Hsslive.co.in: Kerala Higher Secondary News, Plus Two Notes, Plus One Notes, Plus two study material, Higher Secondary Question Paper.

Saturday, June 18, 2022

BSEB Class 12 Chemistry Haloalkanes and Haloarenes Textbook Solutions PDF: Download Bihar Board STD 12th Chemistry Haloalkanes and Haloarenes Book Answers

BSEB Class 12 Chemistry Haloalkanes and Haloarenes Textbook Solutions PDF: Download Bihar Board STD 12th Chemistry Haloalkanes and Haloarenes Book Answers
BSEB Class 12 Chemistry Haloalkanes and Haloarenes Textbook Solutions PDF: Download Bihar Board STD 12th Chemistry Haloalkanes and Haloarenes Book Answers


BSEB Class 12th Chemistry Haloalkanes and Haloarenes Textbooks Solutions and answers for students are now available in pdf format. Bihar Board Class 12th Chemistry Haloalkanes and Haloarenes Book answers and solutions are one of the most important study materials for any student. The Bihar Board Class 12th Chemistry Haloalkanes and Haloarenes books are published by the Bihar Board Publishers. These Bihar Board Class 12th Chemistry Haloalkanes and Haloarenes textbooks are prepared by a group of expert faculty members. Students can download these BSEB STD 12th Chemistry Haloalkanes and Haloarenes book solutions pdf online from this page.

Bihar Board Class 12th Chemistry Haloalkanes and Haloarenes Textbooks Solutions PDF

Bihar Board STD 12th Chemistry Haloalkanes and Haloarenes Books Solutions with Answers are prepared and published by the Bihar Board Publishers. It is an autonomous organization to advise and assist qualitative improvements in school education. If you are in search of BSEB Class 12th Chemistry Haloalkanes and Haloarenes Books Answers Solutions, then you are in the right place. Here is a complete hub of Bihar Board Class 12th Chemistry Haloalkanes and Haloarenes solutions that are available here for free PDF downloads to help students for their adequate preparation. You can find all the subjects of Bihar Board STD 12th Chemistry Haloalkanes and Haloarenes Textbooks. These Bihar Board Class 12th Chemistry Haloalkanes and Haloarenes Textbooks Solutions English PDF will be helpful for effective education, and a maximum number of questions in exams are chosen from Bihar Board.

Bihar Board Class 12th Chemistry Haloalkanes and Haloarenes Books Solutions

Board BSEB
Materials Textbook Solutions/Guide
Format DOC/PDF
Class 12th
Subject Chemistry Haloalkanes and Haloarenes
Chapters All
Provider Hsslive


How to download Bihar Board Class 12th Chemistry Haloalkanes and Haloarenes Textbook Solutions Answers PDF Online?

  1. Visit our website - Hsslive
  2. Click on the Bihar Board Class 12th Chemistry Haloalkanes and Haloarenes Answers.
  3. Look for your Bihar Board STD 12th Chemistry Haloalkanes and Haloarenes Textbooks PDF.
  4. Now download or read the Bihar Board Class 12th Chemistry Haloalkanes and Haloarenes Textbook Solutions for PDF Free.


BSEB Class 12th Chemistry Haloalkanes and Haloarenes Textbooks Solutions with Answer PDF Download

Find below the list of all BSEB Class 12th Chemistry Haloalkanes and Haloarenes Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:

Bihar Board Class 12 Chemistry Haloalkanes and Haloarenes Intext Questions and Answers

Question 1.
Write structures of the following compounds:
(i) 2-Chloro-3-methyl pentane
(ii) l-Chloro-4-ethyl cyclohexane
(iii) 4-tert-Butyl-3-iodoheptane
(iv) 1, 4-Dibromobut-2-ene
(v) l-Bromo-4-sec-butyl-2-methylbenzene.
Answer:

Question 2.
Why is sulphuric acid not used during the reaction of alcohols with KI.
Answer:
Sulphuric acid (H2SO4) cannot be used along with KI in the conversion of an alcohol to an alkyl iodide as it converts KI to corresponding HI and then oxidizes it to I2.
(i) KI + H2SO4 → KHSO4 + HI
(ii) H2SO4 → H2O + SO2 + O.
2HI + O → H2O +I2

Question 3.
Write structures of different dihalogen derivatives of propane.
Answer:

(ii)ClCH2CH2CH2Cl
(iii) Cl2CHCH2CH3
(iv) CH3CCl2CH3

Question 4
Among the isomeric alkanes of molecular formula C5H12, identify the one that on photochemical chlorination yields:
(i) A single monochloride,
(ii) Three isomeric monochlorides,
(iii) Four isomeric monochlorides.
Answer:

All the hydrogen atoms of 4 methyl groups are equivalent and replacement of any hydrogen will give the same product.

The equivalent hydrogens are grouped as a, b and c. The replacement of equivalent hydrogens will give the same product. Thus, three isomeric products possible.

Similarly, the equivalent hydrogens are grouped as a, b, c and d. Thus, four isomeric products are possible.

