NCERT Solutions for Class 11 Chemistry Chapter 12 Organic Chemistry – Some Basic Principles and Techniques

C–1 is sp² hybridised.

C–2 is sp hybridised.

(ii)

C–1 is sp³ hybridised.

C–2 is sp² hybridised.

C–3 is sp² hybridised.

(iii)

C–1 and C–3 are sp³ hybridised.

C–2 is sp² hybridised.

(iv)
C–1 is sp² hybridised.

C–2 is sp² hybridised.

C–3 is sp hybridised.

(v) C₆H₆

All the 6 carbon atoms in benzene are sp² hybridised.

Question 12.2:

Indicate the σ and π bonds in the following molecules:

C₆H₆, C₆H₁₂, CH₂Cl₂, CH₂ = C = CH₂, CH₃NO₂, HCONHCH₃

Answer 12.2:

(i) C₆H₆

There are six C – C sigma bonds, six C–H sigma bonds, and three C=C pi resonating bonds in the given compound.

(ii) C₆H₁₂

There are six C – C sigma bonds and twelve C–H sigma bonds in the given compound.

(iii) CH₂Cl₂

There are two C – Cl sigma bonds and two C–H sigma bonds in the given compound.

(iv) CH₂ = C = CH₂

There are two C–C sigma bonds, four C–H sigma bonds, and two C=C pi bonds in the given compound.

(v) CH₃NO₂

There are three C–H sigma bonds, one C–N sigma bond, one N–O sigma bond, and one N=O pi bond in the given compound.

(vi) HCONHCH₃

There are four C–H sigma bonds, two C–N sigma bond, one N–H sigma bond, and one C=O pi bond in the given compound.

Question 12.3:

Write bond-line formulas for: Isopropyl alcohol, 2, 3–dimethyl butanal, Heptan–4–one.

Answer 12.3:

Isopropyl alcohol

2, 3–dimethyl butanal

Heptan–4–one

Question 12.4:

Give the IUPAC names of the following compounds:

(a)

 (b)

 (c)

 (d)

 (e)

(f) Cl₂CHCH₂OH

Answer 12.4:

(a)

Propylbenzene

(b)

3–methylpentanenitrite

(c)

2, 5–dimethyl heptane

(d)

3–bromo–3–chloroheptane

(e)

3–chloropropanal

(f) Cl₂CHCH₂OH

2, 2–dichloroethanol

Question 12.5:

Which of the following represents the correct IUPAC name for the compounds concerned?

(a) 2,2-Dimethylpentane or 2-Dimethylpentane

(b) 2,4,7-Trimethyloctane or 2,5,7-Trimethyloctane

(c) 2-Chloro-4-methylpentane or 4-Chloro-2-methylpentane

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

Answer 12.5:

(a) The prefix di shows that there are two methyl groups in the chain. Thus, the correct IUPAC name would be 2,2-Dimethylpentane.

(b) The locant number should start from the minimum. Here, 2,4,7 is lower than 2,5,7. Thus, the correct IUPAC name would be 2,4,7-Trimethyloctane.

(c) If the substituents in the chain are in equivalent positions, then the lower number is given to the substituent group in alphabetical order. Thus, the correct IUPAC name would be 2-Chloro-4-methylpentane.

(d) Out of the two functional groups present in the given compound, the alcoholic group is the principal functional group. Thus, the parent chain will have an –ol suffix. Since the alkyne group is in C–3, the IUPAC name would be But–3–yn–1–ol.

Question 12.6:

Draw formulas for the first five members of each homologous series beginning with the following compounds

(a) H–COOH

(b) CH₃COCH₃

(c) H–CH=CH₂

Answer 12.6:

The first five members of each homologous series beginning with the given compounds are

(a)

H–COOH: Methanoic acid

CH₃–COOH: Ethanoic acid

CH₃–CH₂–COOH: Propanoic acid

CH₃–CH₂–CH₂–COOH: Butanoic acid

CH₃–CH₂–CH₂–CH₂–COOH: Pentanoic acid

(b)

CH₃COCH₃: Propanone

CH₃COCH₂CH₃: Butanone

CH₃COCH₂CH₂CH₃ : Pentan-2-one

CH₃COCH₂CH₂CH₂CH₃: Hexan-2-one

CH₃COCH₂CH₂CH₂CH₂CH₃ : Heptan-2-one

(c)

H–CH=CH₂: Ethene

CH₃–CH=CH₂: Propene

CH₃–CH₂–CH=CH₂: 1-Butene

CH₃–CH₂–CH₂–CH=CH₂: 1-Pentene

CH₃–CH₂–CH₂–CH₂–CH=CH₂: 1-Hexene

Question 12.7:

Give condensed and bond line structural formulas and identify the functional group(s) present, if any, for :

(a)2,2,4-Trimethylpentane

(b)2-Hydroxy-1,2,3-propanetricarboxylic acid

(c) Hexanedial

Answer 12.7:

(a) 2, 2, 4–trimethylpentane

Condensed formula

(CH₃)₂CHCH₂C (CH₃)₃

Bond line formula :

(b) 2–hydroxy–1, 2, 3–propanetricarboxylic acid

Condensed Formula

(COOH)CH₂C(OH) (COOH)CH₂(COOH)

Bond line formula:

Functional groups:

Carboxylic acid (-COOH) and Hydroxyl (-OH) groups

(c) Hexanedial

Condensed Formula

(CHO)(CH₂)₄(CHO)

Bond line formula:

Functional groups:

Aldehyde (-CHO)

Question 12.8:

Identify the functional groups in the following compounds.

