Haloalkanes And Haloarenes - Class 12 Engineering Chemistry - Extra Questions

t-butylbromide 

$$CH_3-CH_2-Br$$

Halogenation of Alkanes.

$$\textbf{(i)}$$

The complete mechanism is shown:

• More the nucleophilic activity of the attacking reagent more will be the $$SN_2$$ path.
• Since in $$SN_1$$ mechanism the reagent does not enter into the rate-determining step of ionization. However, it can also be expected that nucleophilic activity changes the mechanism from $$SN_2$$ to $$SN_1$$
  

So answer is true.

The first compound in the given pair undergoes solvolysis in aqueous ethanol more quickly as it is primary alkyl halide. So the statement is True
The second compound is tertiary alkyl halide and undergoes elimination reaction preferentially.

Reactivity towards $$SN_{1}$$ reaction depends upon the stability of carbocation.

This is $$SN_{1}$$ reaction.

In both the compounds, a tertiary carbocation is formed so it depends on the tendency of leaving the group.

$$\text{Option B is correct.}$$

In presence of polar a protic solvent such as acetone and nucleophilic such as $$NaSH ,SN^2$$ reaction.Occur attacks will occur from equitorial position because vacant is at equitorial position.

The elimination reaction occurs by $$E_{2}$$ mechanism.

$$E_{2}$$ reaction.

The elimination occurs by an $$E_{2}$$ mechanism.

The elimination reaction is $$E_{2}$$ reaction.

The given reaction follows $$SN_{2}$$ mechanism.

(A) Base catalyzed dehydrohalogenation of alkyl halide to form alkene is E1 reaction when carbocation intermediate is involved.
(B)  Base catalyzed dehydrohalogenation of alkyl halide to form alkene is  E2 reaction when no carbocation intermediate is involved.
(C) Base catalyzed conversion of halohydrin to epoxide is an intramolecular $$SN^2$$ reaction.
(D) The one step replacemet of the leaving group L with nucleophile Nu is $$SN^2$$ reaction
(E)  Two step replacemet of the leaving group L with nucleophile Nu is $$SN^1$$ reaction.

There are two different $$\beta$$ point ions so,product will be two from two different $$\beta$$ elimination.

An alkyl halide of formula $$C_{6}H_{13}Br$$ on treatment with potassium t-butoxide gives two isomerism alkenes $$(C_{6}H_{12})$$. Both alkenes on hydrogenation give 2,3-dimethyl butane.

Wurtz reaction involves the reduction of an alkyl halide with Na in ether. Wurtz reaction is a convenient method for the preparation of symmetrical alkanes (R-R)  i.e. alkane containing even the number of C atoms.

Answer: c
Explanation: A nucleophile is a chemical species that donates an electron pair to an electrophile to form a chemical bond in relation to a reaction, it is an electron rich species.

(i) Since, iodide ion is better leaving group than bromide ion, methyl iodide will react faster than methyl bromide in $$SN^2$$ reaction with hydroxide ion.

This is an example of bimolecular nucleophilic substitution reaction $$S_N2$$. Cyanide ion acts as a nucleophile and bromide ion acts as leaving group. It is one step reaction and the mechanism is as shown above.

The reactions are given above:

In SN$$^2$$ reaction, steric factors determine the reactivity. more reactive alkyl halides have less steric hinderance. Hence, the decreasing order of the reactivity of alkyl halides is $$\displaystyle 1^o > 2^o > 3^o$$.
(i)  1-Bromopentane >  2-Bromopentane > 2-Bromo-2-methylbutane

(ii) 1-Bromo-3-methylbutane >  2-Bromo-3-methylbutane >  2-Bromo-2-methylbutane

(iii) 1-Bromobutane >  1-Bromo-3-methylbutane >  1-Bromo-2-methylbutane > 1-Bromo-2, 2-dimethylpropane

The products of various reactions are as shown.

