Sponsor Area
KI produces HI, In the presence of sulphuric acid (H2SO4).
2KI + H2SO4 --> 2KHSO4 +2HI
Since H2SO4 is an oxidizing agent, it oxidizes HI( Produced in the reaction to I2).
2HI + H2SO4 --> I2 +SO2 +H2O
As a result, the reaction between alcohol and HI to produce alkyl iodide cannot occur. Therefore, sulphuric acid is not used during the reaction of alcohols with KI. Instead, a non-oxidiznig acid such H3PO4 is used.
To have a single monochloride, there should be only one type of H-atom in the isomer of the alkane of the molecular formula C5H12. This is because, replacement of any H-atom leads to the formation of the same product. The isomer is neopentane.
Neopentane
To have three isomeric monochlorides, the isomer of the alkane of the molecular formula C5H12 should contain three different types of H-atoms
The equivalent hydrogens are grouped as a, b and c. The replacement of equivalent hydrogens will give the same product.The equivalent hydrogens are grouped as a, b and c and d. Thus, four isomeric products are possible.
Arrange each set of compounds in order of increasing boiling points:
Bromomethane, Bromoform, Chloro-methane, Dibromomethane.
Alkyl halides containing same alkyl group,the boiling point increases with an increase in the atomic mass of the halgoen. Since the atomic mass bromine is greater than chlorine, the boiling point of bromomethane is higher than that of chloromethane. Also, for alkyl halides containing the same alkyl group, the boiling point inrease in as number of halide group increase. Therefore, the boiling point of dibromomethane is higher than chloromethane and bromomethane, but lower than bromoform. Such as,
Chloromethane<bromomethane<dibromomethane<bromoform.
Sponsor Area
Arrange each set of compounds in order of increasing boiling points:
1-chloropropane, Isopropyl chloride, 1-chlorobutane.
Alkyl halides containing same alkyl group,the boiling point increases with an increase in the atomic mass of the halgoen. 1-chlorobutane is higher than that of isopropyl chloride and 1-chloropropane. Also, the boiling point decrease with an increase in branching in the chain. Thus, the boiling point of isopropyl alcohol is lower than chloropropane, i.e.
Isopropyl chloride < 1- chloropropane < 1- chlorobutane
Arrange each set of compounds in order of increasing boiling points:
1-bromobutane is a primary alkyl halide whereas 2-bromobutane is secondary alkyl halide. The nucleophile approaching is more hindered in 2- bromobutane than in 1-bromobutane. Therefore, 1- bromobutane reacts more rapidly than 2- bromobutane by an SN2 mechainsm. 
Which alkyl halide from the following pairs would you expect to react more rapidly by an SN2 mechanism? Explain your answer.
2-bromo-2-methylpropane is tertiary alkyl halide whereas 2-bromobutane is secondary alkyl halide also, 2-bromo-2-methylpropane has number of substituents are present. Hence, 2- bromobutane reacts more rapidly than 2-bromo-2-methylpropane by an SN2 mechanism.
Arrange each set of compounds in order of increasing boiling points:
Both of the alkyl halides are primary. However, the substituent CH3 is at a greater distance to the carbon atom linked to Br in 1- bromo-3-methylbutane than in 1-bromo-2-methylbutane. Therefore, the approaching nucleophile is less hindered in case of the 1-bromo-3-methylbutane than in case of the 1-bromo-2-methylbutane. Hence, 1-bromo-3-methylbutane reacts faster than the latter by SN2 mechanism.
In the following pairs of halogen compounds, which compound undergoes faster SN1 reaction?
and 
Greater the stability of the carbocation, faster is the rate of SN1 reaction. Since tertiary carbocation 2-chloro-2-methylpropane is more stable than secondary carbocation 3-chloropentane. Hence SN1 reaction proceed via tertiary cation such as 2-chloro-2-methylpropane.
In the following pairs of halogen compounds, which compound undergoes faster SN1 reaction?
Greater the stability of the carbocation, faster is the rate of SN1 reaction. Since secondary carbocation 2-chloroheptane is more stable than primary carbocation 1-chlorohexane. Hence SN2 reaction proceed via secondary cation such as 2-chloroheptane.
Since D of D2O gets attached to the carbon atom to which MgBr is attached, C is isopropylmagnesium bromide
Therefore, the compound R-Br is 2- bromopropane
When an alkyl halide is trated with Na in the presence of ether, a hydrocarbon containing double bond the number of carbon atom as present in the original halide is obtained as product. This is known as Wurtz raction.
Thionyl chloride is preferred in the preparation of chloro alkanes from alcohol. Give reason.
What is the function of anhydrous zinc chloride in the reaction between alcohols and hydrogen halide?
Name one reagent that is used to convert alcohols into chloroalkanes.
What is the difference in the molar masses of alkyl halides and corresponding parent haloalkanes?
Which will have higher boiling point CH3—CH2—CH2—CH2—Br or CH3—CH(CI)— CH3? Give reason.
Give reason: The order or decreasing boiling points is
CH3I > CH3Br > CH3CI > CH3F.
Why is the boiling point of iodo-benzene higher than chlorobenzene?
When isopropyl chlorode is boiled with alcoholic potassium hydroxide solution, 1- propene is form.
In haloarenes the lone pair of electron on halogen atom is delocalized on the benzene ring. Since aryl halides are stabilizes by resonance hence the energy of acitvation for displacement of halogen form aryl halides is much greater than alkyl halides thus they not under go nucleophillic substiution reaction.
Sponsor Area
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.
In the SN2 mechanism, the reactivity of halides for same halides group increase down the group. Because increase in size increase, the halide becomes a better leaving group. Therefore, CH3I will react faster than CH3Br in SN2 reactions with OH-.
In SN2 mechanism nucleophile attack occur at the atom bearing the leaving group. In case of CH3Cl there is no bulky substituents on the carbon atom or in case of (CH3)3CCl, the attack of the nucleophile at the carbon atom is hindered because of the presence of bulky group on the carbon atom. Hence CH3Cl reacts faster than (CH3)3CCl in SN2 reaction with OH-.
What happens when bromoethane is treated with aqueous potassium hydroxide?
When bromoethane is treated aqueous potassium hydroxide, it form ethanol.
What are polyhalogen compounds? Give two examples of such compounds.
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.
Name the products formed when acetone is warmed with sodium hypoiodite solution.
Why are chlorine derivates of organic compounds used as solvents in industry?
Give one reason why the organic halogen compounds used as solvents in industry are chlorides rather than bromides.
Iodoform gives precipitate with silver nitrate on heating while chloroform does not. why?
Write the names of compounds which gives iodoform tests.
What happnes when iodoform is heated with potassium hydroxide?
Write a chemical test to distinguish CHCl3 and CCl4.
How can you obtain C6H5Cl from benzene?
What happens when chlorobenzene reacts with sodium hydroxide at high temperature and pressure?
Wurtz- fittig reaction: when chloro benzene treated with sodium and chloromethane in dry ether, it gives toluene as main product.
A hydrocarbon C5H12 gives only one chlorination product. Identify the hydrocarbon.
Which compound in each of the following pairs will react faster in SN2 reaction with HO–?
(a) CH3—Br or CH3I
(b) (CH3)3 Cl or CH3Cl2
(c) CH2=CHBr or CH2 = CH—CH2Br.
(a) CH3I (Due to its bigger size thus better leaving group.
(b) CH3Cl (As the size of the alkyl group increases the reactivity decreases and CH3Cl have less number of alkyl group.
(c) CH2 = CHBr (Due to lesser number of alkyl group).
(ii) CH3CHCH2CH3
(iii) CH3CH2CH2CH2Br
(iv) (CH3)3 C—Cl
In the 1-bromo-1-methylcyclohexane, all b -hydrogen atom are equivalent. Thus, dehydrohalogenation of this compound gives only one alkene as follows:
In the given compound, there are two different sets of equivalent b -hydrogen atoms. Thus, dehydrohalogenation of the compound yields two alkenes.
Saytzeff’s rule implies that in dehydrohalogenation reaction, the alkene having a greater number of alkyl groups attached to doubly bonded carbon atoms is preferably produced.
Thus, 2methylbut-2-ene is the major product in this reaction.
In the given compound, there are two different sets of equivalent b -hydrogen atoms. Thus, dehydrohalogenation of the compound yields two alkenes.
According to saytzeff’s rule in dehydrohalogenation reaction, the alkene having a greater number of alkyl groups attached to the doubly bonded carbon atom is preferably formed. Hence, alkene 3,4,4-trimethylpent-2-ene is the major product in this reaction.
p-methoxy benzyl bromide reacts faster than p-nitrobenzyl bromide because NO2group at p-position is an electron withdrawing, the electron density reduced at o- and p-position. Nitro group is resonance hybrid so it produces steric hindrance.
So p-methoxy benzyl bromide reacts faster than p-nitro benzyl bromide. The p-methoxy is a primary substituted.
How will you distinguish between ethyl chloride and vinyl chloride?
Write IUPAC names of isomers 
.
IUPAC names of isomers are:
(i) 1, 2-dichloro ethane.
(ii) 1, 1-dichloro ethane.
Sponsor Area
Haloalkanes undergo nuclephilic substitution reaction. Give reason.
When alkyl halide is treated with AgCN, it form alkyl isocyanide such as
Further reaction of alkyl isocyanide with LiAlH4, which act as reducing agent form secondary amines.
How is acetone converted into iodoform?
