Test: Nucleophilic Substitution Reaction - Chemistry MCQ

α-halocarbonyl are the best S_N2 substrate.
Here nucleophilic substitution reaction will happen and the above product will be formed. Hence this reaction is correct.

Here (PrO) and (CH₃) will be at same carbon atom and (OH) will be at adjacent carbon atom. Hence this reaction is wrong.

n-PrO^– will attack the carbon atom attached to oxygen above the plane and the above product will be formed. Hence the reaction is correct.

Attack of Bromine ion (Br^–) will be at that carbon atom where less substitution is present, the above product will form. Hence this reaction is correct.

Least hindered halide give fastest S_N2 reaction as the hindrance increases. As the hindrance increases, the occurrence of S_N2 reaction decreases.

When ether reacts with the reactant nucleophilic substitution conjugate base reaction (i.e. S_NCB) will take place and there will be retention of configuration and it will be R product. When C₅H₅NH reacts with reactant S_N2 reaction will take place and configuration inversion will take place and it will be S product.

As we know rate order in S_N2 is CH₃X > 1^o > 2^o > 3^o halide and rate order in S_N1 is CH₃X < 1^o < 2^o < 3^o halide. So according to the following representation.

Here the rate of reaction is directly proportional to the concentration of RI and (131_I-), so statement is correct.
If any molecule undergoes substitution, then it forms inversion product due to SN₂ reaction. Hence statement is correct.
For statement, below reaction one (+) molecule converts into (–) molecular and it also converts one optical reaction of (+) molecule.

Final solution contains both Iodine (I¹³¹) and normal (I). So, we can say this is the only incorrect statement about the above reaction.

I & II reaction undergo via S_N2 mechanism. III & IV reaction undergo via S_N1 mechanism. As we know the rate of reaction in S_N2 reaction is directly proportional to the concentration of nucleophile.

The solvation of above compound can be compared by this trend, i.e. Solvation trend: 3^O halide > 2^O halide > 1^O halide > vinylic halide. So, (III) molecule is an example of 3^O halide, (II) molecule is an example of 2^O halide, (IV) molecule is an example of 1^O halide and (I) molecule is an example of vinylic halide.

(1) Formation of carbocation is RDS, more stable the cation more easily it will form. Hence rate order is t-BuX > iso-PrX > EtX > MeX
(2) Formation of carbocation CH₃ is most difficult hence ionization energy is highest. Similarly, EtX has lesser ionization energy than MeX and so on.
(3) Order of energy of activation is MeX > EtX > iso-PrX > t-BuX.

Alpha-halo carbonyl undergoes fastest S_N2 reaction due to the coupling of π* of CO bond and π* of C–X bond.

Rate of formation of II and IV similar because H & D are approximately of same size.
C–H & C–D bond breaking is the part of RDS hence rate in (I) is slower than (III)
It undergoes E2 not via E1 because carbocation is not stable.