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Page No 144:

Question 1:

Which of the following is the correct IUPAC name?
(i) 3-Ethyl-4, 4-dimethylheptane
(ii) 4,4-Dimethyl-3-ethylheptane
(iii) 5-Ethyl-4, 4-dimethylheptane
(iv) 4,4-Bis(methyl)-3-ethylheptane

Answer:

While naming a compound according to IUPAC nomeclature, substituents are written in alphabetical order. Also, prefixes like di, tri, tetra, etc. are not considered in alphabetical order.

Hence, the correct answer is option (i).

Page No 144:

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:

When more than one functional group is present in a carbon chain, functional group higher in priority order (carboxylic acid here) is given lowest number and its suffix (oic acid) is used. The name of the second functional group (ketone here) is written as prefix (oxo).

Hence, the correct answer is option (iv).

Page No 144:

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:

In case of tri and higher substituted benzene derivatives, lowest locant rule is followed and name of substituents is written in alphabetical order.

Hence, the correct answer is option (ii).



Page No 145:

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) CH3 – CH2 *CH2 – CH3
(ii) CH3 *CH = CH – CH3
(iii) CH3 – CH2 – C ≡ *CH
(iv) CH3 – CH2 – CH = *CH2

Answer:

The change in hybridization is directly related to the change in the electronegativity of carbon atoms. As the s-character increases in hybridization, electronegativity also increases. sp hybrid orbital has maximum s-character (50%).

Hence, the correct answer is option (iii).

Page No 145:

Question 5:

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

Answer:

Alkyl halides can not show any other functional group with same molecular formula.

Hence, the correct answer is option (iii).

Page No 145:

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:

Steam distillation is a suitable method for the extraction of steam volatile compounds which are immiscible with water such as essential oils from flowers. The mixture boils just below 373 K and oil is later separated from water by separating funnel.

Hence, the correct answer is option (iv).

Page No 145:

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 components of different inks can be separated on the basis of their degree of adsorption on thin layer adsorbent coated glass plate. This technique is known as thin layer chromatography.

Hence, the correct answer is option (iv).

Page No 145:

Question 8:

The principle involved in paper chromatography is
(i) Adsorption
(ii) Partition
(iii) Solubility
(iv) Volatility

Answer:

Paper chromatography is based on the principle of separating different components of a mixture using chromatography paper. This special paper has water absorbed in it as stationery phase and carries out differential partitioning of constituents of a mixture.

Hence, the correct answer is option (ii).



Page No 146:

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:

The stability of a carbocation decreases in the order: +R (resonance) effect, +I (inductive) effect, −I (inductive) effect. In option (i), +I effect of two methyl groups stabilizes carbocation while in option (ii), +R effect of OCH₃ is present. Whereas in option (iii), carbocation is destabilised by −I effect of OCH₃.

Hence, the correct answer is option (i).

Page No 146:

Question 10:

Correct IUPAC name for is ___________.
(i) 2- ethyl-3-methylpentane
(ii) 3,4- dimethylhexane
(iii) 2-sec-butylbutane
(iv) 2, 3-dimethylbutane

Answer:

The first rule of IUPAC nomenclature is to select the longest carbon chain in the molecule.

Hence, the correct answer is option (ii).

Page No 146:

Question 11:

In which of the following compounds the carbon marked with asterisk is expected to have greatest positive charge?
(i) *CH3 — CH2 — Cl
(ii) *CH3 — CH2 — Mg+Cl
(iii) *CH3 — CH2 — Br
(iv) *CH3 — CH2 — CH3

Answer:

The presence of electron withdrawing groups increases positive charge in carbon chain due to inductive effect. Among the groups attached to carbon chain in given options, chlorine(Cl) is the most electronegative of all and therefore, imparts greatest positive charge.

Hence, the correct answer is option (i).

Page No 146:

Question 12:

Ionic species are stabilised by the dispersal of charge. Which of the following carboxylate ion is the most stable?
(i)
(ii)
(iii)
(iv)

Answer:

The electronegativity of group attached is directly proportional to the dispersal of charge and stability of carboxylate ion. Fluorine is the most electronegative element and two fluorines are attached to carboxylate ion in option (iv).

