Q: The major product of the following reaction is:

Nucleophile PhS−substitute the Br−through SN2 mechanism with inversion of configuration at α−C

Q: Match the Column-I with Column-II and mark the correct option from the codes given below.

(A) CH3−CHBr−CD3E2⟶ Alc. KOHCH2=CH−CD3E2 reaction is a single-step reaction in which both deprotonation from β−C and loss of leaving group from α−C occur simultaneously in the rate-determining step. C-D bond is stronger than C−H bond. C−H is preferably broken in elimination.(B) Ph−CHBr−CH3 reacts faster than Ph−CHBr−CD3 in E2 reaction because in latter case, stronger C-D bond is to be broken in the rate determining step.(C) Ph−CH2−CH2BrC2H5O−⟶C2H5ODPh−CD=CH2(D) Both PhCH2CH2Br and PhCD2CH2Br will react at same rate in E1 reaction because C−H bond is broken in fast non-determining.Thus, (A) →q;(B)→q;(C)→r; (D) →p,s

Q: on hydrolysis in presence of acetone?

Stabilized by +R effect of −OCH3(I) ⟶H2OK(II) ⟶H2OL

Q: Identify the set of reagents/reaction of the followingalkyl halides for a SN2 reaction is

CH3−CH2−CH2−BrNaOH,80∘Cβ -elimination(x)⟶ Conc. alc. CH3−CH=CH2

Q: In SN2 reaction, the correct order of reactivity for the following compounds: CH3Cl,CH3CH2Cl, (CH3)2CHCl and (CH3)3CCl is:

Reactivity of SN2∝ steric hindrance 1 Reactivity order: 1∘ halide >2∘ halide >3∘ halide.

Q: On heating an aliphatic primary amine with chloroform and ethanolic potassium hydroxide, the organic compound formed is:

R−NH2⟶CHCl3/KOHCH3CH3+KOH⟶R−N≡C+KCl+H2O Alkyl isocyanide CCl¨+KCl+H2O α-elimination (dichloro carbene)

Question 1

The total number of alkenes possible by dehydrobromination of 3-bromo-3-cyclopentylhexane using alcoholic KOH is

Accepted Answer

The substrate has three different types of B–H, therefore, first, three structural isomers of alkenes are expected asThe last two alkenes (II) and (III) are also capable of showing geometrical isomerism; hence, two geometrical isomers for each of them will be counted giving a total of five isomers

Question 2

The major product of the following reaction is:

Accepted Answer

Nucleophile PhS−substitute the Br−through SN​2 mechanism with inversion of configuration at α−C

Question 3

Match the Column-I with Column-II and mark the correct option from the codes given below.

Accepted Answer

(A) CH3​−CHBr−CD3​E2⟶ Alc. KOH​​CH2​=CH−CD3​E2 reaction is a single-step reaction in which both deprotonation from β−C and loss of leaving group from α−C occur simultaneously in the rate-determining step. C-D bond is stronger than C−H bond. C−H is preferably broken in elimination.(B) Ph−CHBr−CH3​ reacts faster than Ph−CHBr−CD3​ in E2 reaction because in latter case, stronger C-D bond is to be broken in the rate determining step.(C) Ph−CH2​−CH2​BrC2​H5​O−⟶C2​H5​OD​​Ph−CD=CH2​(D) Both PhCH2​CH2​Br and PhCD2​CH2​Br will react at same rate in E1 reaction because C−H bond is broken in fast non-determining.Thus, (A) →q;(B)→q;(C)→r; (D) →p,s

Question 4

on hydrolysis in presence of acetone?

Accepted Answer

Stabilized by +R effect of −OCH3​(I) ⟶H2​O​K(II) ⟶H2​O​L

Question 5

Identify the set of reagents/reaction of the followingalkyl halides for a SN​2 reaction is

Accepted Answer

CH3​−CH2​−CH2​−BrNaOH,80∘Cβ -elimination(x)​⟶ Conc. alc. ​​CH3​−CH=CH2​

Question 6

The order of reactivity of the following alkyl halides for a SN​2 reaction is:

Accepted Answer

Reactivity of alkyl halide ∝ power of leaving group. size ↓ Base strength ↑ leaving tendency ↓IΘBrΘClΘFΘ​

Question 7

The synthesis of alkyl fluorides is best accomplished by:

Accepted Answer

Alkyl fluorides can be prepared by action of mercurous fluoride or antimony trifluorides (inorganic fluorides) on corresponding alkyl halide. It is known as Swarts reaction.

Question 8

In a nucleophilic substitution reaction: R−Br+Cl−⟶DMF​R−Cl+Br−, which one of the following undergoes complete inversion of configuration?

Accepted Answer

When alkyl halide having chiral carbon then only complete inversion takes place.In option 3   hence it gives complete inversion of configuration

Question 9

In SN​2 reaction, the correct order of reactivity for the following compounds: CH3​Cl,CH3​CH2​Cl, (CH3​)2​CHCl and (CH3​)3​CCl is:

Accepted Answer

Reactivity of SN​2∝ steric hindrance 1​
Reactivity order: 1∘ halide >2∘ halide >3∘ halide.

Question 10

On heating an aliphatic primary amine with chloroform and ethanolic potassium hydroxide, the organic compound formed is:

Accepted Answer

R−NH2​⟶CHCl3​/KOH​CH3​CH3​+KOH⟶​R−N≡C+KCl+H2​O Alkyl isocyanide CCl¨+KCl+H2​O​
α-elimination (dichloro carbene)

Question 11

A solution of (-)-chloro-1-phenylethane in toluene racemises slowly in the presence of a small amount of SbCl5​, due to the formation of:

Accepted Answer

The given compound, (-)- chloro-1-phenylethane in the presence of SbCl5​ forms a carbocation. Since, the carbocation is a planar species, therefore it can be attacked by SbCl6–​ either from the front or back side of the carbocation with equal ease. As a result, 50:50 mixture of two enantiomers of 1-chloro-1-phenylethane undergoes racemization due to the formation of a carbocation intermediate.

Question 12

Carbylamine forms from aliphatic or aromatic primary amine via which of the following:

Accepted Answer

R−NH2​⟶CHCl3​/KOH​R−N≡CCHCl3​+KOH⟶CCl¨+KCl+H2​O​
α-elimination (dichloro carbene)

Question 13

In Williamson synthesis of mixed ether having a primary and tertiary alkyl group if tertiary halide is used, then:

Accepted Answer

Tertiary alkyl halide mainly give β-elimination reaction so alkene will be formed as major product.

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Solutions for Organic Chemistry II for IIT JEE Main and Advanced by Rajesh Agarwal

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