RBSE Solutions for Class 11 Chemistry Chapter 12 Organic Chemistry: Some Basic Principles and Techniques

Rajasthan Board RBSE Class 11 Chemistry Chapter 12 Organic Chemistry: Some Basic Principles and Techniques

RBSE Class 11 Chemistry Chapter 12 Text Book Questions

RBSE Class 11 Chemistry Chapter 12 Multiple Choice Questions

Question 1.
Which of the following method is used for estimation of nitrogen?
(a) Leibig’s method
(b) Lassaigne’s method
(c) Kjeldahl’s method
(d) Carius method
Answer:
(c) Kjeldahl’s method

Question 2.
The salt obtained by melting sodium metal with nitrogen containing carbon compound:
(a) NaNO₂
(b) NaNO₃
(c) NaCN
(d) NaNH₂
Answer:
(c) NaCN

Question 3.
IUPAC name of iso-butane is :
(a) 2-methyl butane
(b) 2-methyl propane
(c) 2-ethyl butane
(d) 2-butyne
Answer:
(b) 2-methyl propane

Question 4.
Prefix used for —COOH functional group is :
(a) Carbomayl
(b) Carbonyl
(c) Carboxy
(d) Alkoxycarbonyl
Answer:
(c) Carboxy

Question 5.
Which of the following substituent do not show +I effect ?
(a) Br
(b) R—OH
(c) FeCl₃
(d) —CHR₂
Answer:
(a) Br

Question 6.
Which of the following reagent is nucleophilic ?
(a) Br
(b) R—OH
(c) FeCl₃
(d) CO₂
Answer:
(b) R—OH

Question 7.
Which of the following is most stable carbocation ?
(a) ⁺CH₃
(b) ⁺CH₂CH₃
(c)

(d) (CH₃)₃C⁺
Answer:
(d) (CH₃)₃C⁺

Question 8.
Homolytic fission results in formation of:
(a) Carbocation
(b) Carbene
(c) Nitrene
(d) Free radical
Answer:
(d) Free radical

Question 9.
Which of the following group do not give addition reactions ?
(a) C ≡ C
(b) C = C
(c) C = O
(d) CH₃ — CH₃
Answer:
(d) CH₃ — CH₃

RBSE Class 11 Chemistry Chapter 12 Very Short Answer Type Questions

Question 10.
Give two examples of—I effect.
Answer:
— NO₂, —F.

Question 11.
Write the name of two neutral and two negatively charged nucleophiles.
Answer:

Negatively Charged nucleophiles ⇒ Cyanide ion(CN^–) Chloride ion(Cl^–)

Question 12.
Write the name of the method used for estimation of halogens.
Answer:
Carius method is used for estimation of Halogens.

Question 13.
Write the formula for percentage of nitrogen in Kjeldhal’s method.
Answer:
% of Nitrogen

Question 14.
Which method is used for estimation of carbon and hydrogen ?
Answer:
Liebig method is used for estimation of carbon and hydrogen.

Question 15.
Define resonance.
Answer:
Resonance is a method of describing delocalized electrons within certain molecules or polyatomic ions where the bonding cannot be expressed by one single Lewis structure. A molecule or ion with such delocalized electrons is represented by several contributing structures called resonance structures.

Question 16.
What is the hybridisation of carbon atom in carbanion?
Answer:
sp³.

Question 17.
Given the order of stability of primary, secondary and tertiary carbocation.
Answer:
Increasing order of stability :

Question 18.
Define Carbene.
Answer:
Carbene is defined as a neutral species two valence and two unshared pair electron, it is very reactive intermediate.

Question 19.
Write the IUPAC name of trimethyl methane.
Answer:

Question 20.
Write the IUPAC name of OHC—CH₂ —CH₂—COOH
Answer:

Question 21.
Write the prefix and suffix used for —OH group.
Answer:
Prefix of —OH group = hydroxy
Suffix of —OH group = – ol

Question 22.
What is the difference between the mass of two successive members of homologous series ?
Answer:
—CH₂ = 12 + 2 = 14
The difference between the mass of two successive members of homologous series is 14.