Question 5.
Draw the structures of major moon halo products in each of the following reactions:


Answer:

Question 6.
Arrange each set of compounds in order of increasing boiling points.
(i) Bromomet-hane, Bromoform, Chloromethane, Dibromo- methane.
(ii) 1-Chloropropane, Isopropyl chloride, 1-Chlorobutane.
Answer:
(i) Chloromethane, Bromomethane, Dibromomethane Bromoform. Boiling point increases with increase in molecular mass.
(ii) Isopropylchloride < 1-Chloropropane < 1-Chlorbutane. Isopropyl chloride is a branched-chain compound and hence has lowered. p. than l-Chloropropane.

Question 7.
Which alkyl halide from the following pairs would you expect to react more rapidly by SN2 mechanism? Explain your Answer.

Answer:
(i) CH3CH2CH2CH2Br. Being primary halide, there would not be any steric hindrance.

Secondary alkyl halide would react faster than tertiary halide,
 The presence of methyl group close to the halide group will increase the steric hindrance and decrease the rate of the reaction.

Question 8.
In the following pairs of halogen compounds, which compound undergoes faster SN1 reaction?

Answer:

Tertiary halides react faster than secondary halides because of the greater stability of test-carbocation.

Because of the greater stability of the secondary carbocation than primary.

Question 9.
Identify A, B, C, D, E, R and R’ in the following: –

Answer:

Bihar Board Class 12 Chemistry Haloalkanes and Haloarenes Text Book Questions and Answers

Question 1.
Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:
(i) (CH3)2 CHCH(Cl)CH3,
(ii) CH3CH2CH (CH3)CH(C2H5)Cl
(iii) CH3CH2C (CH3)2CH2I
(iv) (CH3)3 CCH2CH(Br) C6H5
(v) CH3CH(CH3)CH(Br)CH3
(vi) CH3C(C2H5)2CH2Br
(vii) CH3C(Cl)(C2H5)CH2CH3
(viii) CH3CH = C(Cl)CH2CH(CH3)2
(ix) CH3CH = CHC(Br) (CH3)2
(x) p-ClC6H4CH2CH(CH3)2
(xi) m-ClCH2C6H4CH2C(CH3)3
(xii) o-Br-C6H4CH(CH3)CH2CH3.
Answer:
(i) 2-chloro-3-methylbutane, 2° alkyl halide.
(ii) 3-Chloro-4-methyl hexane, 2° alkyl halide.
(iii) i-lodo-2,2-dimethylbutane, 1 alkyl halide.
(iv) l-Bromo-2,2-dimethyl-l-phenylbutane,2° benzylic halide.
(v) 2-Bromo-3-methylbutane/ 2° alkyl halide.
(vi) t-Bromo-2-ethyl-2-methyl butane, 1° alkyl halide.
(vii) 3-chloro-3-methylpentane, 3° alkyl halide.
(viii) 3-chloro-5-methyl-hex-2-ene, vinylic halide
(ix) 4-Bromo-4-methylpent-2-ene, allylic halide.
(x) l-(4-Chlorophenyl)-2-methylpropane, aryl halide.
(xi) l-chloromethyl-3-(2,2-dimethyl propyl) benzene, 1° benzylic halide.
(xii) l-Bromo-2-(l-methyl propyl) benzene, aryl halide.

Question 2.
Give the IUPAC names of the following compounds:
(i) CH3CH (Cl)CH (Br)CH3,
(ii) CHF2CBrClF,
(iii) ClCH2C ≡ CCH2Br,
(iv) (CCl3)3CCl,
(v) CH3C (p-ClC6H4)2CH(Br)CH3,
(vi) (CH3)3CCH = ClC6H4I-p
Answer:
(i) 2-Bromo-3-dilorobutane.
(ii) 1-Bromo-l-Chloro-l, 2,2-trifluoroethane.
(iii) l-Bromo-4-chlorobut-2-yne.
(iv) 2-Trichloromethyl-l, 1,1,2,3,3,3-heptachloropropane-
(v) 2,2-Bis (4-Chlorophenyl) butane.
(vi) l-Chloro-l-(4-iodophenyl)-3,3-dimethylbut-l-ene.

Question 3.
Write the structure of the following organic halogen compounds.
(i) 2-Chloro-3-methylpentane,
(ii) p-Bromochlorobenzene,
(iii) 1- Chloro-4-ethylcyclohexane,
(iv) 2-(2-Chlorophenyl)-l-iodooctane,
(v) Perfluorobenene, (vi) 4-tert-Butyl-3-iodoheptane,
(vii) l-Bromo-4- sec-butyl-2-methylbenene,
(viii) 1,4-Dibromobut-2-ene.
Answer:


(viii) Br-CH2CH=CH-CH2Br.

Question 4.
Which one of the following has highest dipole moment?
(a) CH2Cl2,
(b) CHCl3,
(C) CCl4
Answer:
(b) CHCl3.

Question 5.
A hydrocarbon C5H10 does not react with chlorine in dark but gives a single mono chloro compound C5H9 Cl in bright sunlight. Identify the hydrocarbon.
Answer:
C5H10 cannot be an alkene since it does not give addition reaction with Cl2 in dark. Therefore it must be a cyclic hydrocarbon. Moreover, it gives a single mono chloro compound C5H9 Cl with Cl2 in bright sunlight, therefore, it must be symmetrical, i.e., Cyclopentane 

Question 6.
Write the isomers of the compound having formula C4H9Br.
Answer:
(i) CH3-CH2-CH2-CH2Br:1- Bromobutane
2- Bromobutane

2-Bromobutane shows optical isomerism. One of them is dextro or d-form and the other is lalvo or 1-form.

