NCERT Solutions for Class 12 Chemistry Chapter 10 Haloalkanes and Haloarenes are part of NCERT Solutions for Class 12 Chemistry. Here we have given NCERT Solutions for Class 12 Chemistry Chapter 10 Haloalkanes and Haloarenes.
|Chapter Name||Haloalkanes and Haloarenes|
|Number of Questions Solved||31|
NCERT Solutions for Class 12 Chemistry Chapter 10 Haloalkanes and Haloarenes
NCERT IN-TEXT QUESTIONS
Write the structures of the following compounds : (C.B.S.E. Delhi 2010)
(iii) 4-tert. butyl-3-iodoheptane
(v) 1-Bromo-4-sec butyl-2-methylbenzene. (C.B.S.E. Sample paper 2011)
Why is sulphuric acid not used during the reaction of alcohols with KI ?
KI is expected to give HI on reacting with H2SO4 which will convert alcohols (R – OH) to alkyl iodides (R – I). However, H2SO4 is a strong oxidising agent and it oxidises HI formed during the reaction to I2 which does not react with alcohol.
To solve the problem, H2S04 is replaced by phosphoric acid (H3P04) which provides HI for the reaction and does not give I2 as is done by H2S04.
Write the structures of different dihalogen derivatives of propane.
Propane (CH3CH2CH3) has two primary and one secondary hydrogen atoms present. Four isomeric dihalogen derivatives are possible. Let the halogen X be Br.
Among the isomeric alkanes of molecular formula C5H12, identify the one which on photochemical chlorination yields
(i) A single monochloride
(ii) Three isomeric monochlorides
(iii) Four isomeric monochlorides.
The molecular formula C5H12 represents three structural isomers which are chain isomers.
(i) The isomer is symmetrical with four primary (1°) carbon atoms and one quaternary (4°) carbon atom. Since all the hydrogen atoms are equivalent, it will yield only one monochloride upon photochlorination i.e., chlorination carried in the presence of ultra-violet light.
(ii) In the straight chain isomer pentane, there are three groups of equivalent hydrogen atoms. As a result, three isomeric monochlorides are possible.
(iii) The branched chain isomer has four types of equivalent hydrogen atoms present. It will give four isomeric monochlorides upon chlorination.
Draw the structures of the major monohaloproducts in each of the following reactions:
The reaction is carried in the presence of dry acetone upon heating. It is called Finkelstein reaction. In this reaction, I– ion being a stronger nucleophile displaces Br– ion. NaBr formed is insoluble in dry acetone whereas Nal dissolves. This shifts the equilibrium in the forward direction.
Under the reaction conditions allylic halogenation will take place. Addition of bromine can be possible in case the reaction is carried at room temperature.
Arrange each set of compounds in order of increasing boiling points :
(i) Bromomethane, bromoform, chloromethane, dibromomethane
(ii) 1- Chloropropane, isopropylchloride, 1- chlorobutane.
(i) The boiling points of organic compounds are linked with the van der Waals’ forces of attraction which depend upon the molecular size. In the present case, all the compounds contain only one carbon atom. The molecular size depends upon size of the halogen atom and also upon the number of halogen atoms present in different molecules. The increasing order of boiling points is :
CH3Cl(chloromethane) < CH3Br (bromomethane) < CH2Br2 (dibromomethane) < CHBr3 (bromoform)
(ii) The same criteria is followed in this case. We all know that the branching of the carbon atom chain decreases the size of the isomer and this decreases its boiling point as compared to straight chain isomer. The increasing order of boiling point is :
(CH3)2CHCl (isopropylchloride or 2-chloropropane) < ClCH2CH2CH3 (1-chloropropane) < ClCH2CH2CH2CH3 (1-chlorobutane)
Which alkyl halide from the following pairs would you expect to react more rapidly by SN² mechanism ? Explain your answer.
If the leaving group is the same in different isomers of a particular molecular formula, the reactivity of the isomers towards SN² mechanism decreases with the increase in steric hindrance. In the light of above, the reactivity order in different cases is :
(i) CH3CH2CH2CH2Br is a primary alkyl halide (1°). It is more reactive than the other isomer which is a secondary (2°) alkyl halide because less steric hindrance is caused by primary alkyl group as compared to secondary alkyl group.
is a secondary alkyl halide (2°). It is more reactive than the other isomer which is a tertiary alkyl halide (3°). The explanation is the same.
(iii) Here both the isomers are primary alkyl halides (1°). However, the isomer with CH3 group at C2 atom exerts more steric hindrance to the attacking nucleophile at C1 atom as compared to the other isomer in which a CH3 group is attached to C3 atom. It is, therefore, less reactive.
