Chapter 10 Haloalkanes and Haloarenes Solutions
Question - 11 : - Give the IUPAC names of the following compounds:
Answer - 11 : -
(i) CH3CH(Cl)CH(Br)CH3
(ii) CHF2CBrClF
(iii) ClCH2C≡CCH2Br
(iv) (CCl3)3CCl
(v) CH3C(p-ClC6H4)2CH(Br)CH3
(vi) (CH3)3CCH=CClC6H4I-p
Answer
(iv)
2−(Trichloromethyl)−1,1,1,2,3,3,3−heptachloropropane
(v)
2−Bromo−3, 3−bis(4−chlorophenyl) butane
(vi)
1−chloro−1−(4−iodophenyl)−3, 3−dimethylbut−1−ene
Question - 12 : - Write the structures of the following organic halogencompounds.
Answer - 12 : -
(i) 2-Chloro-3-methylpentane
(ii) p-Bromochlorobenzene
(iii) 1-Chloro-4-ethylcyclohexane
(iv) 2-(2-Chlorophenyl)-1-iodooctane
(v) Perfluorobenzene
(vi) 4-tert-Butyl-3-iodoheptane
(vii) 1-Bromo-4-sec-butyl-2-methylbenzene
(viii) 1,4-Dibromobut-2-ene
Answer
(i)
2-Chloro-3-methylpentane
(ii)
p-Bromochlorobenzene
(iii)
1-Chloro-4-ethylcyclohexane
(iv)
2-(2-Chlorophenyl)-1-iodooctane
(v)
Perfluorobenzene
(vi)
4-Tert-Butyl-3-iodoheptane
(vii)
1-Bromo-4-sec-butyl-2-methylbenzene
(viii)
1,4-Dibromobut-2-ene
Question - 13 : - Which one of the following has the highest dipole moment?
Answer - 13 : -
(i) CH2Cl2
(ii) CHCl3
(iii) CCl4
Answer
(i)
Dichlormethane (CH2Cl2)
μ = 1.60D
(ii)
Chloroform (CHCl3)
μ = 1.08D
(iii)
Carbon tetrachloride (CCl4)
μ = 0D
CCl4 is a symmetricalmolecule. Therefore, the dipole moments of all four C−Cl bonds cancel eachother. Hence, its resultant dipole moment is zero.
As shown in the above figure, in CHCl3,the resultant of dipole moments of two C−Cl bonds is opposed by the resultantof dipole moments of one C−H bond and one C−Cl bond. Since the resultant of oneC−H bond and one C−Cl bond dipole moments is smaller than two C−Cl bonds, theopposition is to a small extent. As a result, CHCl3 hasa small dipole moment of 1.08 D.
On the other hand, in case of CH2Cl2,the resultant of the dipole moments of two C−Cl bonds is strengthened by theresultant of the dipole moments of two C−H bonds. As a result, CH2Cl2 hasa higher dipole moment of 1.60 D than CHCl3 i.e., CH2Cl2 hasthe highest dipole moment.
Hence, the given compounds can be arranged in theincreasing order of their dipole moments as:
CCl4 < CHCl3 2Cl2
Question - 14 : - A hydrocarbon C5H10 doesnot react with chlorine in dark but gives a single monochloro compound C5H9Clin bright sunlight. Identify the hydrocarbon.
Answer - 14 : -
A hydrocarbon with the molecular formula, C5H10 belongsto the group with a general molecular formula CnH2n.Therefore, it may either be an alkene or a cycloalkane.
Since hydrocarbon does not react with chlorine in thedark, it cannot be an alkene. Thus, it should be a cycloalkane.
Further, the hydrocarbon gives a single monochlorocompound, C5H9Cl by reacting with chlorine in bright sunlight. Sincea single monochloro compound is formed, the hydrocarbon must contain H−atomsthat are all equivalent. Also, as all H−atoms of a cycloalkane are equivalent,the hydrocarbon must be a cycloalkane. Hence, the said compound iscyclopentane.Cyclopentane (C5H10)
The reactions involved in the question are:
Write the isomers of the compound havingformula C4H9Br.
Answer - 15 : -
There are four isomers of the compoundhaving the formula C4H9Br. These isomers are given below.
(a)
1−Bromobutane
(b)
2−Bromobutane
(c)
1−Bromo−2−methylpropane
(d)
2−Bromo−2−methylpropane
Question - 16 : - Write the equations for the preparation of 1−iodobutanefrom
Answer - 16 : -
(i) 1-butanol
(ii) 1-chlorobutane
(iii) but-1-ene.
Answer
(i)
(ii)
(iii)
Question - 17 : - What are ambident nucleophiles? Explain with an example.
Answer - 17 : -
Ambident nucleophiles are nucleophiles having twonucleophilic sites. Thus, ambident nucleophiles have two sites through whichthey can attack.
For example, nitrite ion is an ambident nucleophile.
Nitrite ion can attack through oxygen resulting in theformation of alkyl nitrites. Also, it can attack through nitrogen resulting inthe formation of nitroalkanes.
Question - 18 : - Which compound in each of the followingpairs will react faster in SN2 reaction with OH−?
Answer - 18 : -
(i) CH_3Br or CH_3I
(ii) (CH_3)_3CCl or CH_3Cl
Answer
(i) In the SN2 mechanism, thereactivity of halides for the same alkyl group increases in the order. Thishappens because as the size increases, the halide ion becomes a better leavinggroup.
R−F << R−Cl < R−Br < R−I
Therefore, CH3Iwill react faster than CH3Br in SN2 reactions with OH−.
(ii)
The SN2 mechanism involvesthe attack of the nucleophile at the atom bearing the leaving group. But, incase of (CH3)3CCl, the attack of the nucleophile at the carbon atom ishindered because of the presence of bulky substituents on that carbon atombearing the leaving group. On the other hand, there are no bulky substituentson the carbon atom bearing the leaving group in CH3Cl.Hence, CH3Cl reacts faster than (CH3)3CClin SN2 reaction with OH−.
Question - 19 : - Predict all the alkenes that would be formed bydehydrohalogenation of the following halides with sodium ethoxide in ethanoland identify the major alkene:
Answer - 19 : -
(i) 1-Bromo-1-methylcyclohexane
(ii) 2-Chloro-2-methylbutane
(iii) 2,2,3-Trimethyl-3-bromopentane.
Answer
(i)
In the given compound, there are two typesof β-hydrogen atoms are present. Thus, dehydrohalogenation of thiscompound gives two alkenes.
(ii)
Question - 20 : - How will you bring about the following conversions?
Answer - 20 : -
(i) Ethanol to but-1-yne
(ii) Ethane to bromoethene
(iii) Propene to 1-nitropropane
(iv) Toluene to benzyl alcohol
(v) Propene to propyne
(vi) Ethanol to ethyl fluoride
(vii) Bromomethane to propanone
(viii) But-1-ene to but-2-ene
(ix) 1-Chlorobutane to n-octane
(x) Benzene to biphenyl.
Answer
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
