Chemistry For HSC Part 2 - Chapter No.7 - Shorts Answers, Scientific Reasons And Text Book Exercise

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Chapter No.6
Introduction To Organic Chemistry
Shorts Answers, Scientific Reasons And Text Book Exercise

Short Answers

Q.1: Draw and labelled the orbital Structure of Ethane, Ethene and Ethyne.

Q.2: Give the mechanism of the chlorination of methane in the presence of sunlight.

OR

Describe halogenation reaction in alkane.

OR

Describe free radical chain reaction of alkanes with halogen.

Scientific Reasons

Q.1: Why alkanes are called "Parafins"?
Ans: In alkanes all the four valencies of carbon atoms are fully satisfied, they can take up no more hydrogen or any other atom. Thus "fully saturated" in respect of chemical combination, these hydrocarbons are classed as "saturated hydrocarbons.
On account of their feeble chemical activity, they are also called "PARAFFINS" (Latin - Parum affinis = little affinity).

Q.2: Why alkenes or alkylene are called "OLEFINES"?
Ans: Alkenes are unsaturated hydrocarbons, containing a double bond. Since all the valencies of carbon of the carbon atom in alkenes are not fully satisfied, they can take up more hydrogen atoms are called unsaturated hydrocarbons.
The hydrocarbons of this class are frequently called olefines (olefiant = oil forming) as the first member of the class, ethylene reacted with chlorine to form an oily product.

Q.3: What is the qualitative test for unsaturation in ethene (alkene)?
Ans: The addition of Br₂ to ethene (alkene) is frequently used as a qualitative test for unsaturation.
Bromine itself is a dark red-brown liquid, where as both the ethene (alkene) and the addition product are colourless. Thus to test whether a substance is an alkene, we add to it a solution of bromine in CCl₄. If the substance is an alkene or an alkyne the bromine solution will be rapidly decolourized.

Text Book Exercise 

6. Describe simple chemical tests that would distinguish between:
(a) An alkene and an alkane
(b) An alkene and an alkyne
(c) An alkane and an alkyl — halide

Answer:
(a) Chemical tests which are used to distinguish the alkane and alkene
The presence of the double bond in alkenes makes them more reactive than alkanes. This higher reactivity of the alkenes over alkanes distinguish them by following simple tests:

1. WITH BAEYER’S REAGENT
2. WITH BROMINE WATER TEST

Experiment No.1: WITH BAEYER’S REAGENT (Hydroxylation):
Reaction with alkane: No reaction.
Reaction with alkene: In case of ALKENE, dilute and alkaline KMnO₄ solution on addition at room temperature gives a dihydroxy alcohol or glycol. The PINK color of KMnO₄ is changed to the brown precipitated MnO₂ which show the unsaturaion in the molecules of alkene.

This reaction is called the Baeyer's test for unsaturation.

Experiment No.2: WITH BROMINE WATER TEST:
Reaction with alkane: No reaction.
Reaction with alkene: When ethene is treated with a brown solution of bromine water in CCl₄ at room temperature, the bromine solution will be decolorized due to formation of vicinal dihalide (1,2-dibromoethane).

this reaction will occur for unsaturated compounds containing carbon-carbon double bonds.

(b) Chemical tests which are used to distinguish the alkene and alkyne
Terminal alkynes may be distinguished from other alkynes or alkenes because they react rapidly with:

1. Cuprous ions and
2. Silver ions

in alkaline solution to produce copper acetylide and silver acetylide respectively.

Experiment No.1: With Ammonical (Cu₂Cl₂)
Reaction with alkene: No Reaction.
Reaction with alkyne: When acetylene is passed through ammonical solution of Cuprous chloride, Reddish brown precipitate of copper acetylide are formed.

Experiment No.2: With Tollen's Reagent: (Ammonical AgNO₃)
Reaction with alkene: No Reaction.
Reaction with alkyne: When acetylene is passed through ammonical solution of Silver nitrate, a white insoluble silver acetylide are formed.

