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) Crystallization

(i) On dissolving in solvent, the solid which is insoluble is separated out by filtration and the soluble solid is separated by crystallisation. It is a common and useful laboratory technique. It can be used to purify substances. (ii) Distillation under reduced pressure is done, because at low pressure vapours are formed below its boiling point and decomposition does not take place. (iii) A compound is said to be steam volatile if it distills under the conditions of steam distillation. Liquid is converted into vapours by passing steam followed by condensation. (iv) The vapours of such liquids are formed within the same temperature range and are condensed simultaneously. 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 bottom flask. (v) Liquids having different boiling points vaporise at different temperatures. The vapours are cooled and the liquids so formed collected separately. 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

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 bond with empty p -orbital present at adjacent positively charged carbon. (iii) Hyperconjugation (c) sp 2 hybridised carbon with the empty p orbital (iv) Isomers (d) Ethyne (v) sp hybridisation (e) Species that can revive a pair of electrons (vi) Electrophile (f) Species that can supply a pair of electrons

The hybridisation of carbocation is sp 2 . For a carbocation to be formed, the carbon atom must lose one electron. Thus, the carbon atom has three bond pair of electrons and no lone electrons. In CH 3 + , for example, the carbon has only three sigma bonds. Thus, it is sp 2 hybridised. Since the fourth pair of electrons is nonexistent, it is impossible to be sp 3 hybridised. A nucleophile is a chemical species that donates an electron pair to form a chemical bond in relation to a reaction. All molecules or ions with a free pair of electrons or at least one pi bond can act as nucleophiles. Because nucleophiles donate electrons. Hyperconjugation -The σ electrons of C - H bond of the alkyl group enter into partial conjugation with the attached unsaturated system or with the unshared p - orbital. Hyperconjugation is a permanent effect. An isomer is a molecule with the same molecular formula as another molecule, but with a different chemical structure. That is, isomers contain the same number of atoms of each element but have different arrangements of their atoms. For example, cyclohexane and 1 - hexene has molecular formula C 6 H 12 but properties are different. In C 2 H 2 (ethyne), Carbon is sp hybridised, because C has 2 σ and 2 π bonds. An electrophile is an atom or molecule that accepts an electron pair to make a covalent bond. It is electron loving i.e., positive charge species or electron deficient species and it will act as Lewis acid. So they can receive a pair of electrons. (i) → (c); (ii) → (f); (iii) → (b); (iv) → (a); (v) → (d); (vi) → (e) Column I Column II (i) Carbocation (c) sp 2 hybridised carbon with the empty p orbital (ii) Nucleophile (f) Species that can supply a pair of electrons (iii) Hyperconjugation (b) Conjugation of electrons of C - H 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 revive a pair of electrons

Match Column I with Column II. Column I Column II (i) Dumas method (a) AgNO 3 (ii) Kjeldahl's method (b) Silica gel (iii) Carius method (c) Nitrogen gas (iv) Chromatography (d) Free radicals (v) Homolysis (e) Ammonium sulphate

(i)Dumas method -----(c)Nitrogen gas In Dumas method, Compounds containing Nitrogen are estimated. (ii)Kjeldahl’s method ------(e)Ammonium sulphate In Kjeldahl’s method, Nitrogen present in compound is converted into ( NH 4 ) 2 SO 4 . (iii)Carius method-----(a) AgNO 3 In Carius method, halogens are estimated as a silver halide by using AgNO 3 . (iv)Chromatography-----(b)Silica gel In chromatography, silica gel is used as a fixed phase and acts as an adsorbent. (v)Homolysis -----(d)Free radicals As a result of homolytic fission of bond, shared pair of electrons equally distributed and free radicals are formed. (i) → (c); (ii) → (e); (iii) → (a); (iv) → (b); (v) → (d) Column I Column II (i) Dumas method (c) Nitrogen gas (ii) Kjeldahhl's method (e) Ammonium sulphate (iii) Carius method (a) AgNO 3 (iv) Chromatography (b) Silica gel (v) Homolysis (d) Free radicals

HARD

11th CBSE

IMPORTANT

Earn 100

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) ${CH}_{3} CH = {CHCH}_{2} {NH}_{2}$

Important Questions on Organic Chemistry – Some Basic Principles and Techniques

HARD

11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 12 - Organic Chemistry – Some Basic Principles and Techniques>Short Answer Type>Q 45

Which of the following compound will not exist as resonance hybrid. Give reason for your answer:

${CH}_{3} CH = {CHCH}_{2} {NH}_{2}$

MEDIUM

11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 12 - Organic Chemistry – Some Basic Principles and Techniques>Short Answer Type>Q 46

Why does

{SO}_{3}

act as an electrophile?

HARD

11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 12 - Organic Chemistry – Some Basic Principles and Techniques>Short Answer Type>Q 47

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

${CH}_{2} — CH = CH — O \leftrightarrow \overset{\oplus}{C} H_{2} - CH = CH - \overset{⊝}{O} I II$

HARD

11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 12 - Organic Chemistry – Some Basic Principles and Techniques>Short Answer Type>Q 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.

HARD

11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 12 - Organic Chemistry – Some Basic Principles and Techniques>Short Answer Type>Q 49

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

$(A)$ ${CH}_{3} COOH$ and $(B)$ ${CH}_{3} CO \overset{⊝}{O}$

MEDIUM

11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 12 - Organic Chemistry – Some Basic Principles and Techniques>Matching Type>Q 50

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) Crystallization

HARD

11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 12 - Organic Chemistry – Some Basic Principles and Techniques>Matching Type>Q 51

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 $C - H$ bond with empty $p$ -orbital present at adjacent positively charged carbon.
(iii)	Hyperconjugation	(c)	${sp}^{2}$ hybridised carbon with the empty $p$ orbital
(iv)	Isomers	(d)	Ethyne
(v)	$sp$ hybridisation	(e)	Species that can revive a pair of electrons
(vi)	Electrophile	(f)	Species that can supply a pair of electrons

MEDIUM

11th CBSE

IMPORTANT

NCERT Exemplar Chemistry - Class 11>Chapter 12 - Organic Chemistry – Some Basic Principles and Techniques>Matching Type>Q 52

Match Column I with Column II.

	Column I		Column II
(i)	Dumas method	(a)	${AgNO}_{3}$
(ii)	Kjeldahl's method	(b)	Silica gel
(iii)	Carius method	(c)	Nitrogen gas
(iv)	Chromatography	(d)	Free radicals
(v)	Homolysis	(e)	Ammonium sulphate

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

Important Questions on Organic Chemistry – Some Basic Principles and Techniques

Learn and Prepare for any exam you want !

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) ${CH}_{3} CH = {CHCH}_{2} {NH}_{2}$