SN2 Reaction Basics - Free MCQ Practice Test with solutions, NEET Chemistry

CH₃ Br is more reactive termed and   is stronger nucleophile.

Answer: D - All of the above.

Rate law: An S_N2 reaction is bimolecular in the rate-determining step and follows rate = k [nucleophile][substrate]. Therefore, increasing the strength or concentration of the nucleophile increases the rate.

Solvent effect: Polar protic solvents hydrogen-bond and solvate anionic nucleophiles, lowering their reactivity. Polar aprotic solvents do not solvate anions strongly, so nucleophiles remain more available and the S_N2 reaction is faster in polar aprotic solvents.

Leaving-group ability: A better leaving group stabilises the negative charge after bond breaking and lowers the activation energy. Thus, substrates with better leaving groups react faster by the S_N2 pathway; for example, I⁻ > Br⁻ > Cl⁻ as leaving groups.

All three statements are therefore correct, so the correct option is D.

S_N2 reaction is a one step (concerted) reaction that involes a single transition state.

Allyl bromide (I) is most reactive among the given halides as pi bonds from allylic position stabilises the transition state. Vinyl iodide (II) is least reactive due to partial double bond character. Electron withdrawing inductive effect of CH₃O^- increases reactivity of (IV) over (III)

 (IV) is most reactive as it is benzylic as well as electron withdrawing effect of — NO₂ further increases the reactivity. (Ill) is least reactive due to resonance effect resulting in partial double bond character between carbon and chlorine. (I) is less reactive in S_N2 reaction than II, due to greater steric hindrance in (I).

Substrates are same as they are at same potential energies. Nucleophiles are different as magnifested by different activation energies and different products with different potential energies,

HO^- is stronger nucleophile than CH₃COO^- . Hence, lower activation energy of (I) than (II).

The transition state in S_N2 reaction is pentavalent as indicated here.
 

In a S_N2 reaction, Rate = k[substrate] [nucleophile]
Hence, rate depends both on concentration of substrate as well as nucleophile. Rate of reaction also depends on nature of leaving group. A better leaving group gives lower activation energy hence, greater rate of reaction. Number of lone pair has no role to play in nucieophilicity hence reactivity.

Increasing concentration of nucleophile (NaOH) increases the reactivity. Adding some Nal also catalyse reaction because of following equilibrium.

Now reaction occur with CH₃I which has better leaving group. Heavier isotope forms stronger covalent bond. Hence, replacing CH₃Br by ¹⁴CH₃Br stabilises substrate, decreases rate whereas replacing NaOH by NaO¹⁸H stabilises transition state increases rate of reaction.

Both nucleophile and leaving group are electron rich species, nucleophile attacks from backside (most remote position) to have minimum repulsion with leaving group. Also, due to the presence of leaving group on front side there is less physical space for attack of nucleophile.

As indicated in the given reaction, α-carbon is less hindered in 1- chloropropane. Also + I effect of three methyl groups in 2- chloro-2-methyl propane decreases electropositive character of α-carbon, decreases reactivity further in S_N2 reaction.

Steric hindrance plays the most important role in S_N2 reaction. Hence, although Cl is poorer leaving group than I, Cl is substituted predominantly in the above reaction due to less steric hindrance at α-carbon. Addition of Nal replace Cl by I and substitution becomes easier.

Stronger bond formation at α-carbon affects the stability of product (thermodynamics) not the rate (kinetics) of reaction. Change of isotope at α-position or in halogens affect the rate, hence show kinetic isotopic effect. Presence of electron withdrawing group in substrate increases electrophilicity of α-carbon hence, increases its reactivity towards nucleophiles.

Products are same in both reactions, hence same potential energies of products are shown. Also I^- is better leaving group, has lower activation energy in S_N2 reaction as indicated by curve-l in diagram.

Since I^- is better leaving group, k_I > k_II

Following neutralisation reaction is preferred over S_N2 reaction :

Rate = A [C₂H₅I] [OH^-] Rate is linear function of both alkyl halide and nucleophile concentration both will decrease by same factor. Hence, if after 1.0 h concentration of iodoethane decreases to  concentration of hydroxide will also decrease by the same  factor and rate by  , i.e. by 9 times.

Compounds (II), (III), (IV), (V), (VIII) and (X) react at faster rate than 2-bromobutane in S_N2 reaction.

Only the circled groups are substituted in S_N2,