Haloalkanes and Haloarenes

(i) Freon 12:- the chlorofluorocarbon compounds of methane and ethane are collectively known as freons. They are extremely stable, unreactive, non-toxic and non-corrosive. Example of Freon 12 is CCl₂F₂. It is mainly used in refrigeration and eventually makes its way into the atmosphere where it diffuses unchanged into the stratosphere. In stratosphere, Freon is able to initiate the radical chain reactions that can upset the natural ozone balance.

• DDT: - it stands for p, p'-Dichlorodiphenyltrichloroethane (DDT). It is mainly used as an
• Iodoform: - it was earlier used as an antiseptic but the antiseptic properties are due the liberation of free iodine and not due to the iodoform itself. Due to its objectionable smell, it has been replaced by other formulations containing iodine.
• Carbon tetrachloride (CCl₄):- it is used as an industrial solvent for oils, fats, resins etc. and also in dry Used as a cleaning fluid both in industries, as a degreasing agent and in the home, as a spot remover and as a fire extinguisher.

Let us explain each question from Haloalkane and Haloarenes chapter one by one. Students who are currently in school and plan to take the CBSE board exam for class 12th soon, need to prepare all these questions. Let us get started.

(i) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.

Sp² hybrid carbon (s-character=33.33%) in chlorobenzene is more electronegative than a sp³-hybrid carbon (s-character=25%) in cyclohexylchloride, due to greater s-character. Thus, Carbon atom of chlorobenzene has less tendency to release electrons to Cl than carbon atom of cyclohexylchloride.

Chlorine atom in chlorobenzene is attached to the benzene ring. Here, the carbon-chlorine bond is polar. However, electron delocalization is possible in the benzene ring. Due to the electron delocalization, the overall dipole moment reduced since the polarity is distributed over the entire ring structure.

On the other hand, the chlorine atom in cyclohexyl chloride is attached to a saturated carbon atom in the cyclohexane ring. Because of this, there is no electron delocalization in the saturated structure. Hence, the dipole moment is more localized and stronger as compared to chlorobenzene.

 (ii) Alkyl halides, though polar, are immiscible with water.

Alkyl halides are polar because of the difference in electronegativity between carbon and halogen atoms. Here, there is not enough polarity to make them soluble in water.

The alkyl part of molecule is non-polar as well as hydrophobic in nature. This portion is large enough to dominate the solubility characteristics of the molecule. The entire molecule becomes insoluble in water due to this.

Water molecules create extensive hydrogen bonds with each other. However, Alkyl halides cannot form hydrogen bonds with water, which reduces their solubility in water. Haloalkane and Haloarene NCERT excercise discusses this question in further detail.

(iii) Grignard reagents should be prepared under anhydrous conditions.

Grignard reagents are very reactive organometallic compounds and they react readily with water to produce hydrocarbons. This destroys the Grignard reagent and makes it ineffective for its original purpose.

The reaction with water not only destroy Grignard reagent but also it produces by products that can complicate the synthetic pathway.

Preparing Grignard reagents under anhydrous conditions makes sure that they remain stable and effective for use in further reactions. Moisture-free conditions are important for the formation and utilization of reagents.

R-Mg-X + H-OH → R-H + Mg (OH)X

Do note that questions based on such equations from organic chemistry are also be asked in the CUET exam which makes it important for students to remember it.

Thus, Grignard reagents must be prepared under anhydrous conditions.

(i) Since the conversion of ethanol to but-1-yne involves the addition the two extra carbons that is why the conversion is carried out in 3 steps.

1. In the first step ethanol is treated with thionyl chloride (SOCl₂) in the presence of pyridine to give chloroethane

• In the second step the acetylene is treated with sodamide(NaNH₂) in the presence of liquid ammonia to give sodium acetylide

• The last step involves the reaction between chloroethane and sodium acetylide to give the final product as the But-1-yne and NaCl as the by-product.

• The bromination of ethane (Br₂ in the presence of light at 520-670K) gives bromoethane which on further treatment with alcoholic KOH results in the alkene called ethane which again on bromination with Br₂/CCl₄ gives dibromide(vicinal bromide) called 1,2-dibromoethane which on treatment with alcoholic KOH gives bromoethene as the final product.

• Propene on treatment with HBr in the presence of peroxide (anti-markovnikoff's rule which states the negative part e. Br^- goes to the carbon which has more number of hydrogens and positive part i.e. H⁺ goes to the carbon which has lesser number of hydrogens) gives 1-bromopropane which on treatment with silver nitrite in the presence of alcohol or water gives 1-nitropropane as the final product.