(a) 

(b)

(c)

Answer 12.8:

(a) Hydroxyl (–OH), Aldehyde (–CHO), Methoxy (–OMe),

C=C double bond

(b) Ester -(O = C – O)-, 1^oAmino (–NH₂) (aromatic), Diethylamine (N(C₂H₅)₂ ) – 3^oAmino

(c) Nitro (–NO₂),

C=C Ethylenic double bond

Question 12.9:

Which of the two: O₂NCH₂CH₂O^– or CH₃CH₂O^– is expected to be more stable and why?

Answer 12.9:

Since NO₂ belongs to the electron-withdrawing group, it shows –I effect. NO₂ tries to decrease the negative charge on the compound by withdrawing the electrons toward it. This stabilizes the compound whereas the ethyl group belongs to the electron-releasing group and shows +I effect. This results in an increase in the negative charge on the compound thus destabilizing the compound. Hence, I would expect O₂NCH₂CH₂O^–to be more stable than CH₃CH₂O^–.

Question 12.10:

Explain why alkyl groups act as electron donors when attached to a π system.

Answer 12.10:

Due to hyperconjugation, an alkyl group behaves as an electron-donor group when attached to a π system. For example, look at propene.

The sigma electrons of the C-H bond get delocalized due to hyperconjugation. The alkyl is attached directly to an unsaturated system. The delocalization happens due to the partial overlap of an sp³-s sigma bond orbital with an empty p orbital of the n bond of an adjacent carbon atom.

This process can be shown as:

The overlap as shown above results in delocalization making the compounds more stable.

Question 12.11:

Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation.

(a) C₆H₅OH

(b) C₆H₅NO₂

(c) CH₃CH = CH – CHO

(d) C₆H₅CHO

(e) C₆H₅-CH₂

(f)CH₃CH=CHCH₂

Answer 12.11:

(a) The structure of C₆H₅OH is: 

Resonating structures: 

(b) The structure of C₆H₅NO₂ is: 

Resonating structures: 

(c) CH₃CH = CH – CHO

Resonating structures: 

(d) The structure of C₆H₅CHO is: 

Resonating structures: 

(e) C₆H₅-CH₂ 

Resonating structures:

(f) CH₃CH=CHCH₂

Resonating structures: 

Question 12.12:

What are electrophiles and nucleophiles? Explain with examples.

Answer 12.12:

A nucleophile is a reagent that has an electron pair and is willing to donate it. It is also known as a nucleus-loving reagent. Ex: NC^–, OH^–, R₃C^– (carbanions), etc.

An electrophile is a reagent which is in need of an electron pair and is also known as an electron-loving pair. Ex: Carbonyl groups, CH₃CH₂⁺(Carbocations), Neutral molecules (due to the presence of a lone pair).

Question 12.13:

Identify the reagents shown in bold in the following equations as nucleophiles or electrophiles:

(a) CH₃COOH + HO^–$ightarrow$ CH₃COO^– + H₂O

(b) CH₃COCH₃+ C^–N $ightarrow$ (CH₃)₂C(CN) + (OH)^–

(c) C₆H₅ + CH₃C⁺O $ightarrow$ C₆H₅COCH₃

Answer 12.13:

A nucleophile is a reagent that has an electron pair and is willing to donate it. It is also known as a nucleus-loving reagent.

An electrophile is a reagent which is in need of an electron pair and is also known as an electron-loving pair.

(a) CH₃COOH + HO^–$ightarrow$ CH₃COO^– + H₂O

It is a nucleophile since HO- is electron rich in nature.

(b) CH₃COCH₃+ C^–N $ightarrow$ (CH₃)₂C(CN) + (OH)^–

It is a nucleophile since C^–N is electron rich in nature.

(c) C₆H₅ + CH₃C⁺O $ightarrow$ C₆H₅COCH₃

It is an electrophile since CH₃C⁺O is electron-deficient in nature.

Question 12.14:

Classify the following reactions in one of the reaction type studied in this unit.

(a) CH₃CH₂Br + HS^–$ightarrow$CH₃CH₂SH + Br^–

(b) (CH₃)₂ C=CH₂ + HCl$ightarrow$(CH₃)₂ClC^–CH₃

(c) CH₃CH₂Br + HO^–$ightarrow$CH₂=CH₂ + H₂O + Br^–

(d) (CH₃)₃C–CH₂ OH + HBr$ightarrow$(CH₃)₂ CBrCH₂CH₃ + H₂O

Answer 12.14:

(a) Substitution (nucleophilic) reaction since bromine group gets substituted by –SH group.

