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Chapter 1 · Haloalkanes and Haloarenes

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#1 SUB 2M 🖼 1

Question

What are haloalkane and haloarenes? Explain with examples.

Answer

“The replacement of hydrogen atom(s) in an aliphatic or aromatic hydrocarbon by halogen atom(s) results in the formation of alkyl halide (haloalkane) and aryl halide (haloarene), respectively.”
Haloalkanes contain halogen atom(s) attached to the sp3 hybridised carbon atom of an alkyl group whereas haloarenes contain halogen atom(s) attached to sp2 hybridised carbon atom(s) of an aryl group.
#2 SUB 2M

Question

Write down uses of halogen containing organic compounds.

Answer

These classes of compounds find wide applications in industry as well as in day-to-day life.
They are used as solvents for relatively non-polar compounds and as starting materials for the synthesis of wide range of organic compounds.
Chlorine containing antibiotic, chloramphenicol, produced by microorganisms is very effective for the treatment of typhoid fever.
Synthetic halogen compounds, viz. chloroquine is used for the treatment of malaria; halothane is used as an anaesthetic during surgery.
Certain fully fluorinated compounds are being considered as potential blood substitutes in surgery.
Our body produces iodine containing hormone thyroxine the deficiency of which causes a disease called goiter.
#3 SUB 2M 🖼 1

Question

Explain classification of haloalkane and haloarene on the basis of number of halogen atoms.

Answer

These may be classified as mono, di, or polyhalogen (tri-,tetra-, etc.) compounds depending on whether they contain one, two or more halogen atoms in their structures.
Examples:
#4 SUB 4M 🖼 3

Question

Explain classification of monohalogen compounds on the basis of sp3 (C – X) bonds.

Answer

These are the types:
(a) Alkyl halides or haloalkanes (R – X)
(b) Allylic halides
(c) Benzylic halides
(a) Alkyl halides or haloalkanes (R—X)
“In alkyl halides, the halogen atom is bonded to an alkyl group (R).”
They form a homologous series represented by CnH2n+1X.
They are further classified as primary, secondary or tertiary according to the nature of carbon to which halogen is attached.
If halogen is attached to a primary carbon atom in an alkyl halide, the alkyl halide is called primary alkyl halide or alkyl halide.
Similarly, if halogen is attached to secondary or tertiary carbon atom, the alkyl halide is called secondary alkyl halide (2°) and tertiary (3°) alkyl halide, respectively.
(b) Allylic halides:
“These are compounds in which the halogen atom is bonded to an sp3-hybridised carbon atom adjacent to carbon-carbon double bond.”
(c) Benzylic halides:
“There are the compounds in which the halogen atom is bonded to an sp3-hybridised carbon atom attached to an aromatic ring.”
#5 SUB 2M 🖼 2

Question

What are Vinylic halides and Aryl halides Explain with examples.
OR
Explain classification of halogen compound containing sp2 C – X bonds.

Answer

(a) Vinylic halides: “These are compounds in which the halogen atom is bonded to a
sp
2-hybridised carbon atom of a carbon-carbon double bond (C = C).”
(b) Aryl halides: “These are compounds in which the halogen atom is directly bonded to the sp2-hybridised carbon atom of an aromatic ring.”
#6 SUB 2M 🖼 8

Question

Explain the Common and IUPAC nomenclature of haloalkane and haloarene compounds
with examples.

Answer

The common names of alkyl halides are derived by naming the alkyl group followed by the name of halide.
In the IUPAC system of nomenclature, alkyl halides are named as halosubstituted hydrocarbons.
For mono halogen substituted derivatives of benzene, common and IUPAC names are the same.
For dihalogen derivatives, the prefixes o-, m-, p- are used in common system but in IUPAC system, the numerals 1,2; 1,3 and 1,4 are used.
Common name: n-Propyl bromide Isopropyl chloride Isobutyl chloride
IUPAC name: 1-Bromopropane 2-Chloropropane 1-Chloro-2-methylpropane
Common name: Bromobenzene m-Dibromobenzene sym-Tribromobenzene
IUPAC name: Bromobenzene 1,3-Dibromobenzene 1,3,5-Tribromobenzene
Common name: Neopentylchloride isopropylbromide
IUPAC name: 1-Chloro-2,2-dimethylpropane 2-Bromopropane
#7 SUB 2M 🖼 2

Question

Explain geminal and vicinal dihalide with examples.

Answer

When both the halogen atoms are present on the same carbon atom, it is known as geminal dihalide.
In common name system, gem-dihalides are named as alkylidene halides.
When halogen atoms are present on adjacent carbon atoms, it is known as vicinal dihalides.
In common name system, vic-dihalides are named as alkylene dihalides.
Common name: Ethylidene chloride (gem-dihalide) Ethylene dichloride (vic-dihalide)
IUPAC name: 1, 1-Dichloroethane 1, 2-Dichloroethane

Common and IUPAC Names of some Halide Compounds

Structure

Common Name

IUPAC Name

CH3CH2CH(Cl)CH3

sec-Butyl chloride

2-Chlorobutane

(CH3)3CCH2Br

neo-Pentyl bromide

1-Bromo-2,2-dimethylpropane

(CH3)3CBr

tert-Butyl bromide

2-Bromo-2-methylpropane

CH2 = CHCl

Vinyl chloride

Chloroethene

CH2 = CHCH2Br

Allyl bromide

3-Bromopropene

o-Chlorotoluene

1-Chloro-2-methylbenzene

OR

2-Chlorotoluene

Benzyl chloride

Chlorophenylmethane

CH2Cl2

Methylene chloride

Dichloromethane

CHCl3

Chloroform

Trichloromethane

CHBr3

Bromoform

Tribromomethane

CCl4

Carbon tetrachloride

Tetrachloromethane

CH3CH2CH2F

n-Propyl fluoride

1-Fluoropropane

#8 SUB 2M 🖼 2

Question

Write a note on nature of C-X bond.
OR
Explain polarity of C – X bond in Haloalkane.

Answer

Halogen atoms are more electronegative than carbon, therefore, carbon-halogen bond of alkyl halide is polarised.
The carbon atom bears a partial positive charge whereas the halogen atom bears a partial negative charge.
As we go down the group in the periodic table, the size of halogen atom increases. Fluorine atom is the smallest and iodine atom is the largest.
Consequently the carbon-halogen bond length also increases from C–F to C–I.
Bond Length: C – F < C – Cl < C – Br < C – I
Bond enthalpy: C – F > C – Cl > C – Br > C – I
Dipole Moment: CH3 – Cl > CH3 – F > CH3
– Br > CH3 – I
#9 SUB 4M 🖼 2

Question

Write down the preparation of alkyl halides from alcohols.

Answer

The hydroxyl group of an alcohol is replaced by halogen on reaction with concentrated halogen acids, phosphorus halides or thionyl chloride etc.
(i) Reaction with thionyl chloride (SOCl2):
Thionyl chloride is first preferred because in this reaction alkyl halide is formed along with gases SO2 and HCl. The two gaseous products are escapable, hence, the reaction gives pure alkyl halides.
R–OH + SOCl2 R–Cl + SO2 + HCl
(ii) Reaction with halogen acids or Lucas Test:
The reactions of primary and secondary alcohols with HCl require the presence of a catalyst, anhy ZnCl2. With tertiary alcohols, the reaction is conducted by simply shaking the alcohol with concentrated HCl at room temperature.
Constant boiling with HBr (48%) is used for preparing alkyl bromide.
Good yields of R—I may be obtained by heating alcohols with sodium or potassium iodide in 95% orthophosphoric acid.
The order of reactivity of alcohols with a given haloacid is 3° > 2° > 1°.
R–OH + HCl R–Cl + H2O
(iii) Reaction with Phosphorus halides:
Phosphorus tribromide and triiodide are usually generated in situ (produced in the reaction mixture) by the reaction of red phosphorus with bromine and iodine respectively.
3R–OH + PX3 3R–X + H3PO3(X = Cl, Br)
R–OH + PCl5 R–Cl + POCl3 + HCl
(iv) Reaction with NaBr and H2SO4:
R – OH + NaBr + H2SO4 R–Br + NaHSO4 + H2O
(v) Reaction with halogen:
R–OH R–X
The above methods are not applicable for the preparation of aryl halides because the carbon-oxygen bond in phenols has a partial double bond character and is difficult to break being stronger than a single bond.
#10 SUB 3M 🖼 5

Question

Explain the preparation of haloalkanes from hydrocarbon.

Answer

(I) From alkanes by free radical halogenation :
Free radical chlorination or bromination of alkanes gives a complex mixture of isomeric mono- and polyhaloalkanes, which is difficult to separate as pure compounds. Consequently, the yield of any single compound is low.
Example:
(II) From alkenes:
(a) Addition of hydrogen halides:
An alkene is converted to corresponding alkyl halide by reaction with hydrogen chloride, hydrogen bromide or hydrogen iodide.
Propene yields two products, however only one predominates as per Markovnikov’s rule.
(b) Addition of halogens:
In the laboratory, addition of bromine in CCl4 to an alkene resulting in discharge of reddish brown colour of bromine constitutes an important method for the detection of double bond in a molecule.
The addition results in the synthesis of vic-dibromides, which are colourless.
#11 SUB 2M 🖼 1

Question

Explain Finkelstein and Swarts reaction.
OR
Explain preparation of haloalkane by halogen exchange method.

Answer

(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 Finkelstein reaction.
R–X + Nal R–I + NaX
X = Cl, Br
Example:
(i) CH3 - Cl + NaI CH3 - I + NaCl
(ii) CH3 - CH2 - Br + NaI CH3 - CH2 - I + NaBr
NaCl or NaBr thus formed is precipitated in dry acetone.
(2) Swarts reaction:
The synthesis of alkyl fluorides is best accomplished by heating an alkyl chloride/bromide in the presence of a metallic fluoride such as AgF, Hg2F2, CoF2 or SbF3.
The reaction is termed as Swarts reaction.
H3C - Br + AgF H3C–F + AgBr
#12 SUB 2M

Question

Explain preparation of haloarene compounds from hydrocarbons.

Answer

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 like iron or iron(III) chloride.
The ortho and para isomers can be easily separated due to large difference in their melting points.
Reactions with iodine are reversible in nature and require the presence of an oxidising agent (HNO3, HIO4) to oxidise the HI formed during iodination.
Fluoro compounds are not prepared by this method due to high reactivity of fluorine.
#13 SUB 2M 🖼 3

Question

Explain preparation of haloarene from amine compound. OR
Preparation of hyloarene by Sandmayer reaction.

Answer

When a primary aromatic amine, dissolved or suspended in cold aqueous mineral acid, is treated with sodium nitrite, a diazonium salt is formed.
Mixing the solution of freshly prepared diazonium salt with cuprous chloride or cuprous bromide results in the replacement of the diazonium group by –Cl or –Br.
This reaction is known as Sandmayer reaction.
Replacement of the diazonium group by iodine does not require the presence of cuprous halide and is done simply by shaking the diazonium salt with potassium iodide.
#14 SUB 2M

Question

Write down physical state, colour and odour of halogen compound.