Non-Superimposible image of the object,

Question 7.
Write the equations for the preparation of 1-iodobutane from (a) 1-butanol, (b) l-chlorobutanol, (c) but-l-ene.
Answer:
(a) Preparation of 1-iodobutane from 1-butanol

(b) From 1-chloro butane
CH3 -CH2-CH2,-CH2Cl + KI → CH3-CH2-CH2-CH2I + KCl
(c) From but-l-ene

Question 8.
What are ambident nucleophiles? Explain with an example.
Answer:
Ambident nucleophiles are the reagents which have two nucleophilic centres. Groups like  which can attack either from carbon or nitrogen. Actually, cyanide group is a hybrid of two contributing structures  and can act as a nucleophile in two different ways, i.e., linking through carbon atom resulting in the formation of alkyl cyanides and through nitrogen atom leading to the formation of isocyanide.

Question 9.
Which compound in each of the following pairs will react faster in SN2 reaction with OH?
(i) CH3Br or CH3I,
(ii) (CH3)3 CCl or CH3Cl.
Answer:
(i) CH3I
(ii) CH3Cl.

Question 10.
Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene.
(i) 1-Bromo-l-methylcyclohexane,
(ii) 2-Chloro-2-methylbutane,
(iii) 2,2,3-Trimethyl-3-bromopentane.
Answer:

Question 11.
How will you bring the following conversions?
(i) Ethanol to but-l-yne,
(ii) Ethane to bromoethene,
(iii) Propene to 1-nitro-propane,
(iv) Toluene to benzyl alcohol,
(v) Propene to propyne,
(vi) Ethanol to ethyl fluoride,
(vii) Bromomethane to propanone,
(viii) But-l-ene to but-2-ene,
(ix) 1-Chlorobutane to n-octane,
(x) Benzene to biphenyl.
Answer:
(i) Ethanol to but-l-yne

(ii) Ethane to bromoethene

(iii) Propene to 1-nitro-propane

(iv) Toluene to benzyl alcohol

(v) Propene to propyne

(vi) Ethanol to ethyl fluoride

(vii) Bromomethane topropanone

(viii)But-l-enetobut-2-ene

(ix) 1-Chlorobutane to n-octane

(x) Benzene to biphenyl

Question 12.
Explain why (i) the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride (ii) alkyl halides, though polar, are immiscible with water, (iii) Grignard’s reagents should be prepared under anhydrous conditions?
Answer:
(i) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride  because C attached to Cl in Chlorobenzene is sp2 hybridized and so more electronegative than C attached to Cl is cyclohexyl chloride which is sp3 hybridized and so less electronegative. Therefore C of C—Cl bond in cyclohexyl chloride is more willing to release electrons to chlorine and thus is more polar.

(ii) Alkyl halides though polar (2.05-2.15 D) are insoluble in water because they can neither form hydrogen bonds with water nor can they break the hydrogen bonds already existing between water molecules.

(iii) Grignard reagents are highly reactive and react with any source of proton to give hydrocarbons. Even water is sufficiently acidic to convert them to corresponding hydrocarbons.
R-Mg-X + H2O → R-H + Mg (OH) X
Therefore, Grignard reagents should be prepared under only anhydrous conditions like in ether.

Question 13.
Give the uses of Freon 12, DDT, Carbon tetrachloride and iodoform.
Answer:
Uses of Freon 12-It is widely used as:(i) a refrigerant (cooling agent) in refrigerators and air conditions.
(ii) a propellant in aerosols and foams (i.e, hair sprays, deodorants, shaving creams, cleansers, insecticides etc.)
Uses of DDT-It is a cheap, but powerful insecticide. It is widely used for sugarcane and fodder crops and to kill mosquitoes and other insects. Through its use malaria has virtually been minimised in India. However, its use has been stopped in several advanced countries of the world.

Uses of Carbon tetrachloride-

  • It is used as an industrial solvent for oils, fats, resins, lacquers etc. and also in dry-cleaning. —
  • as a fire extinguisher under the name of Pyrene.
  • in the industrial preparation of chloroform.
  • as a medicine for hookworms.

Uses of iodoform-

  1. It is used as an antiseptic for dressing wounds.
  2. It is used in the preparation of certain pharmaceuticals.

Question 14.
Write the structure of the major organic product in each of the following reactions:


(v) C6H5ONa+C2H5Cl →
(vi) CH3CH2CH2OH+SOCl2

(viii) CH3CH = C(CH3)2+HBr →
Answer:
(i) CH3 CH2CH2 I,
(ii) (CH3)2 C = CH2,
(iii) CH3 CH (OH) CH2CH3
(iv) CH3CH2CN,
(v) C6H5OC2H5,
(vi) CH3CH2CH2Cl,
(vii) CH3 CH2 CH2 CH2 CH2 Br,
(viii) CH3 CH2 CBr (CH3)2.