The reagents used for the given conversions are as follows:

Two primary alkyl halides with molecular formula $$C_4H_9Br$$ are possible. They are n-butyl bromide and isobutyl bromide.
When (a) is reacted with sodium metal it gives compound (d), $$C_8H_{18}$$ which is different from the compound formed when n-butyl bromide is reacted with sodium. Hence, compound a is isobutyl bromide and compound d is 2,5-dimethylhexane.
$$\displaystyle 2CH_3CH_2CH_2CH_2Br + 2Na \xrightarrow {Wurtz\ reaction} CH_3CH_2CH_2CH_2CH_2CH_2CH_2CH_3 (n-octane)$$
$$\displaystyle 2CH_3CH(CH_3)CH_2Br + 2Na \xrightarrow {wurtz\ reaction} CH_3CH(CH_3)CH_2CH_2CH(CH_3)CH_3 (2,5-dimethylhexane)$$
(a) reacted with alcoholic KOH to give compound (b).
$$\displaystyle CH_3CH(CH_3)CH_2Br \xrightarrow {alc KOH} CH_3C(CH_3)=CH_2 (2-methyl-1-propene)$$
Compound (b) is reacted with $$HBr$$ to give (c) which is an isomer of (a).
$$\displaystyle CH_3-C(CH_3)=CH_2 \xrightarrow [Markownikoff's\ rule] {HBr} CH_3-CBr(CH_3)CH_3 (tert-butylbromide)$$

The isomer of compound $$\displaystyle C_4H_9Br $$ which is most reactive towards $$S_N1$$ reaction is tert butyl bromide. In general, tertiary alkyl halides have maximum rate for $$S_N1$$ reaction.

The IUPAC name of the given compound is $$4-Bromo-4-methyl pent-2-ene$$

Chlorobenzene is extremely less reactive towards a nucleophilic substitution reaction.

The name is of the compound is 3-bromo-2-methylpropene or 3-bromo-2-methylprop-1-ene.

2-Bromo-4-chloro pentane is the IUPAC name of the compound.

In the $$SN_1$$ reaction, the big barrier is carbocation stability. Since, the first step of the $$SN_1$$ reaction is loss of a leaving group to give a carbocation, the rate of the reaction will be proportional to the stability of the carbocation. Carbocation stability increases with increasing substitution of the carbon ($$tertiary > secondary > primary$$)  as well as with resonance.

REF.Image

The IUPAC name of the above compound is 2 - Iodo - 2 - methylpropane.

The compounds in which two same halogen atoms are attached to same carbon are called Geminal halides.

(1) SN$$\displaystyle ^1$$ reaction:
Consider the alkaline hydrolysis of tert butyl bromide to give tert butanol.
$$\displaystyle CH_3C(CH_3)_2-Br+NaOH \xrightarrow {\Delta } CH_3C(CH_3)_2-OH+NaBr$$
The rate of the reaction is proportional to the concentration of tert butyl bromide only and is independent of the concentration of hydroxide ion.
Rate $$\displaystyle = K[ CH_3C(CH_3)_2-Br] $$
The order of the reaction is 1. In the rate determining step, carbon to Br bond breaking is followed by the carbon to hydroxide ion bond formation. Hence, the reaction is unimolecular as only tert butyl bromide is involved in the rate determining step.

Thus SN$$\displaystyle ^1$$ reaction is unimolecular nucleophilic substitution reaction.

(2)  SN$$\displaystyle ^2$$ reaction
Consider the alkaline hydrolysis of methyl bromide to give methanol.
$$\displaystyle CH_3-Br+NaOH \xrightarrow {\Delta } CH_3-OH+NaBr$$
The rate of the reaction is proportional to the concentration of methyl bromide as well as the concentration of hydroxide ion.
Rate $$\displaystyle = K[CH_3-Br][OH^-]$$
The order of the reaction is 2. In the rate determining step, the carbon to hydroxide ion bond formation and carbon to Br bond breaking are simultaneous processes. Hence, the reaction is bimolecular as both methyl bromide and NaOH are involved in the rate determining step.

Thus SN$$\displaystyle ^2$$ reaction is bimolecular nucleophilic substitution reaction.

$$\displaystyle SN^1$$ reaction corresponds to unimolecular nucleophilic substitution reaction.  The order of reaction is one.  The hydrolysis of tert-butyl bromide with aqueous $$NaOH$$ solution is an example of $$\displaystyle SN^1$$ reaction.  The rate of the reaction depends on the concentration of tert butyl bromide but it is independent of the concentration of $$NaOH.$$ Hence, the rate determining step only involves tert-butyl bromide. In $$\displaystyle SN^1$$ reaction, racemic mixture is obtained.

The mechanism of $$\displaystyle SN^1$$ reaction involves two steps. First step is the dissociation of $$\displaystyle C-Br$$ bond to from tertiary carbocation. In the second step, the nucleophile $$\displaystyle X^-$$ attacks the carbocation to form $$\displaystyle C-X$$ bond.