Answer

Alkyl halides are colourless when pure.
However, bromides and iodides develop colour when exposed to light.
Many volatile halogen compounds have sweet smell.
Methyl chloride, methyl bromide, ethyl chloride and some chlorofluoromethanes are gases at room temperature. Higher members are liquids or solids.
#15 SUB 3M 🖼 1

Question

Explain boiling point of alkyl halides.

Answer

Molecules of organic halogen compounds are generally polar.
Due to greater polarity as well as higher molecular mass as compared to the parent hydrocarbon, the intermolecular forces of attraction (dipole-dipole and van der Waals) are stronger in the halogen derivatives.
That is why the boiling points of chlorides, bromides and iodides are considerably higher than those of the hydrocarbons of comparable molecular mass.
The pattern of variation of boiling points of different halides is depicted in Fig.
For the same alkyl group, the boiling points of alkyl halides decrease in the order: RI > RBr > RCl > RF.
This is because with the increase in size and mass of halogen atom, the magnitude of van der Waal forces increases.
The boiling points of isomeric haloalkanes decrease with increase in branching (1° > 2° > 3°).
For example, 2-bromo-2-methylpropane has the lowest boiling point among the three isomers.
#16 SUB 2M 🖼 1

Question

Explain boiling point and melting point of dihalo-benzene.
OR
Give reason why melting point of p-Dichloro benzene is more than ortho and meta isomers.

Answer

Boiling points of isomeric dihalobenzenes are nearly the same.
However, the para-isomers are high melting as compared to their ortho- and meta-isomers.
It is due to symmetry of para-isomers that fits in crystal lattice better as compared to ortho- and meta-isomers.
#17 SUB 2M

Question

Explain density of haloalkane compound.

Answer

Bromo, iodo and polychloro derivatives of hydrocarbons are heavier than water.
The density increases with increase in number of carbon atoms, halogen atoms and atomic mass of the halogen atoms.
Some density order of haloalkane compound is given below:
CCl4 > CHCl3 > CH2Cl2 > CH3Cl

Compound

Density

(g/mL)

Compound

Density

(g/mL)

n-C3H7Cl

0.89

CH2Cl2

1.336

n-C3H7Br

1.335

CHCl3

1.489

n-C3H7I

1.747

CCl4

1.595

#18 SUB 3M

Question

Explain solubility of haloalkane.

Answer

The haloalkanes are slightly soluble in water.
In order to dissolve haloalkane in water, energy is required to overcome the attractions between the haloalkane molecules and break the hydrogen bonds between water molecules.
Less energy is released when new attractions are set up between the haloalkane and the water molecules as these are not as strong as the original hydrogen bonds in water.
As a result, the solubility of haloalkanes in water is low.
However, haloalkanes tend to dissolve in organic solvents because the new intermolecular attractions between haloalkanes and solvent molecules have the same strength as the ones being broken in separate haloalkane and solvent molecules.
#19 SUB 3M 🖼 1

Question

What is nucleophilic substitution reaction? Write down different products obtained from alkyl halides.

Answer

Nucleophiles are electron rich species. Therefore, they attack at that part of the substrate molecule which is electron deficient.
The reaction in which a nucleophile replaces already existing nucleophile in a molecule is called nucleophilic substitution reaction.
Haloalkanes are substrate in these reactions.
In this type of reaction, a nucleophile reacts with haloalkane (the substrate) having a partial positive charge on the carbon atom bonded to halogen.
A substitution reaction takes place and halogen atom, called leaving group departs as halide ion. Since the substitution reaction is initiated by a nucleophile, it is called nucleophilic substitution reaction.
It is one of the most useful classes of organic reactions of alkyl halides in which halogen is bonded to sp3 hybridised carbon.
R – X + Nu R – Nu + X

Reagent

Nucleophile (Nu)

Substitution product R – Nu

Class of main product

NaOH (KOH)

HO

ROH

Alcohol

H2O

H2O

ROH

Alcohol

NaOR'

R'O

ROR'

Ether

Nal

I

R-I

Alkyl iodide

NH3

NH3

RNH2

Primary amine

R'NH2

R'NH2

RNHR'

Sec. amine

R'R"NH

R'R"NH

RNR'R"

Tert. amine

KCN

RCN

Nitrile (cyanide

AgCN

Ag-CN

RNC

Isonitrile (isocyanide)

KNO2

O = N – O

R – O – N = O

Alkyl nitrite

AgNO2

R-NO2

Nitroalkane

R'COOAg

R'COO

R'COOR

Ester

LiAlH4

H

RH

Hydrocarbon

R'M+

R'

RR'

Alkane

#20 SUB 🖼 2

Question

What are ambident nucleophiles? Explain with an example.

Answer

Ambident nucleophiles are nucleophiles having two nucleophilic sites. Thus, ambident nucleophiles have two sites through which they can attack.
Groups like cyanides and nitrites possess two nucleophilic centres are called ambident nucleophiles.
Cyanide group is a hybrid of two contributing structures and therefore, it can act as a nucleophile in two different ways .
The Linking through carbon atom resulting in alkyl cyanides and through nitrogen atom leading to isocyanides.
Similarly nitrite ion also represents an ambident nucleophile with two different points of linkage .
The linkage through oxygen results in alkyl nitrites while through nitrogen atom, it leads to nitroalkanes.
#21 SUB 3M 🖼 2

Question

Explain bimolecular nucleophilic substitution (SN2) reaction with example.
OR Explain Mechanism of following reaction : CH3Cl + : OH CH3OH + : Cl

Answer

In the year 1937, Edward Davies Hughes and Sir Christopher Ingold proposed a mechanism for an SN2 reaction.
The reaction between CH3Cl and hydroxide ion to yield methanol and chloride ion follows a second order kinetics.
The rate depends upon the concentration of both the reactants.
Rate =
The incoming nucleophile interacts with alkyl halide causing the carbon-halide bond to break and a new bond is formed between carbon and attacking nucleophile.
Here, it is C-O bond formed between C and -OH. These two processes take place simultaneously in a single step and no intermediate is formed.
As the reaction progresses and the bond between the incoming nucleophile and the carbon atom starts forming, the bond between carbon atom and leaving group weakens.
As this happens, the three C-H bonds of the substrate start moving away from the attacking nucleophile.
In transition state all the three C-H bonds are in the same plane and the attacking and leaving nucleophiles are partially attached to the carbon.
As the attacking nucleophile approaches closer to the carbon, C-H bonds still continue to move in the same direction till the attacking nucleophile attaches to carbon and leaving group leaves the carbon.
As a result configuration is inverted. This process is called inversion of configuration.
In the transition state, the carbon atom is simultaneously bonded to incoming nucleophile and the outgoing leaving group.
Such structures are unstable and cannot be isolated. Thus, in the transition state, carbon is simultaneously bonded to five atoms.
#22 SUB 2M 🖼 2

Question

Explain reactivity order towards SN2 reaction of alkyl halides.

Answer

Reactivity order of primary secondary and tertiary halides toward SN2 reaction is
CH3–X > 1° – halide > 2° – halide > 3° – halide.
The presence of bulky substituents on or near the carbon atom has a dramatic inhibiting effect.
Of the simple alkyl halides, methyl halides react most rapidly in SN2 reactions because there are only three small hydrogen atoms.
Tertiary halides are the least reactive because bulky groups hinder the approach of nucleophiles.
Thus, the order of reactivity followed is: Primary halide > Secondary halide > Tertiary halide.
#23 SUB 3M 🖼 3

Question

Explain unimolecular nucleophilic substitution (SN1) reaction with mechanism.

Answer

SN1 reactions are generally carried out in polar protic solvents (like water, alcohol, acetic acid etc.)
The reaction between tert- butyl bromide and hydroxide ion yields tert-butyl alcohol and follows the first order kinetics,
The rate of reaction depends upon the concentration of only one reactant, which is tert- butyl bromide.
Rate = K[(CH3)3 C Br]
(CH3)3CBr + OH (CH3)3 COH + Br
2-Bromo-2-methylpropane 2-methylpropan-2-ol
It occurs in two steps. In step I, the polarised
C–Br bond undergoes slow cleavage to produce a carbocation and a bromide ion.
Step I:
The carbocation formed is then attacked bynucleophile in step II to complete the substitution reaction.
Step II:
Step I is the slowest and reversible. It involves the C–Br bond breaking, for which the energy is obtained through solvation of halide ion with the proton of protic solvent.
Since the rate of reaction depends upon the slowest step, the rate of reaction depends only on the concentration of alkyl halide and not on the concentration of hydroxide ion.
Further, greater the stability of carbocation, greater will be its ease of formation from alkyl halide and faster will be the rate of reaction. In case of alkyl halides, 3o alkyl halides undergo SN1 reaction very fast because of the high stability of 3o carbocations.
Reactivity order for SN1 reaction
3° – halide > 2° – halide > 1° – halide > CH3 – X
#24 SUB 2M 🖼 1

Question

Explain reactivity order of SN1 and SN2 reaction for alkyl halide, allylic halide and benzylic halide.

Answer

The order of reactivity of alkyl halides towards SN1and SN2 reactions are as follows:
For the same reasons, allylic and benzylic halides show high reactivity towards the SN1 reaction. The carbocation thus formed gets stabilised through resonance as shown below:
For a given alkyl group, the reactivity of the halide, R-X, follows the same order in both the mechanisms R–I > R–Br > R–Cl >> R–F.
#25 SUB 3M 🖼 1

Question

What is stereochemistry? Explain plane polarised light, optical activity and optical isomers.

Answer

Stereochemistry is the branch of chemistry that involves "the study of the different spatial arrangements of atoms in molecules".
Plane polarised light (PPL) is obtained when ordinary light is passed through a nicol prism a light of single wavelength vibrating in a single plane is obtained this is known as plane polarised light.
Optical activity: “When plane polarised light is passed through a solution the light is rotated either in clockwise or in a anticlockwise direction by a certain angle. These types of compounds are known as optical active compound and such a property is known as optical activity.”
The angle by which the plane polarised light is rotated is measured by an instrument called polarimeter.
If the compound rotates the plane of plane polarised light to the right, i.e., in a clockwise direction, it is called dextrorotatory (Greek for right rotating) or the d-form and is indicated by placing a positive (+) sign before the degree of rotation.
If the light is rotated towards left (anticlockwise direction), the compound is said to be laevo-rotatory or the l-form and a negative (–) sign is placed before the degree of rotation.
Such (+) and (–) isomers of a compound are called optical isomers and the phenomenon is termed as optical isomerism.
#26 SUB 3M 🖼 3

Question

Explain molecular asymmetry, chirality and enantiomers.