Question 15.
Write the mechanism of the following reaction:

Answer:

Question 16.
Arrange the compounds of each set in order of reactivity towards SN2 displacement:
(i) 2-Bromo-2-methyl butane, 1-Bromopentane, 2-Bromopentane
(ii) l-Bromo-3-methylbutane, 2-Bromo-2-methylb utane, 3-Bromo-2-methylbutane
(iii) 1-Bromobutane, l-Bromo-2,2-dimethylpropane l-Bromo-2- methylbutane < l-Bromo-3-methyl butane.
Answer:
(i) 2-Bromo-2-methylbutane < 2-Bromopentane < 1- Bromopentane.
(ii) 2-Bromo-2-methyl butane < 3-Bromo-2-methylbutane < 1-Bromo-3-methyl butane.
(iii) l-Bromo-2,2-dimethylpropane < l-Bromo-2-methylbutane < l-Bromo-3-methyl butane.

Question 18.
p-Dichldrobenzene has higher m.p. andsolubility toll those of o- and m-isomers. Discuss.
Answer:
p-Dichlorobehzene has higher m.p. and solubility than those of o- and m-isomers because p-isomer is more symmetrical and hence its molecular forces of attraction are stroriger and hence the p-isomer melts at a higher temperature and has comparatively higher solubility.

Question 19.
How the following conversions can be carried out?
(i) Propene to propah-l-ol
(ii) Ethanol to but-l-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 propanenitrate
(viii) Aniline to chlorobenzene
(ix) 2-chlorobutane to 3,4-dimethyl hexane
(x) 2-Methyl-l-propene to 2-chloro-2-methyl propane
(xi) Ethyl chloride to propanoic acid
(xii) But-l-ene to n-butyliodide
(xiii) 2-chloropropane to 1-propanol
(xiv) Isopropyl alcohol to iodoform
(xv) Chlorobenzene to p-nitrophenol
(xvi) 2-Bfombpropane to 1-bromopropane
(xvii) Chloroethane to butane
(xviii) Benzene to Diphenyl
(xix) tert-Butyl bromide to isobutyl bromide
(xx) Aniline to phenyl isocyanide.
Answer:
(i) Propene to propan-l-ol

(ii) Ethanol to but-l-yne

(v) Benzene to 4-bromonitrobenzene

(vi) Benzyl alcohol to 2-phenylethanoic acid

(vii) Ethanol to propane nitrile

(viii) Aniline to chlorobenzene

(ix) 2-Chlorobutane to 3,4-dimethyl hexane

(x) 2-Methyl-l-propene to 2-chloro-2-methyl-propane

(xi) Ethyl chloride to propanoic acid

(xii) But-l-ene ton-butyl iodide ‘

(xiii) 2-chloropropane to 1-propanol

(xiv) Isopropyl alcohol to iodoform

(xv) Chlorbenzene to p-nitrophenol

(xvi) 2-Bfombpropane to 1-bromopropane

(xvii) Chloroethane to butane

(xviii) Benzene to Diphenyl

(xix) tert-Butyl bromide to isobutyl bromide

(xx) Aniline to phenyl isocyanide.

Question 20.
The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols, but in the presence of alcoholic ] KOH alkenes are major products. Explain.
Answer:
In aqueous solution KOH is almost completely ionised to give OH ions which being a strong nucleophile brings about a; substitution reaction on alkyl chloride to form alcohols. Further in the aq. solution OH ions are highly solvated (hydrated). This solvation reduces the basic character of OH ions which, therefore, fail to abstract a hydrogen from the β-carbon of the alkyl chloride to form an alkene.

In contrast, an alcoholic solution of KOH contains alkoxide  ions which being a much stronger base than OH ions preferentially eliminates a molecule of HC1 from an alkyl halide to form alkenes.
R-Cl + KOH (aq) → R-OH + KCl [Substitution]
CH3CH2CH2Cl + KOH (ale.) → CH3CH = CH2 [Elimination].

Question 21.
Primary alkyl halide C4H9Br (a) reacted with alcoholic, KOH to give compound (b). Compound (b) reacted with HBr to give (c) which is an isomer of (a). When (a) was reacted with sodium metal it gives a compound (d), C8H18 which is different from the compound I when n-butyl bromide was reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions. s
Answer:
C4H9Br has two primary alkyl halides:

n-Butyl bromide gives a different product (C8H18) on treatment with sodium than that given by isobutyl bromide (C8H18).

Therefore (a) is isobutyl bromide. It is confirmed from the following reactions:

Evidently (c) is an isomer of (a).
Therefore compound (a) is isobutyl bromide and not n-Butyl bromide.

Question 22.
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?
Answer:
(i)

(ii) When bromobenzene is treated with Mg in the presence of dry ether, phenyl magnesium bromide (Grignard’s reagent) is formed.

(iii) When chlorobenzene is subject to hydrolysis with NaOH at 300°C and 200 atm. pressure, phenol is formed.

(iv) When ethyl chloride is treated with aq. KOH, ethyl alcohol is formed.

(v) When methyl bromide is treated with sodium in the presence of dry ether, ethane is formed. [Wurtz reaction]

(vi) When methyl chloride is treated with KCN, ethanenitrile is formed.