Hence, the correct answer is option (iv).

Page No 146:

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.
H3C — HC = CH2 + H→?
(i) 2° Carbanion
(ii) 1° Carbocation
(iii) 2° Carbocation
(iv) 1° Carbanion

Answer:

The attack of electrophile (H⁺) can generate either 1⁰ or 2⁰ carbocation. But 2⁰ carbocation is more stable.

Hence, the correct answer is option (iii).



Page No 147:

Question 14:

Covalent bond can undergo fission in two different ways. The correct representation involving a heterolytic fission of CH3 — Br is

Answer:

During the heterolytic fission, shared pair of electrons will move towards more electronegative bromine (Br) atom and it will acquire negative charge. The carbon atom will be left with positive charge.

Hence, the correct answer is option (ii).

Page No 147:

Question 15:

The addition of HCl to an alkene proceeds in two steps. The first step is the attack of H+ ion to portion which can be shown as
(i)

(ii)

(iii)

(iv) All of these are possible

Answer:

In this reaction, transfer of electrons of double bond of alkene to attacking electrophile (H⁺) takes place. This is correctly represented by a curved arrow originating from the site of electron density and ending at the site where electrons will be shifted to (H⁺ here).

Hence, the correct answer is option (ii).

Page No 147:

Question 16:

In the given question two or more options may be correct.
Which of the following compounds contain all the carbon atoms in the same hybridisation state?
(i) H — C ≡ C — C ≡ C — H
(ii) CH3 — C ≡ C —  CH3
(iii) CH2 = C = CH2
(iv) CH2 = CH — CH = CH2

Answer:

In option (i), all the carbon atoms are in sp hybridisation while in option (iv), all the carbon atoms are in sp2 hybridisation.

Hence, the correct answers are options (i) and (iv).

Page No 147:

Question 17:

In the given question two or more options may be correct.
In which of the following representations given below spatial arrangement of group/ atom different from that given in structure ‘A’?

(i)
(ii)
(iii)
(iv)

Answer:

To find the spatial arrangement of groups/ atoms, the lowest priority group/atom (H here) should be on the dashed wedge. Structure A has a clockwise spatial arrangement of groups/ atoms which is similar to the structure given in option (ii) only.

Hence, the correct answers are options (i), (iii) and (iv).



Page No 148:

Question 18:

In the given question two or more options may be correct.
Electrophiles are electron seeking species. Which of the following groups contain only electrophiles?
(i) BF3, NH3, H2O
(ii) AlCl3, SO3, NO2+
(iii) NO2+, CH3+, CH3-C+=O
(iv) C2H5-, C2H5, C2H5+

Answer:

The examples of electrophiles are positively charged species, electron deficient neutral molecules and lewis acids.

Hence, the correct answers are options (ii) and (iii).

Page No 148:

Question 19:

In the given question two or more options may be correct.
Which of the following pairs are position isomers?
(i)

(ii)

(iii)

(iv)

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

Answer:

When two or more compounds differ in position of substituent or functional group (ketone here) on the carbon skeleton, they are called position isomers.

Hence, the correct answer is option (ii).



Page No 149:

Question 20:

In the given question two or more options may be correct.
Which of the following pairs are not functional group isomers?
(i)

(ii)


(iii)

(iv)

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

Answer:

Functional group isomers are compounds with same molecular formula but different functional groups. I and IV contain aldehyde functional group while II and III contain ketone functional group.

Hence, the correct answers are options (i) and (iii).

Page No 149:

Question 21:

In the given question two or more options may be correct.
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:

Nucleophile (nucleus loving) is a species that can be negatively charged or neutral with lone pair of electrons for donation to an electrophile (electron seeking species).

Hence, the correct answers are options (i) and (iii).