Answer

The observation of Louis Pasteur (1848) that crystals of certain compounds exist in the form of mirror images laid the foundation of modern stereochemistry.
He demonstrated that aqueous solutions of both types of crystals showed optical rotation, equal in magnitude (for solution of equal concentration) but opposite in direction.
He believed that this difference in optical activity was associated with the three dimensional arrangements of atoms in the molecules (configurations) of two types of crystals.
Dutch scientist, J. Van’t Hoff and French scientist, C. LeBel in the same year (1874), independently argued that the spatial arrangement of four groups (valencies) around a central carbon is tetrahedral and if all the substituents attached to that carbon are different, the mirror image of the molecule is not superimposed (overlapped) on the molecule; such a carbon is called asymmetric carbon or stereocentre.
The resulting molecule would lack of symmetry and is referred to as asymmetric molecule.
Chirality: The symmetry and asymmetry are also observed in many day to day objects: a sphere, a cube, a cone, are all identical to their mirror images and can be superimposed.
However, many objects are non superimposable on their mirror images. For example, your left and right hand look similar but if you put your left hand on your right hand by moving them in the same plane, they do not coincide.
The objects which are non-superimposable on their mirror image (like a pair of hands) are said to be chiral and this property is known as chirality.
Chiral molecules are optically active, while the objects, which are, superimposable on their mirror images are called achiral. These molecules are optically inactive.
The above test of molecular chirality can be applied to organic molecules by constructing models and its mirror images or by drawing three dimensional structures and attempting to superimpose them in our minds.
Let us consider two simple molecules propan-2-ol and butan-2-ol and their mirror images.
As you can see very clearly, propan-2-ol (A)
does not contain an asymmetric carbon, as all the four groups attached to the tetrahedral carbon
are not different.
We rotate the mirror image (B) of the molecule by 180° (structure C) and try to overlap the structure (C) with the structure (A), these structures completely overlap. Thus propan-2-ol is an achiral molecule.
Butan-2-ol has four different groups attached to the tetrahedral carbon and as expected is chiral.
Some common examples of chiral molecules such as 2-chlorobutane, 2, 3-dihyroxypropanal, (OHC–CHOH–CH2OH), bromochloro-iodomethane (BrClCHI), 2-bromopropanoic acid (H3C–CHBr–COOH), etc.
The stereoisomers related to each other as non-superimposable mirror images are called enantiomers . A and B and D and E are enantiomers.
Enantiomers possess identical physical properties namely, melting point, boiling point, refractive index, etc.
They only differ with respect to the rotation of plane polarised light. If one of the enantiomer is dextro rotatory, the other will be laevo rotatory.
#27 SUB 2M

Question

What is racemic mixture?

Answer

"A mixture containing two enantiomers in equal proportions will have zero optical rotation, as the rotation due to one isomer will be cancelled by the rotation due to the other isomer. Such a mixture is known as racemic mixture or racemic modification."
A racemic mixture is represented by
prefixing dl or (±) before the name, for example (±) butan-2-ol. The process of conversion of enantiomer into a racemic mixture is known as racemisation.
#28 SUB 2M

Question

What is retention? Explain with examples.

Answer

Retention of configuration is the preservation of the spatial arrangement of bonds to an asymmetric centre during a chemical reaction or transformation.
In general, if during a reaction, no bond to the stereocentre is broken, the product will have the same general configuration of groups around the stereocentre as that of reactant.
Such a reaction is said to proceed with retention of the configuration.
Consider as an example, the reaction that takes place when (–)-2-methylbutan-1-ol is heated with concentrated hydrochloric acid.
It is important to note that configuration at a symmetric centre in the reactant and product is same but the sign of optical rotation has changed in the product.
This is so because two different compounds with same configuration at asymmetric centre may have different optical rotation. One may be dextrorotatory (plus sign of optical rotation) while other may be laevorotatory (negative sign of optical rotation).
#29 SUB 2M

Question

Explain inversion, retention and racemisation.

Answer

There are three outcomes for a reaction at an asymmetric carbon atom, when a bond directly linked to an asymmetric carbon atom is broken.
Consider the replacement of a group X by Y in the following reaction;
If (A) is the only compound obtained, the process is called retention of configuration.
If (B) is the only compound obtained, the process is called inversion of configuration. Configuration has been inverted in B.
If a 50:50 mixture of A and B is obtained then the process is called racemisation.
The product is optically inactive, as one isomer will rotate the plane polarised light in the direction opposite to another.
#30 SUB 4M 🖼 2

Question

Explain stereochemistry of SN2 and SN1 reaction.
SN2 reaction:

Answer

In case of optically active alkyl halides, the product formed as a result of SN2 mechanism has the inverted configuration as compared to the reactant.
This is because the nucleophile attaches itself on the side opposite to the one where the halogen atom is present.
When (–)-2-bromooctane is allowed to react with sodium hydroxide, (+)-octan-2-ol is formed with the –OH group occupying the position opposite to what bromide had occupied.
Thus SN2 reactions of optically active halides are accompanied by inversion of configuration.
SN1 reaction:
In case of optically active alkyl halides, SN1 reactions are accompanied by racemisation.
Actually the carbocation formed in the slow step being sp2 hybridised is planar (achiral).
The attack of the nucleophile may be accomplished from either side of the plane of carbocation resulting in a mixture of products,
One having the same configuration (the OH attaching on the same position as halide ion) and the other having opposite configuration (the OH attaching on the side opposite to halide ion).
This may be illustrated by hydrolysis of optically active 2-bromobutane, which results in the formation of (±)-butan-2-ol.
#31 SUB 3M ▦ 1

Question

Write down the difference between SN1 and SN2 reaction.

Answer

SN2 Reaction

SN1 Reaction

  • SN2 reaction means it is bimolecular nucleophilic substitution reaction.
  • SN1 reaction means it is unimolecular nucleophilic substitution reaction.
  • The rate of SN2 reaction depends on the concentration of Substrate and Nucleophiles both. Rate = K [S]1. [Nu]1
  • The rate of SN1 reaction depends on the concentration of Substrate only.
    Rate = K [S]1
  • SN2 reaction follow second order kinetics.
  • SN1 reaction follows first order kinetics.
  • The reaction of chloromethane and hydroxide ion gives methanol and chloride ion is example of SN2 reaction.

  • The reaction of tert-butyl bromide with hydroxide ion gives tert-butyl alcohol is example of SN1 reaction.

  • SN2 reaction complete in a single step without formation of an intermediate.
  • SN1 reaction complete in two steps.
    Step 1: Formation of carbocation by cleavage

    of C X in haloalkane

Step 2: Attack of nucleophiles on carbocation

to form product.

  • Reactivity order:
  • Reactivity order:
  • SN2 reaction shows inversion of configuration.
  • SN1 reaction shows racemic mixture.
#32 SUB 3M 🖼 2

Question

Explain β – Elimination reaction with example.
OR
Explain Dehydrohalogenation reaction of Haloalkane.
OR
Explain Zaitsev or Saytzeff rule with example.

Answer

When a haloalkane with β-hydrogen atom is heated with alcoholic solution of potassium hydroxide, there is elimination of hydrogen atom from β carbon and a halogen atom from the α-carbon atom.
As a result, an alkene is formed as a product. Since β-hydrogen atom is involved in elimination, it is often called β-elimination.
If there is possibility of formation of more than one alkene due to the availability of more than one
β-hydrogen atoms, usually one alkene is formed as the major product.
These form part of a pattern was first observed by Russian chemist, Alexander Zaitsev (also pronounced as Saytzeff).
A rule which can be summarised as in dehydrohalogenation reactions, the preferred product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms.
Thus, 2-bromopentane gives pent-2-ene as the major product.
#33 SUB 3M 🖼 2

Question

What is organo metallic compound? Explain Grignard reagent.

Answer

Most organic chlorides, bromides and iodides react with certain metals to give compounds containing
carbon-metal bonds. Such compounds are known as organo-metallic compounds.
An important class of organo-metallic compounds discovered by Victor Grignard in 1900 is alkyl
magnesium halide, RMgX, referred as Grignard Reagents.
These reagents are obtained by the reaction of haloalkanes with magnesium metal in dry ether.
In the Grignard reagent, the carbon-magnesium bond is covalent but highly polar, with carbon pulling electrons from electropositive magnesium; the magnesium halogen bond is essentially ionic.
Grignard reagents are highly reactive and react with any source of proton to give hydrocarbons. Even water, alcohols, amines are sufficiently acidic to convert them to corresponding hydrocarbons.
RMgX + H2O RH + Mg(OH)X
It is therefore necessary to avoid even traces of moisture from a Grignard reagent. That is why reaction is carried out in dry ether. On the other hand, this could be considered as one of the methods of converting halides to hydrocarbons.
#34 SUB 3M 🖼 6

Question

Write a note on Wurtz reaction and its limitations.

Answer

Alkyl halides react with sodium in dry ether to give hydrocarbons containing double the number of carbon atoms present in the halide.
This reaction is known as Wurtz reaction.
2RX + 2Na R – R + 2NaX
Limitation:
(1) This method is suitable to produce an alkane only with even number of carbon.
(2) If alkyl halide is tertiary, the alkene is major product.
#35 SUB 4M 🖼 1

Question

Why haloarenes are less reactive towards nucleophilic substitution reactions. Give reasons.

Answer

Aryl halides are extremely less reactive towards nucleophilic substitution reactions due to the following reasons:
(i) Resonance effect: In haloarenes, the electron pairs on halogen atom are in conjugation with π-electrons of the ring and the following resonating structures are possible:
C-CI bond acquires a partial double bond character due to resonance.
As a result, the bond cleavage in haloarene is more difficult than haloalkane and therefore, they are less reactive towards nucleophilic substitution reaction.
(ii) Difference in hybridization of carbon atom in C-X bond: In haloalkane, the carbon atom attached to halogen is sp3 hybridised while in case of haloarene, the carbon atom attached to halogen is sp2 hybridised.
The sp2 hybridised carbon with a greater s-character is more electronegative and can hold the electron pair of C-X bond more tightly than sp3 hybridised carbon in haloalkane with less s-character.
Thus, C-CI bond length in haloalkane is 177 pm while in haloarenes is 169 pm.
Since it is difficult to break a shorter bond than a longer bond, therefore, haloarenes are less reactive than haloalkanes towards nucleophilic substitution reaction.
(iii) Instability of phenyl cation: In case of haloarenes, the phenyl cation formed as a result of self-ionization will not be stabilised by resonance and therefore, SN1 mechanism is ruled out.
(iv) Repulsion: Because of the possible repulsion, it is less likely for the electron rich in nucleophile to approach electron rich arenes.
#36 SUB 4M 🖼 3

Question

Give the reaction of chlorobenzene with hydroxyl group.

Answer

Chlorobenzene can be converted into phenol by heating in aqueous sodium hydroxide solution at a temperature of 623K and a pressure of 300 atmospheres. Because atomatic compounds undergo nucleoplhic reactions on High Temp and High presure.
The presence of an electron withdrawing group (-NO2) at ortho- and para- positions increases the reactivity of haloarenes.
The effect is pronounced when (-NO2) group is introduced at ortho- and para positions. However, no effect on reactivity of haloarenes is observed by the presence of electron withdrawing group at meta-position.
#37 SUB 3M 🖼 1

Question

Why electron withdrawing groups such as – NO2 show their effect only at o and p – position and not at m – position.

