Updated By SaraNextGen

On April 21, 2024, 11:35 AM

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:

The first five members of each homologous series beginning with the given compounds are shown as follows:

(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:

(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:

The functional groups present in the given compound are carboxylic acid (–COOH) and alcoholic (–OH) groups.

(c) Hexanedial

Condensed Formula:

(CHO) (CH₂)₄ (CHO)

Bond line Formula:

The functional group present in the given compound is aldehyde (–CHO).

Question 12.8:

Identify the functional groups in the following compounds

(a)

(b)

(c)

Answer:

The functional groups present in the given compounds are:

(a) Aldehyde (–CHO),

Hydroxyl(–OH),

Methoxy (–OMe),

C=C double bond 

(b) Amino (–NH₂); primary amine,

Ester (-O-CO-),

Triethylamine (N(C₂H₅)₂); tertiary amine

(c) Nitro (–NO₂),

C=C double bond 

Question 12.9:

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

Answer:

NO₂ group is an electron-withdrawing group. Hence, it shows –I effect. By withdrawing the electrons toward it, the NO₂ group decreases the negative charge on the compound, thereby stabilising it. On the other hand, ethyl group is an electron-releasing group. Hence, the ethyl group shows +I effect. This increases the negative charge on the compound, thereby destabilising it. Hence, O₂NCH₂CH₂O^– is expected to be more stable than CH₃CH₂O^–_.

Question 12.10:

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

Answer:

When an alkyl group is attached to a π system, it acts as an electron-donor group by the process of hyperconjugation. To understand this concept better, let us take the example of propene.

In hyperconjugation, the sigma electrons of the C–H bond of an alkyl group are delocalised. This group is directly attached to an atom of an unsaturated system. The delocalisation occurs because of a partial overlap of a sp³ –s sigma bond orbital with an empty p orbital of the π bond of an adjacent carbon atom.

The process of hyperconjugation in propene is shown as follows:

This type of overlap leads to a delocalisation (also known as no-bond resonance) of the π electrons, making the molecule 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)  (f) 

Answer:

(a) The structure of C₆H₅OH is:

The resonating structures of phenol are represented as:

(b) The structure of C₆H₅NO₂ is:

The resonating structures of nitro benzene are represented as:

(c) CH₃CH = CH – CHO

The resonating structures of the given compound are represented as:

(d) The structure of C₆H₅CHO is:

The resonating structures of benzaldehyde are represented as:

(e) C₆H₅CH₂^⊕

The resonating structures of the given compound are:

(f) CH₃ CH = CH CH₂^⊕

The resonating structures of the given compound are:

Question 12.12:

What are electrophiles and nucleophiles? Explain with examples.

Answer:

An electrophile is a reagent that takes away an electron pair. In other words, an electron-seeking reagent is called an electrophile (E⁺). Electrophiles are electron-deficient and can receive an electron pair.

Carbocations and neutral molecules having functional groups such as carbonyl group ( ) are examples of electrophiles.

A nulceophile is a reagent that brings an electron pair. In other words, a nucleus-seeking reagent is called a nulceophile (Nu:).

For example: OH^–, NC^–, carbanions (R₃C^–), etc.

Neutral molecules such as H₂Ö and ammonia also act as nulceophiles because of the presence of a lone pair.

Question 12.13:

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

Answer:

Electrophiles are electron-deficient species and can receive an electron pair. On the other hand, nucleophiles are electron-rich species and can donate their electrons.

Here, HO^– acts as a nucleophile as it is an electron-rich species, i.e., it is a nucleus-seeking species.

Here, ^–CN acts as a nucleophile as it is an electron-rich species, i.e., it is a nucleus-seeking species.

Here,

acts as an electrophile as it is an electron-deficient species.

Question 12.14:

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

(a) CH₃CH₂Br + HS^– → CH₃CH₂SH + Br^–

(b) (CH₃)₂ C = CH₂ + HCl → (CH₃)₂ ClC–CH₃

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

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

Answer:

(a) It is an example of substitution reaction as in this reaction the bromine group in bromoethane is substituted by the –SH group.

(b) It is an example of addition reaction as in this reaction two reactant molecules combine to form a single product.

(c) It is an example of elimination reaction as in this reaction hydrogen and bromine are removed from bromoethane to give ethene.

(d) In this reaction, substitution takes place, followed by a rearrangement of atoms and groups of atoms.

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:

(a) Compounds having the same molecular formula but with different structures are called structural isomers. The given compounds have the same molecular formula but they differ in the position of the functional group (ketone group).

In structure I, ketone group is at the C-3 of the parent chain (hexane chain) and in structure II, ketone group is at the C-2 of the parent chain (hexane chain). Hence, the given pair represents structural isomers.

(b) Compounds having the same molecular formula, the same constitution, and the sequence of covalent bonds, but with different relative position of their atoms in space are called geometrical isomers.

In structures I and II, the relative position of Deuterium (D) and hydrogen (H) in space are different. Hence, the given pairs represent geometrical isomers.

(c) The given structures are canonical structures or contributing structures. They are hypothetical and individually do not represent any real molecule. Hence, the given pair represents resonance structures, called resonance isomers.