Page No 149:

Question 22:

In the given question two or more options may be correct.
Hyperconjugation involves delocalisation of ___________.
(i) electrons of carbon-hydrogen σ bond of an alkyl group directly attached to an atom of unsaturated system.
(ii) electrons of carbon-hydrogen σ bond of alkyl group directly attached to the positively charged carbon atom.
(iii) π-electrons of carbon-carbon bond
(iv) lone pair of electrons

Answer:

Hyperconjugation is a permanent stabilizing interaction. In this effect, delocalization of σ electrons of C―H bond of an alkyl group directly attached to an atom of unsaturated system or carbocation takes place.

Hence, the correct answers are options (i) and (ii).



Page No 150:

Question 23:

I.

II.
III.

IV.

V.

VI.

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

Answer:

Metamerism is a type of isomerism in which compounds have same molecular formula but different alkyl groups on either side of functional group. The compounds, V, VI and VII can be called metamers because they have different alkyl groups on either side of ether functional group.

Page No 150:

Question 24:

I.

II.
III.

IV.

V.

VI.

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

Answer:

Alcohols are functional group isomers of ethers of same molecular formula. Compounds I, II, III and IV are alcohols while compounds V, VI and VII are ethers of same molecular formula. Hence, the following pairs of compounds are functional group isomers–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 VII.

Page No 150:

Question 25:

I.

II.
III.

IV.

V.

VI.

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

Answer:

The isomerism in which compounds differ in the position of substituent atom or functional group on the carbon skeleton is called position isomerism. The pairs of compounds that represents position isomerism from given list are I and II, III and IV, V and VI.

Page No 150:

Question 26:

I.

II.
III.

IV.

V.

VI.

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

Answer:

The phenomenon of compounds with similar molecular formula but different carbon skeletons is called chain isomerism. The pairs of compounds that represents chain isomerism are I and II, I and III, I and IV, II and III, II and IV, III and IV.

Page No 150:

Question 27:

For testing halogens in an organic compound with AgNO3 solution, sodium extract (Lassaigne’s test) is acidified with dilute HNO3. What will happen if a student acidifies the extract with dilute H2SO4 in place of dilute HNO3?

Answer:

On acidifying the sodium extract with dilute H2SO4, a white precipitate of Ag2SO4 will be formed due to reaction of Ag⁺ ions from AgNOsolution with SO42− ions from dilute H2SO4. This may give a false impression to the student that white precipitate of AgCl is formed due to presence of chlorine.

Page No 150:

Question 28:

What is the hybridisation of each carbon in H2C = C = CH2.

Answer:

The hybridisation of central carbon is sp due to formation of two π bonds while the hybridisation of two carbons at the two ends of molecule is sp2 due to formation of single π bond and two sigma bonds.

Page No 150:

Question 29:

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

Answer:

The change in electronegativity of carbon atoms is directly related to change in hybridisation. As the s-character increases in hybridisation, electronegativity also increases. The sp hybrid orbital has maximum s-character (50%) followed by sp2 hybrid orbital which has 33% s-character and sp3 hybrid orbital has the least s-character i.e. 25%. Therefore, the order of electronegativity is also the same–sp > sp2sp3.

Page No 150:

Question 30:

Show the polarisation of carbon-magnesium bond in the following structure.

CH3 — CH2 — CH2 — CH2 — Mg — X

Answer:

CH3 — CH2 — CH2 — CH2δ−—← Mgδ+ — X.
The bond between carbon and magnesium is polar covalent bond due to difference in electronegativity of the two atoms. The electron density is shifted toward more electronegative carbon. As a result, carbon bears partial negative (δ−) charge and magnesium bears partial positive (δ+) charge.
 

Page No 150:

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:

The above two compounds show position isomerism because they have same molecular formula but differ in position of thioether (S) functional group on carbon skeleton.

Page No 150:

Question 32:

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

Answer:

The four carbon chain is the correctly selected chain to name the given compound according to IUPAC system. This is because it is the longest possible carbon chain accomodating both the functional groups–carboxylic acid and alcohol.



Page No 151:

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:

No, Kjeldahl method can not be used for the estimation of nitrogen present in DNA and RNA. This is because nitrogen present in the ring system does not convert to ammonium sulphate and then, ammonia gas can not be liberated.