Answer

As shown, the presence of nitro group at ortho- and para-positions withdraws the electron density from the benzene ring and thus facilitates the attack of the nucleophile on haloarene.
The carbanion thus, formed is stabilised through resonance.
The negative charge appeared at ortho- and
para-positions with respect to the halogen substituent is stabilised by –NO2 group while in case of
meta-nitrobenzene, none of the resonating structures bear the negative charge on carbon atom bearing the –NO2 group.
Therefore, the presence of nitro group at meta- position does not stabilise the negative charge and no effect on reactivity is observed by the presence of –NO2 group at meta-position.
#38 SUB 3M 🖼 1

Question

Why Haloarenes undergo electrophilic substitution reactions at o and p – position? Why it is less reactive than benzene?
OR
Although chlorine is an electron withdrawing group, yet it is ortho, para directing in electrophilic aromatic substitution reaction, Explain.

Answer

Haloarenes undergo the usual electrophilic reactions of the benzene ring such as halogenation, nitration, sulphonation and Friedel-Crafts reactions.
Halogen atom besides being slightly deactivating is o, p- directing; therefore, further substitution occurs at ortho- and para- positions with respect to the halogen atom.
The o, p-directing influence of halogen atom can be easily understood if we consider the resonating structures of halobenzene as shown:
Due to resonance, the electron density increases more at ortho- and para-positions than at meta-positions.
Further, the halogen atom because of its -I effect has some tendency to withdraw electrons from the benzene ring.
As a result, the ring gets somewhat deactivated as compared to benzene and hence the electrophilic substitution reactions in haloarenes occur slowly and require more drastic conditions as compared to those in benzene.
#39 SUB 3M 🖼 5

Question

Explain Halogenation, Nitration, Sulphonation and Friedel crafts reaction of chlorobenzene.

Answer

(i) Halogenation
(ii) Nitration
(iii) Sulphonation
(iv) Friedel-Crafts reaction
#40 SUB 4M 🖼 2

Question

Write down reaction of halogens with metals.
OR
Give Fittig, Wurtz – Fittig and Grignard reaction of haloarene.

Answer

(1) Fittig reaction:
2 mole of Aryl halides react with sodium in presence of dry ether to give diphenyl
(2) Wurtz-Fittig reaction:
A mixture of an alkyl halide and aryl halide gives an alkylarene when treated with sodium in dry ether is called Wurtz-Fittig reaction.
(3) Grignard reaction:
Aryl halide compounds react with Mg Metal in presence of dry ether & gives Aryl Magnesium halide.
#41 SUB 3M

Question

What are polyhalogen compounds? Write a note on dichloromethane (Methylene chloride)

Answer

“Carbon compounds containing more than one halogen atom are usually referred to as polyhalogen compounds.”
Uses:
Dichloromethane is widely used as a solvent as a paint remover, as a propellant in aerosols, and as a process solvent in the manufacture of drugs.
It is also used as a metal cleaning and finishing solvent.
Harmful effects:
Methylene chloride harms the human central nervous system.
Exposure to lower levels of methylene chloride in air can lead to slightly impaired hearing and vision.
Higher levels of methylene chloride in air cause dizziness, nausea, tingling, and numbness in the fingers and toes.
In humans, direct skin contact with methylene chloride causes intense burning and mild redness of the skin.
Direct contact with the eyes can burn the cornea.
#42 SUB 2M 🖼 1

Question

Write a note on trichloromethane (Chloroform).

Answer

Uses : Chloroform CHCl3 is employed as a solvent for fats, alkaloids, iodine and other substances.
The major use of chloroform today is in the production of the freon refrigerant R-22.
It was once used as a general anaesthetic in surgery but has been replaced by less toxic, safer anaesthetics, such as ether.
Harmful effects:
As might be expected from its use as an anaesthetic, inhaling chloroform vapours depresses the central nervous system.
Breathing about 900 parts of chloroform per million parts of air (900 parts per million) for a short time can cause dizziness, fatigue and headache.
Chronic chloroform exposure may cause damage to the liver (where chloroform is metabolised to phosgene) and to the kidneys, and some people develop sores when the skin is immersed in chloroform.
Chloroform is slowly oxidised by air in the presence of light to an extremely poisonous gas, carbonyl chloride, also known as phosgene.
It is therefore stored in closed dark coloured bottles completely filled so that air is kept out.
#43 SN 2M

Question

Write a short note on triiodomethane (Iodoform).

Answer

It was used earlier as an antiseptic but the antiseptic properties are due to the liberation of free iodine and not due to iodoform itself.
Due to its objectionable smell, it has been replaced by other formulations containing iodine.
#44 SN 3M

Question

Write short note on tetra chloromethane (CCl4) carbon tetra chloride.

Answer

Uses:
It is produced in large quantities for use in the manufacture of refrigerants and propellants for aerosol cans.
It is also used as feedstock in the synthesis of chlorofluorocarbons and other chemicals, pharmaceutical manufacturing, and general
solvent use.
Until the mid 1960s, it was also widely used as a cleaning fluid, both in industry, as a degreasing agent, and in the home, as a spot remover and as fire extinguisher.
Harmful effects:
There is some evidence that exposure to carbon tetrachloride causes liver cancer in humans.
The most common effects are dizziness, light headedness, nausea and vomiting, which can cause permanent damage to nerve cells.
In severe cases, these effects can lead rapidly to stupor, coma, unconsciousness or death.
Exposure to CCl4 can make the heart beat irregularly or stop. The chemical may irritate the eyes on contact.
When carbon tetrachloride is released into the air, it rises to the atmosphere and depletes the ozone layer.
Depletion of the ozone layer is believed to increase human exposure to Depletion of the ultraviolet rays, leading to increased skin cancer, eye diseases and disorders, and possible disruption of the immune system.
#45 SUB 2M

Question

Write note on freons.

Answer

The chlorofluorocarbon compounds of methane and ethane are collectively known as freons.
They are extremely stable, unreactive, non-toxic, non- corrosive and easily liquefiable gases.
Freon 12 (CCl2F2) is one of the most common freons in industrial use.
It is manufactured from tetrachloromethane by Swarts reaction.
These are usually produced for aerosol propellants, refrigeration and air conditioning purposes.
By 1974, total freon production in the world was about 2 billion pounds annually.
Most freon, even that used in refrigeration, eventually makes its way into the atmosphere where it diffuses unchanged into the stratosphere.
In stratosphere, freon is able to initiate radical chain reactions that can upset the natural ozone balance.
#46 SUB 3M 🖼 1

Question

Write a note on DDT.

Answer

DDT, the first chlorinated organic insecticides, was originally prepared in 1873, but it was not until 1939 that Paul Muller of Geigy Pharmaceuticals in Switzerland discovered the effectiveness of DDT as an insecticide.
Paul Muller was awarded the Nobel Prize in Medicine and Physiology in 1948 for this discovery.
The use of DDT increased enormously on a worldwide basis after world war II, primarily because of its effectiveness against the mosquito that spreads malaria and lice that carry typhus.
However, problems related to extensive use of DDT began to appear in the late 1940s.
Many species of insects developed resistance to DDT, and it was also discovered to have a high toxicity towards fish.
The chemical stability of DDT and its fat solubility compounded the problem.
DDT is not metabolised very rapidly by animals; instead, it is deposited and stored in the fatty tissues.
If ingestion continues at a steady rate, DDT builds up within the animal over time.
The use of DDT was banned in the United States in 1973, although it is still in use in some other parts of the world.
#47 SUB

Question

Give the uses of freon 12, DDT, carbon tetrachloride and iodoform.

Answer

(1) Uses of Freon – 12
Freon–12 (dichlorodifluoromethane, CCl2F2) is commonly known as CFC. It is used as a refrigerant in refrigerators and air conditioners. It is also used in aerosol spray propellants such as body sprays, hair sprays, etc. However, it damages the ozone layer. Hence, its manufacture was banned in the United States and many other countries in 1994.
(2) Use of DDT:
DDT (p, p-dichlorodiphenyltrichloroethane) is one of the best known insecticides. It is very effective against mosquitoes and lice. But due its harmful effects, it was banned in the United States in 1973.
(3) Uses of carbontetrachloride (CCI4)
It is used for manufacturing refrigerants and propellants for aerosol cans.
It is used as feedstock in the synthesis of chlorofluorocarbons and other chemicals.
It is used as a solvent in the manufacture of pharmaceutical products.
(4) Uses of iodoform (CHI3)
Iodoform was used earlier as an antiseptic, but now it has been replaced by other formulations-containing iodine-due to its objectionable smell. The antiseptic property of iodoform is only due to the liberation of free iodine when it comes in contact with the skin.
#48 SUB 2M PYQ 🖼 2

Question

Write down chemical equations to prepare following substances from 1-Chloropropane.
[March 2023] [2 Marks]

Answer

(i)
(ii)
#49 SUB 2M PYQ 🖼 1

Question

Give conversion: Ethene into butane.
[July 2022] [2 Marks]

Answer

#50 SUB 2M

Question

State the reaction equation of ethyl chloride with the following compounds:
(i) Aqueous KOH (ii) Alcoholic KOH
[June 2024] [2 Marks]

Answer

(i) CH3CH2Cl + KOH CH3CH2OH + KCl
Aqueous Ethanol
(ii) CH3CH2Cl + KOH CH2=CH2 + KCl + H2O
Alcoholic Ethane
Class 12 Chemistry (Part 2) 002
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#51 SUB 🖼 4

Question

Write structures of the following compounds:
(i) 2-Chloro-3-methylpentane
(ii) 1-Chloro-4-ethylcyclohexane
(iii) 4-tert. Butyl-3-iodoheptane
(iv) 1,4-Dibromobut-2-ene
(v) 1-Bromo-4-sec. butyl-2-methylbenzene

Answer

(i) 2-Chloro-3-methylpentane:
(ii) 1-Chloro-4-ethylcyclohexane:
(iii) 4-tert. Butyl-3-iodoheptane:
(iv) 1,4-Dibromobut-2-ene:
Br – CH2 – CH = CH – CH2 – Br
(v) 1-Bromo-4-sec. butyl-2 methylbenzene:
#52 SUB 🖼 1

Question

Why is sulphuric acid not used during the reaction of alcohols with KI?

Answer

H2SO4 cannot be used along with KI in the conversion of an alcohol to an alkyl iodide as it converts KI to corresponding acid, HI which is then oxidised by it to I2.
#53 SUB 🖼 1

Question

Write structures of different dihalogen derivatives of propane.