Acetone is treated with iodine and sodium carbonate forming triodo acetone then the resulting compound treated with sodium hydroxide forming sodium acetate and iodoform.
An alkyl halide having molecular formula C4H9Cl is optically active. What is its structure?
Out of chlorobenzene and chloro-methane which is more reactive towards nucleophilic substitution reactions?
Hydrogen atom of chloroform is definitely acidic, but that of methane is not. Why?
What is the role of phosphorus in the preparation of ethyl iodide from ethyl alcohol and iodine?
tetra - Butyl iodide undergoes following reactions:
(CH3)3 Cl+H2O → (CH3)3COH + HI
(CH3)3CI + OH– → (CH3)2 C = CH2 + H2O + I–
Give reasons.
RBr reacts with AgNO2 to give both RNO2 and RONO. Give reason.
The nitrite ion, has two nucleophilic sites N and O. Reaction with the unshared electron pair on N gives RNO2 and reaction with the unshared pair of electrons on O gives R—O—N = O.
Structure of nitrite ion,
Why a small amount of ethyl alcohol is added to chloroform stored for use as an anaes-thatic?
To convert any carbonyl chloride formed into harmless ethyl carbonate.
2C2H5OH + COCl2 → (C2H5)2CO3 + 2HCl
After using carbon tetrachloride as a fire extinguisher inside a closed space, the space is thoroughly ventilated. Why?
Unlike the chlorine atom in CH3CI, that in chlorobenzene, C6H5Cl, is not easily replaced by the —OH group. Give reason.
When a iodoform is heated with silver nitrate solution, a precipitate is formed. No precipitate is formed if chloroform is used in place of iodoform. Why?
Iodoform gives a precipitate with silver nitrate on heating while chloroform does not.Carbon –iodide bond is quite weak as compared to carbon – chlorine bond .Therefore, when Iodoform is heated with AgNO3 solution, C-I bond gets cleaved easily and iodide ions react with AgNO3 to give precipitate of AgI. On the other hand, C-Cl bond does not get cleaved.
bonding. Hence, the smaller the halogen atom, the more extensive is delocalization.
What happens when chloroform reacts with concentrated nitric acid?
When chloroform reacts with concentrated nitric acid, it forms chloropicrin.
How would you bring about the following conversion: 1-butane to 1-chloro butane?
1- chloro butane can be obtaine by reacting 1- butane with chlorine in presence of sunlight.
The use of chloroform as anaesthesia is decreasing. why?
Write the structural formula of 1-bromo-2, 2-dimethyl propane.
Structural formula of 1-bromo-2, 2-dimethyl propane.
Write a convenient method of preparation of iodoalkanes.
Give the name of the catalyst used in Friedel-Crafts reaction.
Name one reagent that is used to convert alcohols into chloroalkanes.
What is Wurtz reaction?
Wurtz reaction: 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.
RX + 2Na + XR’ → R—R’ + 2NaX
Arrange the following halides in order of increasing SN2 reactivity:
CH3Cl, CH3Br, CH3CH2Cl, (CH3)2CHCl.
Identify the products A, B and C in the following:
When ethyl iodide is treated with alc. KOH. It gives ethene which reacting with bromine gives 1,2 dibromoethane and further reacting with KCN gives ethlene dicyanide.
Give chemical equation for the reaction by which propene can be obtained from 1-bromopane.
Propene can be obtain by reacting 1-bromopane with alcoholic potassium hydroxide.
CH3CH2CH2Br + KOH (alc.) → CH3CH = CH2 + KBr + H2O
i) Aluminium carbide react wity water to form aluminium hydroxide and methane.
ii) Methane react with chlorine in presence of light (hv) to form methyl chloride.
Sponsor Area
Write the structure of all the possible isomes of dichloro ethene. Which of these will have zero dipole moment?
Which effect should the following resonance of vinyl chloride have on its dipole moment?
CH2 = CH—Cl ↔ CH2–— CH = Cl+
How would you convert bromo ethane to n-butane?
Bromo ethane react with sodium in dry ether to give n-butane. This reaction is known as Wurtz reaction.
Iodination of benzene is carried in the presence of oxidising agent like HIO3. Give reason.
The iodination of benzene is a reversible reaction. Therefore, yield of C6H5I is very poor because HI combines with C6H5I and forms back the reactants.
C6H5 + I2 ⇌ C6H5I + HI
In the presence of oxidizing agent like HIO3 or HNO3, the co-product HI is oxidised to iodine and iodination proceeds favourably in the forward direction.
5HI + HIO3 + 3I2 + 3H2O
Out of C6H5CH2Cl and C6H5CHCIC6H5 which is more easily hydrolysed by aqueous KOH.
Hydrolysis by aqueous KOH proceed through the formation of carbocation. If carbocation is stable, then the compound is easily hydrolyzed by aqueous KOH. In the given compounds benzyl chloride forms primary carbocation, while chlorodiphenylmethane forms secondary carbocation, which is more stable than primary carbocation, hence C6H5CHClC6H5 is more easily hydrolysed by aqueous KOH.
Structural isomers of C5H11Br are:
i) CH3CH2CH2CH2CH2Br =1-Bromopentane (1°)
ii)CH3CH2CH2CH(Br)CH3 =2-Bromopentane (2°)
iii)CH3CH2CH(Br)CH2CH3 =3-Bromopentane (2°)
iv)(CH3)2CHCH2CH2Br=1-Bromo-3-methylbutane(1°)
v) (CH3)2CHCHBrCH3 =2-Bromo-3-methyl butane (2°)
vi) (CH3)2CBrCH2CH3 =2-Bromo-2-methyl butane (3°)
vii) CH3CH2CH(CH3)CH2Br= 1-Bromo-2-methyl butane (1°)
viii) (CH3)3CCH2Br =1-Bromo-2, 2-dimethyl propane (1°)
Write the structural formula of the following:
2-chloro-2-methyl propane
2-chloro-2-methyl propane 
Write the structural formula of the following:
1-Bromo-2, 2-dimethyl propane
1-Bromo-2, 2-dimethyl propane
Write the structural formula of the following:
2-Bromo-2, 3-dichloro butane.
2-Bromo-2, 3-dichloro butane
Write the products of the following reactions:
Write the products of the following reactions:
Write the products of the following reactions:
i) There is increase in the boiling points with increasing molar masses of the members of the same family of compounds. Since, there is increase in the molar masses from R—Cl to R—I, therefore, there is increase in their boiling points also. Thus, the boiling point of R—I is greater than the boiling point of R—Br and the boiling point of R—Br is greater than that of R—Cl. Thus, boiling points follow the order R—I > R—Br > R—Cl.
ii) The Van der Waals interactions between the molecules of R—I are stronger than that between R—Br molecules. Therefore, more energy and thus high temperature is required to cause the boiling point of R—I as compared to R—Br. Again, the molecular interactions due to Van der Waals forces in R—Br are stronger as compared to that in R—Cl. Therefore, the boiling point of bromoalkane is higher than the boiling point of chloroalkane.
What are nucleophiles? Give some examples of nucleophiles.
Nucleophile is molecule or ion which have ability to donate it electron. A nucleophile has affection for a positively charged centre. The nucleophile is contained in nucleophilic reagent (nucleo = + ve centre, philic = liking, love affection).
Examples of nucleophiles and nucleophilic reagents.
|
Reagent |
KOH |
KI |
KCN |
NH3 |
NaSR |
AgNO2 |
KNO2 |
|
Nucleophile |
OH- |
I– |
CN- |
NH3 |
SR- |
NO2- |
ONO- |
Write the cause of chemical reactivity of haloalkanes.
+) and the halogen atom acquires an equivalent small negative charge ( 
-). This is represented as
Because of a such rearrangement of the electrons in the C—X bond the haloalkanes (R—X, alkyl halides) are reactive molecules. Consequently, R—
X can be converted into different compounds with different functional groups under proper experimental conditions.
Nucleophilic Substitution Reaction : A chemical reaction in which a stronger nucleophilic (electron rich group which loves +ve centre) substitutes a weaker nucleophile is called the nucleophilic substitution reaction. For example, CH3Br + OH→ CH3OH + Br is a nucleophilic substitution reaction.
In haloalkanes (R—X) the halogen is more electronegative than carbon. Thus, C—X bond is polarized as C 
+—X 
-. Therefore, a strong nucleophile (electron rich species) such as Z attacks the positively charged carbon atom and replaces weak nucleophile halogen from C—Z. Thus a new product is formed.
When haloalkane treated with aqueous KOH, it form alcohol.
Alkyl iodides are often prepared by the reaction of alkyl chlorides or bromides with NaI in dry acetone. This reaction is known as Finkelstein reaction.
Haloalkane treated with ammonia to get primary amine.
When haloalkane is treated with mercaptile, it form thioether.
Grignard reagents are highly reactive and react with haloalkane give alkane.
What is meant by dehydrohalogenation?
Grignard reagents: Alkyl magnesium halides (RMgX) are called Grignard reagents.
Preparation of Grignard reagents: When a solution of alkyl halide (R—X) in ether is allowed to stand over magnesium turning for some time then metal gradually dissolves. In this reaction alkyl magnesium halide is formed. This compound is called Grignard reagent.
When bromoethane react with sodium in dry ether to give butane.
When bromoethane react with Alcoholic potassium hydroxide, it give ethene as main product.
On the reaction of bromoethane with magnesium metal in dry ether Grignard Reagents is form. Grignard reagents are highly reactive and react with ethanol to give butane.
