TN Board 11th Chemistry Important Questions Chapter 14 Haloalkanes and Haloarenes

TN State Board 11th Chemistry Important Questions Chapter 14 Haloalkanes and Haloarenes

Question 1.
Give an example each for:

(i) a primary alkyl halide
Answer:

(ii) a secondary alkyl halide
Answer:

(iii) a tertiary alkyl halide
Answer:

Question 2.
The common names of certain alkyl halides are given below. Write their structure and their IUPAC names.

(i) n-propyl fluoride
Answer:
Structure:CH₃CH₂CH₂F
IUPAC Name: 1 -Fluoropropane

(ii) A-butyl chloride
Answer:
Structure: CH₃ — CH₂ — CH₂ — CH₂ — Cl
IUPAC Name: 1-Chlorobutane

(iii) sec-butyl chloride
Answer:
Structure:
IUPAC Name: 2 – Chlorobutane

(iv) tert-butyl chloride
Answer:
Structure:
IUPAC Name: 2 – Chloro – 2 – methyl propane

(v) neo pentyl bromide
Answer:
Structure:
IUPAC Name: 1 – Bromo – 2, 2 – dimethyl propane

(vi) vinyl chloride
Answer:
Structure: CH₂ = CH — Cl
IUPAC Name: Chloroethene

(vii) allyl bromide
Answer:
Structure: CH₂ = CH — CH₂ — Br
IUPAC Name: 3 – Bromopropene

Question 3.
Classify the following alkyl halides as primary, secondary and tertiary or vinyl / allyl halides.
(i) Bromoethane:
Answer:

(ii) 2 – chloropropane:
Answer:

(iii) 2-iodo-2-methyl propane:
Answer:

(iv) 2-fluoro propane:
Answer:

(v) 1 – chloro – 2 – methyl propane:
Answer:

(vi) 2 – chloro butane
Answer:

(vii) 1-bromo-2, 2, dimethyl propane:
Answer:

(viii) Chloro ethane:
Answer:

CH₂ = CH – Cl – Vinyl halide

(ix) 3-bromo propen e:
CH₂ = CH — CH₂ – Br – Allyl halide

Question 4.
Give examples for:

(i) a dichloroalkane
Answer:

1, 1, dichloroethane

(ii) a tribromoalkane.
Answer:

1, 2, 3 tribromopropane

Question 5.
Write the IUPAC names of the following:

(i)

Answer:
1, 2 dichloroethane

(ii) CH₃CHCl₂
Answer:
1, 1 dichloroethane

(iii)

Answer:
1 – bromo – 2, 3, dichlorobutane

(iv)

Answer:
2-bromo – 3 – chloro – 2, 4 dimethyl pentane

Question 6.
What is Lucas reagent? How does it react with tertiary butyl alcohol? Give equation for the reaction.
Answer:
Mixture of con.HCl and anhydrous ZnCl₂ is called Lucas reagent.

Question 7.
Complete the following equations.

(i) CH₃CH₂OH (Ethanol) + PCl₅ →
Answer:
CH₃CH₂Cl (Chloroethane) + POCl₃ + HCl

(ii) 3 CH₃CH₂OH (Ethanol) + PCl₃ →
Answer:
3 CH₃CH₂Cl (Chloroethane) + H₃PO₃

(iii)

Answer:
CH₃CH₂I (Iodo ethane) + NaBr

(iv)

Answer:
CH₃CH₂F (Flouoroethane) + AgBr

(v)

Answer:
CH₃CH₂Br (Bromoethane) + CO₂ + AgBr

Question 8.
Give equations for: (i)Finkelstein reaction, (ii) Hunsdiecker reaction, (iii) Swartz reaction.
Answer:
(i) Finkelstein reaction:
Chloro or bromoalkane on heating with a concentrated solution of sodium iodide in dry acetone gives iodo alkanes. This reaction is called Finkelstein reaction, (S_N² reaction).

(ii) Hunsdiecker reaction:
Silver salts of fatty acids when refluxed with bromine in CCl₄ gives bromo alkane.

(iii) Swartz reaction:
Chloro or bromo alkanes on heating with metallic fluorides like AgF, SbF₃ or Hg₂F₂ gives fluoroalkanes. This reaction is called Swarts reaaction.

Question 9.
Chlorination of methane gives different products. How are they Separated?
Answer:
Chlorination of methane gives different products,
(i) Chloromethane,
(ii) Dichloromethane,
(iii) Trichloro methane,
(iv) Tetraohloro methane which have differences in their boiling points. Hence, these can be separated by fractional distillation.

Question 10.
Identify the products, (major organic product)

(i) CH₃CH = CH₂ + HBr →
Answer:

(ii)

Answer:

(iii)

Answer:

(iv)

Answer:

Question 11.
Starting from propene, how will you prepare:

(i) 1 – bromopropane
Answer:
CH₃ CH = CH₂ + HBr CH₃CH₂CH₂Br
(1 – Bromopropane)

(ii) 2 – bromopropane
Answer:
CH₃ CH = CH₂ + HBr →

(iii) 1 – propanol
Answer:
CH₃ CH = CH₂ CH₃CH₂CH₂Br CH₃CH₂CH₂OH (1 – Propanol)

(iv) propyl amine
Answer:
CH₃ CH = CH₂ + HBr CH₃CH₂CH₂Br CH₃CH₂CH₂NH₂ (Propylamine)

(v) propyl cyanide
Answer:
CH₃ CH = CH₂ + HBr CH₃ CH₂ CH₂ Br CH₃CH₂CH₂CN (Propyl cyanide)

Question 12.
Identify the reagent used in the following conversions.