Answer

#54 SUB 🖼 4

Question

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

(i) A single monochloride:
All the hydrogen atoms are equivalent and replacement of any hydrogen will give the same product.
(ii) Three isomeric monochlorides:
CaH3CbH2CcH2CbH2CaH3
The equivalent hydrogens are grouped as a, b and c. The replacement of equivalent hydrogens will give the same product.
(iii) Four isomeric monochlorides:
Similarly the equivalent hydrogens are grouped as a, b, c and d. Thus, four isomeric products are possible.
#55 SUB 2M

Question

Arrange each set of compounds in order of increasing boiling points.
(i) Bromomethane, Bromoform, Chloromethane, Diabromomethane.
(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-Chlorobutane.
Isopropylchloride being branched has lower b.p. than 1-Chloropropane.
#56 SUB 🖼 9

Question

Draw the structures of major monohalo products in each of the following reactions:
(i)
(ii)
(iii)
(iv)
(v) CH3CH2Br + NaI
(vi)

Answer

(i)
(ii)
(iii)
(iv)
(v) CH3CH2Br + NaI CH3CH2I + NaBr
(vi)
#57 SUB 🖼 5

Question

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

(ii)
(iii)

Answer

(i) CH3CH2CH2CH2Br Being primary halide, there won't be any steric hindrance.
(ii) Secondary halide reacts faster than tertiary halide.
(iii) The presence of methyl group closer to the halide group will increase the steric hindrance and decrease the rate.
#58 SUB 🖼 4

Question

In the following pairs of halogen compounds, which compound undergoes faster SN1 reaction?
(i) and
(ii) and

Answer

(i) Tertiary halide reacts faster than secondary halide because of the greater stability of tert-carbocation
(ii) Because of greater stability of secondary carbocation than primary.
#59 SUB 🖼 5

Question

Identify A, B, C, D, E, R and R1 in the following:
(i) (ii)
(iii)

Answer

(i)
(ii)
(iii)
Mg(OH) X +
Class 12 Chemistry (Part 2) 003
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#60 SUB 1M 🖼 12

Question

Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:
(i) (CH3)2CHCH(Cl)CH3
(ii) CH3CH2CH(CH3)CH(C2H5)Cl
(iii) CH3CH2C(CH3)2CH2I
(iv) (CH3)3CCH2CH(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)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
(xi)
(xii)
#61 SUB 🖼 6

Question

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 = CClC6H4I – p

Answer

(i)
(ii)
(iii)
(iv)
(v)
(vi)
#62 SUB 🖼 7

Question

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)-1-iodooctane
(v) 2-Bromobutane
(vi) 4-tert-Butyl-3-iodoheptane
(vii) 1-Bromo-4-sec-butyl-2-methylbenzene
(viii) 1,4-Dibromobut-2-ene

Answer

(i) 2-Chloro-3-methylpentane
(ii) p-Bromochlorobenzene
(iii) 1-Chloro-4-ethylcyclohexane
(iv) 2-(2-Chlorophenyl)-1-iodooctane
(v) 2-Bromobutane
(vi) 4-tert-Butyl-3-iodoheptane
(vii) 1-Bromo-4-sec-butyl-2-methylbenzene
(viii) 1,4-Dibromobut-2-ene
Br – 1CH22CH = 3CH – 4CH2 – Br
#63 SUB

Question

Which one of the following has the highest dipole moment?
(i) CH2Cl2 (ii) CHCl3 (iii) CCl4

Answer

CH2Cl2 has the highest dipole moment.
Order of dipole moment: CCl4 < CHCl3 < CH2Cl2
#64 SUB 3M 🖼 2

Question

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

Answer

A hydrocarbon with the molecular formula, C5H10 belongs to the group with a general molecular formula CnH2n. Therefore, it may either be an alkene or a cycloalkane.
Since hydrocarbon does not react with chlorine in 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.
The reactions involved in the question are:
#65 SUB 🖼 4

Question

Write the isomers of the compound having formula C4H9Br.

Answer

There are four isomers of the compound having the formula C4H9Br. These isomers are given below.
(i)
(ii)
(iii)
(iv)
#66 SUB 🖼 5

Question

Write the equations for the preparation of 1-iodobutane from (i) 1-butanol, (ii) 1-chlorobutane, (iii) but-1-ene.

Answer

(i) 1-Iodobutane from 1-butanol
(ii) 1-Iodobutane from 1-chlorobutune
(iii) 1-Iodobutane from but-1-ene
#67 SUB

Question

What are ambident nucleophiles? Explain with an example.

Answer

Refer Que. no. 20 Page no. 18
#68 SUB 🖼 1

Question

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

Answer

(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–l
Therefore, CH3I will react faster than CH3Br in SN2 reactions with OH.
(ii)
The SN2 mechanism involves the attack of the nucleophile at the atom bearing the leaving group. But, in case of (CH3)3CCl, the attack of the nucleophile at the carbon atom is hindered because of the presence of bulky substituents on that carbon atom bearing the leaving group. On the other hand, there are no bulky substituents on the carbon atom bearing the leaving group in CH3CI. Hence, CH3CI reacts faster than (CH3)3CCI in SN2 reaction with OH.
#69 SUB 🖼 2

Question

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

(i) 1-Bromo-l-methylcyclohexane:
(ii) 2-Chloro-2-methylbutane:
(iii) 2,2,3-Trimethyl-3-bromopentane:
#70 SUB 🖼 3

Question

How will you bring about the following conversions:
(i) Ethanol to but-l-yne (ii) Ethane to bromoethene (iii) Propene to l-nitropropane
(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) l-Chlorobutane to n-octane
(x) Benzene to biphenyl.

Answer

(i) Ethanol to but-l-yne:
(ii) Ethane to bromoethene:
(iii) Propene to l-nitropropane:
(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:[June 2025]
(ix) l-Chlorobutane to n-octane:
(x) Benzene to biphenyl: [March 2020, March2022, April 2022]
#71 SUB 🖼 2

Question

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 reagents should be prepared under anhydrous conditions?

Answer

(i) 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. Therefore, the density of electrons of C-Cl bond near the Cl-atom is less in chlorobenzene than in cyclohexyl chloride.
Moreover, the -R effect of the benzene ring of chlorobenzene decreases the electron density of the C-Cl bond near the Cl-atom. As a result, the polarity of the C-Cl bond in chlorobenzene decreases. Hence, the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.
(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 polar molecules and so held together by
dipole-dipole interactions. Similarly, strong H- bonds exist between the water molecules.
The new force of attraction between the alkyl halides and water molecules is weaker than the alkyl halide-alkyl halide and water-water forces of attraction. Hence, alkyl halides (though polar) are immiscible with water.
  • (iii) Grignard reagents are very reactive. In the presence of moisture, they react to give alkanes.
Therefore, Grignard reagents should be prepared under anhydrous conditions.
#72 SUB

Question

Give the uses of freon 12, DDT, carbon tetrachloride and iodoform.

Answer

Refer Que. no. 47 Page no. 31
#73 SUB 🖼 20

Question

Write the structure of the major organic product in each of the following reactions:
(i) CH3– CH2 – CH2 – Cl + NaI
(ii) (CH3)3CBr + KOH
(iii) CH3CH(Br)CH2CH3 + NaOH
(iv) CH3CH2Br + KCN
(v) C6H5ONa + C2H5Cl
(vi) CH3CH2CH2OH + SOCl2
(vii) CH3CH2CH = CH2 + HBr
(viii) CH3CH = C(CH3)2 + HBr

Answer

(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
#74 SUB 🖼 4

Question

Write the mechanism of the following reaction: n BuBr + KCN n BuCN.

Answer

The given reaction is: n BuBr + KCN n BuCH
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.
#75 SUB 🖼 3

Question

Arrange the compounds of each set in order of reactivity towards SN2 displacement:
(i) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane
(ii) 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 2-Bromo-3-methylbutane
(iii) 1-Bromobutane, l-Bromo-2,2-dimethylpropane, l-Bromo-2-methylbutane, 1-Bromo-3-methylbutane.

Answer

(i)
SN2 reactivity order: 1o > 2o > 3o.
(ii)
(iii) 1-Bromobutane > 1-Bromo-3-Methylbutane >
1-Bromo-2-Methylbutane > 1-Bromo-2,2-dymethylbutane
The steric hindrance to the nucleophile in the SN2 mechanism increases with decrease in the distance of the substituents from the atom containing the leaving group. Further, the steric hindrance increases with an increase in the number of substituents.
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-3- methylbutane < 1- Bromobutane
#76 SUB 🖼 1

Question

Out of C6H5CH2CI and C6H5CI(Cl)–C6H5, which is more easily hydrolysed by aqueous KOH.

Answer

2° – carbocation is more stable than 1°– carbocation, so C6H5CH(CI)C6H5 gets easily hydrolysed.
#77 SUB 🖼 1

Question

p-Dichlorobenzene has higher m.p. than those of o- and m-isomers. Discuss it.

Answer

Three isomers of p-Dichlorobenzene
p-Dichlorobenzene is more symmetrical than o-and m-isomers. 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.
#78 SUB 🖼 12

Question

How the following conversions can be carried out?
(i) Propene to propan-l-ol (ii) Ethanol to but-l-yne
(iii) l-Bromopropane to 2-bromopropane (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-l-propene to 2-chloro-2-methylpropane
(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-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) Propene to propan-l-ol
(ii) Ethanol to but-l-yne
(iii) l-Bromopropane to 2-bromopropane
(iv) Toluene to benzyl alcohol
(v) Benzene to 4-bromonitrobenzene[July 2023]
(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-l-propene to 2-chloro-2-methylpropane
(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-Bromopropane to 1-bromopropane
(xvii) Chloroethane to butane
(xviii) Benzene to diphenyl[March 2024, March/April-2022, March 2020]
(xix) tert-Butyl bromide to isobutyl bromide
(xx) Aniline to phenylisocyanide
#79 SUB 🖼 2

Question

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 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.
On the other hand, an alcoholic solution of KOH contains alkoxide (RO) ion, which is a strong base. Thus, it can abstract a hydrogen from the b-carbon of the alkyl chloride and form an alkene by eliminating a molecule of HCI.
OH ion is a much weaker base than RO ion. Also, OH ion is highly solvated in an aqueous solution and as a result, the basic character of OH ion decreases. Therefore, it cannot abstract a hydrogen from the β-carbon.
#80 SUB 🖼 5

Question

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 - butyl bromide and isobutyl bromide.
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.
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-Methylpropane.
Also, compound (b) reacts with HBr to give compound (c) which is an isomer of (a).
Hence, compound (c) is 2-bromo-2-methylpropane.
#81 SUB 🖼 7

Question

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 aqueous KOH,
(v) Methyl bromide is treated with sodium in the presence of dry ether,
(vi) Methyl chloride is treated with KCN?