Page No 151:

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 method will be chosen for purification of the liquid compound from water. This is because the liquid will vaporise below its boiling point owing to its low pressure. Then, it can be condensed back and separated from water by separating funnel.

Page No 151:

Question 35:

“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.”

Draw the possible resonance structures for CH subscript 3 minus straight O with bold. bold. below and bold. bold. on top minus straight C with plus on top straight H subscript 2 and predict which of the structures is more stable. Give reason for your answer.

Answer:

There are two possible resonance structures for the given compound.
text (I)  end text C H subscript 3 minus O with times times below and times times ↷ on top minus C with plus on top H subscript 2 left right arrow C H subscript 3 minus O with times times below and plus on top equals with ↶ on top C H subscript 2 space text (II)  end text
Structure (II) is more stable because all the atoms have their octet complete in this structure.

Page No 151:

Question 36:

“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.”

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

Answer:

Ion (A) is more stable because double bond is more stable inside the ring and this structure is planar. Moreover, it undergoes delocalisation of π electrons i.e. resonance.

Page No 151:

Question 37:

“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.”

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:

The high stability of this cation can be attributed to the delocalisation of positive charge of carbon over three phenyl rings. This leads to three resonating structures by each phenyl ring and overall nine resonating structures.

Page No 151:

Question 38:

“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.”

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

Answer:

Four different carbocations can be obtained from 2-methylbutane.
(I) C+H2-C|H-CH2-CH3               CH3(II) CH3-C+|-CH2-CH3                CH3(III) CH3-C|H-C+H-CH3                 CH3(IV) CH3-C|H-CH2-C+H2                  CH3
Increasing order of stability of carbocations: II > III > I > IV
This is because of  hyperconjugation effect in which the stability of carbocation increases in the order−3⁰ > 2⁰ > 1⁰.
                                  

Page No 151:

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 FeSO4 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:

Ramesh used less sodium and thus, obtained red colour due to formation of [Fe(SCN)]2⁺ as both N and S are present in given organic compound.

N a plus C plus N plus S rightwards arrow N a S C N
F e to the power of 3 plus end exponent plus S C N to the power of minus rightwards arrow left square bracket F e left parenthesis S C N right parenthesis right square bracket to the power of 2 plus end exponent

Manish and Rajni used excess of sodium in preparation of Lassaigne’s extract and therefore, thiocyanate decomposed to give cyanide and sulphide.

N a S C N plus 2 N a rightwards arrow N a C N plus N a subscript 2 S

Cyanide ion on reaction with solid FeSOâ‚„ and dilute sulphuric acid gave Feâ‚„[Fe(CN)₆]₃.xHâ‚‚O due to which Manish and Rajni obtained prussian blue colour.

6 C N to the power of minus plus F e to the power of 2 plus end exponent rightwards arrow left square bracket F e left parenthesis C N right parenthesis subscript 6 right square bracket to the power of 4 minus end exponent space
3 left square bracket F e left parenthesis C N right parenthesis subscript 6 right square bracket to the power of 4 minus end exponent plus 4 F e to the power of 3 plus end exponent rightwards arrow with X H subscript 2 O on top F e subscript 4 left square bracket F e left parenthesis C N right parenthesis subscript 6 right square bracket subscript 3. x H subscript 2 O


 



Page No 152:

Question 40:

Name the compounds whose line formulae are given below :
(i)

(ii)

Answer:

(i) 3-Ethyl-4-methylhept-5-en-2-one

(ii) 3-Nitrocyclohex-1-ene

Page No 152:

Question 41:

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

Answer:

text (a)  end text C H subscript 3 minus C H subscript 2 minus C H subscript 2 minus C H subscript 2 minus C H subscript 2 minus C H subscript 2 minus C H subscript 2 minus B r
text (b)  end text C H subscript 3 minus C H subscript 2 minus C with vertical line below H minus C H subscript 2 minus C H subscript 2 minus C H subscript 2 minus C with parallel to below minus O H
space space space space space space space space space space space space space space space space space space space space space space space space B r space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space O
space space space space space space space space space space space space space

