(iv) benzene Q1

Name the following compounds according to IUPAC system. (i) (CH3)3CCH2CH(Br)C6H5 (ii) CH3C(C2H5)2CH2Br (iii) CH3CH=CHC(Br)(CH3)2 (iv) o-Br-C6H4CH(CH3)CH2CH3 (v) m-ClCH2C6H4CH2C(CH3)3 (vi) CH3C(p-ClC6H4)2CH(Br)CH3 (vii) (CH3)3CCH=CClC6H4I-p

(v)

1−Chloromethyl−3−(2, 2−dimethylpropyl)

1−Bromo−2−(1−methylpropyl) benzene

Answer

(vi) (i)

2−Bromo−3, 3−bis(4−chlorophenyl) butane

1−Bromo−3, 3−dimethyl−1−phenylbutane

(vii) 1−chloro−1−(4−iodophenyl)−3,3−dimethylbut−1−ene (ii)

1−Bromo−2−ethyl−2−methylbutane

Q2

Write the structures and IUPAC name of the following organic halogen compounds (i)isobutyl bromide

(ii)tertiarybutyl bromide (i) 2−Bromo−2−methylpropane (iii)

4−Bromo−4−methylpent−2−ene

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Q4

(ii)

1−Bromo−2−methylpropane

Write the structures of the following organic halogen compounds (i) 2-(2-Chlorophenyl)-1-iodooctane (ii) 1-Bromo-4-sec-butyl-2-methylbenzene

Answer

Q3 Name the following compounds according to IUPAC system. (i)

(i) (ii) (ii)

Q5

Write IUPAC names of the following compounds:

(iii) (i)

(iv) (v)

(ii) (vi)

(iii)

Answer (i) 3-Chloromethyl-2-isopropylpentan-1-ol (ii) 2, 5 Dimethylhexane-1, 3-diol (iii) 3-Bromocyclohexanol (iv) Hex-1-en-3-ol (v) 2-Bromo-3-methylbut-2-en-1-ol (vi) 4-Bromo-3-methylpent-2-ene

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(iv)

(v)

(v)

Answer1

(vi) (vii) Answer

(i)

(i) 2, 2, 4-Trimethylpentan-3-ol (ii) 5-Ethylheptane-2, 4-diol (iii) 1-Methoxy-2-methylpropane (iv) 2, 6-Dimethylphenol (v) 2-Ethoxybutane (vi) 1-Phenoxyheptane (vii) Ethoxybenzene

Q1

(ii)

Draw the structures of major monohalo products in each of the following reactions:

(iii)

(iv) (i)

(ii) (v)

(iii)

(iv)

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(vi)

(vi)

Q2

Identify A, B, C, D, E, R and R1 in the following:

When an alkyl halide is treated with Na in the presence of ether, a hydrocarbon containing double the number of carbon atoms as present in the original halide is obtained as product. This is known as Wurtz reaction. Therefore, the halide, R1 −X, is Therefore, compound D isAnd, compound E is

Answer2

Since D of D2O gets attached to the carbon atom to which MgBr is attached, C is Q3

Write the products of the following reactions:

Therefore, the compound R − Br is

Answer 3

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Q4

Predict the products of the following reactions: (i)

(ii)

Q5 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-1-methylcyclohexane (ii) 2-Chloro-2-methylbutane (iii) 2,2,3-Trimethyl-3-bromopentane.

(iii)

Answer 5 (iv) Answer 4

(i) 1−bromo−1−methylcyclohexane In the given compound, all β-hydrogen atoms are equivalent. Thus, dehydrohalogenation of this compound gives only one alkene.

(i)

(ii)

(ii) (iii) In the given compound, there are two different sets of equivalent β-hydrogen atoms labelled as a and b. Thus, dehydrohalogenation of the compound yields two alkenes.

(iii)

(iv)

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Saytzeff’s rule implies that in dehydrohalogenation reactions, the alkene having a greater number of alkyl groups attached to a doubly bonded carbon atoms is preferably produced.

According to Saytzeff’s rule, in dehydrohalogenation reactions, the alkene having a greater number of alkyl groups attached to the doubly bonded carbon atom is preferably formed.

Therefore, alkene (I) i.e., 2-methylbut-2-ene is the major product in this reaction. (iii) 1 How will you bring about the following conversions?

2,2,3-Trimethyl-3-bromopentane In the given compound, there are two different sets of equivalent βhydrogen atoms labelled as a and b. Thus, dehydrohalogenation of the compound yields two alkenes.