(a) Wurtz reaction
Alkyl halides react with sodium in presence of dry ether to form higher alkanes. For example, methyl bromide reacts with sodium in presence of dry ether to form ethane.
$$\displaystyle CH_3-Br  + 2Na + Br-CH_3 \xrightarrow {\text {dry ether}} CH_3-CH_3 +2NaBr$$

(b) Wurtz-Fittig reaction
Aryl halides react with alkyl halides in presence of sodium to form substituted aromatic compounds. For example, bromobenzene reacts with methyl iodide in presence of sodium to form toluene.
$$\displaystyle C_6H_5-Br+ 2Na+ I-CH_3 \rightarrow C_6H_5-CH_3+ NaBr+NaI$$

The alkaline hydrolysis to tert-butyl bromide with aqueous alkali such as NaOH or KOH is as follows.
$$\underset{\underset{\displaystyle bromide}{\displaystyle tert-butyl}}{(CH_3)_3C}-Br+NaOH(aq)\rightarrow \underset{\underset{\displaystyle alcohol}{\displaystyle tert-butyl}}{(CH_3)_3C}-OH+NaBr$$.
The rate of this reaction depends only on the concentration of the tert-butyl bromide and is independent of the concentration of alkali added.
Rate $$\alpha[(CH_3)_3C-Br]$$
Rate $$=K[(CH_3)_3C-Br]$$
This is a first order reaction because rate of hydrolysis of $$(CH_3)_3-Br$$ is independent of the concentration of alkali or $$OH^-$$ ions. This can be explained by two-step mechanism shown below. Each step is an elementary reaction with its own rate constant, step $$1$$ proceeds much more slowly than step $$2$$.
Step $$1$$. $$(CH_3)_3C-Br\overset{slow}{\underset{k_1}{\rightarrow}}(CH_3)C^++Br^-$$
Rate of reaction $$=k_1[(CH_3)_3C-Br]$$
The first step consists of breaking of C-Br bond and it determines the rate of overall reaction. So, step $$1$$ is called the rate-determining step. The rate determining step in this reaction involves only a single molecule, therefore, it is said to be unimolecular. Also, this type of mechanism is known as $$SN^1$$ mechanism(substitution, nucleophilic, unimolecular).
Step $$2$$. $$(CH_3)_3C^++OH^-\overset{fast}{\underset{k_2}{\rightarrow}}(CH_3)C-OH$$
Rate of reaction$$=k_2[(CH_3)_3C^+][OH^-]$$
The second step involves the attack of $$OH^-$$ ion. This is the fast step, since it is the bond formation step.
Energy profile diagram of $$SN^1$$ mechanism shows that rate of a reaction is independent of the concentration of nucleophile. The first step requires larger activation energy $$(\Delta E_1)$$ than the second step $$(\Delta E_2)$$. The first step to form carbocation determines the rate of overall reaction. The second step, which is the attack of nucleophile on carbocation is exothermic i.e., it is a lower energy transition state. The intermediate carbocation appears at a low point in the diagram. The conditions and reagents which favour the formation of carbocation will accelerate the $$SN^1$$ reaction. The energy difference between products and reactants is $$\Delta$$H, i.e., Heat of reaction.

The conversion of ethyl methanoate into propan-$$2$$-ol is as follows.
$$2CH_3MgX+HCOOCH_2CH_3\overset{H_3O^+}{\rightarrow}CH_3CH(OH)CH_3$$.

The lesser reactivity of haloarenes can be explained as follows :

(1) Stabilization by resonance : The electron pair of halogen atom in haloarene, is in conjugation with the $$\pi$$ electrons of the benzene ring. Thus, haloarene is a resonance hybrid of following structures :-
Thus, due to resonance, the contributing structures II, III and IV indicate that $$C - X$$ bond has partial double bond character. As a result of this, $$C - X$$ bond in haloarene is shorter and hence stronger than that of haloalkane. So it is difficult to break $$C - X$$ bond in haloarenes compared to haloalkanes, therefore, it is less reactive

(2) Polarity of carbon-halogen bond : The dipole moment $$C - X$$ bond in chlorobenzene is $$1.7D$$ while that in chloroalkanes it is in the range $$2.0 - 2.2 D$$. Greater the dipole moment, greater is the polarity and more is the reactivity. Thus, $$C - X$$ bond of haloalkane is more polar than the $$C - X$$ bond of haloarene. Hence, haloarenes are less reactive than haloalkanes.