Write the equations for the following reactions:
Alkyl halide reacts with alcoholic solution of potassium cyanide.
Write the equations for the following reactions:
Bromoethane reacts with alcoholic ammonia.
Write the equations for the following reactions:
Haloalkane reacts with silver cyanide.
Write the equations for the following reactions:
Ethyl chloride reacts with sodium lead alloy.
When ethyl chloride react with sodium lead alloy tetraethyl lead obtaine.
How would you bring about the following conversions?
Bromoethane to ethyne
Conversion of 1-Butane to 1-chlorobutane
What happens when:
2-Bromopropane is treated with HI in the presence of red P?
When 2- bromopropane is treated with HI in the presence of red Phosphorous. It forms 2- Iodopropane.
What happens when:
2-Bromopropane is treated with H2, in the presence of Ni catalyst?
When 2- bromopropane is treated with Hydrogen in presence of nickel catalyst, it forms propane as major product. 
What happens when:
2-Bromoprane is treated with silver cyanide?
Propyl cynaide is formed by this reaction.
What happens when:
2-Bromopropane is treated with ammonia?
When 2-bromopropane is treated with ammonia, it forms propamine-2.
What happens when:
1-Bromopropane reacts with metallic sodium?
When 1-bromopropane reacts with metallic sodium, it forms 1-hexane and sodium bromide.
What happens when:
Iodoethane is heated with alcoholic potassium hydroxide?
When iodoethane is heated with alcoholic potassium hydroxide, it forms ethene as major product.
What happens when:
Iodomethane is treated with ammonia?
When Iodomethane is treated with ammonia, it forms methanamine.
What happens when:
Bromoethane is treated with caustic potash?
When Bromoethane is treated with caustic potash, it forms ethanol.
Haloalkanes react with KCN to form alkyl cyanides as main product while AgCN form isocyanides as the chief product. Give reason.
As there will be lesser steric hind– rance for the approaching nucleo– phile from the back side in (i).
As iodine is a better leaving group because of its large size, it will be released at a faster rate in the presence of incoming nucleophile.
Predict the order of reactivity of the following compounds in SN1 and SN2 reactions:
(a) The four isomeric bromobutanes.
(b) C6H5 H2Br, C6H5CH (C6H5Br, C6H5CH (CH3)Br, C6H5C(CH3)(C6H5)Br.
(a) CH3CH2CH2CH2Br < (CH3)2CHCH2Br < CH3CH2CH(Br)CH3 < (CH3)3CBr (SN1).
CH3CH2CH2CH2Br > (CH3)2CHCH2Br > CH3CH2CH(Br)CH3 > (CH3)3CBr (SN2).
Of the two primary bromides, the carbocation intermediate derived from (CH3)2CHCH2Br is more stable than derived from CH3CH2CH2CH2Br because of greater electron donating inductive effect of (CH3)2CH-group.Therefore, (CH3)2CHCH2Br is more reactive than CH3CH2CH2CH2Br in SN1 reactions. CH3CH2CH(Br)CH3 is a secondary bromide and (CH3)3CBr is a tertinry bromide. Hence the above order in SN1.
The reactivity in SN2 reaction follows the reverse order as the steric hindrance around the electrophilic carbon increases in that order.
(b) C6H5C(CH3)(C6H5)Br > C6H5CH(C6H5)Br > C6H5CH(CH3)Br > C6H5CH2Br (SN1).
C6H5C(CH3)(C6H5)Br > C6H5CH(C6H5)Br > C6H5CH(CH3)Br > C6H5CH2Br (SN2).
Of the two secondary bromides, the carbocation intermediate obtained from C6H5CH(C6H5)Br is more stable than obtained from C6H5CH(CH3)Br because it is stabilized by two phenyl groups due to resonance. Therefore, the former bromide is more reactive than the latter in SN1 reactions. A phenyl group is bulkier than a methyl group. Therefore, C6H5CH(C6H5)Br is less reactive than C6H5CH(CH3)Br in SN2 reactions.
Why (-NO2) group shows its effect only at ortho- and para-position and not at meta–position?
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.
Through resonance, halogen tends to stabilise the carbocation and the effect is more pronounced at ortho- and para- positions. The
inductive effect is stronger than resonance and causes net electron withdrawal and thus causes net deactivation. The resonance effect tends to oppose the inductive effect for the attack at ortho- and parapositions and hence makes the deactivation less for ortho- and paraattack.
Reactivity is thus controlled by the stronger inductive effect and orientation is controlled by resonance effect.
It is a chiral molecules.
It is achiral molecule.
Explain the reason:
The reactivity order of alkyl bromides is tert alkyl bromide > sec alkyl bromide > primary alkyl bromide.
i) The given order is follow the SN1 mechanism. In which carbon halogen bond is break and form carbocation and halogen ion.
ii) The carbocation is formed is then attacked by nucleophile in step to complete the substitution reaction.
According to SN1 mechanism more is stability of carbocation easy is formation of compound. Hence the order of alkyl bromides are given as;
30 > 20 > 10
Since tertiary cation is more stable than secondary and secondary is more stable than primary.
1-bromo-1phenyl-3,3 dimethyl butane
(secondary benzyl halide)
2-(trichloromethyl )-1,1,1,2,3,3,3-heptachloropropane
1-chloro-1-(4-iodophenyl)-3,3-dimethylbut-1-ene
p-Bromochlorobenzene
1-chloro-4-ethylcyclohexane
1-Bromo-4-sec-butyl-2-methylbenzene
1, 4–Dibromobut-2-ene.
Arrange the nucleophilicity (rate of SN2 reactivity) of
(i) H2O, OH–, CH3COO- and CH3O-
(ii) NH3 and PH3.
A hydrocarbon C5H10 does not react with chlorine in dark but gives a single monochloro compound C5H9Cl in bright sunlight. Identify the hydrocarbon.
The hydrocarbon can be either a cycloalkane or an alkene since it does n’t react with Cl2 in the dark, it can not be an alkene but must be cycloalkane since the cycloalkane reacts with Cl2 in the presence of sunlight to give a single monochloro derivative. It shows that it is cyclic hydrocarbon in which hall the hydrogen atoms are identical. Thus, it can be only cyclopentane.
There are four isomers possible for the C7H7Cl
How will you bring about the following conversions?
Propene to 1-nitropropane
How will you bring about the following conversions?
Toluene to Benzyl alcohol
How will you bring about the following conversions?
Ethanol to ethyl fluoride
How will you bring about the following conversions?
Bromomethane topropanone
How will you bring about the following conversions?
But-1-ene to but-2-ene
But-1-ene to but-2-ene:
How will you bring about the following conversions?
1-chlorobutane to n-octane
What happens when?
Chlorobenzene is subjected to hydrolysis.
Chlorobenzene does not undergo hydrolysis under normal conditions. However, it undergoes hydrolysis when heated in an aqueous sodium hydroxide solution at a temperature of 623K and a pressure of 300 atm to form phenol.
What happens when?
Ethyl chloride is treated with aqueous KOH.
When ethyl chloride is treated with aqueous KOH, it forms ethanol and potassium chloride
What happens when?
methyl Bromide is treated with sodium in the presence of dry ether.
When methyl bromide treated with sodium in presence of dry ether, ethane is form.
What happens when?
Methyl chloride is treated with KCN.
Haloalkanes undergo nucleophillic substitution reaction due to high electronegativity of the halogen atom, the C—X bond in haloalkanes (alkyl halides) is slightly polar, thereby the C-atom acquires a slight positive charge (≡ C+δ—X–δ). Hence, C-atom is a good target for attack by nucleophiles (electron easily rich species). Therefore, the X-atom of the halo-alkane is replaced by a nucleophile easily.
Nu : + R+δ—Xδ → R—Nu + X–
On the other hand, in haloarenes the halogen atom releases electron to the benzene nucleus relatively electron-rich with respect to halogen atom. As a result, the electrophile attacks at ortho and para position. Hence, haloarenes undergo electrophilic substitution reactions.
Toluene reacts with bromide in the presence of light to give benzyl bromide, while in presence of FeBr3 it givesp-bromo toluene. Give explanation for the above observation.
Arrange the following compounds in increasing order of SN1 , respectively.
ClCH2=CHCH2CH3, CH3(Cl)=CHCH2CH3, CH3 = CHCH2 CH2Cl, CH3CH = CH. CHCl CH3
CH3CH = CHCH2CH2Cl < CH3CH = CH.CH(Cl)CH3 ClCH2CH = CH.CH2CH3 < CH3C(Cl) = CHCH2CH3< CH3CH = CH.CH(Cl)CH3< CH3CH = CH CH2CH2Cl.
This is due to the stability of carbonium intermediate.
Arrange the following compounds in increasing order of SN1 , respectively.
CH3CH2Br, CH2 = CHCH(Br)CH3, CH3CH(Br) CH3
CH2 = CH.CH(Br) CH3 > CH3CH(Br)CHCH3 >CH3CH2(Br)
(The reason is that primary halide is less reactive than other halide such as secondary and tertiary).
Arrange the following compounds in increasing order of SN1 , respectively.
(CH3)3CCI, C6H5(CH3)2Cl, CH3CH2CH2Cl
What are the structure of the possible isomes of dichloroethane? Which of them will have zero dipole movement?
Arrange the following alkyl halides in the decreasing order of SN2 reactivity.