(i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) (i)

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Ethanol to but-1-yne Ethane to bromoethene Propene to 1-nitropropane Toluene to benzyl alcohol Propene to propyne Ethanol to ethyl fluoride Bromomethane to propanone But-1-ene to but-2-ene 1-Chlorobutane to n-octane Benzene to biphenyl.

(v) (ii)

(vi)

(iii)

(vii)

(iv)

(viii)

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(xiii) 2-Chloropropane to 1-propanol (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 phenylisocyanide Answer (i) (ix)

(x) (ii)

2

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

(iii)

(iv) Toluene to benzyl alcohol (v) Benzene to 4-bromonitrobenzene (vi) Benzyl alcohol to 2-phenylethanoic acid (vii) Ethanol to propanenitrile (viii) Aniline to chlorobenzene (ix) 2-Chlorobutane to 3, 4-dimethylhexane (x) 2-Methyl-1-propene to 2-chloro-2-methylpropane (xi) Ethyl chloride to propanoic acid (xii) But-1-ene to n-butyliodide

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(iv)

(v) (ix)

(vi)

(x)

(xi)

(vii)

(xii)

(viii)

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(xvi) (xiii)

(xiv)

(xvii)

(xviii) (xv)

(xix)

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(iii) When ethyl magnesium chloride is treated with methanal, an adduct is the produced which gives propan-1-ol on hydrolysis.

(xx)

Q3

How are the following conversions carried out? (i) Propene → Propan-2-ol (ii)Benzyl chloride → Benzyl alcohol (iii) Ethyl magnesium chloride → Propan-1-ol. (iv) Methyl magnesium bromide → 2-Methylpropan-2-ol.

Answer (i) If propene is allowed to react with water in the presence of an acid as a catalyst, then propan-2-ol is obtained.

(ii) If benzyl chloride is treated with NaOH (followed by acidification) then benzyl alcohol is produced.

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(iv) When methyl magnesium bromide is treated with propane, an adduct is the product which gives 2-methylpropane-2-ol on hydrolysis.

note :- always write few lines about reaction conditions and remarkable point 1 Finkelstein Reaction Alkyl iodides are often prepared by the reaction of alkyl chlorides/ bromides with NaI in dry acetone. This reaction is known as 4 Fittig reaction :

Finkelstein

reaction. 2 Swarts Reaction Preparation of alkyl fluoride by reaction of metallic fluoride such as AgF, Hg2 F2, CoF2, or SbF3 with alkyl halides

3 Friedel-Crafts reactions [1]Friedel−Crafts Alkylation

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5 Wurtz-Fittig reaction:-

6 Sandmeyer’s reaction Replacement of group of benzene diazonium chloride by halide or cyanide ion is called Sandmeyer reaction

[ii]Friedel-Crafts Acylation

12

reactions of aryl halides with sodium

DPS JODHPUR

Q2

Q1

A hydrocarbon C5H10 does not react with chlorine in dark but gives a single monochloro compound C5H9Cl in bright sunlight. Identify the hydrocarbon.

Answer hydrocarbon with the molecular formula, C5H10 belongs to the group with a general molecular formula C nH2n. Therefore, it may either be an alkene or a cycloalkane.

Primary alkyl halide C4H9Br (a) reacted with alcoholic KOH to give compound (b).Compound (b) is reacted with HBr to give (c) which is an isomer of (a). When (a) is reacted with sodium metal it gives compound (d), C8H18 which is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions. Answer There are two primary alkyl halides having the formula, C4H9Br. They are n − bulyl bromide and isobutyl bromide.

Since hydrocarbon does not react with chlorine in the dark, it cannot be an alkene. Thus, it should be a cycloalkane. Further, the hydrocarbon gives a single monochloro compound, C5H9Cl by reacting with chlorine in bright sunlight. Since a single monochloro compound is formed, the hydrocarbon must contain H−atoms that are all equivalent. Also, as all H−atoms of a cycloalkane are equivalent, the hydrocarbon must be a cycloalkane. Hence, the said compound is cyclopentane.

Therefore, compound (a) is either n−butyl bromide or isobutyl bromide. Now, compound (a) reacts with Na metal to give compound (b) of molecular formula, C8H18, which is different from the compound formed when n−butyl bromide reacts with Na metal. Hence, compound (a) must be isobutyl bromide.

Cyclopentane (C5H10) The reactions involved in the question are:

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Step 1 – removal of leaving group and formation of carbocation

Step 2- attack of nucleophile

Thus, compound (d) is 2, 5−dimethylhexane. It is given that compound (a) reacts with alcoholic KOH to give compound (b). Hence, compound (b) is 2−methylpropene.