(i) Aryl halides are less reactive in nucleophilic substitution reactions. This is due to following two reasons.
(1) $$\displaystyle C-X$$ bond in aryl halides is $$\displaystyle sp^2$$ hybridised. s character is more. Bond length is shorter. Bond is strong and requires greater energy. Hence, aryl halides are less reactive.

 $$\displaystyle C-X$$ bond in alkyl halides is $$\displaystyle sp^3$$ hybridised. s character is less. Bond length is longer. Bond is weak and requires lesser energy. Hence, alkyl halides are more reactive.

(2) The lone pair of electrons of halogen atom in aryl halides is in conjugation with pi electrons of the ring. This resonance imparts partial double bond character to $$\displaystyle C-X$$ bond.

(ii) Example for nucleophilic substitution reactions of aryl halides.
$$\displaystyle Chlorobenzene$$ on heating with $$\displaystyle NaOH$$ at 623 K under pressure gives phenol. Here, $$\displaystyle -Cl$$ is replaced with $$\displaystyle -OH$$

$$\displaystyle C_6H_5-Cl + OH^- \xrightarrow [H^+]{NaOH \: 623 \: K, \: pressure } C_6H_5-OH+Cl^-$$

(i) (Refer to Image 1)

It is an example of nucleophilic aromatic substitution reaction. $$ \displaystyle -Cl$$ group is replaced with $$ \displaystyle -CN$$ group. $$ \displaystyle CN^-$$ is the nucleophile.

More stable the carbocation formed after removal of $$H$$ more will be the ease of removal of H as shown above.

Order of stability of carbocation $$:1>3>2$$

Order of ease of removal of $$H: H_1>H_3>H_2$$

Propan-2-ol on reaction with iodine and sodium hydroxide gives yellow precipitate and the reaction is called Iodoform test.

Here, $$OH^-$$ & $$Cl^-$$ both are ortho & para directing groups. The relativity of $$OH^-> Cl^-$$ and to avoid the steric hinderance the new $$cl^-$$ ( from $$KCl$$) is placed at $$4^{th}$$- position from $$OH$$.

3-chloro-prop-1-ene

a) In order for a haloalkane to dissolve in water, energy is required to overcome the attraction 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.

b) Grignard reagents are very reactive. In the presence of moisture, they react to give alkanes. Therefore, Grignard reagents should be prepared under anhydrous conditions.

$$RMgX + H_2O \rightarrow R-H + Mg(OH)X$$

c) Refer Image: In vinyl chloride, the chlorine atom is attached to the double-bonded carbon. So, it forms resonance and gets the partial double bond character and thus difficult to break. Whereas the ethyl chloride does not have the property of resonance and hence it shows nucleophilic substitution reaction easily.

$$CH_3-CH_2-Br \xrightarrow []{KCN}  CH_3-CH_2-CN$$

Substitution reaction happens in which hydrogen atom in the Methane are replaced one at a time by chloride Ion or chlorine atom. a mixture of Methane and chlorine is exposed to a flame  that explodes producing carbon and hydrogen , chlorine 

The most stable radical is allylic radical.

The carbocation is more stable at (I) so iodine attacks at (I).

Due to chemical unreactivity and heat-absorbing properties, a wide variety of applications of chlorofluro carbons $$\left( { CFC }_{ S } \right) $$ are there. For example $$\left( { CFC }_{ S } \right) $$ have been used as refrigerants in air conditioners and refrigerated, in aerosol spray cans, in manufacturing foams as industrial solvents, and as cleaning agents in the manufacture of electronics.

$$1^o$$ isomers:-                                                                             $$CH_3$$

$$CH_3CH_2CH_2Br \xrightarrow[(as \,alcholic \,so - H_2O)]{KOH(alc)} CH_3 - CH = CH_2$$
                                                                                  $$\downarrow{HBr}$$
                                                                   $$CH_3 - \underset{Br}{\underset{|}{CH}} - CH_3$$
                                                                                  $$\downarrow{KOH}(aq)$$
                                                                    $$CH_3 - \underset{OH}{\underset{|}{CH}} - CH_3$$

$$t-Buo^-$$ in $$t-Bu-OH$$ it act as a base abstract the proton and $$Cl^-$$ would leave forming a double or $$\alpha-\beta$$ insaturated ketone.

In chloro benzene, chlorine is other and para directing in nature. Hence Nitration account para position of chlorine. 

$$i.$$ From ethyl alcohol:

$$\rightarrow$$ The reaction between Haloalkanes and $$Na$$ in presence of dry ether leads to the formation of alkanes. This reaction is known as Wartz reaction.