Page No 152:

Question 42:

Draw the resonance structures of the following compounds;
(i)

(ii) CH2 = CH — CH = CH2

(iii)

Answer:

text (i) end text space C H subscript 2 equals C H minus stack C l with bullet bullet below and bullet bullet on top bullet bullet left right arrow C with circled dash on top H subscript 2 minus C H equals stack stack C l with bullet bullet below with circled plus on top bullet bullet
text (ii)  end text C H subscript 2 equals C H minus C H equals C H subscript 2 left right arrow C with circled plus on top H subscript 2 minus C H equals C H minus C with circled dash on top H subscript 2
space space space space space space space space space space space space space space space space space space space space space space space space space space O space space space space space space space space space space space space space space space space space space space space space space bullet bullet O with circled dash on top bullet bullet
text (iii)  end text C H subscript 2 equals C H minus C with parallel to on top H left right arrow C with circled plus on top H subscript 2 minus C H equals C with vertical line on top H

Page No 152:

Question 43:

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

i C+H3, C+H2Br, C+HBr2, C+Br3ii CH3, CH2Cl, CHCl2, CCl3

Answer:

(i) C+H3 is the most stable species in this set of ions because replacement of H by more electronegative Br will destabilise the carbocation as Br is electron withdrawing group and will show −I (inductive) effect.
(ii) CCl3 is the most stable species in this set of ions because replacement of H by more electronegative Cl will aid in dispersing the negative charge on carbon atom as Cl is electron withdrawing group and will show −I (inductive) effect.

Page No 152:

Question 44:

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

Answer:

S.No. Inductive effect Resonance effect
1. Polarisation of σ-bond caused by polarisation of adjacent σ-bond. Polarisation in conjugated system due to interaction between 2 π-bonds or between a π-bond and lone pair of electrons on adjacent atom.
2. It vanishes after three to four carbon atoms. It is transmitted throughout the chain.
3. It develops partial positive and negative charges as electron density is slightly shifted towards an atom. It develops complete positive and negative charges as complete transfer of electrons takes place.

Page No 152:

Question 45:

Which of the following compounds will not exist as resonance hybrid. Give reason for your answer :
(i) CH3OH
(ii) R – CONH2
(iii) CH3CH = CHCH2NH2

Answer:

CH₃OH  and CH₃CH = CHCHâ‚‚NHâ‚‚ will not exist as resonance hybrid because both of them do not involve conjugated system which is the basic requirement for resonance to occur.

Page No 152:

Question 46:

Why does SO3 act as an electrophile?

Answer:

Sulphur is electron deficient due to attachment of three electronegative oxygen atoms to it which pull electron density towards them. This creates a positive charge over sulphur in resonating structures. Therefore, SO₃ acts as an electrophile.

Page No 152:

Question 47:

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

CH2=CH-CH=OCH2-CH=CH-OIII

Answer:

Structure I is more stable due to greater number of covalent bonds, complete octet of all atoms and no charges in the structure.

Page No 152:

Question 48:

By mistake, an alcohol (boiling point 97°C) was mixed with a hydrocarbon (boiling point 68°C). Suggest a suitable method to separate the two compounds. Explain the reason for your choice.

Answer:

The given two compounds can be separated by simple distillation because they have a difference of more than 25 °C in their boiling points.

Page No 152:

Question 49:

Which of the two structures (A) and (B) given below is more stabilised by resonance? Explain.
CH3COOHandCH3COOAB

Answer:

Structure (B) is more stabilised by resonance as it does not involve separation of opposite charges and negative charge is always present on more electronegative oxygen atom.



Page No 153:

Question 50:

In the given question more than one correlation is possible between options of Column I and Column II. Make as many correlations as you can.
Match the type of mixture of compounds in Column I with the technique of separation / purification given in Column II.
 