(B) and (C) are primary and are very reactive, but bromide is more reactive than chloride.
(D) and (E) are both secondary but nucleophile attack on (D) is hindered because of stearic hindrance by two —CH3 group.
(A) is tertiary and therefore not reactive under SN2 conditions ;
(C) > (B) > (E) > (D) > (A).
Give reasons:
Iodoform is obtained by the reaction of acetone with hypoidodide ion and not with iodide ion.
How will you prepare 1, 2-ethane diol from 1-chloro ethane?
When 1-chloro ethane is treated with alc. KOH, it produce ethene which on further reaction with potassium permangnet gives 1,2 ethane diol.
How will you prepare 1-bromo propane from 2-bromo propane?
2-bromo propane is treated with alc. KOH, it forms propene which on further reaction with hydrogen bromide in presence of peroxide forms 1-bromo propane.
How will you prepare 1-bromo propane from 1-chloro propane?
1-chloro propane is treated with alc. KOH giving propene which on further reaction with hydrogen bromide in presence of peroxide gives 1- bromopropane.
Chloro benzene on reaction with chloroethane in presence of dry ether forms toluene.
when tetrachloromethane is treated with Antimony trifluoride, it gives freons.
Conversion of chlorobenzene to p-nitrophenol.
The mixture of the products is separated by fractional crystallisation and the para iosmer is used in the next step as follows:
Conversion of deuterobenzene from benzene.
Alcohol have ability to form hydrogen bond therefore it forms bond with hydrogen and thus strengthen the bond. thus, boiling point of alcohol is more then haloalkane.
Since molecule (iii) has two hydroxy group therefore it forms more stronger hydrogen bond than (iv). In case of haloalkane bromine is larger in size thus vander waal forces is more as campare to chlorine thus order of increasing boiling point is given in order;
Secondary alkyl halides are more reactive than primary alkyl halides and tertiary alkyl halides are more reactive than the secondary ones. Explain.
The C—X bond in aryl halides is shorter in comparison to alkyl halides. Explain.
Explain giving reason although haloalkanes are polar in character, yet they are insoluble in water.
Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene.
(i) 1-Bromo-1-methyl cyclohexane.
(ii) 2-Chloro-2-methyl butane.
(iii) 2, 2, 3-Trimethyl-3-bromopentane.
i) In 1-bromo-1-methylcyclohexane compound, all 
atoms are equivalent. Thus, dehydrohalogenation of this compound gives only one alkene.
ii) In 2-chloro- 2-methyl butane compound, there are two different sets of equivalent 
atoms. Thus, dehydrohalogenation of the compound yeilds two alkenes.
Saytzeff’s rule implies that in dehydrohalgenation reaction, the alkene having a greater number of alkyl groups attached to a doubly bonded carbon atoms is perferably produced. Hence, alkene (I) is the major product in this reaction.
iii) In the 2, 2, 3-Trimethyl-3 bromopentane compounds, there are two different 
atoms. Thus, dehydrohalgenation of the compounds yields two alkenes.
Saytzeff’s rule implies that in dehydrohalgenation reaction, the alkene having a greater number of alkyl groups attached to a doubly bonded carbon atoms is perferably produced. Hence, alkene (I) is the major product in this reaction.
The given reaction is SN2 reaction. The nucleophile, Nu– approaches the carbon atom on the side opposite from the halogen and forms new covalent bond with the carbon. This is because KCN is an ionic compound, K+ [C ≡ N]–. Since carbon carrying a lone pair of electrons is more reactive than nitrogen carrying a lone pair, the transition state is formed by the carbon of the cyanide ion forming band with the carbon halogen bond thus,
Explain why the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride?
Explain why alkyl halides, though polar, are immiscible with water.
Explain why Grignard reagents should be prepared under anhydrous conditions?
Give the uses of Freon 12, DDT, carbon tetrachloride and iodoform.
Freon-12: The chlorofluorocarbon compounds of methane and ethane are collectively known as ferons.
Dichloro difluoro methane (Freon 12) is the most common freons. It is manufactured from tetra-chloromethane by the action of antimony trifluoride in the presence of antimony pentafluoride.
Uses:
(a) It is used as a refrigerant (cooling agent) in refrigerators and air conditioners.
(b) It is also used as a propellant in aerosols and foams to spray out deodorants, cleaners, hair sprays, shaving creams.
(c) It is also used as insecticides.
(ii) DDT (p, p’ - Dichlorodiphenyl trichloro ethane):
Preparation: It is manufactured by the condensation of chlorobenzene with trichloro acetaldehyde (chloral) in the presence of sulphuric acid.
Uses:
(a) It is a powerful insecticide. It is highly stable and not easily decomposed.
(b) It is used for killing insects and mosquitoes.
(iii) Carbon tetrachloride (CCl4) : Preparation: It is prepared industrially by chlorination of methane and by the action of chlorine on carbon disulphide in the presence of aluminium chloride as catalyst.
Uses:
(a) It is used as a solvent for oils, fats and waxes.
(b) It is used as a fire extinguisher under the name pyrene.
(c) It is used as dry cleaning.
(d) It is used for the manufacture of freon.
(iv) Iodoform (CHI3):
Preparation: It is prepared by using ethanol or acetone with sodium hydroxide and iodine or sodium carbonate and iodine in water.
Uses:
(a) It is used as an antiseptic and this nature is due to iodine that it liberates. However, because of its very unpleasant smell, it has now been replaced by better antiseptics.
(b) It is used in the manufacture of pharmaceuticals.
Benzene to 3, 4-Dibromo nitrobenzene:
Conversion of benzyl alcohol to -2-phenyl ethanoic acid.
A hydrocarbon (Y) decolourises bromine water. On ozonolysis it gives 2-methyl butanol and formaldehyde. Give the name of the compound.
A hydocarbon (Z) has molecular formula C8H10. It does not decolourise bromine water and is oxidised to benzoic acid on heating with K2Cr2O7 . It can also have the three other isomers A, B and C. Write the structure of A, B and C.
A.
Ethylchloride is the most reactive halide of ethyl series.B.
Ethyl chloride may be converted to ethyl iodide by the action of sodium iodide.C.
Butene-1, CH3CH2CH = CH2 is more stable than its isomer butene-2, CH3CH = CHCH3.D.
Tert butyl bromide reacts with aq KOH to give tertiary butyl alcohol as the major product.E.
Iodide is a stronger electrophile than boromide.A.
1-Bromo-2-buteneOnly two isomeric monochloro derivatives are possible for
n-hexane
D.
2-methyl propane.Which of the following will have zero dipole moment?
C.
trans-1, 2-dichloro ethyleneB.
transition stateButane nitrile may be prepared by heating
D.
propyl chloride with KCNGive one example of Wurtz-Fitting reaction.
When chloro propane react with bromo ethane in the presence of ether, it forms bromo propane and chloro propane.
Structural formula of 4-chloro-2-pentene.
What happens when CH3-CH2CH2-Cl is boiled with alcoholic KOH?
When chloro propane is boiled with alcoholic KOH then, it forms propene
CH3-CH2CH2-Cl + alc. KOH --> CH3-CH2=CH2
Arrange CH3F, CHI, CH3Br and CH3Cl in order of their reactivity in nucleophilic substitution reaction.
Reactivity order given compounds;
CH3F < CH3Cl <CH3Br <CHI
How is ethyl iodide prepared from ethanol?
When ethanol is treated with phosphorus triiodide then, it forms ethyl iodide.
3C2H5OH +PI3 --> 3C2H5I + H3PO3
How will you convert isopropyl bromide to n-propyl bromide?
Conversion of n-propyl bromide from isopropyl bromide.
Give one chemical test to distinguish between C2H5Br and C6H5Br.
Hydrolysis of C2H5Br with refluxing aqueous KOH followed by acidification with dil HNO3 and subsequent treatment with AgNO3 gives light yellow ppt. of AgBr whereas C6H5Br does not gives this test.
Why is chloroethane insoluble in water?
Chloro ethane does not have ability to form hydrogen bond with water. Therefore, chloroethane is insoluble in water.
Starting from ethyl bromide, how will you get (i) ethanol (ii) ethylene.
i) When ethyl bromide react with aq. KOH it forms ethanol.
CH3 -CH2 -Br + KOH (aq.) ---> CH3 -CH2 -OH
ii) When ethyl bromide react with alc. KOH it forms ethylene.
CH3 -CH2 -Br +KOH (alc.) ---> CH2 =CH2
What happens when trimethylene dibromide is heated with zinc?
When 1,3-Dibromopropane or trimethylene dibromide is heated with zinc, it forms cyclopropane.
Alkyl halide has the higest density then water because density increases with increase in number of carbon atoms and halogen atom as atomic mass increase.
How will you obtain 2-bromopropane from propene?
Addition of HBr to propene gives 2-bromopropane. The hydrogen of HBr adds to carbon 1 of propene and bromine adds to carbon 2.
The boiling point of ethyl bromide is higher than that of ethyl chloride because bromine is bigger in size and vander waal forces are maximum. Hence, it have higher boiling point.
Account for the following observation:
n-propyl chloride boils at a high temperature than methyl chloride.
n- propyl chloride boils at a high temperature than methyl chloride because size of n- propyl chloride is bigger than that of methyl chloride. Hence, greater the size of molecular larger the molecular mass of molecule thus higher the boiling point.
Account for the following observation:
Haloalkanes are denser than the corresponding hydrocarbons.