Question 23.
Write the name and structure of two aromatic hetrocyclic compounds.
Answer:
Pyridine

RBSE Class 11 Chemistry Chapter 12 Short Answer Type Questions

Question 24.
Which method is used to test nitrogen, sulphur and halogens ?
Answer:
Lassaigne test method is used to test nitrogen, sulphur and halogens. This test involves two steps

• preparation of sodium fusion extract
• Detection of elements using sodium fusion extract.

Question 25.
Write the structural formula and IUPAC name of the following compounds :
(i) Formic acid
(ii) Ethyl acetate
(iii) Ethyl methyl ether
Answer:

Question 26.
What is the difference between homelytic and heterolytic cleavage ?
Answer:

Question 27.
Write two examples of molecules showing—I effect.
Answer:

In these molecules, —NO₂ and —Cl atoms or groups exhibit —I effect. Because these are electron withdrawing group or atoms.

Question 28.
Write four examples of neutral electrophiles.
Answer:
BF₃, AlCl₃, SO₃ and ZnCl₂.

Question 29.
Write the following in decreasing order of + I effect:
(CH₃ )₂ CH—, CH₃—, (CH₂)₃ C —, — CH₂CH₃
Answer:
The decreasing order of +1 effect is given as below:
(CH₃)₃ C➝(CH₃)₂ CH➝ -CH₂CH₃ CH₃ —

Question 30.
Write the name and structure of third homologous member of alkane, alkene and alkanone.
Answer:
Third homologous member of
Alkane ➝ Propane (C₃H₈)
Alkene ➝ Butene C₄ H₈
Alkanone ➝ Pentanone (CH₃CH₂CH₂COCH₃)

Question 31.
Why alkanes are called paraffins ?
Answer:
Paraffins is a Latin word which means (Parum = little and affinis = reactivity). Alkanes are called paraffins because they have a little affinity towards a general reagent.

Question 32.
Write the IUPAC name of chloroform, formic acid and iso-pentane
Answer:

Question 33.
Write the formula of isobutyl alcohol and butyl chloride.
Answer:

Question 34.
Classify the following into electrophiles and nucleophiles:
NH₃, BF₃, H₂O, FeCl₃, OH^–, H₃O⁺, SO₃, CCl₂
Answer:
Electrophiles : Electron pair acceptors :
BF₃, FeCl₃, H₃O⁺, SO₃, CCl₂
Nucleophiles : Electron pair donors :

Question 35.
In which compounds, electrophilic addition reactions are generally found? Explain.
Answer:
Electrophilic addition reaction are generally found in organic compounds containing multiple bonds. For example, ethene reacts with bromine by an electrophilic addition reaction. In this reaction all parts of adding reagent appear in the product.

Question 36.
Explain rearrangement reactions with examples.
Answer:
In a rearrangement reaction, a molecule undergoes a reorganization of its constituent parts. For example, alkene on heating with strong acid form another isomeric alkene.

Question 37.
Write the difference between homolytic and heterolytic cleavage.
Answer:

Question 38.
Explain the free radical reactions.
Answer:
Free radicals are chemical species that contain a singly occupied orbital. They are neutral and tend to be highly reactive. These are formed by homolytic cleavage of covalent bonds. These have high tendency to pair the unpaired electron. These readly combine with each other or with other molecules to pair their electrons. In a free radical reaction, free readily is formed as reaction intermediate. It is the reaction which involves free radicals. These reactions take place mainly in the presence of UV light.

Question 39.
Write the IUPAC name of following:

Answer:

Question 40.
Give the structural formula of 2-methoxy- 2-methyl propane.
Answer:
Structural formula of 2-methoxy-2-methyl propane

Question 41.
Write the odd term in Thiophene and Pyridine.
Answer:

Question 42.
Which of the following will have more hyperconjugation ? Explain with reason
\(\left(\mathbf{C} \mathbf{H}_{3}\right)_{2}^{\oplus} \mathbf{C H},^{\oplus} \mathbf{C H}_{2} \mathbf{C H}_{3}\)
Answer:
\(\left(\mathrm{CH}_{3}\right)_{2} \stackrel{\oplus}{\mathrm{C}} \mathrm{H}\)
carbocation will have more hyperconjugation than \(^{\oplus} \mathrm{CH}_{2} \mathrm{CH}_{3}\) because it is a secondary carbocation due to presence of two methyl groups on the carbon atom having positive charge. Greater the number of alkyl groups in a carbocation, greater will be the hyperconjugation and greater will be the stability of that carbocation.