• Chlorination of toluene (Cl₂ at 773K) gives benzyl chloride with the removal of HCl. Benzyl Chloride on further treatment with dilute KOH in the presence of heat gives benzyl alcohol as the final (dilute KOH removes Cl from the benzyl chloride and replaces it by OH)

• Propene is treated with Br₂/CCl₄e. bromination of alkene takes place to give dibromides, in this case 1,2-dibromopropane which on further treatment with alcoholic KOH gives propyne.(two times removal of KBr)

• Ethanol in treatment with thionyl chloride(SOCl₂) in the presence of pyridine gives ethyl chloride with the evolution of SO₂ The ethyl chloride on treatment with mercury fluoride gives ethyl fluoride (the Cl of ethyl chloride is being replaced by F from Hg₂F₂).

• Bromoethane on treatment with alcoholic KCN forms Acetonitrile with the removal of KBr. CN is added in the case where there is a need to increase the number of Further the acetonitrile is treated with CH₃MgBr(Grignard reagent) in the presence of ether, the intermediate so formed is hydrolysed, and the product formation takes place i.e. Propanone and NH₃ is obtained as the by- product.

• But-1-ene reacts with HBr to form 2-bromobutane(acc. To markovnikoff's rule) and further 2- bromobutane is treated with alcoholic KOH in the presence of heat to form But-2-ene with the removal of HBr.(alcoholic KOH removes H from the carbon which has less number of hydrogens just next to the carbon to which halide is attached and gives the major product e. Saytzeff rule)

• The conversion of 1-chlorobutane to n-octane in the presence of Na metal and dry ether is called Wurtz Two equivalent Chlorobutane reacts to give n-octane as a final product and NaCl as the by-product.

 

The conversion of benzene to biphenyl using Na and dry ether is called FITTIG reaction.

Firstly, the benzene is treated with Br₂/FeBr₃ results in the formation of bromobenzene. It is an electrophilic substitution. Further the bromobenzene is treated with Na metal in presence of dry ether to form biphenyl as the final product and NaBr as the by-product.

 

If instead of Na metal, bromobenzene is treated with Cu metal then the reaction is called Ullmann's reaction. The final product obtained in this reaction will also be biphenyl.

Carbon due to its chemical property of catenation, tetravalency and the ability to form stable covalent bonds form hydrocarbon and halo-alkanes easily.

The process of removal of hydrogen and halogen from adjacent carbon atoms of haloalkens is called dehydrohalogenation. NaOEt ( Sodium ethoxide) used in Dehydrohalogenation.  It is a strong base and helps remove a β-hydrogen from the haloalkane.

The chemical reaction for

Reaction:

R–CH2 –CHX–R′+NaOEt   ethanol > R–CH=CHR'+NaX+EtOH

(i) 1-Bromo-1-methylcyclohexane: 

      Br
       |
 CH3–C1–cyclohexane  + NaOEt/EtOH → CH3–C=cyclohexane (double bond at C1-C2) + NaBr + EtOH

Major Product: 1-Methylcyclohexene

(ii) 2-Chloro-2-methylbutane

      CH3
           |
 CH3–C–CH2–CH3   + NaOEt/EtOH → CH3–C=CH–CH3 + NaCl + EtOH
           |                                
          Cl                             CH3

Major Product: 2-Methyl-2-butene (major), 2-Methyl-1-butene (minor)

(iii) 2,2,3-Trimethyl-3-bromopentane

       CH3         CH3
         |            |
 CH3–C–C–CH2–CH3  + NaOEt/EtOH → CH3–C=C–CH2–CH3 + NaBr + EtOH
         |   |                           
        CH3  Br                        CH3

Major Product: 2,2,3-Trimethyl-2-pentene

(a) 1-butanol is treated with KI in the presence of H₃PO₄ where the OH is being replaced by the iodine and gives H₂O and KH₂PO₄ as the by-product with 1-iodobutane as the final

(b) The conversion of 1-chlorobutane to 1-iodobutane simply takes place by treating the reactant with KI in the presence of acetone. The iodine from KI normally replaces the chlorine from the reactant and gives 1-iodobutane as the final product with KCl as the by product.

(c) The conversion of but-1-ene to 1-iodobutane takes place according to anti-markovnikoff (positive charged ion H⁺ goes to the carbon which has less number of hydrogens and negative part Br- goes to the carbon which has more number of hydrogens.) i.e. the attack of HBr on but-1-ene in the presence of peroxide gives 1-bromobutane as the intermediate which on treatment with NaI+Acetone gives 1- iodobutane as the final product and NaBr as the by-product.

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.