Ncert Solutions Chemistry Class 12th

As per the molecular formula $C_5{H}_{10}$ , one thing can be confirmed; the hydrogen is either cycloalkane or it is an alkene. However, alkenes readily react with chlorine in even dark because of the double bond. Therefore, it is possible that the hydrocarbon is cycloalkane. 

Now, let us consider the reactivity. Cycloalkanes are the saturated hydrocarbons with no double bonds. These do not react with Chlore in dark because such a reaction needs UV light for initiating the process of free radical substitution. Haloalkane and Haloarenes NCERT solutions cover this as well as other questions in further detail.

In bright sunlight, the UV light starts free radical substitution reaction. In cyclopentane, the substitution will result in single monochloro product because all hydrogen atoms are equivalent in the molecule. Therefore, by replacing any hydrogen with chlorine atom will result again in the same product i.e. $C_5{H}_9\mathrm{Cl}.$ This confirms that the hydrocarbon is cyclopentane.

Dipole moment of a molecule depends on the electronegativity difference between atoms bonded covalently and geometry of the molecule (how far the atoms are from each other). Dipole moment is important to understand the polarity of a molecule.

The three dimensional structures of the three compounds along with the direction of dipole moment in each of their bonds are given below:-

CCl₄ being symmetrical has zero dipole moment. In CHCl₃, the resultant of the two C-Cl dipole moments is opposed by the resultant of C-H and C-Cl bonds. Since the dipole Moment of latter resultant is expected to be smaller than the former, CHCl₃ has a finite dipole moment (1.03D)

In CH₂Cl_2, the resultant of two C-Cl dipole moments is reinforced by resultant of two C-H dipoles. Therefore, CH₂Cl₂ (1.62D) has a dipole moment higher than that of CHCl₃. Thus, CH₂Cl₂ has the highest dipole moment.

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