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(i) 2-chloro-3-methylbutane, (2^? alkyl halide)

Explanation: The Cl atom is attached with two alkyl groups therefore 2^? alkyl halide.

(ii)  3-chloro-4-methylhexane (2^? alkyl halide)

Explanation: The Cl atom is attached with two alkyl groups therefore 2^? alkyl halide.

(iii)  1-iodo-2,2-dimethylbutane (1^? alkyl halide)

Explanation: The atom I is attached with one alkyl group therefore 1^? alkyl halide.

(iv)  1-bromo-3,3-dimethyl-1-phenylbutane (2^? benzylic halide)

Explanation: The atom Br is attached with two alkyl groups with the benzene group therefore 2^? benzylic halide.

(v)  2-bromo-3-methylbutane (2^? alkyl halide)

Explanation: The atom Br is attached with two alkyl groups therefore 2^? alkyl halide.

(vi)  1-bromo-2-ethyl-2-methylbutane (1^? alkyl halide)

Explanation: The atom Br is attached with one alkyl group therefore 1^? alkyl halide.)

(vii)  3-chloro-3-methylpentane (3^? alkyl halide)

Explanation: The atom Cl is attached with three alkyl groups therefore 3^? alkyl halide.

(viii)  3-chloro-5-methylhex-2-ene (vinylic halide)

Explanation: The atom Cl is attached with the double bond i.e. at vinylic position ∴ vinylic halide.)

(ix)  4-bromo-4-methylpent-2-ene (allylic halide)

Explanation: The Br atom is attached with the carbon next to the double bond i.e. at allylic position therefore allylic halide.

(x)  1-chloro-4- (2-methylpropyl)benzene (aryl halide)

Explanation: The Cl atom is attached with the benzene i.e. at aryl position therefore aryl halide.

(xi)  1-chloro-3- (2,2-dimethylpropyl)benzene (1^? benzylic halide)

Explanation: The Cl is attached with one alkyl group at the benzene carbon therefore 1^? benzylic halide.

(xii)  1-bromo-2- (1-methylpropyl)benzene (aryl halide)

Explanation: The Br is attached with the benzene i.e. at aryl position therefore aryl halide.

(i) Bromocyclohexane is reacting with the metal Mg to give organo-metallic compound/Grignard reagent

i.e. Cyclohexyl Magnesium Bromide (A).

Grignard reagents are highly reactive with hydrocarbons to give alkanes. Cyclohexyl Magnesium Bromide reacts with water to cyclohexane (B) and Mg (OH)Br.

(ii) As the above reaction, haloalkane with metal Mg react to form the Grignard reagent, alkyl magnesium halide, RMgX, and further with hydrocarbon gives alkane.

∴ C is CH₃CH (MgBr)CH₃, a Grignard Reagent, R is CH₃CH (Br)CH₃ and D ≅ H

(iii) In the Wurtz Reaction, alkyl halide in the presence of sodium in dry ether gives the double number of

Now, in the reaction 2,2,3,3-tetramethyl butane is formed after the alkyl halide reacting in the presence of sodium in dry ether.

∴ Number of carbons will be 3 in R¹ –X is CH₃C (X) (CH₃)CH₃

And the remaining reaction is same as the above i.e. alkyl halide reacting with Magnesium to give Grignard Reagent (D) and further with hydrocarbon H₂O to give alkane (E)

∴ D is CH₃C (CH₃) (CH₃)MgBr E is CH₃CH (CH₃)CH₃

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A 10.5 1. Cyclohexanol will react with thionyl chloride to form Chlorocyclohexane with the evolution of sulphur dioxide and hydrogen chloride gas as shown below :

2. 4-Ethylnitrobenzene will undergo benzylic bromination in presence of heat or light. Now, as benzylic radicals are more stable therefore benzylic hydrogen is abstracted. Hence, the reaction yields 4-(1- Bromoethyl)nitrobenzene as a product

3. 4-Hydroxymethylphenol will react with hydrochloric acid under thermal conditions to yield 4- Chloromethylphenol a shown in the reaction below:

4. 1-Methylcyclohexene will react with hydrogen iodide via markovnikov’s addition mechanism to give 1- Iodo1-methylcyclohexane as shown in the reaction below: -

5.Bromoethane will react with sodium iodide in a Finkelstein reaction manner to yield Iodoethane in presence of dry acetone and heat as shown in the reaction below : 

6. Cyclohexene reacts with bromine in presence of heat and light to undergo allylic bromination to yield 3-bromocyclohex-1-ene as the major product as shown in the reaction below : 

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To have a single monochloride, there should be only one type of H-atom in the isomer of the alkane of the molecular formula C₅H₁₂. This is because of the fact that replacement of any H-atom leads to the formation of the same product. Therefore the isomer is 2,2-dimethylpropane.

To have three isomeric monochlorides, the isomer of the alkane of the molecular formula C₅H₁₂ should contain three different types of H-atoms. Therefore, the isomer is n-pentane. It can be observed that there are three types of H atoms labelled as a, b and c in n-pentane as shown below : -

To have four isomeric monochlorides, the isomer of the alkane of the molecular formula C₅H₁₂ should contain four different types of H-atoms in it. Therefore, the required isomer is 2-methylbutane. It can be observed that there are four different types of H-atoms labelled as a, b, c, and d in 2-methylbutane as shown below: -

Here is the reasoning that you can learn more while practicing the NCERT Solutions for Haloalkanes And Haloarenes. 

 

This outcome is directly attributed to the free radical halogenation mechanism. Usually, in photochemical chlorination, a chlorine atom can replace any hydrogen atom in the alkane. The number of unique products, or isomeric monochlorides, depends on the number of distinct types of hydrogen atoms in the starting isomeric alkane. 

 

For instance, the high symmetry of 2,2-dimethylpropane (neopentane) means all its hydrogen atoms are equivalent. That itself yields only a single monochloride. But, in contrast, the different primary and secondary hydrogens in n-pentane and 2-methylbutane create multiple products.

In the presence of sulphuric acid (H₂SO₄), KI produces HI as follows : -

2KI + H₂SO₄ → 2KHSO₄ + 2KI

Since H₂SO₄ is an oxidising agent, it oxidises HI (produced in the reaction to I₂)

2HI + H₂SO_{4 →} I₂ + SO₂ +;

As a result, the reaction between an alcohol and HI to produce alkyl iodide cannot occur. Therefore, sulphuric acid is not used during the reaction of alcohols with KI. Instead, a non-oxidising acid such as H3PO4 is used in the reaction to get the desired product.

A few things you can remember here, while solving NCERT Solutions for Haloalkanes And Haloarenes.

• Sulphuric acid is a powerful oxidising agent. We know this because the sulphur atom in sulphuric acid is in its highest possible oxidation state (+6). It can only be reduced. And it can readily accept electrons from other substances.

• It oxidises the necessary intermediate, hydrogen iodide (HI), to iodine (I2). We know this because iodide ions are strong reducing agents, and they can simply give up electrons. So when sulphuric acid accepts electrons, it can convert the iodide to elemental iodine.

• The alcohol cannot be converted to alkyl iodide. The reason for this is that the side reaction consumes the iodide ions necessary to act as a nucleophile and replace the alcohol's -OH group. This is the desired alkyl iodide.

• A non-oxidising acid like phosphoric acid (H3PO4) is used instead to produce the desired alkyl iodide. Phosphoric acid is a weaker oxidising agent than sulphuric acid and does not react with the iodide ions. That allows the intended reaction with the alcohol to proceed.