In the following pairs of halogen compounds, which compound undergoes reaction faster ? (C.B.S.E. Delhi 2008, Outside Delhi 2010, 2013)
The reactivity of a particular halogen compound towards SN¹ reaction depends upon the stability of the carbocation formed as a result of ionisation. This is a slow step and is called rate determining step. The order of relative stabilities of different carbocations is in the order : tertiary > secondary > primary. In the light of this, the order of reactivity in the two cases is explained.
- The isomer (a) is a tertiary alkyl chloride while the other isomer (b) is a secondary alkyl chloride. The isomer (a) is more reactive towards S i reaction since the tertiary carbocation formed in this case is more stable than the secondary carbocation which is likely to be formed in the other case.
- The isomer (a) is a secondary alkyl chloride while the other isomer (b) is primary in nature. The secondary alkyl chloride (a) is expected to react faster since the secondary carbocation formed is more stable than the primary carbocation which is likely to be formed in the other case.
Identify A, B, C, D, E, R and R’ in the following :
Name the following compounds according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary) vinyl or aryl halides.
(i) (CH3)2CHCH(C1)CH3 (C.B.S.E. Delhi 2013)
Give the IUPAC names of the following compounds :
Write the structures of the following compounds :
(iv) 4-tert. butyl -3-iodooctane
(v) 1, 4-Dibromobut-2-ene
Which one of the following has highest dipole moment ?
- Tetrachloromethane (CCl4) is a symmetrical molecule and has zero dipole moment.
- In chloroform (CHC13), the resultant dipole moment of two C- Cl bonds is opposed by resultant dipole moment of C – Cl and C- H bonds. Since the latter resultant dipole moment is smaller, the molecule as a whole has dipole moment (μ) = 103 D).
- In dichloromethane (CH2Cl2), the resultant dipole moments of two C – Cl and two C – H bonds reinforce one another. The molecule has maximum dipole moment (μ) = 1.62 D.
CH2Cl2 has the maximum dipole moment value.
A hydrocarbon C5H10 does not react with chlorine in dark but gives a single monobromo compound in bright sun light. Identify the hydrocarbon.
(i) The hydrocarbon with molecular formula C5H10 is either an alkene or cycloalkane.
(ii) Since the hydrocarbon does not react with chlorine (Cl2) in the dark, it is not an alkene.
(iii) As it forms only single monochloroderivative in bright sun light, it is cyclopentane which is symmetrical. In this, all the ten hydrogen atoms are identical.
Write the isomers of the compound having the formula C4H9Br. (Haryana Board 2013)
The compound has the following structural isomers.
2-Bromobutane has a chiral carbon and it is expected to exhibit optical isomerism.
Write equations for the preparation of 1-Iodobutane from :
(a) Butan-1- ol
What are ambident nucleophiles ? Explain with an example.
Nucleophiles which can attack through two different sites, are called ambident nucleophiles. For example, cyanide ion exists as a hybrid of following two structures. It can attack either
through carbon to form cyanides (or nitriles) or through nitrogen to form isocyanides (or carbyl amines). For more details, consult section 11.7.
Which compound in the following pairs will react faster in Sn2 reaction ?
(a) CH3Br or CH3I
(b) (CH3)3CCl or CH3Cl (C.B.S.E. 2008)
(a) CH3– I will react faster than CH3– Br because I ion is a better leaving group than Br– ion. Alternatively, bond dissociation entbalpy of C- I bond is less than that of C – Br bond.
(b) CH3Cl will react faster because of less steric hindrance as compared to (CH3)3CCl
Predict all the alkenes that would be formed by dehydrohalogenation of following alkyl halides with sodium ethoxide in ethanol.
(iii) 3-Bromo-2, 2, 3-trimethylpentane.
(i) 1-Bromo-l-methylcyclohexane has two β-bydrogen atoms. This will give a mixture of two alkenes as a result of dehydrohalogenation. Since alkene (B) is more substituted according to SaytzefFs rule, it is more stable and will be the major product. The same rule applies to the other alkyl halides also.
(ii) The compound has two sets of β-hydrogen atoms. Therefore, two elimination products are formed. However, more substituted alkene is formed in greater proportion as compared to less substituted alkene.
The explanation is similar. More substituted alkene is formed in preference to less substituted alkene.
How will you bring about the following conversions? (Haryana Board 2011)
(i) Ethanol to but-1-yne
(ii) Ethane to bromoethane
(iii) Propene to 1-nitropropane
(iv) Toluene to benzyl alcohol
(v) Propene to propyne
(vi) Ethanol to ethyl fluoride
(vii) Bromometbane to propanone
(viii) But-1-ene to but-2-ene
(ix) 1-Chiorobutane to n-octane
(x) Benzene to biphenyl
(i) Dipole moment of chlorobenzene is lower than that of cyclohexylchloride (C.B.S.E 2016)
(ii) Alkyl halides though polar, are immiscible with water.