Bihar Board Class 12 Chemistry Haloalkanes and Haloarenes Additional Important Questions and Answers

Very Short Answer Type Questions

Question 1.
Name the alkyl halide which can be used to prepare methane and ethane in single steps.
Answer:
Methyl iodide (CH3I)

Question 2.
What happens when chlorobenzene is treated with soda mide in liquid ammonia. (W.B. JEE 2001)
Answer:

Question 3.
Iodoform gives precipitate with silver nitrate, but chloroform does not. Explain.
Answer:
C-I bond is much weaker than C-Cl bond. Therefore when CHI, is heated, it gives T ions which give a yellow ppt of Agl on reaction with AgNO3 solution. On the other hand C-Cl bond of CHCl3 does not break on heating to give Cl ions and hence ppt. of AgCl is not formed.

Question 4.
Which alkyl halide has the highest density and why?
Answer:
CH3I. Because of its smallest carbon content and heaviest halogen, i.e., I.

Question 5.
Arrange the following in order of their increasing reactivity towards sulphonation with fuming sulphuric acid:
Benzene, toluene, methoxybenzene, chlorobenzene.
Answer:
Chlorobenzene < benzeoe < toluene < methoxy benzene.

Question 6.
Which of the following compounds would show optical isomerism and why?
(i) 1-Bromobutane,
(ii) 2-Bromobutane.
Answer:
2-Bromobutane  because it contains a chiral carbon.

Question 7.
Arrange the following in order of their increasing reactivity in nucleophilic substitution reactions: CH3F, CH3I, CH3Br, CH3Cl.
Answer:
CH3F < CH3Cl < CH3Br < CH3I.

Question 8.
Under what conditions, 2-methyl propene can be converted into isobutyl bromide by hydrogen bromide?
Answer:
In the presence of peroxides.

Question 9.
Explain why thionyl chloride method is preferred for preparing alkyl chlorides from alcohols?
Answer:
Because the by-products of the reaction i.e,, SO2 and HCl being gassed escape into the atmosphere leaving alkyl chlorides in almost pure state.

Question 10.
Which compounds responds to iodoform test?
Answer:
Compounds containing CH3 CH(OH)- or CH3-C = O groups.

Question 11.
Name an aldehyde which will respond to iodoform test.
Answer:
Acetaldehyde, (CH3-CHO).

Question 12.
Give one chemical test to distinguish between C2H5Br and C6H5Br.
Answer:
Hydrolysis of C2H5Br with KOH followed by acidification with dil. HNO3 and subsequent treatment with AgNO3 gives light yellow, ppt. of AgBr whereas C2H5Br does not give this test.

Question 13.
What happens when chlorine is passed through boiling toluene in the presence of sunlight?
Answer:
Benzyl chloride  is formed.

Question 14.
For a given alkyl group, what is the trend in the b.pts of the following halides RBr, RC1, RF, RI?
Answer:
B. Pts. of RI > RBr > RC1 > RF.

Question 15.
Among the isomeric dichlorobenzenes, p-isomer has the highest melting point. Why?
Answer:
Because of the symmetry of its structures .

Question 16.
Which of the SN1 and SN2 reactions proceeds with complete stereochemical inversion of configuration and which proceeds with racemisation?
Answer:
SN2 reactions proceed with complete stereochemical inversion of configuration and SN1 reaction proceed with racemisation.

Question 17.
Write a chemical reaction to illustrate Saytzeff’s rule.
Answer:

Question 18.
Arrange the following in order of increasing ease towards nucleophilic substitution. 4-nitrochlorobenzene, Chlorobenzene, 2,4, 6-trinitrochlorobenzene, 2,4-dinitrochlorobenzene.
Answer:
Chlorobenzene < 4-nitrochlorobenzene < 2, 4, dinitrochlorobenzene < 2,4 6-trinitrochlorobenzene.

Question 19.
A hydrocarbon C5H12 gives only one monochlorination product. Identify the hydrocarbon.
Answer:
It must be neopentane (CH3)4C in which all the hydrogen atoms attached to 4 methyl groups are identical.

Question 20.
An alkyl halide having molecular formula C4H9Cl is optically active. What is its structure?
Answer:
2-Chlorobutane. CH3-CHCl-CH2-CH3.

Question 21.
What type of isomerism is shown by 1,2-dichloroethene?
Answer:
Cis-trans or geometrical isomerism.

Question 22.
Out of chloromethane and chlorobenzene which is more reactive towards nucleophilic substitution reactions.
Answer:
Chloromethane (CH3Cl) because it is an alkyl halide.

Question 23.
Which is a better nucleophile, a bromide ion or an iodine ion?
Answer:
Iodide ion (I ).

Question 24.
Write the structure of the main product obtained by the action of cone. H2SO4 on 2-methylbutan-l-ol.
Answer:
2-Methylbut-2-ene. The 1° carbocation initially formed img being less stable rearranges to more stable 2° carbocation img (CH3)2 through 1, 2-hydride shift during dehydration to give more stable alkene in keeping with Saytzeff rule.

Question 25.
A hydrocarbon C5H10 does not react with chlorine in dark but gives a single mono chloro compound C5H9Cl in bright sunlight. Identify the hydrocarbon.
Answer:
C5H10 is a saturated compound and all hydrogen atoms are same. Therefore C5H10 is a symmetrical molecule having cyclic structure: Cyclopentane.