(i) CH₃CH₂OH CH₃CH₂I
Answer:
X = HI

(ii)

Answer:
X = HI

(iii) CH₃CH₂OH CH₃CH₂Cl
Answer:
X = PCl₅ or PCl₃or SOCl₂

(iv) CH₃CH₂Br CH₃CH₂I
Answer:
X = NaI

(v) CH₃CH₂Br CH₃CH₂F
Answer:
X = AgF

Question 13.
Account for the following:

(i) Ethyl bromide has a higher boiling point than ethane.
Answer:
Haloalkanes have higher boiling point than those of corresponding alkanes because of the polarity of C – X bond. As a result, strong dipole-dipole attractions exist in alkyl halides. Such kind of attractions are absent in alkanes.

(ii) The boiling poinj of methyl halides decrease in the order CH₃I > CH₃Br > CH₃Cl > CH₃F
Answer:
This is because, with increase in size and mass of halogen atom, the magnitude of Vander Waals force of attraction increases.

(iii) The boiling point of CCl₄ is greater than CCl₃.
Answer:
With the increase in number of halogen atoms, the Vander Waals force of attraction increases. Hence, the alkyl halide which has more number of halogen atoms has a higher boiling point.

(iv) n – propyl chloride has a higher boiling point in methyl chloride.
Answer:
For the same halogen atom, boiling points of haloalkanes increase with the size of alkyl groups. Larger the size of the alkyl group, greater is the Vander Waals force of attraction.

Question 14.
Tertiary butyl chloride has the lowest boiling point compared to its isomers. Explain why?
Answer:
This is because of branching of the chain makes the molecule more compact and therefore decreases the surface area of haloalkanes and so the magnitude of Vander Waals force of attraction decreases, n – butyl chloride > isobutyl chloride > tert – butyl chloride.

Question 15.
Explain why?

(i) Ethyl bromide is insoluble in water.
Answer:
Ethyl bromide is insoluble in water because it cannot form hydrogen bonding with water.

(ii) The densities of haloalkanes are of the order RI > RBr > RCl > RF.
Answer:
The density increases with increasing number and atomic mass of halogen.

Question 16.
Name the reagents used to convert bromo ethane into:

(i) ethyl nitrate –
Answer:
KNO₂

(ii) nitro ethane –
Answer:
Ag NO₂

(iii) propane nitrite –
Answer:
KCN

(iv) ethyl carbyl amines –
Answer:
AgCN

(v) ethane thiol –
Answer:
NaSH or KSH

(vi) ethoxy ethane –
Answer:
Ag₂O or C₂H₅ONa

(vii) diethyl amine –
Answer:
C₂H₅ NH₂

(viii) tri-ethyl amine –
Answer:
(C₂H₅)₂NH

(ix) ethyl acetate –
Answer:
CH₃COOAg

(x) 1 – butyne –
Answer:
HC ≡ C.Na

(xi) butane –
Answer:
Na/ether

(xii) ethene –
Answer:
alc.KOH

Question 17.
Give equations for the conversions in:

(i) CH₃CH₂Br + KNO₂ →
Answer:
CH₃CH₂ONO + KBr

(ii) CH₃CH₂Br + AgNO →
Answer:
CH₃CH₂NO₂ (nitroethane) + AgBr

(iii) CH₃CH₂Br + KCN →
Answer:
CH₃CH₂CN (propane nitrite) + KBr

(iv) CH₃CH₂Br + AgCN →
Answer:
CH₃CH₂CN (ethyl carbylatnines (or) ethyl isocyanide) + AgBr

(v) CH₃CH₂Br + NaSH →
Answer:
CH₃CH₂SH (Sodium ethoxide) + NaBr

(vi) →

Answer:
CH₃CH₂OCH₂CH₃ (ethoxy ethane) + NaBr

(vii) →

Answer:

(viii) →

Answer:

(ix) →

Answer:

(x) →

Answer:
NaBr + CH₃CHC≡CH (1 – butyne)

(xi) →

Answer:

2 NaBr + CH₃CH₂CH₂CH₃ (n – butane)

(xii) CH₃CH₂Br + alc KOH →
Answer:
CH₂ = CH₂ + KBr + H₂O

Question 18.
What is ammonolysis? Why this method is not recommended for the preparation of mono alkyl amines?
Answer:
Reaction of an alkyl halide with alcoholic ammonia to give a mixture of amines is known as ammonolysis.

This method is not recommended because the product is a mixture of amines.

Question 19.
What are ambident nucleophiles? Give two examples.
Answer:
Nucleophiles which attack the nucleophilic sites from two sides are called ambident nucleophiles. For example, cyanide ion, CN^– ion can attack the reaction site through carbon or nitrogen atom depending upon the reagent and reaction condition, eg: CN^– and NO₂^–

Question 20.
When ethyl bromide is treated with KCN, the product is ethyl cyanide while, when treated with AgCN, the product is ethyl isocyanide. Explain why?
Answer:
When ethyl bromide is treated with KCN, the nucleophile is ^–CN (the negative charge on carbon) and hence ethyl cyanide is formed. KCN is an ionic compound. It ionizes to give ^–CN as a nucleophile.

whereas AgCN is a covalent compound. It gives ^–NC (negative charge on nitrogen) as a nucleophile. Hence ethyl isocyanide is formed.