  Column I   Column II
(i) Two solids which have different solubilities in a solvent and which do not undergo reaction when dissolved in it. (a)  Steam distillation
(ii) Liquid that decomposes at its boiling point (b) Fractional distillation
(iii) Steam volatile liquid (c) Simple distillation
(iv) Two liquids which have boiling points close to each other (d) Distillation under reduced pressure
(v) Two liquids with large difference in boiling points. (e) Crystallisation

Answer:

  Column I Column II
(i) Two solids which have different solubilities in a solvent and which do not undergo reaction when dissolved in it. (e) Crystallisation 
(ii) Liquid that decomposes at its boiling point (d) Distillation under reduced pressure
(iii) Steam volatile liquid (a) Steam distillation
(iv) Two liquids which have boiling points close to each other (b) Fractional distillation
(v) Two liquids with large difference in boiling points. (c) Simple distillation

Page No 153:

Question 51:

In the given question more than one correlation is possible between options of Column I and Column II. Make as many correlations as you can.
Match the terms mentioned in Column I with the terms in Column II.
 

  Column I   Column II
(i) Carbocation (a) Cyclohexane and 1- hexene
(ii) Nucleophile (b) Conjugation of electrons of C – H σ bond with empty p-orbital present at adjacent positively charged carbon.
(iii) Hyperconjugation (c) sp2 hybridised carbon with empty p-orbital
(iv) Isomers (d) Ethyne
(v) sp hybridisation (e) Species that can receive a pair of electrons
(vi) Electrophile (f) Species that can supply a pair of electrons

Answer:

  Coulmn I Column II
(i) Carbocation (c) sp2 hybridised carbon with empty p-orbital
(ii) Nucleophile (f) Species that can supply a pair of electrons
(iii) Hyperconjugation (b) Conjugation of electrons of C – H σ bond with empty p-orbital present at adjacent positively charged carbon
(iv) Isomers (a) Cyclohexane and 1- hexene
(v) sp hybridisation (d) Ethyne
(vi) Electrophile (e) Species that can receive a pair of electrons

Page No 153:

Question 52:

In the given question more than one correlation is possible between options of Column I and Column II. Make as many correlations as you can.
Match Column I with Column II.
 

  Column I   Column II
(i) Dumas method (a) AgNO3
(ii) Kjeldahl’s method (b) Silica gel
(iii) Carius method (c) Nitrogen gas
(iv) Chromatography (d) Free radicals
(v) Homolysis (e) Ammonium sulphate

Answer:

  Column I Column II
(i) Dumas method (c) Nitrogen gas
(ii) Kjeldahl's method (e) Ammonium sulphate
(iii) Carius method (a) AgNO₃
(iv) Chromatography (b) Silica gel
(v) Homolysis (d) Free radicals
 



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Question 53:

In the given question more than one correlation is possible between options of Column I and Column II. Make as many correlations as you can.
Match the intermediates given in Column I with their probable structure in Column II.
 

  Column I   Column II
(i) Free radical (a) Trigonal planar
(ii) Carbocation (b) Pyramidal
(iii) Carbanion (c) Linear

Answer:

  Column I Column II
(i) Free radical (a) Trigonal planar
(ii) Carbocation (a) Trigonal planar
(iii) Carbanion (b) Pyramidal

Page No 154:

Question 54:

In the given question more than one correlation is possible between options of Column I and Column II. Make as many correlations as you can.
Match the ions given in Column I with their nature given in Column II.
 

  Column I   Column II                  
(i) (a) Stable due to resonance
(ii) (b) Destabilised due to inductive effect
(iii) (c) Stabilised by hyperconjugation
(iv) (d) A secondary carbocation

Answer:

  Column I Column II
(i) (a) Stable due to resonance
(ii) (b) Destabilised due to inductive effect
(iii) (b) Destabilised due to inductive effect
(iv) (c) Stabilised by hyperconjugation
(d) A secondary carbocation
 

Page No 154:

Question 55:

In the given question 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) : Simple distillation can help in separating a mixture of propan-1-ol (boiling point 97°C) and propanone (boiling point 56°C).
Reason (R) : Liquids with a difference of more than 20°C in their boiling points can be separated by simple distillation.