Question 43.
Classify the following reactions :
(a) CH₃ CH₂Br + HS^– ➝ CH₃ CH₂SH + Br^–
(b) (CH₃ )₂ C = CH₂ + HCl ➝ (CH₃ )₂ —CCl—CH₃
(c) CH₃ CH₂Br + HO^– ➝ CH₂ = CH₂ + H₂O + Br^–
(d) (CH₃ )₃ C—CH₂OH + HBr ➝ (CH₃)₃C—CH₂Br + H₂O
Answer:
(a) CH₃CH₂Br + HS^– ➝ CH₃CH₂SH + Br^–
This reaction is called nucleophilic substitution reaction.
(b) (CH₃ )₂ C = CH₂ + HCl ➝ (CH₃)₂C(Cl)CH₃
This reaction is called electrophilic addition reaction.
(c) CH₃CH₂Br + HO^– ➝ CH₂ = CH₂ + H₂O + Br^–
This reaction is called elimination reaction.
(d) (CH₃)₃C—CH₂OH + HBr ➝ (CH₃ )₃ C—CH₂ Br + H₂O
This reaction is called nucleophilic substitution reaction.

Question 44.
Explain substitution reactions with examples.
Answer:
Substitution reactions are those reactions in which one atom or group of atoms of one molecule substitutes one atom or group of atoms of another molecule.

Question 45.
Explain Beldstein test.
Answer:
Beldstein Test—This is a qualitative chemical test for halides. In this test, heat the tip of a copper wire in a burner flame until there is no further colouration of the flame. Let the wire cool slightly, then dip it into the unknown solid or liquid and again, heat it in the flame. A green flash is indicative of chlorine, bromine and iodine. Fluorine is not detected because copper fluoride is not volatile.

Question 46.
Write the IUPAC name of glycerol and crotonic acid.
Answer:

RBSE Class 11 Chemistry Chapter 12 Long Answer Type Questions

Question 47.
Write short notes on :
(a) Tetravalency Principle of Kekule
(b) Van’t Hoff and Le Bel’s Principle.
(c) Homologous Series.
Answer:
(a) Tetravalency Principle of Kekule : The main points of this principle are given as below :

• Carbon atom shows tetravalency i. e., its valency is four.
• Carbon atom combines with other carbon atoms to form different open chained closed chained compounds.
• Carbon can form single bond, double bond and triple bond with other atoms.

(b) Van’t Hoff and Le Bel’s Principle : According to this principle, “In an organic compound, four valency of carbon is present at the centre of the structure”. Four covalent bonds of carbon form 109°28′ with each other. All the four valency of carbon are same. This theory shows that all the four valencies of carbon are not present in one plane. This theory helps in understanding the study of three dimensional shape of carbon.

(c) Homologous Series : All organic compounds are made up of a progressively building chain of carbon atoms with a number of compounds having the same functional groups. Such a series of similarly constituted compounds are called a homologous series.

Members of a homologous series are similar in structure and exhibit similar chemical characteristics. The two consecutive members of the series differ in their molecular formula by a —CH₂— group.

Question 48.
Explain the following :
(a) Inductive effect.
(b) Electromeric effect.
(c) Mesomeric effect.
(d) Hyperconjugation.
Answer:
(a) Inductive Effect—The polarization of a σ-bond due to electron withdrawing or electron donating
effect of adjacent groups or atoms is called inductive effect. It is permanent effect. It is represented by I.
It is of two types.
(i) —I effect—The electron withdrawing nature of groups or atoms is known as —I effect.
e.g., —NO₂, —CN, —SO₃H, —F, —Cl etc.