Question 26.
What is the name of polyhaloalkaneTesponsible for depletion ozone layer?
Answer:
It is Freon-12 (dichlorodifluoro methane (CCl2F2).

Question 27.
Wurtz reaction fails in the case of tert-alkyl halides. Explain.
Answer:
Only 1°, 2° alkyl halides respond to Wurtz reaction to form higher alkanes, while 3° alkyl halides prefer to undergo dehydrohalogenation to form alkenes.

Question 28.
Why alkyl halides are insoluble in wafers?
Answer:
Because they do not form hydrogen bonds with water.

Question 29.
What name would you give to the formation of phenol from chlorobenzene reaction?
Answer:
Nucleophilic Substitution reaction.

Question 30.
Which insecticide is used to eradicate malaria?
Answer:
DDT.

Short-Answer Type Questions

Question 1.
p-Dichlorobenzene has higher melting point and lower solubility than those of o- and m-isomers. Explain why?
Answer:
Tire para isomer is more symmetrical and fits closely in the crystal lattice and thus has stronger intermolecular forces of attraction than o- and m- isomers. Since during melting or dissolution, the crystal lattice breaks, therefore larger amount of heat is required to melt or dissolve the Midsomer In other words, its melting point is higher and solubility lower.

Question 2.
Haloarenes are insoluble in water but soluble in benzene.
Answer:
Haloarenes are insoluble in water because neither they can form H-bonds with water nor can they break the H-bonds already existing in water. (Like dissolves like). Due to larger hydrocarbon part in haloarenes, they are soluble in organic solvents like benzene, petroleum ether etc.

Question 3.
Explain why alkyl halides are generally not prepared in the laboratory by free radical halogenation of alkanes?
Answer:
Free-radical halogenation is not a suitable method for laboratory synthesis of alkyl halides because of the following reasons :
(i) It gives a mixture of isomeric mono halogenated products where boiling points are so close that they cannot be easily separated from each other.

(ii) Polyhalogenation may also occur to some extent thereby making the mixture mere complex and hence more difficult to separate.

Question 4.
An Alkyl naide, X, of formula C6H13Cl treatment with potassium tert butoxide gives two isomeric alkenes Y and Z (C6H12). Both alkenes on hydrogenation give 2,3 dimethylbutane. Predict the structures of X, Y and Z.
Answer:
Since the two isomeric alkenes Y and Z (C6H12) on catalytical hydrogenation give the same 2, 3 dimethyl butane, therefore, Y and Z must differ in the position of the double bond, i.e.,

If this is so, then the alkyl halide X (C6H13Cl) must have a H-atom on either side of Cl atom, i.e., X must be 2-chloride-2,3-dimethyl butane

Question 5.
Optically active 2-iodobutane on treatment with Nal in acetone gives a product which does not show a optical activity. Explain.
Answer:
Optically active 2-iodobutane as treatment with Nal in acetone undergoes racemisation and hence the product does not show optical activity as explained :

I undergoes Walden Inversion (SN2) to give 2-iodobutane (II) which is the enantiomer of I. No II undergoes Walden Inversion to give enantiomer I. As a result of this two Walden Inversion, a 50:50 mixture of 1 and II is obtained. In other words active 2-iodobutane undergoes racemization.

Question 6.
Give the structures of the major organic products from 3- ethyl pent-2-ene under each of the following conditions:
(a) HBr in presence of peroxide
(b) Br2/H2O.
Answer:

Question 7.
Predict the main product in each of the following reactions:

Answer:
CCl3 is m-directing

Question 8.
Give the names of the following compounds.

Answer:
(i) 3-Chloro-5-Fluoro-3, 5-dim ethyiheptane.
(ii) 3-Bromo-5-chloro-3, 5-diniethyiheptarie.

Question 9.
Out of the various possible isomer of C7H7Cl containing a benzene ring, suggest the structure with the weakest C—Cl bond.
Answer:
Four isomers are possible

The C-Cl bond in o, m- and p-chlorotoluene has some double-bond character due to resonance. In contrast, in benzyl chloride, the C- Cl bond is a pure single bond. Hence out of 4 isomers the C-Cl bond in benzyl chloride is the weakest.

Question 10.
Account for the following: Haloalkanes undergo nucleophilic substitution reactions whereas haloarenes undergo electrophilic substitutions.
Answer:
Haloalkanes are more polar than haloarenes. Thus the C atom carrying halogen in haloalkanes is more electrons. Haloalkanes do not possess any benzene ring.

Long Answer Type Questions

Question 1.
(a) Draw the structures of all eight isomers that have the molecular formula C5H11Br. Name each isomer according to IUPAC system and classify them as primary, secondary or tertiary bromide. Point out if anyone is optically active.
(b) Amongst the aromatic compounds having the molecular formula C7H7Cl, how many isomers are possible. Write them structures and IUPAC names.
Answer:
(a) The structures and IUPAC names of all 8 isomers are given.
(i) CH3-CH2-CH2-CH2-CH2Br [1-Bromopentane] (1°)



The 2nd C atom is chiral and will show optical activity.




2nd C atom is chiral and this isomer will show optical activity.