Question 21.
Give equation for the reaction between ethyl bromide with
(i) KNO₂ and
(ii) AgNO₂ and write their nucleophiles.
Answer:
(i)

(ii)

Nucleophiles:
If potassium nitrite is used as a reagent, the nucleophile is ^–ONO with the negative charge on the oxygen atom. Hence alkyl nitrite is formed. If silver nitrite is used as a reagent, ^–NO₂ (the negative charge on nitrogen) is the nucleophile and the product is nitroalkane.

Question 22.
What is Williamson’s ether synthesis? Explain with an equation.
Answer:
Haloalkane, when boiled with sodium alkoxide gives corresponding ethers.
eg: This method can be,used to prepare mixed (unsymmetrical) ethers also.
CH₃CH₂Br (Bromoethane ) + NaOCH₂CH₃ (Sodiumethoxide) CH₃CH₂OCH₂CH₃ (diethylether) + NaBr

Question 23.
Briefly outline S_N² mechanism with an example.
Answer:
(i) The reaction follows second order kinetics and the rate equation is
rate = R₂ [RX] [Nucleophile].
Generally primary alkyl halides undergo nucleophilic substitution reaction by S_N² mechanism.

(ii) This reaction involves the formation of a transition state in which both the reactant molecules are partially bonded to each other. The attack of nucleophile occurs from the back side (i.e., opposite to the side in which the halogen is attacked). The carbon at which substitution occurs has inverted configuration during the course of reaction just as an umbrella has tendency to invert in a wind storm. This inversion of configuration is called Walden inversion; after paul walden who first discovered the inversion of configuration of a compound in S_N² reaction.

(iii) When 2 – Bromo octane is heated with sodium hydroxide, 2 – Octanol is formed with inversion of configuration. (-) 2 – Bromo octane is heated with sodium hydroxide (+) 2 – Octanol is formed in which ^–OH group occupies a position opposite to what bromine had occupied,

(a) (-) 2 – Bromo octane
(b) Transition State
(c) (+) 2 – Octanol (product)

Question 24.
Explain S_N¹ mechanism with an example.
Answer:
(i) The rate of the following S_N¹ reaction depends upon the concentration of
alkyl halide (RX) and is independent of the concentration of the nucleophile (OH^–).
Hence Rate of the reaction = k [alkyl halide]

R — Cl + OH^– → R — OH + Cl^–

This S_N¹ reaction follows first order kinetics and occurs in two steps.

We understand S_N¹ reaction mechanism by taking a reaction between tertiary butyl bromide with aqueous KOH.

This reaction takes place in two steps as shown below:
Step-1:
Formation of carbocation: The polar C – Br bond breaks forming a carbocation and bromide ion. This step is slow and hence it is the rate determining step.

The carbocation has 2 equivalent lobes of the vacant 2p orbital, so it can react equally rapidly from either face.

Step-2:
The nucleophile immediately reacts with the carbocation. This step is fast and hence does not affect the rate of the reactions.

As shown above, the nucleophilic reagent OH^– can attack carbocation from both the sides.

(ii) If haloalkane substrate is optically active then, the product obtained will be optically inactive racemic mixture. As nucleophilic reagent OH- can attack carbocation from both the sides, to form equal proportion of dextro and laevorotatory optically active isomers which results in optically inactive racemic mixture.

Question 25.
Explain Saytzeff ‘s Rule with an example.
Answer:
In a dehydrohalogenation reaction, the preferred product is that alkene which has more number of alkyl groups attached to the doubly bonded carbon (more substituted double bond is formed), eg:

Question 26.
Explain E₂ mechanism with an example.
Answer:
The rate of E₂ reaction depends on the concentration of alkyl halide and base
Rate = k [alkyl halide] [base]
It is therefore, a second order reaction. Generally primary alkyl halide undergoes this reaction in the presence of alcoholic KOH. It is a one step process in which the abstraction of the proton from the |3 carbon and expulsion of halide from the a carbon occur simultaneously. The mechanism is shown below.

Question 27.
Explain E₁ mechanism with an example.
Answer:
E₁ mechanism means elimination unimolecular. It follows first order kinetics conversion oftert-butyl chloride to isobutylene by alcoholic potash is an exertic for E₁ reaction. The mechanism is given below:

Step-1:
Heterolytic fission to yield a carbo – cation.

Step-2:
Elimination of a proton from the β – carbon to produce an alkene.

Question 28.
Complete the following equations:

(i) CH₃CH₂Br (Ethyl bromide)
Answer:
CH₃CH₂MgBr (Ethyl magnesium bromide)

(ii) CH₃CH₂Br (Ethyl bromide) + 2Li
Answer:
CH₃CH₂Li (Ethyl Lithium) + LiBr

(iii) 4CH₃CH₂Br (Ethyl bromide) + 4Na/Pb (Sodium-lead alloy)
Answer:
(CH₃CH₂)₄ Pb (Tetra ethyl lead (TEL)) + 4 NaBr + 3 Pb

(iv) CH₃CH₂Br (Bromoethane) + H₂
Answer:
CH₃ – CH₃ (Ethane) + HBr

(v) CH₃CH₂I (lodoethane) + HI
Answer:
CH₃ — CH₃ (Ethane) + I₂

Question 29.
Mention the uses of
(i) chloroform
(ii) iodoform
(iii) carbon tetrachloride.
Answer:
(i) Chloroform:
It is used as a solvent in pharmaceutical industry and for producing pesticides and drugs as an anesthetic. As a preservative for anatomical specimens.

