Haloalkanes and Haloarenes

CH₃CH (Cl)CH (Br)CH₃: 2-Bromo-3-chlorobutane

1. Write the name of side substituent according to the alphabetical order. Example: bromo is written before chloro as B comes before
2. Always use a hyphen {-} between a number and a Since Br is attached to 2 position and Cl is attached to third position i.e. it is written as 2-Bromo and 3-chloro.
3. Lastly, write the name of the longest hydrocarbon chain e. butane of 4 carbons.

CHF₂CBrClF :1-bromo-1-chloro-1,2,2-trifluoroethane

1. While writing the IUPAC name, write the name of side substituent according to the alphabetical Eg: bromo is written before chloro and fluoro as B comes before C and F.
2. Always use a hyphen {-} between a number and a Since Br is attached to first carbon and Cl Is also attached to first carbon i.e. it is written as 1-Bromo and 1-chloro.
3. Since there are 3 F present so it is written as
4. And at the last write the name of the longest hydrocarbon chain e. ethane of 2 carbons.

ClCHC≡CCH₂Br: 1-bromo-4-chlorobut-2-yne

1. While writing the IUPAC name, write the name of side substituent according to the alphabetical That is why bromo is written before chloro as B comes before C.
2. Always use a hyphen {-} between a number and a Since Br is attached to 1 position and Cl isattached to fourthposition i.e. it is written as 1-Bromo and 4-chloro.
3. Also there is a presence of triple bond at the second carbon so the suffix used for triple bond is “yne” along with the position of the bond i.e. but-2-yne

(CCl₃)₃CCl

2- (trichloromethyl)-1,1,1,2,3,3,3-heptachloropropane

Explanation:

1. While writing the IUPAC name, write the name of side substituent according to the alphabetical
2. Always use a hyphen {-} between a number and a
3. Because of the presence of 3 CCl₃ group tri is used as a
4. Since 7 Cl atoms are present in all therefore hepta is used a prefix with the position of chlorine. And at the last write the name of the longest hydrocarbon chain e. propane)

CH₃C (p-ClC₆H₄)₂CH (Br)CH₃

2-bromo-3,3-bis- (4-chlorophenyl)butane

Explanation:

1. While writing the IUPAC name, write the name of side substituent according to the alphabetical
2. Always use a hyphen {-} between a number and a
3. Because of the presence of 2 (p-ClC₆H₂) group bis is used as a prefix instead of di because C₆H₄ contains itself contains numbers 6 &
4. And at the last write the name of the longest hydrocarbon chain e. butane)

(CH₃)₃CCH=CClC₆H₄I-p

1-chloro-1- (4-iodophenyl)-3,3-dimethylbut-1-ene

Explanation:

1. While writing the IUPAC name, write the name of side substituent according to the alphabetical
2. Always use a hyphen {-} between a number and a
3. Because of the presence of 2 methyl group di is used as a
4. And the presence of double bond has the ending 'ene' and the presence of double bond at the first position is written as but-1-ene.)

(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

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 : 

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.