(iii) Grignard reagents should be prepared under anhydrous conditions.
The polarity of C- Cl bond in chlorobenzene is less than that of same bond in cyclohexyl chloride because of carbon atom involved in chlorobenzene is more electronegative (greater s-character) as compared to the carbon atom in case of cyclohexyl chloride (lesser s-character). Therefore, the dipole moment of chlorobenzene is less with respect to cyclohexyl chloride.
(ii) In water, H2O molecules are linked to each other by intermoleculer hydrogen bonding. Although alkyl halides also contain polar C – X bonds, they cannot break the hydrogen bonding in H20 molecules. This means that there is hardly any scope for association between molecules of alkyl halides and water. They therefore, exist as separate layers and are immiscible with each other. For more details, consult section 11.6.
(iii) Grignard reagents (R – Mg – X) should be prepared under anhydrous conditions because these are readily decomposed by water to form alkanes.
That is why ether used as solvent in the preparation of Grignard reagent is completely anhydrous in nature.
Give the uses of freon-12, D.D.T., carbon tetrachloride and iodoform ?
- Freons are the trade names for the commercially used fluoro chioromethanes with formula CFxCly (x + y = 4). A few example are :
CF4 (Freon-14), CF3C1 (Freon-13), CF2Cl2 (Freon-12), CFCl3 (Freon-11)
Out of the various freons mentioned, Freon- 12 is the most common refrigerant. It is prepared by passing hydrogen fluoride
through carbon tetrachionde in the presence of antimony trichioride catalyst.
in addition to their use as refrigerants in place of highly toxic liquid sulphur dioxide (SO2) and ammonia (NH3), large amount of CFCs are also used in the manufacture of disposable foam products such as cups and plates, as aerosol propellants in spray cans and as solvents to clean freshly soldered electronic circuit boards.
- D.D.T. is the abbreviated form of p, p’-dichlorodiphenyltrichloroethane and its actual IUPAC naine has been given above. It is
prepared by heating chiorobenzene with chlorai (trichioroacetaldehyde) in the presence of conc. H2S04
- Carbon tetrachloride (CC14) is also a colourless oily liquid just like chloroform. It is completely immiscible with water but
dissolves in organic solvents.
Carbon tetrachloride is a very useful solvent for oils, fats, resins etc. Ills used as a cleansing agent both in industry and in home because it can easily dissolve grease and other organic matter. But it mainly finds application for the manufacture of refrigerants, propellants for aerosol cans and some pharmaceuticals.
- lodoform is a yellow crystalline solid with a characteristic unpleasant smell. It is insoluble in water but dissolves in alcohol, ether and other organic solvents.
lodoform can be prepared in the laboratory by treating ethyl alcohol or acetone with sodium hydroxide and iodine. The reaction is known as haloform or iodoform reaction.
Physiological effects : lodoform is used as an antiseptic, particularly for dressing wounds. Actually, on coming in contact with skin (organic mater) it decomposes and slowly loses iodine which accounts for the antiseptic properties of iodoform.
Write the structures of the major products in each of the following reactions :
Explain the following reaction : (C.B.S.E. Delhi 2009 Comptt.)
KCN is a resonance hybrid of two contributing structures :
This shows that the cyanide ion is an ambident nucleophile and the nucleophile attack is possible either through carbon atom or nitrogen atom resulting in cyanides and isocyanides respectively. In this case, in the presence of polar solvent, KCN readily ionises to furnish ions. The nucleophile attack takes place predominandy through carbon atom and not through nitrogen atom as C- C bond is more stable than C -N bond.
Arrange the compounds of each set in order of decreasing reactivity towards (SN²) displacement:
(a) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane
(b) 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 2-Bromo-3-methylbutane
(c) 1-Bromobutane, l-Bromo-2, 2-dimethylpropane, l-Bromo-2-methylbutane, l-Bromo-3-methylbutane. (C.B.S.E. Outside Delhi 2011)
The reactivity of a particular haloalkane towards SN2 reaction is inversely proportional to the steric hindrance around the carbon atom involved in C – X bond. More the steric hindrance, lesser will be the reactivity. In the light of this, the decreasing order of reactivity in all the three cases is as follows :
Out of C6H5CH2Cl and C6H5CH(C1)C6H5 which is more easily hydrolysed by aqueous KOH ?
The compound C6H5CH2Cl is a primary aralkyl halide while C6H5CH(Cl)C6H5 is secondary in nature. The hydrolysis of both these compounds with aqueous KOH (polar) is likely to proceed by SN¹ mechanism due to following reasons.
(a) The carbocations formed in both the cases as a result of ionisation are resonance stabilised due to the presence of phenyl groups at the a-position(s).