(ii) Iodoform:
Iodoform is used as an antiseptic for dressing wounds.

(iii) Carbon tetrachloride:
Carbon tetrachloride is used as dry cleaning agent. It is used as a solvent for oils, fats and waxes. As the vapour of CCl₄ is non-combustible, it is used under the name pyrene for extinguishing the fire in oil or petrol.

Question 30.
What are organo metallic compounds? Give example.
Answer:
Organo metallic compounds are organic compounds in which there is a direct carbon -metal bond.
eg: CH₃ MgI – Methyl magnesium iodide
CH₃CH₂Mg Br – Ethyl magnesium bromide.

Question 31.
Briefly outline the nature of carbon-magnesium bond in alkyl magnesium halide.
Answer:
The carbon-magnesium bond in Grignard reagent is covalent but highly polar, The carbon atom is more, electronegative than magnesium. Hence, the carbon atom has partial negative charge and the magnesium atom has partial positive charge,

R^δ- ………. MgX^δ+

Question 32.
Mention the uses of Grignard reagent.
Answer:
Grignard reagents are synthetically very useful compounds. These reagents are converted to various organic compounds like alcohols, carboxylic acids, aldehydes and ketones. The alkyl group being electron rich acts as a carbanion or a nucleophile. They would attack polarized molecules at a point of low electron density.

Question 33.
Using a suitable Grignard reagent, write equations for the synthesis of:

(i) Ethyl alcohol:
Answer:

(ii) Tertiary butyl alcohol:
Answer:

(iii) Acetaldehyde:
Answer:

(iv) Ethyl methyl ether:
Answer:
CH₃ – 0 – CH₂Cl (Chloro dimethyl ether) + CH₃MgI → CH₃ – O – CH₂CH₃ (Ethyl methyl ether) +

(v) Methane:
CH₃MgI+HO – H → CH₄ + MgI (OH)
CH₃MgI + C₂H₅OH (Ethylalcohol) CH₄ (Methane) + MgI(OC₂H₅)

Question 34.
Briefly outline the nature of carbon-halogen (X) bond in haloarenes.
Answer:
In haloarenes the carbon atom is sp² hybridised. The sp² hybridised orbitals are shorter and holds the electron pair of bond more tightly. Halogen atom contains P – orbital with lone pair of electrons which interacts with π – orbitals of benzene ring to form extended conjugated system of π – orbitals. The delocalisation of these electrons give double bond character to C – X bond. The resonance structure of halobenzene is given as:

Due to this double bond character of C — X bond in haloarenes, the C — X bond is shorter in length and stronger than in haloalkanes.

Question 35.
Haloarenes undergo nucleophilic substitution reactions less readily compared to halo alkanes. Explain why?
Answer:
Because of resonance, the carbon-halogen bond gets a double bond character.

Thus, C — X bond is shorter in haloarenes compared to that of haloalkenes i.e., The halogen atom is more firmly held, by the benzene ring, breaking of C — X bond is difficult. Hence haloarenes undergo nucleophilic substitution reaction under drastic conditions.

Question 36.
How is chlorobenzene prepared from
(i) benzene
(ii) benzene diazonium chloride?
Answer:
(i) Chlorobenzene is prepared by the direct chlorination of benzene in the presence of lewis acid catalyst like FeCl3.

(ii) When aqueous solution of benzene diazonium chloride is warmed with Cu₂Cl₂ in HCl gives chlorobenzene.

Question 37.
How is chlorobenzene prepared by
(i) Sandmeyer reaction
(ii) Gattermann reaction?
Answer:
(i) Sandmeyer reaction:

(ii) Gattermann reaction:

Question 38.
How is (i) iodobenzene and (ii) fluorobenzene prepared?
Answer:
(i) Iodobenzene is prepared by warming benzene diazonium chloride with aqueous KI solution.

(ii) Fluorobenzene is prepared by treating benzene diazonium chloride with fluoro boric acid. This reaction produces diazonium fluoroborate which on heating produces fluorobenzene. This reaction is called Balz-schiemann reaction.

Question 39.
Give equation for Wurtz-Fittig reaction.
Answer:
Haloarenes reacts with haloalkanes when heated with sodium in ether solution to form alkyl benzene. This reaction is called Wurtz Fittig reaction.

C₆H₅Cl + 2 Na + ClC₂H₅ C₆H₅C₂H₅ + 2 NaCl

Question 40.
How are (i) benzene, (ii) ortho and para chloro toluene, (iii) ortho and para nitro toluene prepared from toluene.
Answer:
Toluene to benzene:
C₆H₅CH₃ (chloro benzene) + 2(H) C₆H₆ (Benzene) + HCl

(ii) Ortho and para chloro toluene:

(iii) Ortho and paranitro toluene:

Question 41.
Mention the uses of chlorobenzene.
Answer:
(i) Chlorobenzene is used in the manufacture of pesticides like DDT.
(ii) It is used as high boiling solvent in organic synthesis.
(iii) It is used as fibre – swelling agent in textile processing.