The increasing order of reactivity for SN1 is1° halide < 2° halide < 3° halide Reason − Greater the stability of carbocation, more easily the alkyl halide is formed and hence, faster is the reaction rate. The increasing order of stability of carbocation is 1° < 2° < 3°. Since 1° halide forms 1° carbocation, 2° halide forms 2° carbocation , and 3°

halide forms 3° carbocation. Also, compound (b) reacts with HBr to give compound (c) which is an isomer of (a). Hence, compound (c) is 2−bromo−2−methylpropane.

Allylic and benzylic halides are very reactive towards S N1 reaction because of stabilisation of their carbocations through resonance.

2 (SN2) bimolecular Substitution nucleophilic

1 SN1 unimolecular Nucleophilic Substitution (SN1)

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Inversion of configuration takes place. The increasing order of reactivity FOR SN2 is

3° halide < 2° halide < 1° halide For both SN1 and SN 2 reaction, the order of reactivity is R−F << R−Cl < R−Br < R−I Q.1 write the mechanism of SN2 with the following example

A p-Dichlorobenzene is more symmetrical than o-and misomers. For this reason, it fits more closely than o-and m-isomers in the crystal lattice. Therefore, more energy is required to break the crystal lattice of p-dichlorobenzene. As a result, p-dichlorobenzene has a higher melting point and lower solubility than o-and m-isomers. Q2 Why C-Cl bond is more polar than C-F bomnd A2

higer bond length of C-Cl

Q3

Explain why (i) the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride?

The given reaction is:

(ii) alkyl halides, though polar, are immiscible with water? The given reaction is an SN2 reaction. In this reaction, CN− acts as the nucleophile and attacks the carbon atom to which Br is attached. CN − ion is an ambident nucleophile and can attack through both C and N. In this case, it attacks through the C-atom.

Q1 p-Dichlorobenzene has higher m.p. and lower solubility than those of o- and m-isomers. Discuss.

(iii) Grignard reagents should be prepared under anhydrous conditions? A3 (i) C-Cl bond is shorter in chlorobenzene because of partial double bond character than C−Cl bond In cyclohexyl chloride

In chlorobenzene, the Cl-atom is linked to a sp2 hybridized carbon atom. In cyclohexyl chloride, the Cl-atom is linked to a sp3 hybridized carbon atom. Now, sp2 hybridized carbon has more s-character than sp3 hybridized carbon atom. Therefore, the former is more electronegative than the latter. (ii) To be miscible with water, the solute-water force of attraction must be stronger than the solute-solute and water-water forces of attraction. Alkyl halides are not capable of making H-bond with the water molecules. Hence, alkyl halides (though polar) are immiscible with water.

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(iii) Grignard reagents are very reactive. In the presence of moisture, they react to give alkanes.

(i)

(ii) Therefore, Grignard reagents should be prepared under anhydrous conditions. Q4 Out of C6H5CH2Cl and C6H5CHClC6H5, which is more easily hydrolysed by aqueous KOH?

A4 Hydrolysis by aqueous KOH proceeds through the formation of carbocation. If carbocation is stable, then the compound is easily hydrolyzed by aqueous KOH. Now, while

(iii)

A5 (i) 2-bromobutane is a 2° alkylhalide whereas 1-bromobutane is a 1° alkyl halide. The approaching of nucleophile is more hindered in 2-bromobutane than in 1-bromobutane. Therefore, 1-bromobutane reacts more rapidly than 2-bromobutane by an SN2 mechanism.

forms 1°-carbocation,

forms 2°-carbocation, which is more stable

than 1°-carbocation. Hence, easily than

is hydrolyzed more

(ii)

by aqueous KOH.

Q5 Which alkyl halide from the following pairs would you expect to react more rapidly by an SN2 mechanism? Explain your answer. (iii)

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Q6 In the following pairs of halogen compounds, which compound undergoes faster SN1 reaction?

not through nitrogen atom since C—C bond is more stable than C—N bond. However, AgCN is mainly covalent in nature and nitrogen is free to donate electron pair forming isocyanide as the main product.

(i)

Q8 Which compound in each of the following pairs will react faster in SN2 reaction with OH−?

(ii)

(i) CH3Br or CH3I

A6

(ii) (CH3)3CCl or CH3Cl

A8 (i) In the SN2 mechanism, the reactivity of halides for the same alkyl group increases in the order. This happens because as the size increases, the halide ion becomes a better leaving group. R−F << R−Cl < R−Br < R−I Therefore, CH3I will react faster than CH3Br in SN2 reactions with OH−.

