO_{3}$ act as an electrophile?
Answer:

$SO_{3}$ has total three highly electronegative oxygen atoms, that are attracted to the sulphur atom in $SO_{3}$ thus making sulphur electron deficient, that is it gets +ve and thus acts as an electrophile.

Question 47. Resonance structures of propenal are given below. Which of these resonating structures is more stable? Give reason for your answer.

Answer:

Structure I, propenal, is more stable than structure II because structure I has more covalent bonds in its resonating structures and moreover in structure II has only six electrons on its terminal carbon making it less stable.

Question 48. By mistake, an alcohol (boiling point $97^{\circ}C$) was mixed with a hydrocarbon (boiling point $68^{\circ}C$). Suggest a suitable method to separate the two compounds. Explain the reason for your choice.
Answer:

Because both the components above have a large difference in their boiling points, so simple distillation method is used to separate. The temperature when it rises to the low boiling point the hydrocarbon vapours are only formed without any contamination of alcohol.

Question 49. Which of the two structures (A) and (B) given below is more stabilised by resonance? Explain.
(A) $CH_{3}COOH$ and

Answer:

Structure ‘B’ is more stabilised as it does not involve charge separation.. Hence $CH_{3}COO^{-}$ is more stable.
(A)
$CH_{3}COOH$

Question 51. Match the terms mentioned in Column I with the terms in Column II.

Answer:

(i) →(c);(ii) → (f); (iii) → (b); (iv) → (a); (v) → (d); (vi) → (e)

Question 52. Match Column I with Column II.

Answer:

(i) → (c); (ii) → (e); (iii) → (a); (iv) → (b);(v) → (d)

Question 53. Match the intermediates given in Column I with their probable structure in Column II.

Answer:

(i) → (a); (ii) → (a); (iii) → (b)

Question 54. Match the ions given in Column I with their nature-given in Column II.

Answer:

(i) → (a), (b), (d) (ii) → (b) (iii) → (b) (iv) → (c), (d)

Question 56. In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): Energy of resonance hybrid is equal to the average of energies of all canonical forms.
Reason (R): Resonance hybrid cannot be presented by a single structure.
(i) Both A and R are correct, and R is the correct explanation of A.
(ii) Both A and R are correct, but R is not the correct explanation of A.
(iii) Both A and R are not correct.
(iv) A is not correct but R is correct.
Answer:

The answer is the option (iv) A is not correct, but R is correct.
Explanation: Canonical structures always have more energy as compared to the resonance hybrid, and so the Resonance hybrids are always more stable than any of the canonical structures. The delocalisation of electrons lowers the orbitals energy in case of resonance hybrid and gives stability.

Question 57. In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): Pent- 1- ene and pent- 2- ene are position isomers.
Reason (R): Position isomers differ in the position of functional group or a substituent.
(i) Both A and R are correct, and R is the correct explanation of A.
(ii) Both A and R are correct, but R is not the correct explanation of A
(iii) Both A and R are not correct.
(iv) A is not correct but R is correct
Answer:

The answer is the option (i) Both A and R are correct, and R is the correct explanation of A.
Explanation: When two or more compounds have functional groups or substituent atoms attached at different positions on the carbon skeleton, they are termed as position isomers and this phenomenon is termed as position isomerism

Question 58. In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): All the carbon atoms in $H_{2}C=C=CH_{2}$ are $sp^{2}$ hybridised
Reason (R): In this molecule, all the carbon atoms are attached to each other by double bonds.
(i) Both A and R are correct, and R is the correct explanation of A.
(ii) Both A and R are correct, but R is not the correct explanation of A.
(iii) Both A and R are not correct.
(iv) A is not correct but R is correct.
Answer:

The answer is the option (iv) A is not correct, but R is correct.
Explanation: Hybridisation of C can be found out by counting the number of $\sigma$ bonds and $\pi$ bonds present on the C atom. If C has 3$\sigma$ bonds, it is $sp^2$ hybridised. If C has 2$\sigma$ bonds, it is sp hybridised.