Question 42.
What are ‘gem’ and vicinal dihalides? Give examples.
Answer:
In ‘gem’ dihalides, the halogen atoms attached to the same carbon atom.
eg: CH₃CH₂Cl₂ – Ethylidene chloride (or) 1, 1 dichloro ethane.
In ‘vic’ dihalides, the halogen atoms are attached to adjacent carbon atoms.
eg:

Question 43.
How will you prepare
(i) ethylidene chloride from acetaldehyde,
(ii) ethylidene chloride from acetylene,
(iii) ethylene dichloride from ethylene,
(iv) ethylene dichloride from ethylene glycol.
Answer:
(i) CH₃CHO (Acetaldehyde) + PCl₅ → CH₃CHCl₂ (Ethylidene dichloride) + POCl₃

(ii)

(iii)

(iv)

Question 44.
Give the chemical reaction that is used to distinguish ‘gem’ and ‘vic’ dihalides.
Answer:
Gem-Dihalides, on hydrolysis with aqueous KOH give an aldehyde or a ketone vic-Dihalides, on hydrolysis with aqueous KOH gives glycols.

This reaction can be used to distinguish the gem- Dihalides and vie- Dihalides.

Question 45.
What is the action of zinc dust in methanol on ethylidene chloride and ethylene chloride? Give equation.
Answer:
Gem- Dihalides and vic – Dihalides on treatment with zinc dust in methanol give alkenes.
(i) CH₃ — CHCl₂ (Ethylidene dichloride) + Zn CH₂ = CH₂ (Ethylene) + ZnCl₂

(ii)

Question 46.
How is acetylene prepared from ethylidene chloride and ethylene chloride? Give equation.
Answer:
Gem-Dihalides and vic-Dihalides on treatment with alcoholic KOH give alkynes.

Question 47.
What is the action of (i) zinc and HCl (ii) H₂/Ni on chloroform? Give equation.
Answer:
(i) Reduction of chloroform in the presence of Zn + HCl gives methylene chloride.

CHCl₃ (chloroform) CH₂Cl₂ (methylene chloride) + HCl

(ii) CHCl₃ (chloroform) CH₂Cl₂ (methylene chloride) + HCl

Question 48.
How is methylene chloride prepared from chloroform?
Answer:
Chloroform undergoes reduction with zinc and HC1 in the presence of ethyl alcohol to form methylene chloride.
CHCl₃ (Methylene chloride) + 2(H) CH₂Cl₂ + HCl

Question 49.
Complete the following by writing the structures of (A) and (B).

(i)

Answer:

(ii)

Answer:

(iii)

Answer:

(iv)

Answer:

(v)

Answer:
A = CH₂ = CH₂ (ethylene)
B = Two molecules of HCHO (formaldehyde)

Question 50.
Mention the uses of methylene chloride.
Answer:
Methylene chloride is used as:
(i) Aerosol spray propellant.
(ii) Solvent in paint remover.
(iii) Process solvent in the manufacture of drugs. .
(iv) A metal cleaning solvent.

Question 51.
What is phosgene? How is it prepared?
Answer:
Phosgene is a poisonous gas (COCl₂) and is prepared by the oxidation of chloroform in the presence of light and air.

Choose the correct answer:

Question 1.
The IUPAC name of the compound is:

(a) 1 – fluoro – 4 – methyl – 2 – nitrobenzene
(b) 4 – fluoro – 1 – methyl – 3 – nitrobenzene
(c) 4 – methyl – 1 – fluoro – 2 – nitrobenzene
(d) 2 – fluoro – 5 – methyl – 1 – nitrobenzene
Answer:
(a) 1 – fluoro – 4 – methyl – 2 – nitrobenzene

Question 2.
In the addition of HBr to propene, the first step involves, the addition of
(a) H⁺
(b) Br^–
(c) H^.
(d) Br^.
Answer:
(a) H⁺
Hint:
The reaction is electrophilic addition i.e., H⁺.

Question 3.
The intermediate during the addition of HCl to propene in the presence of a peroxide is:
(a)

(b)

(c)

(d)
Answer:

(b)

Hint:
Anti Markovnikoff‘addition is observed only with HBr. Therefore with HCl, even in the presence of peroxides, the reaction occurs through a more stable carbocation intermediate.

Question 4.
In the presence of peroxides hydrogen chloride and hydrogen iodide do not give anti Markovnikoff’s addition to alkenes because:
(a) both are highly ionic.
(b) one is oxidising and the other is reducing.
(c) one of the steps is endothermic in both cases.
(d) all the steps are endothermic in both reactions.
Answer:
(c) one of the steps is endothermic in both cases.
Hint:
Anti Markovnikoff’s addition is a free radical addition. The first step in generation of free radical, the second is a propagation step and the third is termination. The propagation step is

(i) X^. + CH₂ = CH — CH₃ → XCH₂ — CH — CH₃
(ii) XCH₂ — CH — CH₃ + HX → XCH₂CH₂CH₃ + X^.

with X being bromine, both steps are. exothermic and hence peroxide effect is observed with HI and HCl, the step
(ii) is endothermic with HI, both steps are endothermic i.e., energy has to be supplied. Hence peroxide effect is not observed with HCl and HI.

Question 5.
Identify the set of reagent / reaction conditions X and Y in the following set of transformations. CH₃CH₂CH₂Br Product 

(a) X = dilute aqueous NaOH, 20°C.
Y = HBr/acetic acid, 20°C.

(b) X = cone, alcoholic NaOH, 80°C.
Y = HBr/acetic acid, 20°C.

(c) X = dilute aqueous NaOH, 20°C.
Y = Br₂ / CHCl₃ at 0°C.

(d) X- cone, aqueous NaOH, 80°C.
Y = Br₂ / CHCl₃, 0°C.
Answer:
(b) X = cone, alcoholic NaOH, 80°C.
Y = HBr/acetic acid, 20°C.
Hint:

Question 6.
Predict the product (C) obtained in the following reactions of butyne-1.