(c) Chemical tests which are used to distinguish the alkane and alky-halide
Alkyl halide may be distinguished from alkanes because they react rapidly with base.
Experiment: With warm NaOH and Silver Nitrate:
Reaction with alkane: No reaction.
Reaction with alkyl-halide (R-X): The reactivity of alkyl-halide is due to its nucleophilic nature. As Halogen group is highly electronegative and it develops partial positive charge on Carbon-halogen bond and become slightly acidic in nature. On reaction with base it forms salt.

R-X + NaOH → R-OH + NaX 

Add silver nitrate and a precipitate will form:

NaX + AgNO₃ → AgX ↓   + NaNO₃

If,
X= Cl - you get white ppt.
X= Br - pale tan color ppt.
X= I - pale yellow ppt. will formed

7. Which alkane may be obtained by the reduction of:
(a) 2 — bromo propane
(b) Sec — butyl bromide
(c) Isopropyl bromide

Answer:
(a) 2 — bromo propane

(b) Sec — butyl bromide

(c) Iso propyl bromide {same as (a)}

8. Write equations for the formation of hydrocarbon by the action of N a metal on:
(i) Ethyl chloride
(ii) 2 bromo butane
(iii) Methyl iodide

Answer:
(i) Ethyl chloride
When Ethyl chloride is treated with sodium matel butane is formed.

2CH₃-CH₂-I + 2Na ⟶ C₄H₁₀+ 2NaI

(ii) 2-bromo butane
When 2-bromo butane is treated with sodium matel 3,4-dimethyl hexane is formed.

2CH₃-CHBr-CH₂-CH₃ + 2Na → (CH₃)₂-(CH-CH₂-CH₃)₂ + 2 NaBr

OR

(iii) Methyl iodide
When Methyl iodide is treated with sodium matel ethane is formed.

2CH₃I + 2Na ⟶ C₂H₆+ 2NaI

9. A hydrocarbon, C₄H₆ absorbs two moles of hydrogen in the presence of platinum as a catalyst. The reduction product is inert towards bromine and potassium permanganate. Draw possible structures for C₄H₆. What further information is necessary in order to establish the identity of the hydrocarbon?
Answer: When 1 mole of C₄H₆ reacts with 2 moles of H₂ gas, 4 moles of H atoms are added and then the hydrocarbon becomes C₄H₁₀. C₄H₁₀is the molecular formula of a saturated straight-chained alkane, which is inert towards bromine and potassium permanganate. It can be deduced that C₄H₆ would have 2 C=C double bonds or 1 C≡C triple bond.
Hence, there are 4 possible structures for C₄H₆.

10. What are the principal sources of benzene and its homologous?
Answer: Besides aliphatic compounds, large number of compounds were obtained from natural sources e.g. resins, balsams and "aromatic' oils from plants, whose structures at that time were unknown but had one thing in common, a pleasant smell and were thus classified as aromatic (Greek aroma = pleasant smell). These compounds contained a much higher percentage of carbon content than corresponding aliphatic hydrocarbons and majority of simple aromatic compounds contained at least six carbon atoms.
It was also discovered that when these compounds were subjected to various treatment they often yielded benzene or its derivatives. It was thus concluded that aromatic compounds were related to benzene whose molecular formula was found to be C₆H₆.
Benzene was first isolated by Faraday, in 1825 and was found to be one of the constituents of coal tar by Hofmann in 1845. Thus Coal tar and petroleum are the two main or principal sources of obtaining benzene and its derivative on industrial and commercial scale.

11. Draw the structures, of the following compounds:
(i) Benzene
(ii) Toluene
(iii) Ethyl benzene
(iv) Nitrobenzene
(v) P-bromobenzoic acid
(vi) O-dichlorobenzene
Answer:

12. (a) What do you understand by the term "Aromaticity" and how will you explain the stability of benzene molecule?
(b) Draw resonance structures for benzene.
Answer: (a)

(b)

13. What would be the major monochloro product (or products) formed when each of the following compounds reacts with chlorine in presence of ferric chloride?
(i) Ethyl benzene, C₆H₅-CH₂CH₃
(ii) Trifluoromethyl benzene, C₆H₅-CF₃
(iii) Methyl benzoate, C₆H₅COOCH₃

Answer:
(i) Ethyl benzene, C₆H₅-CH₂CH₃

(ii) Trifluoromethyl benzene, C₆H₅-CF₃

(iii) Methyl benzoate, C₆H₅COOCH₃