(b) As water is a polar solvent, it is expected to favour ionisation of the two halogen substituted compounds leading to SN¹ mechanism.
The carbocations that are formed as a result of ionisation in the slow steps are shown :
The ease of hydrolysis depends upon the relative stability of the carbocation/s that are formed in two cases. The secondary carbocation is more stable since the positive charge on the carbocation is delocalised on two phenyl groups that are present at the a-positions. On the other hand, there is only one phenyl group in primary carbocation available for charge delocalisation.
Thus, we may conclude that C6H5CHClC6H5 is more easily hydrolysed by aqueous KOH as compared to C6H5CH2Cl.
p-dichlorobenzene has higher m.p. and lesser solubility than those of o-and m-isomers. Discuss. (C.B.S.E. Delhi 2013)
The three isomers are position isomers which differ in the relative positions of the chlorine atoms in the ring :
As we know, p-isomer is more symmetrical as compared to the other isomers. This means that in the crystal lattice, molecules of the p-isomers are more closely packed as compared to the other isomers. As a result, it has a higher melting point and lower solubility as compared to ortho and meta isomers.
Haloarenes are less polar than haloalkanes and are insoluble in water. This is because of lack of hydrogen bonding. As a result, the attractive forces in haloarenes—water system remain less than the attractive forces in H20 molecules which are hydrogen bonded. Haloarenes are soluble in organic solvents of low polarity such as benzene, ether, chloroform, carbon tetrachioride etc.
How the following conversions can be carried out ?
(i) Propene to propan-1-ol
(ii) Ethanol to but-1-yne
(iii) 1-Bromopropane to 2-bromopropane
(iv) Toluene to benzyl alcohol
(v) Benzene to 4-bromonitrobenzene
(vi) Benzyl alcohol to 2-phenylethanoic acid
(vii) Ethanol to propanenitrite
(viii) Aniline to chlorobenzene
(ix) 2-Chlorobutane to 3, 4 – dimethylhexane
(x) 2-Methylpropene to 2-chloro-2-methylpropane
(xi) Ethyl chloride to propanoic acid
(xii) But-l-ene to n-butyl iodide
(xiii) 2-Chloropropane to propan-l-ol
(xiv) Isopropyl alcohol to iodoform
(xv) Chlorobenzene to p-nitrophenol
(xvi) 2-Bromopropane to 1-bromopropane
(xvii) Chloroethane to butane
(xviii) Benzene to diphenyl
(xix) tert-Butyl bromide to isobutyl bromide
(xx) Aniline to phenyl isocyanide.
The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in presence of alcoholic KOH, alkenes are major products. Explain. (Pb. Board 2009, Haryana Board 2013)
In aqueous medium i.e., water, KOH will be completely dissociated to give OH– ions. They being a strong nucleophile, will bring about the substitution of alkyl halides to form alcohols. At the same time, the OH” ions will be highly hydrated also. They will not be able to abstract a proton (H+) from the p-carbon atom to form alkenes. In other words, in aqueous medium, OH– ions will behave as weak base and elimination leading to alkenes will not be feasible.
In alcoholic KOH, the solution will also contain ethoxide ions (C2H5O–) in addition to OH– ions. They being a stronger base than OH– ions, will abstract a H+ ion from the β-carbon atom giving alkene as the product as a result of dehydrohalogenation.
Primary alkyl halide (a) C4H9Br was reacted with alcoholic KOH to give compound (b). Compound (b) was reacted with HBr to give (c) which was an isomer of (a). When (a) was reacted with sodium metal, it gave a compound (d) C8H18, that was different than the compound when n-butyl bromide was reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions.
The two primary alkyl bromides are possible from the molecular formula (a) C4H9Br. These are:
According to the available information, the isomer (I) does not represent the correct compound because this on reacting with sodium metal (Wurtz reaction) will give n-octane. (C8H18) which is not acutally formed
What happens when
(i) n-butyl chloride is treated with alcoholic KOH,
(ii) bromobenzene is treated with Mg in the presence of dry ether,
(iii) chlorobenzene is subjected to hydrolysis,
(iv) ethyl chloride is treated with (aq.) KOH,
(v) methyl bromide is treated with sodium in the presence of dry ether,
(vi) methyl chloride is treated with KCN ?
(i) But-l-ene is formed as the product as a result of dehydrohalogenation.
(ii) Phenyl magnesium bromide (Grignard reagent) is formed as a result of the reaction.
(iii) Chlorobenzene will not get hydrolysed on boiling with NaOH. No product will be formed.
(iv) Ethyl alcohol is formed as the product
(v) Ethane is formed as a result of Wurtz reaction
(vi) Methyl cyanide is formed.
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