(i)

The alkyl halide (I) is 3° while (II) is 2°. Therefore, (I) forms 3° carbocation while (II) forms 2° carbocation. (ii) there are no bulky substituents on the carbon atom bearing the leaving group in CH3Cl. Hence, CH3Cl reacts faster than (CH3)3CCl in SN2 reaction with OH−.

(ii) The alkyl halide (I) is 2° while (II) is 1°. 2° carbocation is more stable than 1° carbocation. Therefore, (I), 2−chloroheptane, undergoes faster SN1 reaction than (II), 1-chlorohexane.

Q9 Arrange the compounds of each set in order of reactivity towards SN2 displacement:

Q7 Haloalkanes react with KCN to form alkyl cyanides as main product while AgCN forms isocyanides as the chief product. Explain.

(i) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane (ii) 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 3-Bromo2- methylbutane (iii) 1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2methylbutane, 1-Bromo-3-methylbutane.

A7 KCN is predominantly ionic and provides cyanide ions in solution. Although both carbon and nitrogen atoms are in a position to donate electron pairs, the attack takes place mainly through carbon atom and

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A 9 (i)

(iii)

Due to the presence of substituents, hindrance to the approaching nucleophile increases in the following order. 1-Bromopentane < 2-bromopentane < 2-Bromo-2-methylbutane Hence, the increasing order of reactivity towards S N2 displacement is: 2-Bromo-2-methylbutane < 2-Bromopentane < 1-Bromopentane

1-Bromobutane < 1-Bromo-3-methylbutane < 1-Bromo-2-methylbutane

(ii)

< 1-Bromo-2, 2-dimethylpropane Hence, the increasing order of reactivity of the given compounds towards SN2 displacement is: 1-Bromo-2, 2-dimethylpropane < 1-Bromo-2-methylbutane < 1-Bromo-3methylbutane < 1-Bromobutane

Since steric hindrance in alkyl halides increases in the order of 1° < 2° < 3°, the increasing order of reactivity towards S N2 displacement is 3° < 2° < 1°. Hence, the given set of compounds can be arranged in the increasing order of their reactivity towards S N2 displacement as: 2-Bromo-2-methylbutane < 2-Bromo-3-methylbutane < 1-Bromo-3methylbutane [2-Bromo-3-methylbutane is incorrectly given in NCERT]

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Q10 Although chlorine is an electron withdrawing group, yet it is ortho-, para- directing in electrophilic aromatic substitution reactions. Why? A10 Chlorine withdraws electrons through inductive effect and releases electrons through resonance. Through inductive effect, chlorine destabilises the intermediate carbocation formed during the electrophilic substitution.

A12

Through resonance, halogen tends to stabilise the carbocation and the effect is more pronounced at ortho- and para- positions. The inductive effect is stronger than resonance and causes net electron withdrawal and thus causes net deactivation. The resonance effect Reactivity is thus controlled by the stronger inductive effectand orientation is controlled by resonance effect

Q11

In the following pairs of halogen compounds, which would undergo SN2 reaction faster?

A11 In 1st pair is primary halide and therefore undergoes SN2reaction faster. In 2nd As iodine is a better leaving group because of its large size, it will be released at a faster rate in the presence of incoming nucleophile. Q12

Predict the order of reactivity of the following compounds in S N1 andSN2reactions:

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Q23 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. A23

In an aqueous solution, KOH almost completely ionizes to give OH ions. OH− ion is a strong nucleophile, which leads the alkyl chloride to undergo a substitution reaction to form alcohol.

O n the other hand, an alcoholic solution encourage removal of a hydrogen from the β-carbon of the alkyl chloride and form an alkene by eliminating a molecule of HCl.

−

SHORT ANSWER TYPE 1. Aryl chlorides and bromides can be easily prepared by electrophilic substitution of arenes with chlorine and bromine respectively in the presence of Lewis acid catalysts. But why does preparation of aryl iodides requires presence of an oxidising agent? ans Iodination reactions are reversible in nature. To carry out the reaction in the forward direction, HI formed during iodination is removed by oxidation. HIO4 is used as an oxidising agent. 2. Out of o-and p-dibromobenzene which one has higher melting point and why? ans p-Dibromobenzene has higher melting point than its o-isomer. It is due to symmetry of p-isomer which fits in crystal lattice better than the oisomer. 3. Which of the compounds will react faster in SN1 reaction with the OH ion? CH — CH — Cl or C H — CH — Cl ans C6H5—CH2—Cl 3. Why iodoform has appreciable antiseptic property? ans Due to liberation of free iodine. 4. Haloarenes are less reactive than haloalkanes and haloalkenes. Explain.. ansC-X bond is partially doubly bonsded and resonance stabilized 5. Which of the following compounds (a) and (b) will not react with a mixture of NaBr and H SO . Explain why? 3