CH₃CH₂ C ≡ CH + HCl → B C

(a)

(b)

(c)

(d)

Answer:

(a)

Hint:

Question 7.
The reaction between toluene with chlorine in the presence of FeCl₃ gives ‘X’ and the reaction in the presence of light gives ‘Y’. Thus ‘X’ and ‘Y’ are:
(a) X =benzyl chloride; Y = m – chlorotoluene
(b) X = benzal chloride; Y = o – chlorotoluene
(c) X = – m – chlorotoluene; Y = p – chlorotoluene
(d) X = o^– and p^– chlorotoluene;
Y = trichloromethyl benzene.
Answer:
(d) X = o^– and p^– chlorotoluene;
Y = trichloromethyl benzene.
Hint:

Question 8.
Which of the following in a nucleophilic substitution reaction?
(a) 2RX + 2Na → R – R + 2 NaX
(b) RX + H₂ → RH + HX
(c) RX + Mg → RMgX
(d) RX + KOH → ROH + KX
Answer:
(d) RX + KOH → ROH + KX

Question 9.
For the following (a) I^– (b) Cl^– (c) Br^–, the increasing order of nucleophilicity is:
(a) Br^– < Cl^– < I^–
(b) I^– < Br^– < Cl^–
(c) Cl^– < Br^– < I^–
(d) I^– < Cl^– < Br^–
Answer:
(c) Cl^– < Br^– < I^–

Question 10.
Consider the following Bromides.

The correct order of S_N1 reactivity is:
(a) C > B > A
(b) A > B > C
(c) B > C > A
(d) B > A > C
Answer:
(c) B > C > A
Hint:
The reactivity of S_N1 reactions depends upon the stability of the intermediate carbocation. The stability of carbocations derived from A, B, C follows the order

the reactivity follows the order B > C > A.

Question 11.
The major product formed in the following reaction is:

(a)

(b)

(c)

(d)

Answer:

(c)

Hint:
Dehydrohalogenation. Strong bases like -OCH₃ also used for the reaction.

Question 12.
The correct prder of increasing reactivity of C – X bond towards nucleophile in the following compounds are:

I.

II.

III. (CH₃)₃ C – X

IV. (CH₃)₂ CH – X

(a) III < II < I < IV
(b) I < II < IV < III
(c) II < III < I < IV
(d) IV < III < I < II
Answer:
(b) I < II < IV < III
Hint:
Alkyl halides are more reactive than aryl halides. Among alkyl halides 3° alkyl halide (III) are more reactive than 2° alkyl halide (IV). Among aryl halides having electron with drawing groups at ortho and para positions are more reactive than sirnplp aryl halides (I). Thus the overall reactivity increases in the order.
I < II < III < IV

Question 13.
The reagent used to convert ethyl bromide to ethyl fluoride is/are.:
(a) cone NaOH ; HF
(b) HF
(c) AgF
(d) F₂
Answer:
(c) AgF
Hint:
CH₃CH₂Br + AgF → CH₃CH₂F + AgBr

Question 14.
Among the following, the alkyl halide which has the highest boiling point is:
(a) CH₃Cl
(b) CH₃CH₂Cl
(c) CH₃CH₂CH₂Cl
(d) CH₃F
Answer:
(c) CH₃CH₂CH₂Cl
Hint:
Between CH₃Cl and CH₃F, CH₃Cl has a higher boiling point. Among CH₃Cl, CH₃CH₂Cl, CH₃CH₃CH₃Cl, the boiling point increases with increasing number of carbon atoms.

Question 15.
Choose the correct statement.
I. Among methyl chloride, ethyl chloride . and butyl chloride, methyl chloride has the least boiling point.
II. Among the methyl halides, methyl iodide has the highest boiling point.
III. Both I and II are correct.
IV. Both I and II are wrong.
(a) I
(b) II
(c) III
(d) IV
Answer:
(c) III
Hint:
(i) The boiling point and melting point of mono haloalkane increase with the increase in the number of carbon atoms.
(ii) The boiling point and melting point of haloalkanes decreases with respect to the halogen in the following order.

Question 16.
Which of the following are ambident nucleophiles? .
(a) ^–OH
(b) ^–SH
(c) ^–CN
(d) Br^–
Answer:
(c) ^–CN
Hint:
^–CN and CN^– have two nucleophilic sites, where the carbon atom of the cyanide ion attacks the reaction site and CN^– ion, the nitrogen atom attacks the reaction site. Such nucleophiles are ambident nucleophiles.

Question 17.

The products A and B are:
(a) A = CH₃CH₂CN; B = CH₃CH₂NC
(b) A = CH₃CH₂NC ; B = CH₃CH₂CN

(c)

(d) A = CH₃CH₂CN; B = C₃CH₂CN
Answer:
A = CH₃CH₂CN; B = CH₃CH₂NC
Hint:

Question 18.
The major product in the following reaction:

is:

(a) CH₂ = CH CH₂CH₃
(b) CH₃CH CH₂ CH₃
(c) CH₃CH = CH CH₃
(d) CH₃CH = CH₂
Answer:
(c) CH₃CH = CH CH₃
Hint:
The product follows Saytzeff’s rule: In a dehydrohalogenation reaction, the preferred product is that alkene which has more number of alkyl groups attached to the doubly bonded carbon (more substituted double bond is formed).