Answer

(i) When n-butyl chloride is treated with alcoholic KOH, the formation of but-l-ene takes place. This reaction is a dehydrohalogenation reaction.
(ii) When bromobenzene is treated with Mg in the presence of dry ether, phenylmagnesium bromide is formed.
(iii) Chlorobenzene does not undergo hydrolysis under normal conditions. However, it undergoes hydrolysis when heated in an aqueous sodium hydroxide solution at a temperature of 623 K and a pressure of 300 atm to form phenol.
(iv) When ethyl chloride is treated with aqueous KOH, it undergoes hydrolysis to form ethanol.
(v) When methyl bromide is treated with sodium in the presence of dry ether, ethane is formed. This reaction is known as the Wurtz reaction.
(vi) When methyl chloride is treated with KCN, it undergoes a substitution reaction to give methyl cyanide.
S match 100% type: 1 Q ⤓ Export ZIP
#82 MCQ ⚠ needs answer review 1M 🖼 104 ▦ 1

Question

(i)
(ii) CH3CH2CH2CH2Br
(iii)
9. In which of the following molecules carbon atom marked with asterisk (*) is asymmetric ?
10. Which of the following structure is enantiomeric with the molecule (A) given below:
Class 12 Chemistry (Part 2) 004
11. Which of the following is an example of
vicinal-dihallide?
12. The position of – Br in the compound
CH3CH = CHC(Br)(CH3)2, can be classified as:
13. Chlorobenzene is formed by reaction of chlorine with benzene in the presence of AlCl3. Which of the following species attacks the benzene ring in this reaction?
14. What is A in the following reaction?
(A)
(B)
(C)
(D)
15. Ethylidene chloride is a/an ________.
16. A primary alkyl halide would prefer to undergo:
17. Which of the following alkyl halides will undergo SN1 reaction most rapidly?
18. Which is the correct IUPAC name of ?
19. What should be the correct IUPAC name for diethylbromomethane?
20. The reaction of toluene with chloride in the presence of iron and in the absence of light yields.
21. Chloromethane on treatment with excess of ammonia yields mainly:
22. Molecules whose mirror image is non superimposable over them are known as chiral. Which of the following molecule is chiral in nature?
23. Reaction of C6H5CH2Br with aqueous sodium hydroxide follows:
24. Which of the carbon atoms present in the molecule given below are asymmetric?
25. Which of the following compounds will give racemic mixture on nucleophilic substitution by OH ion?
(i) (ii)
(iii)
26. Arrange the following compounds in increasing order of rate of reaction towards nucleophilic substitution reaction.
(i) (ii) (iii)
27. Arrange the following compounds in increasing order of rate of reaction towards nucleophilic substitution reaction.
(i) (ii) (iii)
28. Arrange the following compounds in increasing order of rate of reaction towards nucleophilic substitution reaction.
(i) (ii) (iii)
29. Arrange the following compounds in increasing order of rate of reaction towards nucleophilic substitution reaction.
(i) (ii) (iii)
30. Which is the correct increasing order of boiling points of the following compounds?
Butane, 1-Iodobutane, 1-Bromobutane, 1-Chlorobutane
< 1-Bromobutane < 1-Iodobutane
< 1-Chlorobutane < Butane
< 1-Bromobutane < 1-Chlorobutane
< 1-Iodobutane < 1-Bromobutane
31. Which is the correct increasing order of boiling points of the following compounds?
1-Bromoethane, 1-Bromopropane, 1-Bromobutane, Bromobenzene.
< 1-Bromopropane < 1-Bromoethane
< 1-Bromopropane < 1-Bromobutane
< 1-Bromoethane < Bromobenzene
< 1-Bromobutane < Bromobenzene
1. What is the hybridization of carbon atom having halogen atom in haloarene compounds ?
2. Which structure represents vicinal dihalide?
3. Which structure represents vinyl halide?
4. Which compound represents allylic halide?
5. Which structure represents allyl halide?
6. Which structure represents geminal dihalides?
7. Propyl iodide and isopropyl iodide are:
8. Which sentence indicates alkylidine compounds?
9. Which compound represents secondary halide?
10. Which substance is a primary halide?
11. From which substance trihalogen haloform compound is obtained?
12. How many π-bonds are present in benzene hexa chloride (B.H.C.)?
13. Which molecular formula represents benzylic halide?
14. Identify structure of benzylic halide:
15. Identify ethylidene dibromide:
16. Formula of geminal dihalide is _______:
17. Which of the following is the structure of alkylene dihalide?
18. Match the Column A with Column B.
19. Which halogen element present in thyroxine?
20. Give the IUPAC name of
21. What is the IUPAC name of
?
22. Give the IUPAC name of
Br – CH2 – C C – CH2Br
23. Indicate the structural formula of
1-chloro-4-secbutyl-2-methyl benzene.
(A)
(B)
(C)
(D)
24. The IUPAC name of the compound, (CH3)2CHCH2CH2Br is:
25. What is common name of 3-bromopropene?
26. The order of polarity of CH3I, CH3Br and CH3Cl molecules follows the order:
27. Which substance has highest dipole moment?
28. Which substance has highest melting point?
29. Which of the following substances has the highest bond enthalpy?
30. Which of the following substance has the weakest bond enthalpy?
31. Methyl bromide reacts with AgF to give methyl fluoride and silver bromide. This reaction is called:
32. The following reaction is known by which name?
C2H5OH + SOCl2
C
2H5Cl + SO2(g) + HCl(g)
33. Which product is obtained by reaction between 2-methyl propene and HBr?
34. Mention the main product obtained by reaction between silver acetate and bromine water in presence of carbon disulphide.
35. Which products are obtained by reaction between ethanol and thionyl chloride in presence of catalyst pyridine?
36. Which reaction gives the product 2,2-dibromo propane?
37. Indicate the decreasing order of reactivity of HX for reaction R – OH + HX R – X + H2O.
38. CH2 = CH – CCl3 + HBr _______ product is obtained.
39. Mention the type of reaction for the reaction between propane and chlorine gas in presence of sunlight.
40. Identify correct method to prepare methyl fluoride?
41. Which reagent is not appropriate to prepare alkyl halide from alcohol?
42. Identify industrial method to prepare ethyl bromide.
43. X C2H5Cl; Y CH3COCl, identify X and Y.
44. An alkyl halide may be converted into an alcohol by:
45. Preparation of alkyl halides in laboratory is least preferred by:
46. Ethyl alcohol gives ethyl chloride on treatment with:
47. Number of monochloro derivatives obtained when neo-pentane is chlorinated, is
48. The reaction RCl + Nal R – I + NaCl is known as:
49. Which of the following can be obtained by halide exchange method?
50. Bromination of methane in presence of sunlight is a:
51. Anti-Markovnikov addition of HBr is not observed in.
52. In the preparation of chlorobenzene from aniline, the most suitable reagent is.
53. In the chemical reactions,
The compounds 'A' and 'B' respectively are:
54. + Cl2 M Diphenyl
In this reaction identify molecule (M) and Reagent (R).
55. Which of the following is possible at lower temperature?
56. On reaction of aniline with NaNO2 and HCl at 273K temperature, which product is obtained.
57. How many s and p bonds are present in benzene diazonium chloride?
58. The correct order of melting and boiling points of the primary (1°), secondary (2°) and tertiary (3°) alkyl halides is:
59. Which substance has highest boiling point?
60. For a given alkyl group, the densities/b. p. are in the order:
61. Arrange the following compounds in the decreasing order of their boiling points.
(i) CH3Br (ii) CH3CH2Br
(iii) CH3CH2CH2Br (iv) CH2CH2CH2CH2Br
62. Which of the following has the highest boiling point?
63. Which of the following has the highest density?
64. Identify density order of following compounds.
65. Dehydrohalogenation in haloalkanes produces:
66. The greater the ionic character of the carbon metal bond:
67. The order of reactivities of methyl halides in the formation of Grignard reagent is:
68. On treating a mixture of two alkyl halides with sodium metal in dry ether, 2-methyl propane was obtained. The alkyl halides are:
69. The given reaction is an example of,
C2H5Br + KCN(aq) C2H5CN + KBr:
70. Aryl halides are less reactive towards electrophiles than alkyl halides due to:
71. In Wurtz reaction, alkyl halide reacts with
72. X + KCN CH3CN CH3CH2NH2, what is (X)?
73. Reaction of alkyl halides with aromatic compounds in presence of anhy, AlCl3 is known as:
74. 1, 2-dibromo cyclohexane on dehydrohalogenation gives
75. Which one of the following is not true for the hydrolysis of t-butyl bromide with aqueous NaOH?
76. Grignard reagent is prepared by the reaction between:
77. A mixture of sodium acetate and sodalime is heated and the product treated with excess of chlorine in presence of bright sunlight. The product is:
78. 1-Chlorobutane on reaction with alcoholic KOH gives:
79. Which halide does not get hydrolysed by sodium hydroxide?
80. Optically active compound is:
81. Which one is the most reactive towards SN1 reactions?
82. Which of the following applies in the reaction,
CH3CH(Br)CH2CH3
(i) CH3CH = CHCH3 (major product)
(ii) CH2 = CHCH2CH3 (minor product)
83. Reaction of t-butyl bromide with sodium methoxide produces
84. An alkyl iodide on standing darkens, due to:
85. X compound reacts with Na to give CH3CH2CH2CH3, then compound X is:
86. Mention the reaction conditions to convert ethyl bromide into ethyl alcohol.
87. Which two alkyl chloride on heating with Na metal in presence of dry ether give isobutane?
88. Which substance gives fast SN2 reaction?
89. What is obtained by reaction between tertiary butyl bromide and CH3ONa?
90. C6H5Cl + KCN X Y then X and Y are respectively?
91. Mention the type of dehydrohalogenation reactions of alkyl halide.
92. What is obtained when alkyl halide is heated with Mg metal in presence of dry ether?
93. Identify reaction condition for given reaction?
94.
95. C6H5CH2Br ______ ?
96. Match the Column A with Column B.
97. Identify (Z) in the following reaction series,
C2H5l (X) (Y) (Z):
98. t-butyl chloride preferably undergo hydrolysis by?
99. n-Propyl bromide reacts with ethanolic KOH to form:
100. Compound 'A' reacts with alcoholic KOH to yield compound 'B' which on ozonolysis followed by reaction with Zn/H2O gives methanal and propanal. Compound 'A' is _______.
101. 9.65 C of electric current is passed through fused anhydrous magnesium chloride. The magnesium metal thus, obtained is completely converted into a Grignard reagent. The number of moles of the Grignard reagent obtained is