2

6

5

2

2

2

20

7. Classify the following compounds as primary, secondary and tertiary halides. (i) 1-Bromobut-2-ene (ii) 4-Bromopent-2-ene (iii) 2-Bromo-2-methylpropane ans (i) Primary (ii) Secondary (iii) Tertiary 8 . Compound ‘A’ with molecular formula C4H9Br is treated with aq. KOH solution. The rate of this reaction depends upon the concentration of the compound ‘A’ only. When another optically active isomer ‘B’ of this compound was treated with aq. KOH solution, the rate of reaction was found to be dependent on concentration of compound and KOH both.(i) Write down the structural formula of both compounds ‘A’ and ‘B’. (ii) Out of these two compounds, which one will be converted to the product with inverted configuration.

4

(a) CH CH CH OH ans b), C—O bond is more stable in resonance. 3

–

6. Draw other resonance structures related to the following structure and find out whether the functional group present in the molecule is ortho, para directing or meta directing. ans Ortho-para directing due to increase in the electron density at ortho and para positions. (For resonance structures

ans i) Compound A :

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(b) because of : DPS JODHPUR

Compound B

(ii) Compound ‘B’. 9. Write the structures and names of the compounds formed when compound ‘A’ with molecular formula, C H is treated with Cl in the presence of FeCl . ans. 7

8

2

3

13. A hydrocarbon of molecular mass 72 g mol gives a single monochloro derivative and two dichloro derivatives on photo chlorination. Give the structure of the hydrocarbon. ans. C5H12, pentane has molecular mass 72 g mol–1, i.e. the isomer of pentane which yields single monochloro derivative should have all the 12 hydrogens equivalent. –1

14. Name the alkene which will yield 1-chloro-1-methylcyclohexane by its reaction with HCl. Write the reactions involved. 10. Identify the products A and B formed in the following reaction : (a) CH —CH —CH==CH—CH +HCl → A + B ans 3

2

3

11. Which of the following compounds will have the highest melting point and why?

ansII, due to symmetry of para-positions; it fits into crystal lattice better than other isomers 12. Write down the structure and IUPAC name for neo pentylbromide. ans

; 1-Bromo-2,2-dimethylpropane

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15. Which of the following haloalkanes reacts with aqueous KOH most easily? Explain giving reason. (i) 1-Bromobutane(ii) 2-Bromobutane(iii) 2-Bromo-2methylpropane (iv) 2-Chlorobutane ans (iii); The tertiary carbocation formed in the reaction is stable.

16. Why can aryl halides not be prepared by reaction of phenol with HCl in the presence of ZnCl2? C—O bond in phenols is more stable due to resonance effect and it has double bond character, hence breaking of this bond is difficult. 17. Which of the following compounds would undergo SN1

reaction faster and why? ans (B) Undergoes SN1 reaction faster than (A) because in case of (B), the carbocation formed after the loss of Cl– is stabilised by resonance, whereas, no such stabilisation is possible in the carbocation obtained from (A). 18. Allyl chloride is hydrolysed more readily than npropyl chloride. Why? ans Allyl chloride shows high reactivity as the carbocation formed by hydrolysis is stabilised by resonance while no such stabilisation of carbocation exists in the case of n-propyl chloride. 19. Why is it necessary to avoid even traces of moisture during the use of a Grignard reagent? ans Grignard reagents are highly reactive and react with water to give corresponding hydrocarbons. RMgX + H2O → RH + Mg(OH)X 20. How do polar solvents help in the first step in S 1 mechanism? ans yes because of solvation of carbocation.] 21. Write a test to detect the presence of double bond in a molecule. (1) Unsaturation test with Br2 water (2) Bayer’s test.] 22. Aryl halides are extremely less reactive towards nucleophilic substitution. Predict and explain the order of reactivity of the following compounds towards nucleophilic substitution: N

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ans . III > II > I 23. Predict the major product formed when HCl is added to isobutylene. Explain the mechanism involved.

24. Cyanide ion acts as an ambident nucleophile. From which end it acts as a stronger nucleophile in aqueous medium? Give reason for your answer. ans It acts as a stronger nucleophile from the carbon end because it will lead to the formation of C–C bond which is more stable than the C–N bond. 25 Find which one is chiral 1-chloro-2-methyl butane or 2-chloro -2-methyl butane