Rules for IUPAC Nomenclature:-

The oxidation state of +5 in pentahalides is more as compared to +3 oxidation state in trihalides. Due to the higher positive oxidation state of central atom in pentahalide state, these atoms will have larger polarizing power than the halogen atom attached to them.
But In case of trihalides due to +3 oxidation state the central atom will polarize the halogen atom to a lesser extent as compared to pentahalide state. 
Hence due to larger polarization of bond in pentahalide state as compared to trihalide state, the pentahalides are more covalent than trihalides.

The image formed by a plane mirror is always virtual and erect. The size of the image is equal to that of the object. The image formed is as far behind the mirror as the object is in front of it. The image is laterally inverted.

For the given conversion first, 2-bromopropane undergoes elimination reaction in presence of $$alc. KOH$$ to form propane.

Fine beam of light entering a small hole in a dark room, illuminates the particles in its path because of Scattering of light (Tyndall effect).

Bromoethane can be converted to iodoethane by Finkelstein reaction.

$$\underset{Ethyl \, bromide}{CH_3 CH_2 Br} \xrightarrow{alc. KOH} \underset{Ethene}{CH_2 = CH_2}$$

The Wurtz reaction, named after Charles Adolphe Wurtz, is a coupling reaction whereby two alkyl halides are reacted with sodium metal in dry ether solution to form a higher alkane: 

Alkyl chloride is a reactive compound while vinyl chloride is inert towards nucleophilic substitution.

Vinyl chloride on reaction with dimethyl copper yields Propene-1, with the precipitate of CuCl2.

 Another name of 1,1,2,2-tetrachloroethane is s-Tetrachloroethane Acetylene tetrachloride.

$$CH_3-CH_2Br+NaOH \rightarrow CH_2=CH_2 + NaBr+ H_2O$$

Alkyl iodides are 'less stable and lose free iodine. This iodine makes the remaining iodide darken.

Preparation of benzoyl chloride. Benzoyl chloride could be prepared by reacting thionyl chloride or phosphorus pentachloride with benzoic acid. Dry benzoic acid are treated in a flask, finally pulverized phosphorus pentachloride.

4-Bromopent-2-ene

The property of certain compounds to rotate the plane polarized light in a characteristic way when it is passed through their solutions is called optical activity of compounds.

3-Bromo-2-methylbut-1-ene

1-Bromo-2, 2-dimethyl propane.

4-Bromo-3-methylpent-2-ene

$$CH_2=CH_2+Cl_2 \to CH_2(Cl)-CH_2(Cl)$$

Ethyl$$-4-$$chlorobenzoate

IUPAC name of the given compound is $$2$$-Chloro-$$3$$-methylbutane and it is secondary $$(2^{\circ} )$$ alkyl halide.

Neopentyl bromide ionizes to form first a $$1^\circ$$ carbocation which rearranges to form the more stable $$3^\circ$$ carbocation. This is attacked by weak nucleophile ethanol followed by loss of proton to yield ethyl tert-pentyl ether.

4-Bromo-3-methyl pent-2-ene.

4-Bromo-3-methyl pent-2-ene.

IUPAC name of the given compound is $$3$$-Chloro-$$4$$-methylhexane and it is a secondary $$(2^{\circ}) $$ alkyl halide.

IUPAC name of the given compound is $$1$$-Bromo-$$3,3$$-dimethyl-$$1$$-phenylbutane and it is a secondary $$(2^{\circ} )$$ benzylic halide.

IUPAC name of the given compound is $$3$$-Chloro-$$5$$-methylhex-$$2$$-ene and it is a vinylic halide.

IUPAC name of the given compound is $$1$$-chloromethyl-$$3$$-($$2,2$$-dimethylpropyl)benzene and it is a primary $$(1^{\circ}) $$ benzylic halide.

IUPAC name of the given compound is $$3$$-Chloro-$$3$$-methylpentane and it is a tertiary $$(3^{\circ}) $$ alkyl halide.

IUPAC name of the given compound is $$1$$-Bromo-$$2$$-ethyl-$$2$$-methylpentane and it is a primary $$(1^{\circ}) $$ alkyl halide.

Substitution Reaction

The IUPAC name of the given compound is Chlorobutane.

$$\underset { Methyl\ bromide }{ 2CH_3-Br+2Na } \xrightarrow [  ]{ dry\ ether } \underset { ethane }{ CH_3-CH_3+2NaBr } $$

$$(CH_{3})_{2} CH - CH(Br) CH_{2} CH_{3} \xrightarrow[ethanol/heat]{C_2H_5ONa} (CH_{3})_{2} C = CH CH_{2} CH_{3} $$

Wurtz reaction-

(A) A molecule of HBr is eliminated through E2 mechanism. Ethoxide ion acts as base.
(B) A molecule of HBr is eliminated through E1cb mechanism. Ethoxide ion acts as base.
(C) A molecule of HBr is eliminated through E1 mechanism. Moist silver oxide (silver hydroxide) acts as base.