(b) Four isomers are possible. Their structures and IUPAC names are given below:

Question 2.
(a) How will you prepare Bromoethane from (i) ethanol, (ii) ethene. How will you convert it into (i) n-Butane, (ii) ethyl acetate, (iii) ethene, (iv) Diethyl ether
(b) How will you prepare chlorobenzene from (i) Benzene, (ii) benzene diazonium chloride. How will you convert it into (i) Phenol, (ii) Aniline, (iii) Benzonitrile, (iv) o- and p-chlorotoluene.
Answer:
(a)
(i) Bromoethane from ethanol

(ii) Bromo ethane from ethene:

(i) Conversion of Bromoethane into n-Butane

(ii) Conversion of Bromoethane into ethyl acetate

(iii) Conversion of Bromoethane into ethene

(iv) Conversion of Bromoethane into Diethyl ether


(b)
(i) chlorobenzene from benzene:

(ii) chlorobenzene from benzene diazonium chloride.

(i) Conversion of chlorobenzene into phenol

(ii) Conversion of chlorobenzene into aniline

(iii) Conversion of chiorobenzene into benzonitrile

(iv) o- and p- Cholorotoluene

Question 3.
(a) Explain why Grignard reagent cannot be prepared from Br CH2C ≡ CH?
(b) When phenol reacts with phosphorus pentachloride, minor amount of chlorobenzene is formed. What is the major product? Write down its structure.
(c) If carbon has rectangular planar geometry, how many isomers are possible for CH2Cl2, (ii) If it has square planar geometry how many isomers for H2Cl2 are possible.
Answer:
(a) Initially Mg reacts with Br CH2C ≡ CH to produce the corresponding Grignard reagent (I). Since Br CH2C ≡ CH has acidic acetylenic hydrogen, it immediately reacts with the Grignard reagent (I) to produce (II) and propyne.

These two reactions proceed in tandem till the whole Br CH2C ≡ CH is consumed. Therefore, Grignard reagent cannot be prepared from it.

(b) Because of resonance C-OH bond in phenols is much stronger than C-OH bond in alcohols and hence cannot be displaced by Cl. Instead phenol reacts as nucleophile and brings about a nucleophilic hydrolysis gives triphenylphosphate.

(c) If carbon has rectangular planar geometry then CH2Cl2 will have the following three stereoisomers:

And if carbon has square planar geometry, then CH2Cl2 will have the following two stereoisomers:

Question 4.
(a) Predict the order of reactivity of the following compounds in SN1 reactions:

(b) Predict the order of reactivity of the four isomeric bromobutanes in SN1and SN2 reactions:

(c) Arrange the following halides in the decreasing order of reactivity. CH3CH2Cl (I), CH2 = CH CHClCH3 (II) and CH3CH2 CH Cl CH3 (III).
Answer:
(a) Order of reactivity of towards SN1 reactions:
The first compound is 2° alkyl halide, while all others are tertiary alkyl halides. Since tertiary alkyl halides are more reactive than 2° alkyl halides in SN1 reactions, therefore, first compound is least reactive. Further reactivity increases in the order : Chlorides < bromides < iodide. Titus the order of reactivity is:

(b) In SN1 reactions, the order of reactivity depends upon the stability of the intermediate carbocations. Therefore (CH3)3 C Br which gives a 3° to carbocation, i.e.,  is most reactive, CH3 CH2 CH (Br) CH3 which
gives a 2° carbocation, i.e.,  and hence is less reactive than (CH3)3 CBr. Out of the remaining two 1° alkyl halides, the carbocation (CH3)2CH CH2 is more stable than the carbocation
 -group and hence the alkyl bromide (CH3)2 CHCH2Br is more reactive than CH3CH2CH2CH2Br. Thus the overall increasing reactivity of the four isomeric bromobutanes towards SN1 reactions follows the order:
CH3 CH2CH2CH2 Br < (CH3)2 CH CH2 Br < CH3 CH2 CH (Br) CH3 < (CH3)3 CBr
The reactivity in SN2 reactions, however, follows the reverse order. CH3 CH2 CH2 CH2 Br > (CH3)2 CH CH2 Br > CH3 CH2 CH (Br) CH3 > (CH3)3 C Br.
Since the steric hindrance around the electrophilic carbon {i.e. a- carbon) increases in that order.

(c)
(i) In SN1 reactions, carbocations are the intermediates. Obviously more stable the carbocation, more reactive the alkyl halide. Since alkyl halide (II) gives an allylic carbocation which is stabilized by resonance. Therefore the alkyl halide (II) is the most reactive.

(ii) Out of alkyl halides (I) and (III), (III) gives a more stable 2° carbocation while gives a less stable 1° carbocation, therefore, alkyl halide (III) is more reactive than alkyl halide (I’).

Thus from the above discussion, it follows that the overall reactivity 1 reaction follows the sequence II > III > I.


BSEB Textbook Solutions PDF for Class 12th


Bihar Board Class 12th Chemistry Haloalkanes and Haloarenes Textbooks for Exam Preparations

Bihar Board Class 12th Chemistry Haloalkanes and Haloarenes Textbook Solutions can be of great help in your Bihar Board Class 12th Chemistry Haloalkanes and Haloarenes exam preparation. The BSEB STD 12th Chemistry Haloalkanes and Haloarenes Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 12th Chemistry Haloalkanes and Haloarenes Books State Board syllabus with maximum efficiency.