(ii) +I effect—The electron donating nature of groups or atoms is known as +I effect.
e.g., (CH₃)₃ C—, (CH₃)₂ CH – (CH3 )3 —,
CH₃CH₂ —, (CH₃)₃ —CH₃

(b) Electromeric Effect—It is defined as the complete transfer of a shared pair of π-electrons to one of the atoms joined by a multiple bond on the demand of an attacking reagent. It is temporary effect. It is represented by E and the shifting of the electrons is

shown by a curved arrow
It is of two types :
(i) +E effect—When the transfer of electrons takes place towards the attacking reagent, the effect is called +E effect.
e.g., Addition of acid to alkene.

(ii) -E effect When the transfer of electrons takes place away from the attacking reagent, the effect is called -E effect.
e.g., Addition of cyanide ion to carbonyl compound.

(c) Mesomeric Effect- The electron withdrawing or releasing effect attributed to a substituent through delocalization of π-electrons which can be visualized by drawing various canonical forms, is called mesomeric effect. It is represented by M.

(i) -M effect – It is shown by the substituents that withdraw electrons by delocalisation mechanism from rest of the molecule and are denoted by -M effect.

(ii) +M effect— It is shown by the substituents that release electrons by delocalisation mechanism to rest of the molecule and are denoted by +M effect.
e.g., —OH, —SH, —NH₂, —NR₂ etc.

(d) Hyperconjucation—The delocalization of σ-electrons or lone pair of electrons into adjacent π-orbital or p-orbital is called hyperconjugation. It occurs due to overlapping of σ-bonding orbital or the orbital containing a lone pair with adjacent π-orbital or p-orbital. It is also known as “no bond resonance” or “Baker-Nathan effect”.

Conditions for Hyperconjugation—There must be an a-CH group or a lone pair on atom adjacent to sp² hybrid carbon or other atoms like nitrogen, Oxygen etc.
Example : Alkenes, alkyl carbocations, alkyl free radicals, nitro compounds with α-hydrogen etc.

Question 49.
Write the chemistry for qualitative analysis of nitrogen, sulphur and bromine.
Answer:
Lassaigne’s test is used to detect the presence of nitrogen, sulphur and bromine in an organic compound. It involves following steps :

Step I : Preparation of Sodium Fusion Extract—A small amount of organic substance is fused with small quantity of sodium metal in a fusion tube. The red hot fusion tube is plunged into distilled water and the contents are boiled for a few minutes, then cooled and filtered. The filtrate obtained is called sodium fusion extract or Lassaigne extract. It is usually alkaline.

Step II : Detection of Elements using Sodium Fusion Extract The elements present in the organic compound react with sodium during fusion reaction as follows :
Na + C + N ➝ NaCN (if N is present)
2Na + S ➝ Na₂S (if S is present)
Na + S + C + N ➝ NaSCN
(if both N and S are present)
Na + Br ➝ NaBr (if Br is present)
(i) Test for Nitrogen—The sodium fusion extract is boiled with iron (II) sulphate and then acidified with concentrated H2SO4. The formation of Prussian blue colour confirms the presence of nitrogen

(ii) Test for Sulphur—If sulphur is present in the organic compound, sodium fusion will convert it into sodium sulphide. On treating sodium fusion extract with sodium nitroprusside, appearance of violet colour confirms the presence of sulphur.

(iii) Test for both Nitrogen and Sulphur—On treating sodium fusion extract with Fe³⁺ ions, the formation of blood red colour complex confirms the presence of both nitrogen and sulphur.

(iv) Test for Bromine—Sodium fusion extract is first acidified with nitric acid and then add silver nitrate solution to it. The formation of pale yellow precipitate that is partially soluble in NH₄OH confirms the presence of bromine.

Question 50.
Explain Kjeldahl and Duma’s method. Derive the expression for percentage of nitrogen.
Answer:
Kjeldahl Method—This method is used for estimation of nitrogen. It is used only for those organic compounds that are converted quantitatively to ammonium sulphate on heating strongly with concentrated H₂SO₄ .This method can not be used for the organic compounds having nitrogen in the ring e.g., pyridine and also —NO₂ and —N = N— groups.
This method involves following steps :

(i) Digestion—A known mass (about 0.5 g) of the given organic comound is digested with concentrated H₂SO₄ in the presence of small amount of potassium sulphate and copper sulphate in a Kjeldahl’s flask. Potassium sulphate raises the boiling point of sulphric acid and copper sulphate catalyzes the digestion. In 3 to 4 hrs, the organic compound is completely decomposed to form ammonium sulphate.