Solvolysis reaction:

There are two different positions so from either $$\beta$$ elimination can occur reaction.

((1S,2R)-1-bromopropane-1,2-diyl)dibenzene reacts with ethoxide ion to form (E)-prop-1-ene-1,2-diyldibenzene.
Ethoxide ion is a strong base and affects dehydrobromination to form carbon-carbon double bond.
Anti elimination results in the formation of trans alkene.

The above reaction is an example of solvolysis reaction as $$CH_{3}COOH$$ behaves as both solvent and a nucleophile.

Solvolysis reaction is faster with tertiary halides. 

$$(1S,2S)-1-bromopropane-1,2-diyl)dibenzene$$ reacts with ethoxide ion to form $$(Z)-prop-1-ene-1,2-diyldibenzene.$$
Ethoxide ion is a strong base and affects dhydrobromination to form a carbon carbon double bond.
Anti elimination results in the formation of the cis alkene.

Above reaction is application of wurtz's reaction in which alkane is from by alkyl halide

The treatment of 1,2 dibromodecane with $$KOH$$ in aqueous ethanol gives three isomeric compounds due to dehydrobromination (loss of one $$HBr$$ molecule).
These compounds on treatment with sodium amide in dimethyl sulphoxide gives 1-decyne.

Given reaction is wurtz reaction

Whether an alkyl halide will undergo an $$S_N1$$ or an $$S_N2$$ reaction depends upon a number of factors. Some of the more common factors include the natures of the carbon skeleton, the solvent, the leaving group, and the nature of the nucleophile.
Only those molecules that form extremely stable cations undergo $$S_N1$$ mechanisms. Normally, only compounds that yield $$3$$ (tertiary) carbonications (or resonancestabilized carbocations) undergo $$S_N1$$ mechanisms rather than $$S_N2$$ mechanisms. Carbocations of tertiary alkyl halides not only exhibit stability due to the inductive effect, but the original molecules exhibit steric hindrance of the rear lobe of the bonding orbital, which inhibits $$S_N2$$ mechanisms from occurring. Primary alkyl halides, which have little inductive stability of their cations and exhibit no steric hindrance of the rear lobe of the bonding orbital, generally undergo $$S_N2$$ mechanisms.
Polar protic solvents such as water favor $$S_N1$$ reactions, which produce both a cation and an anion during reaction. These solvents are capable of stabilizing the charges on the ions formed during solvation. Because $$S_N2$$ reactions occur via a concerted mechanism (a mechanism which takes place in one step, with bonds breaking and forming at the same time) and no ions form, polar protic solvents would have little effect upon them. Solvents with low dielectric constants tend not to stabilize ions and thus favor $$S_N2$$ reactions. Conversely, solvents of high dielectric constants stabilize ions, favoring $$S_N1$$ reactions.
$$S_N2$$ and $$E_2$$ reactions require a good nucleophile or a strong base. $$S_N1$$ and $$E_1$$ reactions occur with strong bases with molecules whose -carbon is secondary or tertiary and in the absence of good nucleophiles.
$$S_N1$$ and $$E_1$$ reactions are not synthetically useful because they almost always give a mixture of substitution and elimination products. The proportion of these products does vary with the -carbon branching, however. Generally speaking, greater branching gives more elimination products. These elimination products generally follow Saytzeff's rule. Keep in mind that the -carbon must be secondary or tertiary for $$S_N1$$ or $$E_1$$ to occur at all. Thus the branching effect is much less pronounced than in the $$S_N2$$ and $$E_2$$ reactions.
In the broader context of all substitution and elimination reactions, remember that $$S_N1$$ and $$E_1$$ will not occur in the presence of a strong base or a good nucleophile. In these cases, $$E_2$$ and $$S_N2$$ dominate their unimolecular cousins.
$$E_1cb$$ reactions are quite different from $$E_1$$ reactions. $$E_1$$ reactions require a strong leaving group and a stable cationic intermediate. The leaving group first leaves to create the cationic intermediate. Then, the elimination occurs.
$$E_1cb$$ is the accepted mechanism for the aldol condensation. 