Haloalkane are denser than that of corresponding alkane because density increase with increasing the mass. As the halogen attached to the alkane the mass of the compound increases. Thus, halo alkanes are denser than corresponding alkane.
i) A nucleophile is a chemical species that donates an electron pair to an electrophile to form a chemical bond in realtion to a reaction.
ii) a nucleophilic reagent defined as the molecules or ions with a free pair of electrons or at least one pi bind can act as nucleophiles. For example OH- Cl-, CN- etc
iii) Nucleophilic substitution reaction:- A nucleophile react with haloalkane having a partial positive on the carbon atom bonded to halogen. a substitution reaction take place and atom attached to the substrate departs as leaving atom or ion. Since the substitution reaction is initiated by a nucleophile it is called substitution reaction.For example nucleophillic substitution reaction of alkyl halides.
In vinyl chloride the chloride atom is joined with sp2
hybridised carbon atom of a carbon -carbon double
H2C=CH2
In ethyl chloride the chloride atom attached with ethane.
CH3 -CH2 -Cl
Chlorobenzene on nitration gives 1-chloro-4-nitrobenzene. But cyclohexyl chloride does not give this test.
Ethyl bromide on reaction silver nitrate gives yellowish precipitates and ethyl chloride on reaction with silver nitrate gives whit precipitates.
Uses of the following:
i) BHC (benzene hexachloride) is a important agricultural pesticide mainly used for exterminating white ants, leaf hopper, termite, etc.
ii) Feron are usually used in refrigeration and air conditioning purposes.
iii) D.D.T. (Dichlorophenyltrichloroethane) used as to control pesticides or insect.
How is pure chloroform prepared? How is it oxidised in air or sunlight?
In industry, chloroform is produced by heating a mixture of chlorine or methane. At 400–500 °C, a free radical halogenation occurs, converting these precursors to progressively more chlorinated compounds:
CH4 + Cl2 → CH3Cl + HCl
CH3Cl + Cl2 → CH2Cl2 + HCl
CH2Cl2 + Cl2 → CHCl3 + HCl
Chloroform is slowly oxidised by air in the presence of light to an extremely poisonous gas, carbonyl chloride known as phosgene.
How will you prepare butane from the ethyl iodide?
When ethyl iodide react with sodium in dry ether, it gives butane. This reaction is known as Wurtz reaction.
How will you prepare chlorobenzene from Benzene.
Benzene react with chloride in the presence of ferric chloride to form chlorobenzene.
How will you prepare chlorobenzene from Benzene diazonium chloride ?
Prepartion of chlorobenzne from diazonium chloride.
Why are haloarenes more stable than haloalkanes and undergo electrophilic substitution at o- and p-position?
Haloarenes are more stable because they can donate their lone pair of electrons inside the rings for resonance. Due to resonance, the electron density increase more at ortho and para position. Also, the halogen atom have –I effect and having a tendency to withdraw electrons from the benzene ring. As a result, the ring gets deactivated as compare to benzene and hence the electrophilic substitution reaction occur.
Benzyl chloride will give white precipitate with ethanol followed by silver nitrate whereas chlorobenzene will not.
Explain the following:
Chloroform is chlorine compound but it fails to react with silver nitrate solution.
Chloroform has a stronger bond with the halogen Cl and does not easily hydrolyse to provide Cl-ions. Hence, it fails to react with silver nitrate solution.
Explain the following:
Methyl chloride is hydrolysed more readily than chlorobenzene.
The lone pair of chlorine atom is delocalised over the conjugated double bond ring of benzene which adds a double bond character to the C-Cl bond of the chlorobenzene which makes the bond shorter, so hydrolysed becomes difficult while in alkyl halides methyl chloride there is a single bond and polarity too which is easily hydrolyzed.
This can be explain in the following ways:
When alkyl chloride react with aqueous KOH it forms alcohol. it is subsitution reaction.
C2H5Cl +KOH(aq.) ---> C2H5OH +KCl
When alkyl chloride react with alcoholic KOH it forms alkene. In this case elimination reaction take place.
C2H5Cl +KOH(alc.) ---> H2C=CH2 +KCl +H2O
When ethyl bromide react with KNO2 it forms ethyl nitrite.
C2H5Br +KNO2 --> C2H5ONO
When ethyl bromide react with AgNO2 it forms ethyl nitrite.
C2H5Br + AgNO2 --> C2H5NO2
KCN is predominantly , ionic and therefore , both C and N atoms is available for the electron donation but the reaction occurs through carbon as C-C bonds are stronger than C- N bond, therefore the attack occurs through the carbon atom of the cyanide group forming alkyl cyanides as the main product.
C2H5Br +KCN --> C2H5CN +KBr
What happens when (Give equations):
Chloroform is warmed with aniline in the presence of alc. KOH?
Chloroform on warming with aniline in the presence of KOH gives repelling smell of phenyl isocyanide.
Give reasons:
Vinyl chloride is hydrolysed more slowly than ethyl chloride.
Vinyl chloride is hydrolyzed more slowly than ethyl chloride because the C-Cl bond in vinyl chloride has is sp2 hybridized thus, C-Cl in vinyl chloride has more s character from the C- atom, therefore the bond is stronger and more difficult to break.
The C-Cl bond in vinyl chloride has some double bond and have ability to delocalized the electron to over the molecule. Thus, vinyl chloride is hydrolysed more slowly than ethyl chloride.
1,1 dibromoethane on treatment with KOH (aq) gives acetaldehyde which on heating with tollen’s regents (amm. AgNO3) gives silver mirror test.
1,2 dibromoethane on treatment with KOH(aq) gives ethylene glycol which gives pink colour with ceric ammonium nitrate.
i) Conversion of CHI3 into acetylene
ii) Conversion of chlorobenzene into aniline
How will you prepare n-propyl bromide from isopropyl bromide ?
Conversion of n-propyl bromide from isopropyl bromide.
How will you prepare vinyl bromide from ethyl alcohol?
Conversion of vinyl chloride hydrogen chloride from ethylene chlorine.
Complete the following equation:
CHCl3 + HNO3 →
When chloroform react with nitric acid, nitro chloroform (chloropicrin) and water formed.
CHCl3 + HNO3 → CCl3-NO2 + H2O
Chemical reaction:
C2H5Br +KOH (aq) --> C2H5OH +KBr
When iodoform is heated with silver powder acetylene is formed.
Explain the reason for the following:
A primary alkylhalide, RX, reacts with KCN to give an alkyl cyanide and with AgCN to give an alkyl isocyanide as the major, product.
KCN is ionic in nature so it dissociates completely into K+ and CN- also, C-C bond is more stronger then C-N bond because of this when haloalkanes react with KCN to form alkyl cyanides.
AgCN is convalent in nature thus, it does not dissociate in to Ag+ and CN- and on nitrogen electron density is high thus haloalkane react with AgCN and isocyanide is form.
Explain the reason for the following:
Tertiary butyl iodide forms tertiary butyl alcohol with water, but with an alkali solution, it yields isobutylene (2-methyl propene).
Tertiary butyl iodide forms tertiary butyl alcohol with water due to SN1 mechanism. This reaction is carried out in polar solvents like water and alcohol. The reaction between tert-butyl iodide and hydroxide ion yields tert-butyl alcohol.In the case of alkali, solvent racemisation takes place. Due to the racemisation, the attack of the nucleophile may be accomplished from either side resulting in a mixture of the product. Hence isobutylene is formed.
Explain the reason for the following:
Unlike the chlorine atom in CH3Cl, that in chlorobenzene, C6H5Cl is not easily replaced by the -OH group.
Chlorine in chlorobenzene is not easily replaced by -OH group because of resonace effect. In chlorobenzene the electron pairs are in conjugation with pi-electrons of the ring. C-Cl bond acquires a partial double bond character due to resonace. As a result, the bond cleavage in chlorobenzene is difficult than chloromethane.
Explain the reason for the following:
Halogens delocalize electrons in the benzene ring in the order F > Cl > Br > I.
i)Delocalization of electron in the benzene ring is in the order of F > Cl > Br > I because greater the ability of electronegtivity more is possibilty of electrophillic reaction. Fluorine is most electrongetive element in halogens due to small in size. Hence, it delocalize more electron then others.
ii) As the size increase electronegtivity decrease thus, Iodine is least electronegtive.
Resonance structure of chlorobenzene.
Chlorobenzene are less reactive towards nucleophillic reaction due the following reason:
In chlorobenzene the electron pairs are in conjugation with pi electrons of the ring. Thus, C-Cl bond accquires a partial double bond character due to resonance. As a result, the bond cleavage in haloarene is diffcult than chloromethane.
Which would undergo SN2 reaction faster in the following pair and why?
Sn2 reactions are bimolecular with simultaneous bond-making and bond-breaking steps. Primary alkyl halides prefer to undergo SN2 reactions than tertiary alkyl halides because of less steric hindrance experienced by the approaching nucleophile.
Hence, out of the given pair, (CH3 –CH 2 – Br ) would undergo SN2 reaction faster.
Give reasons:
n-Butyl bromide has higher boiling point than t-butyl bromide.
The boiling point of n-butyl bromide is higher than that of t-butyl bromide because n-butyl bromide is a straight chain molecule having larger surface area and therefore, has stronger intermolecular forces. On the other hand, t-butyl bromide is branched molecule, so it has a smaller surface area. Hence, it has weaker intermolecular force. Thus, n- Butyl bromide has higher boiling point than t-butyl bromide.