Question 19.
Which of the following is used to prepare propanol – 1 from formaldehyde?
(a) CH₃MgI
(b) CH₃CH₂MgI
(c)
(d) CH₃MgBr
Answer:
(b) CH₃CH₂MgI
Hint:
Propanol – 1 contains three carbon atoms. Alkyl magnesium halides teact with HCHO, to give primary alcohol. Thus RMgX, should contain two carbon atoms, i.e., CH₃CH₂MgI.

Question 20.
The correct sequence of reactions involved -for the formation of isopropyl alcohol is/are:
(a) CH₃CHO, CH₃MgI, H₂O/H⁺
(b) HCHO, CH₃MgI, H₂O/H⁺
(c) CH₃Mg I, CH₃CHO, H⁺/ H₂O
(d) CH₃COCH₃, CH₃CHO, H⁺/ H₂O
Answer:
(a) CH₃CHO, CH₃MgI, H₂O/H⁺
Hint:

Question 21.

The product is:

(a)

(b)

(c) CH₃CH₂CH₂CH₂OH

(d)

Answer:

(b)

Hint:

Question 22.
Which of the following reaction is known as Sandmeyer reaotion?
(a) Benzene diazonium chloride Chloro benzene
(b) Benzene diazonium chloride + HBF₄ → Fluoro benzene
(c) Benzene + Chlorine Chloro benzene
(d) Benzene + Nitric acid Nitro benzene
Answer:
(a) Benzene diazonium chloride Chloro benzene
Hint:

Question 23.
Which one of the following is/are ‘gem’ dihalide?

(a)

(b)

(c)

(d)

Answer:

(a)

Question 24.
Chloropicrin is:
(a) trichloro nitro ethane
(b) trichloro nitro methane
(c) nitro methane
(d) nitro ethane
Answer:
(b) trichloro nitro methane

Question 25.
The reagents used to test primary amines are:
(a) chloroform and KOH
(b) mixture of conc H₂SO₄ and HNO₃
(c) aqueous KOH
(d) bromine water
Answer:
(a) chloroform and KOH
Hint:
Chloroform reacts with aliphatic or aromatic primary amine and alcoholic caustic potash, to give foul smelling alkyl isocyanide (carbylamines).

This reaction is used to test primary amine.

Question 26.
SWARTZ reaction is used to prepare:
(a) Chloro benzene
(b) Toluene
(c) DDT
(d) Freon 12
Answer:
(d) Freon 12
Hint:
Freon – 12 is prepared by the action of hydrogen fluoride on carbon tetrachloride in the presence of catalytic amount of antimony pentachloride. This is called swartz reaction.

Question 27.
Assertion:
Chlorination of methane gives different products having differences in boiling points.
Reason:
Chlorination of methane gives a mixture of chloromethane, dichloromethane, tri-chloro methane and carbon tetra chloride.
(a) Both assertion and reason are true and reason is the correct explanation of assertion.
(b) Both assertion and reason are true but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Answer:
(a) Both assertion and reason are true and reason is the correct explanation of assertion.

Question 28.
Assertion:
Bromo ethane reacts with ammonia in the molar ratio 1 : 1 gives ethyl amine.
Reason:
The reaction follows S_N1 mechanism.
(a) Both assertion and reason are true and reason is the correct explanation of assertion.
(b) Both assertion and reason are true but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Answer:
(c) Assertion is true but reason is false.
Hint:
Being a primary alkyl halide bromo ethane undergoes nucleophilic substitution by S_N2 mechanism.

Question 29.
Assertion:
S_N2 reaction of an optically active haloalkane is always accompanied -by inversion of configuration at the asymmetric carbon atom.
Reason:
In the transition state, the nucleophile and the leaving group are attached to the asymmetric carbon atom opposite to each other.
(a) Both assertion and reason are true and reason is the correct explanation of assertion.
(b) Both assertion and reason are true but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Answer:
(a) Both assertion and reason are true and reason is the correct explanation of assertion.

Question 30.
Assertion:
E₁ (elimination, unimolecular) follows first order kinetics.
Reason:
Primary alkyl halides undergo elimination reaction by E₁ mechanism.
(a) Both assertion and reason are true and reason is the correct explanation of assertion.
(b) Both assertion and reason are true but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Answer:
(c) Assertion is true but reason is false.
Hint:
Primary alkyl halides undergo elimination reaction by E₂ mechanism.

Question 31.
Identify a tertiary alkyl halide from the following.
(a) CH₃CH₂Br ‘
(b)

(c)

(d)

Answer:

(d)

Question 32.
Which among the following will have the least boiling point?
(a) CH₃Cl
(b) CH₃ CH₂ Cl
(c) CH₃(CH₂)₂Cl
(d) CH₃ (CH₂)₃ Cl
Answer:
(a) CH₃Cl
Hint:
For the same halogen atom, the boiling points will increase with the size of the alkyl group.

Question 33.
Which of the following pairs, will undergo nucleophilic substitution reaction by S_N2 mechanism only?
(a) CH₃ Br and CH₃ CH₂ Br
(b) (CH₃)₂ CH Br and CH₃ CH₂ Br

(c)

(d)
Answer:
(a) CH₃ Br and CH₃ CH₂ Br
Hint:
Only primary alkyl halides undergo nucleophilic substitution reaction by SN2 mechanism.

Question 34.
Identify the pair of compounds which contain a chiral carbon atoms.
(a)

(b)

(c) (CH₃)₃ C – Br and (CH₃)₂ CH₂ Br
(d) CH₃CH₂Br and CH₃CH₂I
Answer:

Hint:
Chiral carbon atoms are those to which form different atoms or groups attached to Cl.