(b) Addition (electrophilic) reaction since two reactant molecules combines to form a single product.

(c) Elimination (bimolecular) reaction since reaction hydrogen and bromine are removed to form ethene.

(d) Substitution (nucleophilic) reaction since rearrangement of atoms takes place.

Question 12.15:

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

(a) 

(b)

(c)

Answer 12.15:

(a) The given compounds are a pair of structural isomers since they have the same molecular formula but have different structures. These compounds differ in the position of the ketone group. For the first structure, it is in C-3 whereas, for the 2^nd one, it is in C-2.

(b) The given compounds are a pair of geometrical isomers since they have the same molecular formula, sequence of covalent bonds and the same constitution but differ in the relative positioning of the atoms in space. These compounds differ in the positioning of the Deutrium and Hydrogen.

(c) The given compounds are a pair of contributing structures or canonical structures. They do not represent any real molecule and are purely hypothetical. They are also called resonance isomers.

Question 12.16:

For the following bond cleavages, use curved-arrows to show the electron flow and classify each as homolysis or heterolysis. Identify reactive intermediate produced as free radical, carbocation and carbanion.

(a)

(b)

(c) 

(d) 

Answer 12.16:

(a) The bond cleavage can be shown as:

It comes under homolytic cleavage since one of the shared pair in a covalent bond goes with the bonded atom. A free radical is formed as the reaction intermediate.

(b) The bond cleavage can be shown as:

It comes under heterolytic cleavage since the shared remains with the carbon atom of propanone. A carbanion is formed as the reaction intermediate.

(c) The bond cleavage can be shown as:

It comes under heterolytic cleavage since the shared remains with the bromine ion. A carbocation is formed as the reaction intermediate.

(d) The bond cleavage can be shown as:

It comes under heterolytic cleavage since the shared remains with one of the fragments. A carbocation is formed as the reaction intermediate.

Question 12.17:

Explain the terms Inductive and Electromeric effects. Which electron displacement effect explains the following correct orders of acidity of the carboxylic acids?

(a) Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH

(b) CH₃CH₂COOH > (CH₃)₂CHCOOH > (CH₃)₃C.COOH

Answer 12.17:

Electrometric effect

The complete transfer of the shared pair of π electrons to either of the two atoms linked by multiple bonds in the presence of an attacking agent is called the electrometric effect. It can either be – E effect or +E effect.

– E effect: Occurs when electrons are moved away from the attacking agent

+ E effect: Occurs when electrons are moved towards the attacking agent

Inductive effect

Inductive effect involves the permanent displacement of sigma (σ) electrons along a saturated chain, whenever an electron withdrawing or electron donating group is present.

It can either be + I effect or – I effect. When an atom or group attracts electrons towards itself more strongly than hydrogen, it is said to possess – I effect.

When the force with which an atom attracts electrons towards itself is greater than that of hydrogen, it is said to exhibit +I effect.

(a) Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH

The acidity increases with the increase in –I effect which is directly proportional to the number of chlorine atoms.

(b) CH₃CH₂COOH > (CH₃)₂ CHCOOH > (CH₃)₃C.COOH

The acidity increases with the increase in +I effect which is directly proportional to the number of alkyl groups.

Question 12.18:

Give a brief description of the principles of the following techniques taking an example in each case.

(a) Crystallisation

(b) Distillation

(c) Chromatography

Answer 12.18:

(a) Crystallisation

Crystallization is used to purify solid organic compounds.

Principle: The principle on which it works is the difference in the solubility of the compound and impurities in a given solvent. The impure compound is made to dissolve in the solvent at a higher temperature since it is sparingly soluble at lower temperatures. This is continued till we get an almost saturated solution. On cooling and filtering it, we get its’ crystals. Ex: By crystallizing 2-4g of crude aspirin in 20mL of ethyl alcohol, we get pure aspirin. It is heated if needed and left undisturbed until it crystallizes. The crystals are then separated and dried.

(b) Distillation

This method is used to separate non-volatile liquids from volatile impurities. It is also used when the components have a considerable difference in their boiling points.

Principle: The principle on which it works is that liquids having different boiling points vaporise at different temperatures. They are then cooled and the formed liquids are separated.

Ex: A mixture of aniline (b.p = 457 K) and chloroform (b.p = 334 K) is taken in a round bottom flask having a condenser. When they are heated, Chloroform, vaporizes first due to its high volatility and made to pass through a condenser where it cools down. The aniline is left behind in the round bottom flask.

(c) Chromatography

It is widely used for the separation and purification of organic compounds.

Principle: The principle on which it works is that individual components of a mixture move at different paces through the stationary phase under the influence of mobile phase.

Ex: Chromatography can be used to separate a mixture of blue and red ink. This mixture is placed on chromatogram where the component which is less absorbed by the chromatogram moves faster up the paper than the other component which is almost stationary.

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