Give reasons:
Racemic mixture is optically inactive.
The racemic mixture contains two enantiomers (d and l forms) in equal proportions and thus, the rotation due to one isomer is cancelled by the rotation due to another. Therefore, it has zero optical rotation and hence, it is optically inactive.
Give reasons:
The presence of nitro group (–NO2) at o/p positions increases the reactivity of haloarenes towards nucleophilic substitution reactions.
The presence of nitro groups (–NO2) at o/p positions increases the reactivity of haloarenes towards nucleophilic substitution reactions because nitro groups (–NO2) at o/p positions withdraw the electron density from the benzene ring facilitating the attack of the nucleophile. The negative charge in the carbanion formed, at ortho and para positions with respect to a halogen atom, is stabilised through resonance and the presence of nitro groups (–NO2), respectively.
Which alkyl halide from the following pair is chiral and undergoes faster SN2 reaction?
Among the given pair of compounds, alkyl halide (b) has a chiral centre.
The alkyl halide (a) does not contain a chiral centre and it also gives faster SN2 reaction as SN2 is more favourable in primary alkyl halides.
Out of SN1 and SN2, which reaction occurs with
(a) Inversion of configuration
(b) Racemisation
(a) Inversion of configuration takes place in SN2 reaction.
b) 
Racemisation takes place in SN1 reaction.
Draw the structure of major monohalo product in each of the following reactions:
i)
ii)
i)
(ii) Addition in presence of peroxide yields product according to the anti-Markovnikov rule of addition. Anti Markovnikov rule says - in an addition reaction of a generic electrophile HX to an alkene or alkyne, the hydrogen atom of HX becomes bonded to the carbon atom that had the least number of hydrogen atoms in the starting alkene or alkyne.
What happens when CH3-Br is treated with KCN?
It is a nucleophilic substitution reaction. The nucleophile CN- substitutes Br- because cyanide is strong nucleophile than bromide. The reaction is as follows:
CH3-Br+KCN --> CH3-CN+HBr
Chlorobenzene is extremely less reactive towards a nucleophilic substitution reaction. Give two reasons for the same?
1) Resonance effect: The electron pair on chlorine atom is in conjugation with the electrons of the benzene ring which results in the following resonance structures:
This results in delocalization of the electrons of C- Cl bond and a partial double bond character develop in the bond, which makes it difficult for the nucleophile to cleave the C- Cl bond.
2) The nucleophile suffers repulsion from the increased electron density on the benzene ring, as a result, the nucleophile is unable to make a close approach for the attack on the molecule.
Give the IUPAC name of the following compound.
The IUPAC name of compound is 3-Bromo-2-methylpropene.
Although chlorine is an electron withdrawing group, yet it is ortho-, Para-directing in electrophilic aromatic substitution reactions. Explain why it is so?
Chlorine withdraws electrons through inductive effect and releases electrons through resonance. Through inductive effect, chlorine destabilises the intermediate carbocation formed during the Electrophilic substitution
Through resonance, halogen tends to stabilise the carbocation and the effect is more pronounced at ortho-and Para-position. The inductive effect is stronger than resonance and causes net electron withdrawal and thus causes net deactivation. The resonance effect tends to oppose the inductive effect for the attack at ortho-and Para-position and hence makes the deactivation less for ortho- and Para-attack. Reactivity is thus controlled, by the stronger inductive effect and orientation is controlled by a resonance effect.
Write the IUPAC name of the following compound: (CH3)3 CCH2Br
The IUPAC name of the given structure is 2, 2-dimethylbromopropane.
: Answer the following:
(i) Haloalkanes easily dissolve in organic solvents, why?
(ii) What is known as a racemic mixture? Give an example.
(iii) Of the two Bromo derivatives, C6H5CH(CH3)Br and C6H5CH(C6H5)Br, which one is more reactive in Sn1 substitution reaction and why?
i) Haloalkanes can easily dissolve in organic solvents of low polarity because the new forces of attraction set up between haloalkanes and the solvent molecules are of same strength as the forces of attraction being broken.
(ii) A mixture of equal amounts of two enantiomers is known as racemic mixture. For example: When a 3° halide undergoes substitution with KOH, the reaction proceeds through SN 1 mechanism forming the racemic mixture in which one of the products has the same configuration as a reactant, while the other product has an inverted configuration.
(iii) In SN1 reaction mechanism leads through the carbocation pathway.
More stable the carbocation, more reactive will be substrate .the carbocation formed by two compounds are as follows:
Compound (C6H5)2CHBr = carbocation (C6H5)2CH+
Compound C6H5CH(CH3)Br=carbocation C6H5(CH3)CH+
Out of these two carbocations (C6H5)2CH+ is more stable than C6H5(CH3)CH+ because the carbocation(C6H5)2CH+ is resonance stabilised by two benzene rings .therefore (C6H5)2CHBr is more reactive than C6H5CH(CH3)Br. Resonance stabilisation of carbocation with one benzene ring is shown :
Which would undergo SN1 reaction faster in the following pair?
A tertiary alkyl halide tends to undergo the SN1 mechanism because it can form a tertiary carbocation, which is stabilized by the three alkyl groups attached to it. As alkyl groups are electron donating, they allow the positive charge in the carbocation to be delocalized by the induction effect. Hence, out of the given pairs, (CH3)3C-Br would undergo SN1 reaction faster than CH3-CH2-Br.
Order of stability: tertiary>secondary>primary
Give reasons for the following:
(i) Phenol is more acidic than ethanol.
(ii) The boiling point of ethanol is higher in comparison to methoxymethane.
(iii) (CH3)3C - O - CH3 on reaction with HI gives CH3OH and (CH3)3C -I as the main products and not (CH3)3C -OH and CH3I.
An acidic substance is something that can produce hydrogen (H+) in water.
(i) Both ethanol and phenol are weak acids the ethanol is less acid then phenols because when phenol is dissolved in water it produce phenoxide ion by losing of proton and the phenoxide ion is stabilised by resonance.
Ethoxide ion, however, is not stabilised by resonance. On the other hand, it is further destabilised by the positive inductive effect of the alkyl group.
(ii) Ethanol forms intermolecular hydrogen bonds due to the presence of hydrogen attached to the electronegative oxygen atom. As a result, ethanol exists as associated molecules while methoxymethane does not. Hence, the boiling point of ethanol is higher than that of methoxymethane.
(iii) Usually, iodide, being a big nucleophile, attacks on the group with low steric hindrance and the reaction proceeds by the SN2 mechanism.
However, in this case, methanol, on leaving generates a tertiary carbocation, which is more stable. Hence, this reaction proceeds by SN1 mechanism and therefore, we get (CH3)3C-I and CH3-OH as the major products.
Which aerosol depletes ozone layer?
Chlorofluorocarbon, CFC (chemical formula CCl2F2) is an aerosol which depletes ozone layer.
Give reasons for the following:
(i) Ethyl iodide undergoes SN2 reaction faster than ethyl bromide.
(ii) (±) 2-Butanol is optically inactive.
(iii) C -X bond length in halo benzene is smaller than C -X bond length in CH3- X.
(i) Ethyl iodide undergoes SN2 reaction faster than ethyl bromide because in the periodic table size increase if we move down. With an increase in size, basicity decrease, and the ability of the leaving group to leave increase. Iodine is good leaving the group thus it undergoes SN2 reaction faster than ethyl bromide.
(ii) Optically active compounds are those which rotates plane polarised light either left or right direction. In the case of (±) 2-Butanol is optically inactive because it is both dextrorotatory i.e. (+) and laevorotatory i.e. (-) and hence forms a racemic mixture in which the net rotation of plane-polarized light towards the right is cancelled by the left one and so it becomes optically inactive.
(iii) C—X bond length in halobenzene is lower than C—X bond length in CH3—X because in halo- benzene the C—X acquires partial double bond character due to resonance as shown below whereas in CH3—X there is no such resonance. As the bond length of the double bond is smaller than single bond hence C—X bond length in halo- benzene is smaller.
Answer the following question:
(i) What is meant by the chirality of a compound? Give an example.
(ii) Which one of the following compounds is more easily hydrolyzed by KOH and why?
CH3CHCICH2CH3 or CH3CH2CH2Cl
(iii) Which one undergoes S N 2 substitution reaction faster and why?
(i)Chirality: It is a geometrical property of a rigid object ( or molecules ) by which it gets such spatial arrangement of points or atoms that, the molecule becomes non-super imposable of its mirror image.
The chiral molecule of the object does not have any element of symmetry like, a mirror image, centre of inversion (i) etc.
3-bromopent-1-ene is represented as
The centre of chirality is designated as *. It is due to the presence of four different groups.
(ii) Due to +I effect of alkyl groups the 2° carbonium ion CH3—CH+ —CH2—CH3 derived from sec. butyl chloride is more stable than the 1° carbonium ion CH3—CH2—C H2+ derived from n-propyl chloride. Therefore sec. butyl chloride gets hydrolyzed more easily than n-propyl chloride under SN1 conditions.
(iii) 
Undergoes SN2 substitution reaction faster.
Out of 
and 
which is more reactive towards
SN 1 reaction and why?
In the SN1 reaction the formation of carbocation is the rate determining step, and also the stability of carbocation would determine its reactivity.
The order of stability of carbocation is given as Tertiary>Secondary> Primary> methyl.