Question 59. In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): Sulphur present in an organic compound can be estimated quantitatively by Carious method.
Reason (R): Sulphur is separated easily from other atoms in the molecule and gets precipitated as light yellow solid.
(i) Both A and R are correct, and R is the correct explanation of A.
(ii) Both A and R are correct, but R is not the correct explanation of A.
(iii) Both A and R are not correct.
(iv) A is not correct but R is correct.
Answer:

The answer is the option (iii) Both A and R are not correct.
Explanation: Sulphur is estimated by the Carius method in the form of white precipitate of $BaSO_{4}$ on heating with fuming and $BaCl_{2}$. If light yellow solid is obtained it means that impurities are present. It is then filtered, washed, and then dried to get pure $BaSO_{4}$.

Question 60. In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): Components of a mixture of red and blue inks can be separated by distributing the components between stationary and mobile phases in paper chromatography.
Reason (R): The colored components of inks migrate at different rates because paper selectively retains different components according to the difference in their partition between the two phases.
Answer:

The answer is the option (i) Both A and R are correct, and R is the correct explanation of A.
Explanation: In paper chromatography what is used, is a special quality paper known as chromatography paper, which contains water trapped in it, and which acts as the stationary phase. A strip of chromatography paper which is spotted at its base is suspended with the solution of the mixture in a suitable solvent or a mixture of solvents. This solvent acts as the mobile phase. The solvent rises up the paper by the capillary action and flows all over to the spot. The paper now selectively retains different components according to their differing partition in the two phases. The paper strip so developed is now known as a chromatogram.

Question 62. Benzoic acid is an 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:

As per the principle of crystallisation, it is known to us that the solute must be more soluble in hot water and less soluble in the cold water or any other similar solvent.
Now, the solubility of Benzoic acid in water is lesser as it is an organic compound. Moreover, it is less polar, whereas, in comparison, the water molecule is highly polar. But when the temperature of the water is increased, we observed that the solubility of benzoic acid is better in hot water in comparison to the cold water. The other reason being that the impurities are not insoluble in the water. Therefore, they can be filtered from the benzoic acid solution (if not soluble) or they would remain dissolved even after cooling the solution(if completely soluble in cold water). They will not be interfering with recrystallisation of the benzoic acid, as the extra solution can be discarded once the process of crystallisation is over.
The properties of benzoic acid and the impurity which makes this process of purification suitable are: -

• Impurities which are present in benzoic acid are either insoluble in water or soluble in water to such an extent that they remain in solution.

• The solubility of benzoic acid is slightly higher in hot water as compared to cold water.

Question 63. 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:

The two liquids which we are considering above does not have much difference in their boiling points, and so simple distillation cannot just be used for the separation process. So the technique of fractional distillation is used in very such cases, which is a technique where vapours of a liquid mixture are passed through a fractionating column before condensation and the column used for fractionating is fitted over the mouth of the round bottom flask and the liquid with low boiling point distils first.

When vapours of a liquid mixture are passed through a fractionating column, the capours of the low boiling 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.

Question 64. You have a mixture of three liquids A, B and C. There is a large difference in the boiling points of A and rest of the two liquids i.e., B and C. Boiling point of liquids B and C are quite close. Liquid A boils at a higher temperature than B and C and boiling point of B is lower than C. How will you separate the components of the mixture. Draw a diagram showing set up of the apparatus for the process.
Answer:

As per the information available in the question, liquid A boils at a higher temperature than B and C and boiling point of B is lower than C.
Therefore, the correct order of boiling point for all three compounds is:
A > C > B
As it has been provided in the question that there is a significant difference in the boiling points of A and B and C. So, it is possible to separate liquid A from the mixture through the process of simple distillation. The setup for simple distillation is: -



As the remaining two liquids B and C have close boiling points; thus the process of fractional distillation will be used for the separation of the two. The setup for fractional distillation is

Question 65. 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:

The two liquids which we are considering above does not have much difference in their boiling points and so dimple distillation cannot just be used for the separation process. So, the technique of fractional distillation is used in such very cases, which is a technique where vapours of a liquid mixture are passed through a fractionating column before condensation and the column used for fractionating is fitted over the mouth of the round bottom flask and the liquid with low boiling point distills first.

Question 66. 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 different type of separation process for temperature sensitive materials like natural organic compounds. What happens with some organic compounds is that they decompose at higher temperatures and thus normal distillation does not suit the purpose. So, steam or water is added to the apparatus and the temperature of the compounds are depressed as a reason of that, evaporation happens at lower temperatures. Steam distillation is useful for separation if substances that are volatile, insoluble in water and have high vapour pressure at the boiling point of water i.e. $100^{\circ}C$. Then after distillation is over vapours are condensed and hence constituents separate at ease.