Key Requirements:

• There must be an acidic proton. In the case of the one depicted above, the carbonyl acidifies the alpha-carbon proton enough for it to be deprotonated.
• There usually is a bad leaving group. If it were a really good leaving group, either $$E_1$$ or $$E_2$$ would probably occur instead.

The key differences between the two is that $$E_1$$ goes through a cationic intermediate, while $$E_1cb$$ goes through a anionic intermediate. Therefore, never propose a $$E_1cb$$ in acid or a $$E_1$$ in base.Also, the $$cB$$ in $$E_1cb$$ stands for conjugate base meaning the elimination goes through the intermediate that is the conjugate base of my starting material. In this case, the enolate was the conjugate base.

In $$A$$ as all positions are same so only one isomer forms for mono substituted.
In $$B$$, three isomers are forms and only $$1,4$$ disubstituted is optical isomer.
In $$C$$, three isomers are possible and only $$1,2$$ disubstituted is optically inactive
In case of benzene ,$$3$$ isomers will be form and all are optically active.

As $$S_N1$$ and $$E_1$$ are unimolar reactions while $$S_N2$$ and $$E_2$$ are Bimolar reactions so
(a) follows $$S_N2$$ mechanism while b follows $$S_N1$$ mechanism due to resonance.
(c) follows $$E_1cb$$ due to strong base $$PhO^-$$ while 
(d) follows $$E_1$$ mechanism in presence of alc. $$KOH$$ while  
(e) follows $$E_2$$ mechanism in presence of alc. $$KOH$$

The major organic products are shown below:
(i) $$CH_3CH_2CH_2Cl + NaI \xrightarrow [heat ]{acetone} CH_3CH_2CH_2I (1-iodopropane) + NaCl$$

i) Ethanol to but-1-yne

The compound have bromine at 3rd cabon and methyl at 2nd position, double bond(en) at 1st position and three carbons in the longest chain(prop). So, the name of the copmound is 3-bromo-2-methyl propene.

The IUPAC rules for naming alkyl halides

In the $$S_N1$$ reaction, the big barrier is carbocation stability. Since the first step of the $$S_N1$$ reaction is the loss of a leaving group to give a carbocation, the rate of the reaction will be proportional to the stability of the carbocation. Carbocation stability increases with increasing substitution of the carbon ($$tertiary > secondary > primary$$) as well as with resonance.

The conversion of methyl bromide to methanol is through $$ \displaystyle SN^2$$ mechanism. The conversion of tertiary butyl bromide to tertiary butanol is through $$ \displaystyle SN^1$$ mechanism. Tertiary butyl bromide is a tertiary alkyl halide. Tertiary alkyl halides undergo $$ \displaystyle SN^1$$ substitution. Methyl halides and primary alkyl halides  undergo $$ \displaystyle SN^2$$ substitution.

$$ \displaystyle SN^1$$ mechanism	$$ \displaystyle SN^2$$ mechanism
It involves two steps.	It involves only one step.
It involves racemisation.	It involves inversion of configuration.

The chemical equations:
(a) Wurtz reaction:
Alkyl halides react with sodium/ dry ether to form higher alkanes.
$$ \displaystyle \underset {  \displaystyle  \text { methyl bromide } }{ 2CH_3-Br}+2Na \xrightarrow {\text {dry ether }} \underset {  \displaystyle  \text { ethane } }{ CH_3-CH_3}+2NaBr$$

(b) Finkelstein reaction:
The reaction between alkyl chlorides/ bromides and NaI/ dry acetone gives alkyl iodides.
$$ \displaystyle \underset {  \displaystyle  \text { ethyl bromide } }{ CH_3-CH_2-Br}+NaI \rightarrow \underset {  \displaystyle  \text { ethyl iodide } }{ CH_3-CH_2-I}+NaBr$$

1-methylcyclohex-1-ene is formed by dehydrohalogenation of 1-bromo-1-methylcyclohexane with alcoholic KOH. If there is possibility of formation of more than one alkene, then major product is identified by C. According to 1-methylcyclohex-1-ene, during dehydrohalogenation, more substituted alkene is the major product.