Here, 1-chloro-1-methylpropane would form a secondary carbocation, while 1-chloro-2- methylpropane would form a primary carbocation, which is less stable than secondary carbocation. Hence, reactivity towards the SN1 reaction would be higher for 1-chloro-1- methylpropane.
Given reasons:
(i)C–Cl bond length in chlorobenzene is shorter than C–Cl bond length in CH3–Cl.
(ii)The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.
(iii)SN1 reactions are accompanied by racemization in optically active alkyl halides.
(i) The carbon in C-Cl bond in chlorobenzene is sp2 hybridised, while in CH3-Cl is sp3 hybridised. In sp2 hybrid orbitals have more of s -character and hence the carbon if chlorobenzene withdraws the electron pair between C-Cl with greater force. As a result, C-Cl bond is shorter than CH3-Cl.
(ii)Reasons behind the dipole moment of chlorobenzene being lower than that of cyclohexyl chloride are followed:
(a)In chlorobenzene, the C−Cl bond is shorter due to the partial double bond character.
(b)Hybridisations of C in C−Cl in chlorobenzene and cyclohexyl chloride are sp2 and sp3, respectively (sp2 C being more electronegative than sp3 C).
As dipole moment is the product of charge and bond distance, both the factors in chlorobenzene are lower as compared to cyclohexyl chloride. Therefore, the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.
(iii)In SN1 reaction, the formation of carbocation as an intermediate takes place. This carbocation has sp2-hybridised and planar structure. This planar carbocation is attacked by nucleophile from both the sides equally to form d and l isomers in equal proportion. Such products are called racemic mixture. Hence, SN1 reactions are accompanied by racemisation in optically active alkyl halides.
Write the IUPAC name of the following compound: CH2 = CHCH2Br
The IUPAC name of CH2 = CHCH2 Br is 3-bromopropene.
(a) Draw the structures of major monohalo products in each of the following reactions :
(b) Which halogen compound in each of the following pairs will react faster in SN2 reaction:
(i) CH3Br or CH3I
(ii) (CH3)3C−Cl or CH3−Cl
a)
b)
(i) CH3I will react faster in SN2 reaction than CH3Br. This is because I− is a better leaving group, owing to its greater size than Br−. As a result, it will leave at a faster rate in the presence of an incoming nucleophile.
(ii) CH3−Cl will react faster in SN2 reaction than (CH3)3 C−Cl, as CH3−Cl is a primary halide whereas (CH3)3C−Cl is a tertiary halide. Primary halides undergo SN2 reactions faster.
Write IUPAC name of the following compound :
(CH3CH2)2NCH3
IUPAC name of the given compound is N-Ethyl-N-methylethanamine.
Following compounds are given to you :
2-Bromopentane, 2-Bromo-2-methylbutane, 1-Bromopentane
(i) Write the compound which is most reactive towards SN2 reaction.
(ii) Write the compound which is optically active.
(iii) Write the compound which is most reactive towards b-elimination reaction.
(i) 1-bromopropane. Primary alkyl halides are most reactive for SN2.
(ii) 2-bromopentane: It has a unsymmetrical carbon atom. C2 is attached to 4 different groups i.e. methyl, propyl a dihydrogen atom.
(iii) 2-bromo-2-methyl butane: Tertiary alkyl halide is most reactive for elimination.
Out of chlorobenzene and benzyl chloride, which one gets easily hydrolysed by aqueous NaOH and why?
Benzyl chloride gets easily hydrolysed by aqueous NaOH because chlorobenzene a partial double bond develops between carbon and chlorine bond due to which bond become short and strong substitution of chlorine become very difficult.
Aromatic carboxylic acids do not undergo Friedel-Crafts reaction.
In case of aromatic carboxylic acid –COOH attach to the Benzene ring having an electron withdrawing effect and deactivated the benzene ring, hence do not exhibit Friedel craft reaction.
The synthesis of alkyl fluorides is best accomplished by
free radical fluorination
Sandmeyer's reaction
Finkelstein reaction
Swarts reaction
D.
Swarts reaction
Alkyl fluorides can be prepared by the action of mercurous fluorides or antimony trifluoride (inoragnic fluorides) on corresponding alkyl halide.
This reaction is known as Swarts reaction
CH3Br + AgF → CH3F + AgBr
But, when the action of NaI/acetone takes place on alkyl chloride or bromide alkyl iodide forms. This reaction is called 'Finkelstein reaction.
Free radical fluoridation is a highly explosive reaction. So not preferred for the preparation of fluoride.
In SN2 reactions, the correct order of reactivity for the following compounds CH3Cl, CH3CH2Cl, (CH3)2CHCl and (CH3)3CCl is
CH3Cl > (CH3)2CHCl >CH3CH2Cl > (CH3)3CCl
CH3Cl > CH3CH2Cl > (CH3)2CHCl >(CH3)3CCl
CH3CH2Cl > CH3Cl >(CH3)2CHCl >(CH3)3CCl
(CH3)2CHCl > CH3CH2Cl > CH3Cl >(CH3)3CCl
B.
CH3Cl > CH3CH2Cl > (CH3)2CHCl >(CH3)3CCl
The major organic compound formed by the reaction of 1,1,1-trichloroethane with silver powder is
acetylene
ethene
2-butyne
2-butene
C.
2-butyne
The reaction is
An unknown alcohol is treated with the “Lucas reagent” to determine whether the alcohol is primary, secondary or tertiary. Which alcohol reacts fastest and by what mechanism
Secondary alcohol by SN1
Tertiary alcohol by SN1
Secondary alcohol by SN2
Tertiary alcohol by SN2
B.
Tertiary alcohol by SN1
Reaction proceeds through carbocation formation as 30 carbocation is highly stable, hence reaction proceeds through SN1 with 30 alcohol.
What is DDT among the following
Greenhouse gas
A fertilizer
Biodegradable pollutant
Non–biodegradable pollutant
D.
Non–biodegradable pollutant
DDT is P,P' - Dichlorodiphenyl trichloro ethane. It was used as an insecticide. DDT is not metabolised very rapidly by animals. It was also discovered to have a high toxicity towards fish. Therefore, It is a non-bio degradable pollutant.
Consider the following bromides:
The correct order of SN1 reactivity is
A > B > C
B > C > A
B > A > C
C > B > A
B.
B > C > A
Higher the stability of carbocation, faster is the reaction because SN1 reactions involve the formation of carbocation intermediate.
The increasing order of the reactivity of the following halides for the SN1 reaction is
(III) < (II) < (I)
(II) < (I) < (III)
(I) < (III) < (II)
(II) < (III) < (I)
B.
(II) < (I) < (III)
For any SN1 reaction, reactivity is decided by ease of dissociation of alkyl halide
R – X ⇌ R+ + X-
Higher the stability of R+ (carbocation) higher would be reactivity of SN1 reaction.Since the stability of cation follows the order.
3-Methyl-pent-2-ene on reaction with HBr in presence of peroxide forms an addition product.The number of possible stereoisomers for the product is
Six
Two
Zero
Four
D.
Four
3-Methyl-pent-2-ene on reaction with HBr in presence of peroxide forms 2 Bromo-3-methyl pentane.As the molecule is nonsymmetric therefore according to Anti Markownikov rule, due to 2 chiral centre 4 stereo isomers are possible. 
The major product obtained in the following reaction is
C6H5CH (OtBu) CH2C6H5
C6H5CH=CHC6H5
(+)C6H5CH(OtBu)CH2H5
(–)C6H5CH(OtBu)CH2C6H5
B.
C6H5CH=CHC6H5
Elimination reaction is highly favoured if
(a) Bulkier base is used
(b) Higher temperature is used
Hence in given reaction biomolecular ellimination reaction provides major product
Arrange the carbanions 
in order of their decreasing stability
C.
Due to the –I effect of three chlorine atoms and due to pπ - dπ bonding CCl3− is extra stable.Carbanion follow stability order 
Toluene is nitrated and the resulting product is reduced with tin and hydrochloric acid. The product so obtained is diazotised and then heated with cuprous bromide. The reaction mixture so formed contains
mixture of o− and p−bromotoluenes
mixture of o− and p−dibromobenzenes
mixture of o− and p−bromoanilines
mixture of o− and m−bromotoluenes
A.
mixture of o− and p−bromotoluenes
The organic chloro compound, which shows complete stereochemical inversion during a SN2 reaction,is
(C2H5)2CHCl
(CH3)3CCl
(CH3)2CHCl
CH3Cl
B.
(CH3)3CCl
D.
CH3Cl
For SN2 reaction, the C atom is least hindered towards the attack of nucleophile in the case of (CH3Cl).
The treatment of CH3MgX with CH3C≡C−H produces
CH3−CH=CH2
CH3C≡C−CH3
CH4
D.
CH4
CH3-MgX + CH3-C≡C-H→ CH4
Presence of a nitro group in a benzene ring -
activates the ring towards electrophilic substitution
renders the ring basic
deactivates the ring towards nucleophilic substitution
D.
Which of the following is the correct order of decreasing SN2 reactivity?
RCH2X > R3CX > R2CHX
RCH2X > R2CHX > R3CX
R3CX > R2CHX >RCH2X
R2CHX > R3CX >RCH2X
B.
RCH2X > R2CHX > R3CX
More is the steric hindrance at the carbon bearing the halogen, lesser is the SN2 reactivity.
The IUPAC name of
1, 1-diethyl-2-dimethylpentane
4, 4-dimethyl-5, 5-diethylpentane
5, 5-diethyl-4, 4-dimethylpentane
3-ethyl-4, 4 dimethylheptane
D.