102. When 32.25 g of ethyl chloride is subjected to dehydrohalogenation reaction the yield of the alkene formed is 50%. The mass of the product formed is (atomic mass of chlorine is 35.5)
103. SN1 reaction is favoured by:
104. CH3Br + OH CH3OH + Br reaction proceeds by SN2 mechanism. Its rate is dependent on the concentration of:
105. The C–Mg bond in CH3CH2MgBr is:
106. In SN1 reaction, the first step involves the formation of:
107. Which alkyl halide is preferentially hydrolysed by SN1 mechanism?
108. The Mg–Br bond in CH3CH2MgBr is:
109. Which of the following statements about SN2 mechanisms is incorrect?
110. The end product (Q) is in the following sequence of reaction:
111. Which of the following reactions will give the major and minor products?
112. SN1 reaction of alkyl halides leads to:
113. (CH3)3CMgBr on reaction with D2O produces
114. 'A' is:
(A)
(B)
(C)
(D)
115. CH3CH2CH2Cl B C D
In the above reaction, the product D is
116. Identify the end product (C) in the following sequence:
C
117. Fittig reaction can be used to prepare:
118. The antiseptic character of iodoform is due to:
119. Which one of the following compound reacts with chlorobenzene to produce DDT?
120. Chloroform is kept in dark coloured bottles because:
121. Solvent used in dry-cleaning of clothes is:
122. CCl4 is insoluble in water because:
123. Identify IUPAC name of the phosgene.
124. Which substance damages the layer of ozone gas?
125. How many chlorine atoms are present in D.D.T.?
126. Which statement is wrong about chloroform?
127. Which of the following is not inflammable?
128. Chloroform is slowly oxidised by air in the presence of light and air to form:
129. Which of the following have antiseptic property?
130. Assertion : Reaction of CH3–CH(Br)–CH2–CH3 with alcoholic KOH gives
CH3 – CH = CH – CH3.
Reason : The elimination reaction follows Markovnikov's law.
131. Assertion : CH2=CH–CH2–X is an example of an allylic halide.
Reason : In an allylic halide, the halogen atom is bonded to the carbon atom, which has sp2 hybridization.
132. Assertion : Haloalkanes react with KCN to form alkyl cyanide as the major product, while reacting with AgCN to form isocyanide as the major product.
Reason : KCN and AgCN both are ionic compounds.
133. Assertion : If the –NO2 group is present in the ring, it is easy to replace the –Cl group with –OH in chlorobenzene.
Reason : The nitro group leads to strengthening of the C-Cl bond in chlorobenzene.
134. Assertion : The SN2 mechanism proceeds with racemization, whereas the SN1 mechanism proceeds with complete stereochemical inversion.
Reason : SN2 is a two-step reaction, whereas SN1 is a one-step reaction.
135. Assertion : The common name of 1,1-dichloroethane is ethylidene chloride.
Reason : Ethylidene chloride is a geminal dihalide.
136. Assertion : In haloarenes, electrophilic substitution processes occur slowly and require more challenging conditions compared to benzene.
Reason : Halogens are ortho- and para-directing.
137. Assertion : The melting points of symmetrical dihalobenzene are almost the same.
Reason : The molecular weights of symmetrical dihalobenzene are different.
138. Assertion : For the SN1 mechanism, the order of reactivity of alkyl halides is
R-X > 2° R-X > 1° R-X.
Reason : The mechanism follows a carbocation formation pathway.
139. Assertion : Aryl halides are highly reactive towards nucleophilic substitution reaction.
Reason : In the case of aryl halides, the halogen atom is bonded to a carbon atom that has sp hybridization.
140. Assertion : The boiling points of alkyl halide compounds follow the order: R-I > R-Br > R-Cl > R-F.
Reason : Alkyl chloride, bromide, and iodide compounds with almost similar molecular weights have higher boiling points than hydrocarbon compounds.
141. Assertion : The Wurtz reaction of tert-butyl bromide gives 2,2,3, 3-tetramethylbutene.
Reason : In the Wurtz reaction, when an alkyl halide is treated with sodium in dry ether, a hydrocarbon with twice the number of carbon atoms compared to the alkyl halide is obtained.
142. Assertion : The substitution of the –Cl group from chlorobenzene by the -OH group is more difficult compared to chloroethane.
Reason : In chlorobenzene, the C-Cl bond exhibits some characteristics of a partial double bond.
143. Assertion : The reaction of methyl chloride with KCN produces methyl cyanide.
Reason : CN is a strong nucleophile.
144. Assertion : 3° alkyl halides are more reactive for the SN1 mechanism.
Reason : In the SN1 mechanism, the reaction rate depends on the concentration of the alkyl halide.
145. Assertion : The hydrolysis of (-)-2-bromo-octane results in the inversion of configuration.
Reason : This reaction occurs through the formation of a carbocation.
146. Assertion : CHCl3 is filled to the top in a coloured glass bottle.
Reason : CHCl3 forms phosgene when exposed to air.
147. Assertion : Alkyl halides are insoluble in water.
Reason : Although alkyl halides are polar, they do not form hydrogen bonds with water molecules.
148. Assertion : The reaction of styrene (vinyl benzene) with HBr gives 1-bromo-1-phenylethene.
Reason : The benzyl radical is more stable than the alkyl radical.
149. Assertion : The reaction of 2-bromobutane with alcoholic KOH gives but-2-ene.
Reason : The hydrogen on the C3 carbon is more acidic than the hydrogen on the C1 carbon.
150. Assertion : Chlorobenzene is less reactive towards electrophilic substitution reactions than benzene.
Reason : In chlorobenzene, the resonance effect makes the carbocation unstable.
151. Assertion : (CH3)2CH-CH2-CH3 can undergo free radical monochlorination to form four different monochloro isomers.
Reason : (CH3)2CH–CH2–CH3 has four different types of hydrogen atoms.
152. Assertion : The halogen atom cannot replace the -OH group of phenol to form an aryl halide compound.
Reason : Phenol reacts violently with halogen acids.
153. Choose the correct option for the True (T) and False (F) statements given below:
(i) Chloramphenicol is useful in the treatment of malaria.
(ii) Halothane is used as an anesthetic during surgery.
(iii) Chloroquine is used as a medicine to treat typhoid.
(iv) Halobenzene contains a halogen atom attached to a carbon having sp2 hybridization.
154. Choose the correct option for the True (T) and False (F) statements given below:
(i) The reactivity of alcohols for the Lucas test follows the order: 3° > 2° > 1°.
(ii) 3-Bromo cyclohex-1-ene is an allylic halide.
(iii) 1-Bromo-2,2-dimethyl propane is a tertiary alkyl halide.
(iv) 4-Bromo-toluene is a benzylic halide.
155. Choose the correct option for the True (T) and False (F) statements given below:
(i) The dipole moment (polarity) of C–X bonds (where X = F, Cl, Br, I) decreases from C–F to C–I.
(ii) 1,2-Dichloroethane is called a geminal dihalide.
(iii) Free radical monochlorination of 2-methylbutane gives four different structural isomers.
(iv) Alkyl iodide can be obtained from alkyl bromide by the Finkelstein process.
156. Choose the correct option for the True (T) and False (F) statements given below:
(i) Iodobenzene is not a Sandmeyer product.
(ii) Haloarenes can be obtained from aromatic hydrocarbons by the nucleophilic substitution reaction.
(iii) Photochemical chlorination of neopentane gives only one monochloro product.
(iv) The boiling point of the para isomer in a dihaloarene compound is higher than that of the ortho and meta isomers.
157. Choose the correct option for the True (T) and False (F) statements given below:
(i) The boiling point increases from primary to tertiary in isomeric haloalkanes.
(ii) The density increases as the number of equivalent halogens increases in alkyl halides containing similar carbons.
(iii) NO2 is an amphiphilic Nucleophilic reagent.
(iv) Haloalkane compounds react with KCN to form cyanide, whereas with AgCN they form isocyanide.
158. Choose the correct option for the True (T) and False (F) statements given below:
(i) An alkyl halide with a chiral carbon gives a 1:1 mixture of both enantiomers in an SN1 reaction.
(ii) The product obtained from the SN2 reaction of a haloalkane shows inversion of configuration.
(iii) Enantiomers of enantiomeric compounds cannot superimpose on each other.
(iv) A racemic mixture shows optical activity.
159. Choose the correct option for the True (T) and False (F) statements below:
(i) The reactivity of tertiary butyl bromide in an SN2 reaction is the highest.
(ii) If a compound rotates plane-polarized light to the right, it is said to be dextrorotatory.
(iii) 2, 3-Dihydroxypropanal shows the property of chirality.
(iv) The SN2 reaction of an alkyl halide takes place in a single step.
160. Choose the correct option for the True (T) and False (F) statements below:
(i) The nucleophilic substitution process of chlorobenzene cannot occur through SN1 mechanism.
(ii) A racemic mixture rotates plane-polarized light to the right.
(iii) A compound with four different atoms or groups attached to carbon is optically active.
(iv) A reaction involving a Grignard reagent is carried out in dry ether to easily remove side products.
161. Choose the correct option for the True (T) and False (F) statements below:
(i) The reaction of two moles of aryl halides with sodium in dry ether is called the Wurtz-Fitting reaction.
(ii) The production of Freon-1, 2 is done through the Swarts reaction from trichloromethane.
(iii) Freon gas disturbs the natural ozone balance.
(iv) Contact of CH2Cl2 with human eyes can burn the cornea.
162. Choose the correct option for the True (T) and False (F) statements below:
(i) The reactivity of chlorobenzene increases when a -NO2 group is added to the ortho and para positions, but adding a -NO2 group to the meta position has no effect on reactivity.
(ii) Chloroform has pesticidal properties.
(iii) DDT is a chlorine-containing organic pesticide.
(iv) The Wurtz reaction of a haloalkane results in an alkene hydrocarbon with double the carbon count.
163. Choose the correct option for the following true (T) and false (F) statements:
(i) The correct increasing order of reactivity for nucleophilic substitution reactions is: Chlorobenzene < m-Nitrochlorobenzene < o-Nitrochlorobenzene.
(ii) The reaction of aqueous sodium hydroxide with C6H5-CH2Br follows the SN2 mechanism.
(iii) The correct increasing order of boiling points is: 1-Chlorobenzene < 1-Bromobutene < 1-Iodobutene.
(iv) The nucleophilic substitution reaction of with OH ion gives a