The reaction $$H_3C-Br+AgF\rightarrow H_3C-F+AgBr$$ is named as Swartz reaction. It is halogen exchange reaction in which an alkyl bromide is converted into alkyl fluoride. Instead an alkyl bromide, an alkyl chloride can also be used. Instead of $$AgF$$, we can also use $$CoF_2$$, $$SbF_3$$ or $$Hg_2F_2$$.

$$I^−$$ is a good nucleophile as well as a good leaving group. Therefore, if $$(R)-2-iodobutane$$ is treated with NaI, repeated $$S_N^​2$$ reactions occur. As a result, eventually, a $$\text{racemic mixture of (±)-2 iodobutane}$$ is obtained which is optically inactive.

$$CH_3CH_2CH_2CH_2-Cl + KOH \rightarrow CH_3CH_2CH=CH_2+KCl + H_2O$$

Reaction:

Complete Reaction: 

Formation of the $$more \;stable \;carbocation$$ makes $$SN\;1$$ faster.
In all cases, the first molecule reacts $$faster$$ due to formation of more stable cation amongst the two of them. The reasons are
a.  $$3^o$$ cation is more stable than $$2^o$$
b. $$2^o$$ cation is more stable than $$1^o$$
c. $$2^o$$ cation is more stable than $$1^o$$
d. The first molecule stabilises the cation formed by resonance.

$$2C_4H_9Cl \xrightarrow [ Wurtz\ Reaction ]{ 2Na } C_4H_9 - C_4H_9 +2NaCl$$

$$3CH_3CH_2Cl+SbF_3 \xrightarrow [ Swart's \ reaction ]{ \triangle  } 3CH_3CH_2F+SbCl_3$$

i. $$DU$$ in $$A = \dfrac{(2n_C + 2) - (n_H + n_X - n_N)}{2}$$

$$A \rightarrow 3, \ B \rightarrow 1, \ C \rightarrow 6, \ D \rightarrow 2, \ E \rightarrow 4, \ F \rightarrow 5$$

Aryl halides are less reactive towards nucleophilic substitution reaction as compared to alkyl halides due to resonance stabilization.

$$\underset{(O.A\,alcohol)}{A(C_{6}H_{10}O)}\xrightarrow{2H_{2}/Pt}\underset{(O.I.A)}{(B)}\xrightarrow[CrO_{3}]{[O]} Resists\,oxidation$$

This is because ter-alkyl halides prefer to undergo dehydrohalogenation in the presence of sodium metal instead of undergoing Wurtz reaction.

Reaction proceeds by benzene mechanism(elimination-addition ArSN reaction)

Both (F) and (G) are aldehydes lacking $$\alpha-H$$ atom because they give positive Tollens test and undergo cross Cannizzaro reaction. One aldehyde is $$(HCHO)$$, since one product of cross C.R. is $$HCOONa.$$ Total $$C $$ atoms in $$(A)$$ are six ozonolysis product of (A) is $$HCHO$$ and a five $$C$$ aldehyde without $$\alpha-H$$ atom. So another aldehyde must be $$(Me_3C - CHO).$$

Halogen in haloarenes withdraws electrons through-I effect and release electrons through +R effect. The inductive effect is stronger than resonance effect and causes net electron withdrawl. As a result, the electrophilic substitution reactions in haloarness occur slowly.

The available data suggests that compound 'A' is an $$acid.$$ The compound 'B' is an $$alcohol$$ and it gives the acid 'A' upon oxidation with alkaline $$KMnO_4$$. Similarly when the acid 'A' is reduced with $$LiAlH_4$$, alcohol 'B' is formed. The two compounds i.e., 'A' and 'B' react in the presence of $$H_2SO_4$$ to form an $$ester\  'C'$$. The reaction is called the esterification reaction.

3-methyl-1-pentene has molecular formula $$C_6H_{12}$$ and is optically active. On catalytic hydrogenation, it gives 3-methylpentane having molecular formula $$C_6H_{14}$$ which is optically inactive.

$$CH_{3} CH_{2} C \equiv CH + HCl  \xrightarrow[Addition]{Mark.} CH_{3} CH_{2} - \underset{Cl}{\underset{|} {C}} = CH_{2}  $$

No, we will not get any precipitate if we add silver nitrate solution to chloromethane because chloromethane is a covalent compound and hence does not contain $$Cl^{-}$$ ions.

(i) Secondary alkyl halide

(ii) Primary alkyl halide

(iii) Aryl halide

(iv) Allyl halide

As the resulting product is optically inactive the alkene should have same groups attached along one of the $$C =C$$ bond. As the carbon is having similar group, it is having achiral carbon, so optical isomerism is lost in the molecule.

$$ \begin{array}{l} \text { Given compound o-bromotoulene } \\ \end{array} $$

Bromination of Alkyl halides is free radical Substitution reaction n-Butane on brominatian two possibilite (a) $$\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}^{\circ}+\mathrm{H}^{\circ}$$