Only One Option Correct Type
Direction (Q. Nos. 1- 26) This section contains 26 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct.
Which of the following statements about a SN1 mechanism is true?
SN1 reaction proceeds via carbocation intermediate, hence 3° halide (forms most stable tertiary carbocation) is most reactive. Also, polar protic solvent stabilises carbocation, increases reactivity of SN1 reaction.
Place the following alkyl halides in increasing order of reactivity in a SN1 reaction.
Reactivity follow the order of stability of carbocation. Stability of carbocation follows the order 1° < 2° < 3°
Which of the following will form a white precipitate on treatment with AgNO3(ag)?
Both option (b) and option (c) forms resonance stabilised carbocations with Ag+ as
Which of the following is most reactive in SN1 reaction?
It forms the most stable carbocation.
Which of the following is most reactive in SN1 reaction?
Iodide is best leaving group among halides and it forms the most stable carbocation.
A solution of (+)-1 -chloro-1-phenylethane in toluene racemises slowly in the presence of a small amount of SbCI5 due to the formation of
Which of the following compounds is most rapidly hydrolysed by SN1 mechanism?
It forms the most stable, triphenyl methyl carbocation.
Which is the correct order of reactivity towards β-elimination with a strong base?
The order of reactivity in E2 reaction is :
1° < 2° < 3° : Alkyl halide
Consider the elimination reaction given below :
Major product is
It is resonance stabilised, most stable diene
Choose the correct statement about the elimination reaction shown (ignore competing substitution).
Bromide is a better leaving group and in E2 elimination reaction, more substituted, more stable product is formed as major product.
Which of the following undergoes E1 reaction most readily?
E1 reaction proceeds via carbocation intermediate and reactivity follow the stability of carbocation.
What is the major product of the E2 reaction of frans-1-bromo-2-methylcylcohexane with strong base?
In E2 reaction, strong base takes β-H from side anti to the leaving group, in the present case, on tertiary β-carbon, hydrogen atom is not available in anti position to leaving group, hence elimination occur from other β-carbon although it produces less substituted, less stable alkene.
Cyanide anion has two atoms that have lone pair of electrons. Either could act as nucleophile in the reaction. Yet in the vast majority of the cases, carbon acts as nucleophile and forms a bond to the substrate, why?
Carbon, being less electronegative than nitrogen, a better electron donor and stronger nucleophile.
In an aprotic solvent, which relative ordering best describes the nucleophilicity of the halide ions?
In aprotic solvents, nucleophiles are not solvated. Hence, basicity parallels nucleophilicity in case of halides.
Which statement is true about SN2 mechanism?
Rate of SN2 reaction is directly proportional to concentration of both substrate and nucleophile. Also, polar aprotic solvents increases reactivity as they solvates cationic part of nucleophile. A better leaving group increases reactivity of SN2 reaction because leaving group departs in the rate determining step.
Predict the product in nucleophilic substitution reactions?
SN2 reaction leads to inversion of configuration at the chiral α-carbon.
The most reactive in the following in reaction with NaCN is
Electron with drawing nitro group increases electrophilicity at the α-carbon.
Which SN2 reaction would take place most rapidly?
HS- is the strongest amongst the given nucleophiles.
The reactivity of 2-bromo-2-methyl butane (I), 1-bromo pentane (II) and 2-bromo pentane (III) towards SN2 reaction is
In SN2 reaction, the order of reactivity is
1° > 2° > 3° : Alkyl halides
Which of the following is correct increasing rate of SN2 reaction?
Steric hindrance in substrate to the attack of nucleophile decreases reactivity in SN2 reaction.
Chloroethane is treated separately with aqueous NaSCN and NaOCN. Which of the following is true statement regarding the substitution product?
Both SCN- and OCN- are ambident nucleophiles. In SCh-, more nucleophilic 'S’ acts as donor while in OCN-, more nucleophilic nitrogen acts as donor.
Consider the reaction coordinate diagram for a SN2 reaction, that is, for a reaction in which the charges are more distributed in the transition state than in the reactants. How does a change to a more polar solvent affect this reaction coordinate diagram?
A more polar solvents have .greater effect on stability of nucleophiles (due to fully developed negative charge on it) than on transition state (has partially developed negative charge).
The major product in the following reaction is
Only the primary aliphatic bromide is substituted here in SN2 reaction.
What will be the major monobromination product in the following reaction?
Nitro group exert strong .electron withdrawing resonance effect from para position, hence destabilises the stability of free radical formed at para methyl group. Therefore, bromination occur preferably at meta methyl group.
In the following groups :
OAc(I),—OMe(ll),—OSO2Me(lll),—OSO2CF3(IV) the order of leaving gro up ability is
Greater the stability of leaving group as base, better the leaving group.
Grignard's reagent when exposed to moisture
Grignard’s reagent consists of very basic carbanion, reacts with water at room temperature as :
Direction (Q. Nos. 27-30) This section is based on Statement I and Statement II. Select the correct answer from the codes given below.
Statement I : Direct fluorination of ethane usually fails to produce monofluorination product.
Statement II : Fluorination of ethane produces hexafluoro ethane.
Fluorine is highly reactive, it is difficult to control,the reaction at mpnofluorination level.
Statement I : When CH3CH2I is treated with a saturated AgCN solution, CH3CH2NC is formed as the major product.
Statement II : Cyanide is an ambident ligand.
Ag—C bond is covalent in AgCN, hence nucle ophilic attacks occur from nitrogen atom.
Statement I : 1-chloropropane when treated with Nal in acetone, 1-iodopropane is formed.
Statement II : Iodide is stronger nucleophile than chloride ion.
It is the lower solubility of NaCI in acetone that enables substitution to take place successfully.
Statement I : Bimolecular elimination (E2) reaction is stereospecific.
Statement II : Strong base approaches from anti position to leaving group at β-carbon
In E2 reaction, either enantiomer.