racemic mixture.
164. Choose the correct option for the following true (T) and false (F) statements:
(i) SN2 mechanism is not applicable to tertiary alkyl halides.
(ii) The rate of the SN2 mechanism depends on the concentration of the nucleophile.
(iii) The product from the SN2 reaction of chloromethane with OH⁻ shows inversion of configuration.
(iv) The SN1 mechanism occurs in two steps, whereas the SN2 mechanism occurs in one step.

Column-A

Column-B

(1) Benzyl bromide

(M)

(2) Ethylidene bromide

(N) CH3CHBr2

(3) Phenyl bromide

(O) CH2 = CHCH2Br

(4) Allyl bromide

(P)

Options

  1. (A) TFFT
  2. (B) FFTF
  3. (C) FFTT
  4. (D) TFTT

Answer

not detected

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#86 MCQ ⚠ needs answer review 1M

Question

(D)

Answer

not detected

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No questions extracted under this section.

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#104 CS 🖼 6

Question

Alkyl halide nucleophilic substitution reactions primarily follow the SN1 and SN2 mechanisms because the C–X bond is polar. The reaction rate of the SN1 mechanism depends on the stability of the intermediate carbocation, while the reaction rate of the SN2 mechanism depends on steric hindrance. Chirality plays a crucial role in understanding the mechanisms of SN1 and SN2. A chiral alkyl halide undergoes the SN1 mechanism to give a racemic mixture, whereas the SN2 mechanism is characterized by inversion of configuration.
(a) Which of the following compounds is chiral?
(A) 2-methyl-2-chloropropane
(B) 2-chloropropane
(C) 2,2-dimethyl-1-chloropropane
(D) 2-chlorobutane
(b) Which of the following compounds is more reactive for the SN2 mechanism?
(A) CH3–CH2–Br (B)
(C) (D) CH3 – CH2 – Cl
(c)
Arrange the four butyl bromides in the ascending order of their reactivity for the SN1 mechanism.
(A) I < II < III < IV
(B) IV < II < I < III
(C) II < I < III < IV
(D) IV < I < II < III
(d) Which of the following compounds will give a racemic mixture according to the SN1 mechanism?
(A) (CH3)2 CH – CH2 – Br
(B) (CH3)3 CH – Br
(C) CH3 – CH (Br) – CH2 – CH3
(D) (CH3)2 CH – Br

Answer

(a) (D) 2-Chlorobutane
2-Chloro butane: The second carbon is bonded to a hydrogen atom (H), a chlorine atom (Cl), a methyl group (CH3), and an ethyl group (C2H5). Since all four are different, it is chiral.
(b) (A) CH3–CH2–Br
SN2 reactivity depends on steric hindrance. The less crowded the carbon atom attached to the leaving group, the faster the reaction.
The order of reactivity: Primary (1°) > Secondary (2°) > Tertiary (3°).
Bromine is a better leaving group than chlorine because the C–Br bond is weaker and the Br ion is more stable. Therefore, Ethyl bromide is the most reactive.
(c) (B) IV < II < I < III
SN1 reactivity depends on the stability of the carbocation intermediate.
The stability order for carbocations: Tertiary (3°) > Secondary (2°) > Primary (1°).
I: CH3–CH2–CH(Br)–CH3: 2° Carbocation.
II: (CH3)2CH–CH2–Br: 1° Alkyl halide, but it has some branching nearby.
III: (CH3)3C–Br: 3° Carbocation (Most stable).
IV: CH3–CH2–CH2–CH2–Br: 1° Alkyl halide (Least stable).
Order: IV < II < I < III
(d) (C) CH3 – CH (Br) – CH2 – CH3
A chiral alkyl halide undergoes SN1 to produce a racemic mixture.
2-bromobutane: Upon losing the bromine atom, it forms a secondary carbocation. The subsequent attack by a nucleophile results in a 50:50 mixture of the (R) and (S) enantiomers, known as a racemic mixture.
#105 CS 🖼 10

Question

When a haloalkane with a β-hydrogen atom is heated with alcoholic KOH, a dehydrohalogenation reaction occurs, resulting in the formation of an alkene. This process is called β-elimination because the hydrogen atom present at the β-position of the haloalkane is removed.
If a haloalkane contains more than one β-hydrogen, multiple products may be formed. However, the major product is the one in which the alkene has the greatest number of alkyl groups attached. This rule is known as the Zaitsev's rule.
(a)
Identify A and B
(A) (B)
(A)
(B)
(C)
(D)
(b)
Identify the product A.
(A) 2-Methyl propene
(B) But-2-ene
(C) But-1-ene
(D) 2-Methyl-but-2-ene
(c)
Identify the product C.
(A) Propene (B) Propyne
(C) Propan-1-ol (D) Propan-2-ol
(d) The compound (X) C6H13Cl undergoes dehydrohalogenation to give CH3= CH–C2H5. Identify compound X.
(A)
(B)
(C)
(D) None of these
(e)
Identify X and Y.
(X) (Y)
(A) Dilute NaOH HBr / Acetic acid
(B) Alcoholic NaOH HBr / Acetic acid
(C) Aqueous NaOH Br2 / Acetic acid
(D) Alcoholic NaOH Br2 / CHCl3

Answer

(a) (B)
(b) (A) 2-Methyl propene
(c) (D) Propan-2-ol
(d) (C)
(e) (B) Alcoholic NaOH HBr / Acetic acid
#106 CS 🖼 22

Question

Alkyl magnesium halides are Grignard reagents. In the Grignard reaction, the carbon-magnesium bond is covalent but it is more polar because the carbon pulls electrons from the electron-rich magnesium. The magnesium-halogen bond is essentially ionic. The hydrocarbon part of the Grignard reagent acts as a source of carbon anions. Therefore, the Grignard reaction rapidly undergoes nucleophilic addition with aldehydes and ketones, resulting in the formation of nucleophilic products, which upon hydrolysis yield alcohols.
(a) Which of the following Grignard reagents is required to convert propanone to 2-methylpropan-2-ol?
(A) CH3–CH2–Mg Br
(B) CH3–Mg Br
(C)
(D) Above all
(b) CH3 – CH2 – OH
Identify the product B.
(A) CH3–CH2–CH2 –OH
(B) CH3–CH2–CH2–CH2–OH
(C) CH3–CH2–OH
(D)
(c)
Identify the product E.
(A)
(B) (CH3)3 CH
(C) CH3 – CH2 – CH3
(D) CH3 – CH2 – CH2 – CH3
(d)
Identify the product B.
(A) (B)
(C) (D) None of these
(e) Which of the following Grignard reactions with methanal can produce ?
(A)
(B)
(C)
(D)

Answer

(a) (B) CH3–Mg Br
(b) (D)
(c) (B) (CH3)3 CH
It’s a Wortz reaction. The product is formed by dimensation of two alkyl groups.
So, R' = (CH3)3Ic – X (X = Cl/Br)
(d) (A)
(e) (D)
#107 CS

Question

A pharmaceutical company is synthesizing an important intermediate using chloroethane (C2H5Cl) and bromoethane (C2H5Br) as alkylating agents. The reaction conditions (solvent, base strength, temperature) determine whether the reaction proceeds by SN1 or SN2 mechanism.
The following experimental observations were made:

Substrate

Solvent used

Reaction rate

C2H5Br

Stong base in polar aprotic solvent (DMSO)

Very fast

C2H5Cl

Weak base in protic solvent (Ethanol)

Slow

tert-Butyl chloride (+ - Bucl)

Ethanol

Moderate

tert-Butyl chloride (+ - BuCl)

Water

Very fast

Additional data:
C2H5Br bond is weaker than C2H5Cl
Polar protic solvents stabilize carbocations.
Polar aprotic solvents enhance nucleophilicity.
The company wants to understand which reactions follow SN1 or SN2, and which substrate - solvent combination gives maximum yeild.
(a) Identify which reactions above proceeds via SN2 mechanism. Give reason.
(b) Explain why C2H5Br reacts much faster than C2H5Cl in SN2 reaction.
(c) Why does tert-butyl chloride react faster in water than in ethanol?
(d) Predict whether aryl halides (like chlorobenzene) undergo SN1 or SN2. Explain.

Answer

(a) Only C2H5Br in polar aprotic solvent (DMSO) undergoes SN2.
Reason:
Primary halide favors backside attack.
polar aprotic solvent nucleophile becomes stronger.
(b) Because:
C–Br bond is weak (lower bond energy).
Br is better leaving group than Cl.
Transition state stabilizes more effectively.
(c) SN1 reaction Rate depends on carbocation stability.
Water is most polar solvent, stabilizes carbocation and increases ionization rate.
(d) Aryl halides do not undergo SN1 or SN2.
SN1 not possible because, carbocation (aryl+ is highly unstable and positive charge cannot form in benzene ring.
SN2 not possible because, backside attack is blocked by ring structure.
#108 CS

Question

A chemical industry manufactures chlorobenzene, bromobenzene and phenyl chloride derivatives.
They study the effect of substituents on:
Reactivity of haloarenes
Nucleophilic substitution
Electrophilic substitution
Bond stength and stability
Some observations:

Compound

Reactivity toward nucleophiles

(A)

Chlorobenzene

Very low

(B)

p-Nitrochlorobenzene

High

(C)

o-Nitrochlorobenzene

Even higher

(D)

p-Methoxy chlorobenzene

Lowest

Important points:
NO2 group = Strong electron withdrawing activates nucleophilic substitution.
OCH3 group = Electron donating deactivates nucleophilic substitution.
Bromine is better leaving group than chlorine.
(a) Arrange the compounds in increasing order of nucleophilic substitution reactivity.
(b) Why does p-nitrochlorobenzene react faster than chlorobenzene?
(c) Why is o-nitrobromobenzene more reactive than p-nitrochlorobenzene?
(d) Explain why p-methoxychlorobenzene shows the lowest reactivity.

Answer

(a) D < A < B < C (Lowest Highest)
(b) Because of NO2 group strongly electron with drawing, stabilizes Meisen Heimer complex and facilitates nucleophilic attack.
(c) Because, NO2 at ortho position withdraws electrons strongly. Bromine is better leaving group then chlorine and ortho effect increases stabilization.
(d) Because, OCH3 group (i) Electron donating (+M) (ii) Increasing electron density on the ring (iii) Repels nucleophiles, (iv) Decreases Meisen\Heimer complex formation.
Haloalkane: An alkane in which one or more H atoms are replaced by halogen atoms (R–X).
Haloarene (Aryl Halide): An aromatic compound where a halogen is bonded directly to the aromatic ring.
Alkyl Halide: A haloalkane where the halogen is attached to an sp3 carbon (same as haloalkane in practice).
Aryl Halide: Halogen bonded to an sp2 carbon of an aromatic ring (C–X is stronger, less reactive).
Vinylic Halide: A halogen attached to an sp2 carbon of a C=C (halogen on the double bond carbon).
Allylic Halide: A halogen bonded to a carbon next to a double bond (stabilized by resonance).
Benzylic Halide: A halogen attached to the carbon next to an aromatic ring (resonance-stabilized intermediate).
Nucleophile: An electron-rich species that donates an electron pair to form a new bond (Nu,:OH,:CN, etc.).
Electrophile: An electron-deficient species that accepts an electron pair (E+, H+, Br+, carbocations).
SN1 Reaction: Unimolecular nucleophilic substitution; rate [R–X]; proceeds via carbocation intermediate; favours 3° centres.
SN2 Reaction: Bimolecular nucleophilic substitution; rate [R–X][Nu]; backside attack with inversion of configuration; favours 1° centres.
Carbocation: Positively charged carbon intermediate (R–C+), stabilized by alkyl groups via hyperconjugation.
Radical Halogenation: Free-radical substitution (initiation, propagation, termination) commonly used to halogenate alkanes (UV light required).
Grignard Reagent: Organomagnesium halide (R–MgX) formed in dry ether; strong nucleophile/base used for C–C bond formation.
Sandmeyer Reaction: Conversion of aryl diazonium salts to aryl halides (CuX mediated) — key route to introduce Cl/Br/CN/I onto benzene.
Saytzeff (Zaitsev) Rule: In elimination, the more substituted (more stable) alkene is the major product.
Wurtz/Fittig Reactions: Coupling of alkyl/aryl halides with sodium in dry ether to form C–C bonds (Wurtz: alkyl–alkyl; Fittig: aryl–aryl).
Polyhalogen Compounds: Molecules with multiple halogens (e.g., CHCl3, CCl4, freons, DDT) — note uses and environmental/health hazards.
C–X Bond Trends: Bond length ↑ and bond enthalpy ↓ down the halogen group (C–I longest/weakest, C–F shortest/strongest); influences reactivity and ease of substitution.
Reactivity Order (Alkyl Halides): For nucleophilic substitution: SN1 (3° > 2° > 1°); SN2 (methyl > 1° > 2° > 3°).

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