Distillation—The digested reaction mixture, on cooling, is transferred to a round bottomed distillation flask, and distilled with a concentrated alkali solution (NaOH). Ammonia produced is absorbed in a known volume of HCl solution of a known strength.

The unneutralised HCl is then back-titrated against a standard alkali. From the acid consumed, the amount of ammonia produced and hence the mass of nitrogen is calculated.

Calculations : Let,
Weight of the organic compound = W g
Volume of the standard acid required for complete neutralization of the evolved ammonia = V ml
Normality of the standard solution of acid = N
From the law of equivalence (normality equation), 1000 ml of 1 N acid = 1000 ml of 1 N NH₃
= 17 g NH₃ = 14 g Nitrogen
Then,
V ml of N acid = V ml of NH₃
NV milli equivalent of acid = NV milli equivalent of ammonia.
Therefore, mass of nitrogen in the evolved ammonia

Then, percentage of nitrogen in the sample

∴ Percentage of Nitrogen in the sample

Duma’s method—By this method, percentage of nitrogen in all the compounds can be estimated. A known mass of the organic compound is heated with cupric oxide in an atmosphere of carbon dioxide. The carbon and hydrogen in the compound are oxidised to carbon dioxide and water respectively, while nitrogen is set free. Any oxide of nitrogen produced during this process, is reduced back to free nitrogen by heated copper gauze. The gaseous mixture consisting of CO₂, H₂O and N₂ is collected over an aqueous solution of potassium hydroxide. All the gases except nitrogen are absorbed by the solution. The volume of gas (nitrogen) collected is measured. From the volume of nitrogen obtained the percentage of nitrogen in the compound is calculated.

Calculations : Let,
The mass of the organic compound taken be = W g
Volume of nitrogen collected = V₁ ml
Atmospheric pressure = P mm Hg
Temperature at which gas is collected = T₁ K
Therefore,
Pressure of the N₂ gas.P₁ = (P – p) mm Hg

Question 51.
Explain the formation, stability and geometry of the following reaction intermediates : carbanion, carbene and free radicals.
Answer:
(i) Carbanion Formation : It is formed by means of heterolytic fission. The organic compounds containing active hydrogen atom reacts with base to form carbanion.
e.g.,

Stability—There are three types of carbanion : Primary, Secondary and Tertiary carbanion. Greater the +I effect, lesser will be the stability of carbanion. The relative order of stability among carbanions is.

Geometry : In carbanion, carbon atom is sp³ hybridised. Its geometry is pyramidal.

(ii) Carbene : Formation : Carbene is formed by the photolysis of the aliphatic diazo compounds or ketenes.

of carbenes are highly reactive but the singlet state is generally considered as less stable.

Geometry—In singlet state of carbene, carbon is sp² hybridised. The bond angle is 120° and geometry is trigonal planar. In triplet state of carbene, carbon is sp hybridised. The bond angle is 180° and geometry is linear.

(iii) Free Radicals Formation—Free radicals are formed by means of homolytic fission

Stability—Free radicals are classified as primary, secondary and tertiary. Tertiary alkyl free radical is most stable due to greater +I effect.
Greater the +I effect, greater will be the stability of free radical.

Geometry—The carbon in alkyl free radicals is sp2 hybridised. The geometry is trigonal planar and bond angle is 120°.

Question 52.
(a) Write short notes on reactions and stability of free radials.
(b) Write the method for preparation and stability of carbocation.
Answer.
(a) Free Radicals : Free radicals are chemical species that contain a singly occupied orbital. They are neutral and tend to be highly reactive. These have high tendency to pair with the unpaired electron. They readily combine with each other or with other molecules to pair their electrons. Some reactions of free radicals are given below
(i) Recombination into hydrocarbons

(ii) Disproportionation

(iii) Reaction with olefins

The alkyl free radicals are unstable and highly reactive. These are of three types—Primary, Secondary and Tertiary alkyl free radicals. The relative order of stability amongst these free radicals is given as below :

Greater the +I effect, greater will be the stability of free radical.