(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 difference between boiling points of propan-1-ol and propanone is 31°C (97°C−56°C).  Thus, simple distillation can be used to separate them from the mixture.
Hence, the correct answer is option (i).



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Question 56:

In the given question 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 energy of resonance hybrid is lower than that of any of the canonical structures and it is a hybrid of more than one canonical structures.
Hence, the correct answer is option (iv).

Page No 155:

Question 57:

In the given question 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:

Pent-1-ene and pent-2-ene differ in the position of double bond.
Hence, the correct answer is option is (i).

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Question 58:

In the given question 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 H2C = C = CHare sphybridised
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 central carbon atom has two σ bonds and two π bonds. So, it is sp hybridised. The terminal carbon atoms have three σ bonds and one π bond. So, they are sp2 hybridised.
Hence, the correct answer is (iv).

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Question 59:

In the given question 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 Carius 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 percentage of sulphur in an organic compound can be estimated from mass of barium sulphate. Sulphur is oxidised to sulphuric acid and then precipitated as barium sulphate by adding excess of barium chloride solution in water.
Hence, the correct answer is option (iii).

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Question 60:

In the given question 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 coloured components of inks migrate at different rates because paper selectively retains different components according to the difference in their partition between the two phases.

(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:

Paper chromatography is a type of partition chromatography in which components of a mixture can be separated between stationary (chromatography paper) and mobile phases (water).
Hence, the correct answer is option (i).



Page No 156:

Question 61:

What is meant by hybridisation? Compound CH= C = CHcontains sp or sphybridised carbon atoms. Will it be a planar molecule?

Answer:

Hybridisation is the phenomenon of intermixing of atomic orbitals of different energies to form new hybrid orbitals of equivalent energies and shape.
The central carbon atom has two σ bonds and two π bonds. So, it is sp hybridised. The terminal carbon atoms have three σ bonds and one π bond. So, they are sp2 hybridised.
It is not a planar molecule as the two hydrogens attached to each of the terminal carbon atoms are perpendicular to each other.

Page No 156:

Question 62:

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 is sparingly soluble in water at room temperature while it is soluble in hot water. The impurities are soluble in water at room temperature. So, when saturated solution of crude sample of benzoic acid is cooled, pure benzoic acid crystallises out and impurities remain soluble in mother liquor. Hence, crystallisation is an appropriate technique to obtain pure benzoic acid.

Page No 156:

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:

Liquid (A) will come out first in distillate as vapours of liquid of lower boiling point will ascend to top of the column. The top of the fractionating column is richer in more volatile component. Then, vapours of low boiling point liquid i.e. liquid (A) will condense and collected in receiver.

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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:

The components of the mixture can be separated by fractional distillation because the boiling points of liquid B and C are very close. Liquid B with lowest boiling point will be vaporised and collected first in receiver. Then, liquid C will boil and condensed . Liquid A will be enriched in the fractionating flask as it has the highest boiling point.

 

Page No 156:

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:


When the boiling points of two liquids in a mixture is very close, then, we require bubble plate type fractionating column. 
The principle involved in separation of such liquids is that their vapours are passed through a fractionating column. Lower boiling point liquid will vapourise first and ascend to the top of the column. Thus, the top of the column is occupied by vapours of low boiling point liquid. These vapours pass through the condenser and pure liquid is collected in the receiver. After successive distillations, liquid with higher boiling point remains in the distillation flask as vapours of higher boiling point condense first. In this way, both liquids are separated.
This process is very important in petroleum industry for separating different fractions of crude oil. It is also used to separate air components to produce liquid oxygen and nitrogen.

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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:

A liquid with high boiling point can be steam distilled without decomposing it. Steam is passed through the heated flask containing liquid and temperature of the liquid is decreased. The mixture boils when vapour pressure due to liquid and water is equal to atmospheric pressure. Since, vapour pressure of liquid is less than vapour pressure of the mixture, the liquid boils at a temperature below its boiling point. The mixture of steam and liquid is condensed and collected. Then, the liquid which is insoluble in water is separated by separating funnel.



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