Question 35.
Assertion:
Benzyl chloride when kept in acetone-water, produces, benzyl alcohol.
Reason:
The reaction follows S_N2 mechanism.
(a) If both assertion and reason are true and reason is the correct explanation of the assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
(c) If assertion is true, but reason is false.
Hint:
Correct reason: Benzyl halides undergo hydrolysis by S_N1 mechanism.

Question 36.
Assertion:
The presence of nitro group facilitates nucleophilic substitution reactions in aryl halides.
Reason:
The intermediate carbanion is- stabilised due to the presence of the nitro group.
(a) If both assertion and reason are true and reason is the correct explanation of the assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
(a) If both assertion and reason are true and reason is the correct explanation of the assertion.

Question 37.
Assertion:
Nucleophilic substitution reactions on an optically active alkyl halide gives a mixture of enantiomer.
Reason:
The reaction occurs through S_N2 mechanism.
(a) If both assertion and reason are true and reason is the correct explanation of the assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Hint:
The reaction occurs through S_N1 mechanism.

Question 38.
Assertion:
1 – Butene on reaction with HBr in the presence of peroxide produces 1-bromobutane.
Reason:
It involves the formation of primary free radical.
(a) If both assertion and reason are true and reason is the correct explanation of the assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
(c) If assertion is true, but reason is false.
Hint:
The reaction occurs through the formation of more stable 2° free radical.

Question 39.
Assertion:
Chloral reacts with phenyl chloride to form DDT.
Reason:
It is an electrophilic substitution reaction.
(a) If both assertion and reason are true and reason is the correct explanation of the assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
(a) If both assertion and reason are true and reason is the correct explanation of the assertion.

Question 40.
Consider the following reaction:

Which statements are correct about the above reaction?
(a) I and V both are nucleophile.
(b) In III carbon atom is sp³ hybridised.
(c) In IV carbon atom is sp² hybridised.
(d) I and V both are electrophile.
Answer:
(a) I and V both are nucleophile.

Question 41.
Choose the incorrect statement with regard to S_N1 reactions:
(a) The order of reactivity depends on the stability of intermediate carbocations.
(b) Among the isomeric butanes, 1 – bromo butane undergo S_N1 reactions.
(c) Among the isomeric butanes, (CH₃)₃ C — Br undergo S_N1 reactions.
(d) The carbocation (CH₃)₃ C⁺ is most reactive.
Answer:
(b) Among the isomeric butanes, 1 – bromo butane undergo S_N1 reactions.
Hint.
The increasing reactivity of the four isomeric butanes towards S_N1 reaction is
CH₃ CH₂ CH₂ CH₂ Br < (CH₃)₂ CH CH₂ Br < (CH₃)₃ C – Br <

Question 42.
The incorrect statement is the carbon atom is:
(a) The C – Cl bond is chlorobenzene is shorter than C – Cl bond in methyl chloride.
(b) The C – Cl bond in both chloro benzene and methyl chloride is sp³ hybridised.
(c) In chloro benzene, the lone pair of electrons on chlorine enters into conjugation with tho benzene ring.
(d) There is no resonance in methyl chloride, whereas, chloro benzene exhibits resonance.
Answer:
(b) The C – Cl bond in both chloro benzene and methyl chloride is sp³ hybridised.
Hint:
The carbon atom in C – Cl bond in aryl halides is sp² hybridised whereas in methyl chloride it is sp³ hybridised.

Question 43.
Ethylene chloride and ethylidene chloride are isomers. Identify the correct statements.
(a) Both compounds form same products on treatment with alcoholic KOH.
(b) Both compounds form same products on treatment with aqueous KOH.
(c) Both compounds form the same product on reduction.
(d) Both compounds are optically active.

(a) I, II
(b) I, III
(c) II, IV
(d) I, IV
Answer:
(b) I, III

Question 44.
Choose the incorrect statement:

(a) The electrophile in this reaction is NO₂⁺
(b) Chlorine deactivates the benzene ring, by -I effect while activates the benzene ring by + M effect.
(c) Chlorine is an ortho paradirecting group. Nitration is a nucleophilic substitution reaction.
(d) Nitration is a nucleophilic subctitution reaction.
Answer:
(d) Nitration is a nucleophilic subctitution reaction.

Question 45.
Match the entities of column I with appropriate entities of column II.

Column I	Column II
(i) DDT	(A) pesticide
(ii) chloroform	(B) refrigerant
(iii) BHC	(C) insecticide
(iv) Freon – 12	(D) anaesthetic

(a) (i) – (C), (ii) – (D), (iii) – (A), (iv) – (B)
(b) (i) – (D), (ii) – (A), (iii) – (B), (iv) – (C)
(c) (i) – (A), (ii) – (D), (iii) – (B), (iv) – (C)
(d) (i) – (B), (ii) – (C), (iii) – (A), (iv) – (D)
Answer:
(a) (i) – (C), (ii) – (D), (iii) – (A), (iv) – (B)

Question 46.
Match the entities of column I with appropriate entities of column II.

(a) (i)- (B), (ii) – (C), (iii) – (A), (iv) – (D)
(b) (i) – (D), (ii) – (C), (iii) – (A), (iv) – (B)
(c) (i) – (C), (ii) – (A), (iii) – (D), (iv) – (B)
(d) (i) – (A), (ii) – (C), (iii) – (B), (iv) – (D)
Answer:
(b) (i) – (D), (ii) – (C), (iii) – (A), (iv) – (B)

TN Board 11th Chemistry Important Questions