FAQs Regarding Bihar Board Class 12th Chemistry Haloalkanes and Haloarenes Textbook Solutions


How to get BSEB Class 12th Chemistry Haloalkanes and Haloarenes Textbook Answers??

Students can download the Bihar Board Class 12 Chemistry Haloalkanes and Haloarenes Answers PDF from the links provided above.

Can we get a Bihar Board Book PDF for all Classes?

Yes you can get Bihar Board Text Book PDF for all classes using the links provided in the above article.

Important Terms

Bihar Board Class 12th Chemistry Haloalkanes and Haloarenes, BSEB Class 12th Chemistry Haloalkanes and Haloarenes Textbooks, Bihar Board Class 12th Chemistry Haloalkanes and Haloarenes, Bihar Board Class 12th Chemistry Haloalkanes and Haloarenes Textbook solutions, BSEB Class 12th Chemistry Haloalkanes and Haloarenes Textbooks Solutions, Bihar Board STD 12th Chemistry Haloalkanes and Haloarenes, BSEB STD 12th Chemistry Haloalkanes and Haloarenes Textbooks, Bihar Board STD 12th Chemistry Haloalkanes and Haloarenes, Bihar Board STD 12th Chemistry Haloalkanes and Haloarenes Textbook solutions, BSEB STD 12th Chemistry Haloalkanes and Haloarenes Textbooks Solutions,
Share:

0 Comments:

Post a Comment

Plus Two (+2) Previous Year Question Papers

Plus Two (+2) Previous Year Chapter Wise Question Papers, Plus Two (+2) Physics Previous Year Chapter Wise Question Papers , Plus Two (+2) Chemistry Previous Year Chapter Wise Question Papers, Plus Two (+2) Maths Previous Year Chapter Wise Question Papers, Plus Two (+2) Zoology Previous Year Chapter Wise Question Papers, Plus Two (+2) Botany Previous Year Chapter Wise Question Papers, Plus Two (+2) Computer Science Previous Year Chapter Wise Question Papers, Plus Two (+2) Computer Application Previous Year Chapter Wise Question Papers, Plus Two (+2) Commerce Previous Year Chapter Wise Question Papers , Plus Two (+2) Humanities Previous Year Chapter Wise Question Papers , Plus Two (+2) Economics Previous Year Chapter Wise Question Papers , Plus Two (+2) History Previous Year Chapter Wise Question Papers , Plus Two (+2) Islamic History Previous Year Chapter Wise Question Papers, Plus Two (+2) Psychology Previous Year Chapter Wise Question Papers , Plus Two (+2) Sociology Previous Year Chapter Wise Question Papers , Plus Two (+2) Political Science Previous Year Chapter Wise Question Papers, Plus Two (+2) Geography Previous Year Chapter Wise Question Papers, Plus Two (+2) Accountancy Previous Year Chapter Wise Question Papers, Plus Two (+2) Business Studies Previous Year Chapter Wise Question Papers, Plus Two (+2) English Previous Year Chapter Wise Question Papers , Plus Two (+2) Hindi Previous Year Chapter Wise Question Papers, Plus Two (+2) Arabic Previous Year Chapter Wise Question Papers, Plus Two (+2) Kaithang Previous Year Chapter Wise Question Papers , Plus Two (+2) Malayalam Previous Year Chapter Wise Question Papers

Plus One (+1) Previous Year Question Papers

Plus One (+1) Previous Year Chapter Wise Question Papers, Plus One (+1) Physics Previous Year Chapter Wise Question Papers , Plus One (+1) Chemistry Previous Year Chapter Wise Question Papers, Plus One (+1) Maths Previous Year Chapter Wise Question Papers, Plus One (+1) Zoology Previous Year Chapter Wise Question Papers , Plus One (+1) Botany Previous Year Chapter Wise Question Papers, Plus One (+1) Computer Science Previous Year Chapter Wise Question Papers, Plus One (+1) Computer Application Previous Year Chapter Wise Question Papers, Plus One (+1) Commerce Previous Year Chapter Wise Question Papers , Plus One (+1) Humanities Previous Year Chapter Wise Question Papers , Plus One (+1) Economics Previous Year Chapter Wise Question Papers , Plus One (+1) History Previous Year Chapter Wise Question Papers , Plus One (+1) Islamic History Previous Year Chapter Wise Question Papers, Plus One (+1) Psychology Previous Year Chapter Wise Question Papers , Plus One (+1) Sociology Previous Year Chapter Wise Question Papers , Plus One (+1) Political Science Previous Year Chapter Wise Question Papers, Plus One (+1) Geography Previous Year Chapter Wise Question Papers , Plus One (+1) Accountancy Previous Year Chapter Wise Question Papers, Plus One (+1) Business Studies Previous Year Chapter Wise Question Papers, Plus One (+1) English Previous Year Chapter Wise Question Papers , Plus One (+1) Hindi Previous Year Chapter Wise Question Papers, Plus One (+1) Arabic Previous Year Chapter Wise Question Papers, Plus One (+1) Kaithang Previous Year Chapter Wise Question Papers , Plus One (+1) Malayalam Previous Year Chapter Wise Question Papers
Copyright © HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board About | Contact | Privacy Policy