(b) Carbocation : Carbocation are generally formed by means of heterolytic fission. These can be prepared by following methods :
(i) By direct ionization

(ii) By the protonation of unsaturated compounds

Carbocations are highly unstable and reactive species. Alkyl groups directly attached to the positively charged which stabilise the carbocations due to inductive and hyperconjugative effects. The relative order of stability of carbocations is given below—

Question 53.
Explain the Trivial system of nomenclature with example :
Answer:
Trivial System : This is the oldest system of naming organic compounds. The trivial name was generally based on the source, some property or some other reason. Quite frequently, the names chosen had Greek or Latin roots. As given :
(a) Acetic acid derives its name from vinegar of which it is the chief constituent (Latin : acetum = vinegar).
(b) Formic acid was named as it was obtained from red ants. The Greek word for the red ants is formicus.
(c) The names Oxalic acid (oxalus), Malic acid (pyrus malus), Citric acid (citrus) have been derived from botanical sources given in parentheses.
(d) Urea and Uric acid have derived their names from urine in which both are present.
(e) The liquid obtained by the destructive distillation of wood was named as wood spirit. Later on, it was named Methyl alcohol (Greek : methu = spirit; hule = wood).
(f) Names like Glucose (sweet), Pentane (five), Hexane (six), etc. were derived from Greek words describing their properties or structures.
(g) Methane was named as Marsh gas because it was produced in marshes. It was also named as fire damp as it formed explosive mixture with air.

Question 54.
Explain IUPAC system of nomenclature with examples.
Answer:
IUPAC System
(a) The full form of IUPAC system is International Union of Pure and Applied Chemistry.
(b) This sytem of nomenclature was first introduced in 1947 and was modified from time to time.
(c) The most exhaustive rules for nomenclature were first published in 1979 and later revised and updated in 1993.
The systematic IUPAC name of an organic compound consists of four parts.
1. Root word
2. Suffix(es)
3. Prefix(es)
4. Infix
The complete systematic IUPAC name can be represented as :
Prefix + Infix + Root word + 1° suffix + 2° suffix
1. Root Word : It indicates the number of carbon atoms in the longest possible continuous carbon chain also known as parent chain chosen by a set of rules. The root words used for different length of carbon chain (upto 10) are shown below :

2. Suffix : It is again divided into two types—
(i) Primary suffix, and
(ii) Secondary suffix.
(i) Primary Suffix —It is used to indicate the degree of saturation or unsaturation in the main chain. It is added immediately after the root word.

(ii) Secondary Suffix—It is used to indicate the main functional group in the organic compound and is added immediately after the 1° suffix.

The suffixes as well as prefixes used for some important functional groups are shown in the following table in the decreasing order of their priority.

3. Prefix-The prefix is used to indicate the side chains, substituents and low priority functional groups (which are considered as substituents). The prefix may be added immediately before the root word or before the infix.
The prefixes used for some common side chains and substituents are shown below : (The prefixes for functional groups are already given).


Remember that the alkyl groups along with halo, nitro and alkoxy have the same preference. They have lower priority than double and triple bonds.

4. Infix: The infixes like Cyclo, Spiro, Bicyclo are added between the prefix (es) and root word to indicate the nature of parent chain.

• The “Cyclo” infix is used to indicate the cyclic nature of the parent chain.
• The “Spiro” infix is used to indicate the spirocycto compound.
• The “Bicyclo” infix is used to indicate the bicyclic nature of the parent chain.
• The infixes are sometimes called as primary prefixes.

Steps for Writing IUPAC Name

1. The first step in giving IUPAC name to an organic compound is to select the parent chain and assign a root word.
2. Next, the appropriate primary suffix must be added to the root word to indicate the saturation or unsaturation.
3. If the molecule contains functional group or groups, a secondary suffix must be added to indicate the main functional group. This is optional and not necessary if the molecule contains no functional group.
4. Prefix the root word with the infix “cyclo” if the parent chain is cyclic; or with the infix “spiro” if it is a spirocyceo compound; or with the infix “bicyclo” if the compound is bicyclic.
5. Finally add prefix (es) to the name if there are side chains or substituents on the parent chain.
   Example : The IUPAC name of the following compound is arrived in steps mentioned below :

RBSE Solutions for Class 11 Chemistry