3-ethyl-4, 4 dimethylheptane
HBr reacts with CH2 = CH – OCH3 under anhydrous conditions at room temperature to give
CH3CHO and CH3Br
BrCH2CHO and CH3OH
BrCH2 – CH2 – OCH3
H3C – CHBr – OCH3
D.
H3C – CHBr – OCH3
Methyl vinyl ether is a very reactive gas. It is hydrolysed rapidly by dilute acids at room temperature to give methanol and aldehyde. However, under anhydrous conditions at room temperature, it undergoes many addition reactions at the double bond.
Electrophilic addition reaction more favourable.
The IUPAC name of the compound shown below is
2-bromo-6-chlorocyclohex-1-ene
6-bromo-2-chlorocyclohexene
3-bromo-1-chlorocyclohexene
1-bromo-3-chlorocyclohexene
C.
3-bromo-1-chlorocyclohexene
Alkyl halides react with dialkyl copper reagents to give
alkenes
alkyl copper halides
alkanes
alkenyl halides
C.
alkanes
R2CuLi + R'X → R-R' +R-Cu +LiX
Elimination of bromine from 2-bromobutane results in the formation of-
equimolar mixture of 1 and 2-butene
predominantly 2-butene
predominantly 1-butene
predominantly 2-butyne
B.
predominantly 2-butene
Saytzeffs product
Which one of the following has the minimum boiling point?
n-butane
isobutane
1- butene
1- butyne
B.
isobutane
Isobutene (H3C)2 - C=CH2 has the minimum force of attraction (due to steric hindrance).Thus minimum boiling point.
Rate of the reaction
is fastest when Z is
Cl
OCOCH3
OC2H5
NH2
A.
Cl
Cl- is the best leaving group being the weakest nucleophile out of NH2-, Cl-, -OC2H5 and -OOCCH3.
The major product formed in the following reaction is:
C.
It is nucleophilic substitution reaction.
For the following reactions,
i) CH3 CH2 CH2Br + KOH --> CH3CH=CH2 +KBr +H2O
ii) 
iii)
(i) is elimination reaction, (ii) is substitution reaction, and (iii) is addition reaction.
(i) is elimination reaction, (ii) is addition reaction, (iii) is substitution reactions.
(i) is substitution, (ii) addition reaction (iii) is addition reactions
(i) and (ii) are elimination reactions and (iii) is addition reaction.
A.
(i) is elimination reaction, (ii) is substitution reaction, and (iii) is addition reaction.
i) CH3 CH2 CH2Br + KOH --> CH3CH=CH2 +KBr +H2O
[Elimination reaction]
ii) 
[Substitution reaction]
iii)
[Addition reaction]
In which of the following compounds, the C-Cl bond ionisation shall give most stable carbonium ions?
C.
The stability of carbocation follow the order 3> 2> 1> methyl. More the number of alkyl group attached with the carbon atom carrying the positive charge greater would be the tendency to stabilise positive charge via inductive effect, and hence more stable. Due to the presence of benzene group, there is more resonance possibility than others hence it forms a more stable compound.
The reaction of C6H5CH=CHCH3 with HBr produces
C6H5CH2CH2CH2Br
A.
Which of the following is a correct electron displacement for a nucleophilic reaction to take place?
C.
Allylic and benzylic halides show high reactivity towards Sn1 reaction Further, due to greater stabilisation of allyl and benzyl carbocations intermediates by resonance, primary allylic and primary benzylic halides show higher reactivity in Sn1 reactions than other simple primary halides.
Hence, it undergoes nucleophilic reaction readily.
Given,
The enthalpy of hydrogenation of these compounds will be in the order as
I>II>III
I<II<III
II>III>I
II>I>III
B.
I<II<III
the enthalpy of hydrogenation of given compounds is inversely proportional to stability of alkene.
Identity Z in the sequence of reactions.
CH3-(CH2)3-O-CH2CH3
(CH3)2CH2-O-CH2CH3
CH3(CH2)4-O-CH2CH3
CH3CH2-CH(CH3)-O-CH2CH3
A.
CH3-(CH2)3-O-CH2CH3
What products are formed when the following compounds is treated with Br2 in the presence of FeBr3?
C.
CH3 is o/p -directing group, thus.
Which of the following reaction (s) can be used for the preparation of alkyl halides?
I, III and IV
I and II
Only IV
III and IV
A.
I, III and IV
In the equation (I) and (IV) is the presence of Lucas reagent (HCl + anh.ZnCl2) which on reaction alcohols gives alkyl halides. In equation (III alkyl halides is formed due to SN1 reaction.
The order of reactivity of phenyl magnesium bromide (phMgBr) with the following compounds.
III > II> I
II > I > III
I > III > II
I > II > III
D.
I > II > III
Since alkyl group has +I -effect and aryl group has +R - effect, thus, greater the number of alkyl and aryl groups attached to the carbonyl group, its reactivity towards nucleophilic addition reaction. Secondly, as the steric crowding on carbonyl group increases, the reactivity decreases accordingly.
therefore, correct reactivity order for reaction with PhMgBr is
The IUPAC name of the following compound
trans - 2- chloro - 3 - iodo - 2 - pentene
cis - 3- iodo - 4- chloro - 3- pentene
trans - 3- iodo - 4- chloro - 3- pentene
cis - 2- chloro - 3- iodo - 2- pentene
A.
trans - 2- chloro - 3 - iodo - 2 - pentene
Consider the reactions
The mechanisms of reactions (i) and (ii) are respectively
SN1 and SN2
SN1 and SN1
SN2 and SN2
SN2 and SN1
A.
SN1 and SN2
C2H5OH being a weaker nucleophile, when used as a solvent in case, of hindered 1o halide, favours SN1 mechanism while C2H5O- being a strong nucleophile in this reaction favours SN2 mechanism.
In the reaction with HCl, an alkene reacts in accordance with the markownikoff's rule, to give a product 1-chloro -1- methylcyclohexane. The possible alkane is
(a) and (b)
C.
(a) and (b)
For structure A
For structure B,
There is a rearrangement of carbocation occur because 3o carbocation is more stable than 2o- carbocation.
Which one is most reactive towards SN1 reaction?
C6H5CH(C6H5)Br
C6H5CH(CH3)Br
C6H5C(CH3)(C6H5)Br
C6H5CH2Br
C.
C6H5C(CH3)(C6H5)Br
SN1 reaction involves the formation of carbocation intermediate. More the stability of carbocation more is the reactivity of alkyl/aryl halides towards SN1 reaction.
The intermediate carbocations formed by given halides as:
Thus, C6H5C(CH3)(C6H5)Br is most reactive towards SN1 reaction.
Which of the following reactions is an example of nucleophilic substitution reaction?
RX + KOH → ROH + KX
2RX + 2Na → R- R
RX + H2 → RH + HX
RX + Mg → RMgX
A.
RX + KOH → ROH + KX
Nucleophilic substitution reactions involve substitution (replacement) of a group or atom by a nucleophile.
KOH → K+ + OH-
RX + OH- → R - OH + KX
Nucleophiles are either negative charge or a lone pair of electrons bearing species, OH- etc.
Consider the following reaction,
the product Z, is
CH2 = CH2
CH3CH2-O-CH2-CH3
CH3-CH2-O-SO3H
CH3CH2OH
D.
CH3CH2OH
i) PBr3 is a halogenating agent, ie, converts -OH group into -Br
ii) Alc. KOH is a dehydrohalogenation agent.
iii) H2SO4 and H2O converts an olefin into alcohol
Benzene reacts with CH3Cl in the presence of anhydrous AlCl3 to form
toluene
chlorobenzene
benzylchloride
xylene
A.
toluene
CH3Cl in the presence of anhydrous AlCl3 acts as alkylation agent and introduces an alkyl group.
This reaction is known as Friedel-Craft's alkylation of benzene.
In an SN2 substitution reaction of the type which 
one of the followings has the highest relative rate?
CH3-CH2-CH2Br
CH3-CH(CH2)-CH2Br
CH3C(CH3)2CH2Br
CH3CH2Br
D.
CH3CH2Br
The relative reactivity of alkyl halide towards SN N reactions is as follows;
Primary > secondary > Tertiary
However, if the primary alkyl halide or the nucleophile/base is sterically hindered the nucleophile will have difficulty to getting the back side of the alpha carbon as a result of this elimination product will be predominant. Here CH3CH2Br is the least hindered, hence it has the highest relative rate towards SN2 reaction.
B.
The mechanism of electrophilic addition reaction is consistent with the occurrence of rearrangement leading to more stable carbocation.
The order of stability of carbocation is as;
T > S> P > CH3
Consider the following compounds:
The correct decreasing order of their reactivity towards hydrolysis is;
(ii) >(iv) > (iii) > (i)
(i) > (ii) > (iii) > (iv)
(iv) > (ii) > (i) > (iii)
(ii) > (iv)> (i) > (iii)
D.
(ii) > (iv)> (i) > (iii)
Decreasing order of their reactivity toward hydrolysis is:
Identify the major products P, Q and R in the following sequence of reactions:
C.
Mechanism: 
Which of the following is the major product in the reaction of HOBr with propene?
2-bromo, 1-propanol
3-bromo,1-propanol
2-bromo, 2-propanol
1-bromo, 2-propanol
D.
1-bromo, 2-propanol
The reaction through Markownikoff rule.
Sponsor Area
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