Haloalkanes And Haloarenes - Class 12 Medical Chemistry - Extra Questions

When cyclopentyl 4-methylbenzenesulfonate is heated with strong base, it undergoes dehydrobromination to form cyclopentene.  In this reaction, a molecule of 4-methylbenzenesulfonic acid is lost to form carbon carbon double bond.

Firstly $$NH_3$$ attack by $$SN_2$$ mechanism then intramolecular $$SN_2$$ attack occur to give six-membered product as shown above.

When (1-bromopropyl)benzene is heated with strong base, it undergoes dehydrobromination to form (E)-prop-1-enylbenzene. The trans alkene is more stable than cis alkene. Moreover in the given compound, the carbon carbon double bond is in conjugation with aromatic nucleus.

When 2-bromopentanoic acid is heated with strong base, it undergoes dehydrobromination to form (E)-pent-2-enoic acid. The trans alkene is more stable than cis alkene. Moreover in the given compound, the carbon carbon double bond is in conjugation with carboxylic group.

The reaction is major elimination reaction $$(E_{2})$$ some minor product is obtained from $$SN_{2}$$ reaction.

$$E_{2}$$ reaction is anti-elimination reaction.

Major product will be elimination product rather than substitution product because of fact that $$(CH_3)_3CO^-K^+$$ is bulky in nature.

(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.

• The $${ S }_{ N }1$$ reaction is a substitution reaction in organic chemistry. "$${ S }_{ N }$$" stands for nucleophilic substitution and the $$ "1" $$ represents the fact that the rate-determining step is unimolecular. 

The reaction involves a carbocation intermediate and is commonly seen in reactions of secondary or tertiary alkyl halides under strongly basic conditions or, under strongly acidic conditions, with secondary or tertiary alcohols. With primary alkyl halides, 

(1) Methane to ethane

The above reaction proceeds through $$\mathrm{S_N2}$$ mechanism, Br gets replaced with a hydrogen because of greater leaving tendency of Br and the product forms with inversion.

$$ (\pm ) $$ 2-Butanol is a racemic mixture,i.e., there are two enantiomers in equal propertions. The rotation by one enantiomer will be cancelled by the rotation due to the other isomer, making the mixture optically inactive.

In $$SN_1$$ racenisation occur due to formation of planar carbocation,attack of nucleophilic can occur from both above and below the plane.

(2R,3S)-2,3-dibromopentane loses a molecule of bromine to form (E)-pent-2-ene.
Anti elimination results in the formation of trans alkene.

The reaction takes place through $$E_{2}$$ mechanism.

Refer Image,

The reaction will take place through $$SN_{2}$$ mechanism.

Step by Step Reactions:

The reaction between benzene and excess methyl chloride (in presence of aluminum chloride) proves the existence of $$\sigma$$-complex in SE reaction.
The reaction $${ C }_{ 6 }{ H }_{ 6 }+\underset { (Purple) }{ { \quad I }_{ 2 } } \longrightarrow { Brown\\ coloured\\ compound }\xrightarrow [  ]{ \Delta  } { C }_{ 6 }{ H }_{ 6 }+{ \quad I }_{ 2 }$$

proves the existence of $$\pi$$-complex (charge transfer complex).
The reaction $${ C }_{ 6 }{ H }_{ 6 }\xrightarrow [  ]{ Sulphonation } { C }_{ 6 }{ H }_{ 5 }{ SO }_{ 3 }H$$<br><br>$${ C }_{ 6 }{ D }_{ 6 }\xrightarrow [  ]{ Sulphonation } { C }_{ 6 }{ D }_{ 5 }{ SO }_{ 3 }H$$ shows $${1}^{o}$$ kinetic isotope effect. It also shows slightly $${2}^{o}$$ kinetic isotope effect
The reaction $${ C }_{ 6 }{ H }_{ 6 }\xrightarrow [  ]{ Nitration } { C }_{ 6 }{ H }_{ 5 }{ NO }_{ 2 }$$<br><br>$${ C }_{ 6 }{ H }_{ 6 }\xrightarrow [  ]{ Nitration } { C }_{ 6 }{ D }_{ 5 }-{ NO }_{ 2 }$$ does not show $${1}^{o}$$ kinetic effect. It shows slightly $${2}^{o}$$ kinetic isotope effect.

When 1,3-dichloro cyclohexene reacts with alc. KOH, it undergoes elimination reaction to form cyclohex-1,3-diene.

As the product R contains two same group ($$CH_3CH_2$$) group attached to carbon so achiral carbon.

(D) $$Ph-\underset{\displaystyle O}{\underset{\parallel}{C}}-\underset{\,\,\,\,\displaystyle {CH_{2}}}{\underset{\parallel}{\,\,\,\,CD}}-\underset{\displaystyle O}{\underset{\parallel}{C}}-OH\xrightarrow[]{\Delta}Ph-\underset{\displaystyle O}{\underset{\parallel}{C}}-\overset{\overset{\displaystyle D}|}{\underset{\underset{\,\,\,\,\displaystyle {CH_{3}}}|}{C}}-H+Ph-\underset{\displaystyle O}{\underset{\parallel}{C}}-\overset{\overset{\displaystyle H}|}{\underset{\underset{\,\,\,\,\,\,\,\displaystyle {CH_{3}}}|}{C}}-D+CO_{2}\uparrow$$

A-1,3,4,5 ; B- 1,3,4,5 ; C-2,4,5 ; D-2,4,5

2-Bromo-2-methyl butane, because tertiary alkyl halides on dehydrogenation from most substituted alkene which is more stable.

It is due to presence of a chiral carbon in butan -2-ol , which makes it an optically active compound.

2-Bromopentane is optically active as carbon number two is chiral carbon.

When secondary alkyl halide (isopropyl chloride) is treated with benzene in the presence of anhydrous $$AlCl_3$$ isopropylbenzene is obtained.