Test: Reaction Mechanism Level - 1 - Chemistry MCQ

E2 stands for bimolecular elimination. The reaction involves a one-step mechanism in which carbon-hydrogen and carbon-halogen bonds break to form a double bond (C=C Pi bond).

It is nucleophilic substitution reaction which is in first order.

correct ans is 'A'

The conjugate base obtained by the elimination of protons of q -OH group is stabilized by extended conjugation through carbonyl group. And any of the conjugate base is not stabilized by such a factor so the conjugate base corresponds to q -OH group has stability ,so the H corresponding to this OH group is highly acidic. Consequently the H is captured readily by C₂H₂N₂ and readily gets methylated.

Aldehyde is more reactive than ketone, so, E alkene is generated on aldehyde.

Hydrolysis of ether. Lone pair of oxygen captures H atom from HI group that is present on the up gives rise to a stable carbocation(Resonance stabilized as well as well +i effect of 1 methyl group) and the reaction proceeds via SN1 pathway where I- acts as a nucleophile, but for other group, formed carbocation is highly unstable and ends up at alcohol.

It is an example of semipinacol rearrangement. NH₂ and OH group will remain in the diequatorial position in cyclohexane chair form. So after treatment with HNO₂, NH₂ becomes N₂⁺. Then rearrangement takes place through C-C bond migration which is antiperiplanar to N₂⁺. Thus ring contraction takes place and product will be B.

Benzoyl peroxide induces chain growth polymerization.

• This is a classical example of aromatic nucleophilic substitution. Here the reaction proceeds through two steps, in the first step the nucleophile attacks and in the second step the leaving group detaches from the ring.
• The first step is slow, that is rate determining step because when the nucleophile attacks, the ring loses it's aromaticity and a negative charge arises which take part in resonance with NO₂ group.
• So the departure of the leaving group is not the rate determining step (RDS) so it doesn't matter how the leaving group is, it depends on how much the leaving group stabilises the -ve charge. And it follows the order F > Cl > Br >I

OTs is a good leaving group. So after departure of the leaving group (OTs-) a carbocation will form whose geometry is planer so there are two possibilities of attacking the nucleophile ( OAc- furnished by AcOK) :

1) above the plane.

2) down the plane.

So a racemic mixture will be formed which is optically inactive. If the nucleophile attacked the carbon centre from down the group then the configuration of the centre will either of the + or - then the racemic mixture was not obtained.

The base will abstract the most acidic Hydrogen and will attack on C₂H₅  to give A.

In the first step of the reaction a quaternary ammonium salt is formed which is then eliminated to form the alkene. The alkene formation proceeds through the E1cB pathway where the acidic beta hydrogen is abstracted by the Base and the carbanion is stabilised by the positively charged leaving group i. e N⁺(C₂H₅)CH₃ which is a powerful electron withdrawing group and the Hoffman product is predominant product.

• Tropylium - aromatic with 6 pi-electrons (Most stable carbocation)
• Triphenylmethyl carbocation - Positive charge in resonance with two phenyl rings.
• Allylic carbonation - Positive charge delocalized over 3 carbons.
• Methyl carbonation - No stabilizing interactions.

PhS- is a better nucleophile in S_N2 rxn than any other nucleophile given in the question.

• AlCl3 is the strongest Lewis acid due to the presence of vacant p- and d-orbitals.
• FeCl3 is a strong Lewis acid by virtue of its empty d-orbitals.
• BF3 has only p-orbitals that are vacant and is a relatively weak Lewis acid.
• ZnCl2 has only 1 vacant s-orbital and is the weakest.

A polar protic solvent favours SN₁ mechanism because polar solvents has the below properties: 

• It stabilizes the carbocation intermediate. Polar solvents like methanol have a permanent dipole which means that partial negative charge on the molecule will have dipole-dipole interactions with the carbocation, stabilizing it.
• It reduces the reactivity of the nucleophile. The polar solvent can interact electrostatically with the nucleophile. This reduces the reactivity of the nucleophile and enhances the SN₁ reaction.

The SN2 reaction - A Nucleophilic Substitution in which the Rate Determining Step involves 2 components.

SN2 reactions are bimolecular with simultaneous bond-making and bond-breaking steps. -SN2 reactions do not proceed via an intermediate.

In an SN2 reaction there is mostly inversion and little racemisation.

The reaction of bromine with sodium hydroxide forms sodium hypobromite in situ, which transforms the primary amide into an intermediate isocyanate. The formation of an intermediate nitrene is not possible because it also implies the formation of a hydroxamic acid as a byproduct, which has never been observed. The intermediate isocyanate is hydrolyzed to a primary amine, giving off carbon dioxide.

Cannizzaro reaction is given by aldehydes that do not have a α hydrogen atom. Acetaldehyde (CH₃CHO) has 3 α hydrogen atoms and thus does not undergo Cannizzaro reaction.

In SN₂ reaction, inversion of configuration takes place. Since reactant and product are not enantiomers, the sign of the optical rotation may not change, hence a single stereoisomer is obtained.

• The stability of C6H5CH2+ is due to resonance.
• The stability of C6H5CH2CH2+ is due to only inductive effect.
• The stability of C6H5CH+CH3 is due to resonance and +I effect of 1 methyl group.
• The stability of C6H5C+(CH3)2 is due to both resonance and +I effect of 2 methyl groups.

So, the order of stability of carbocation is d>c>a>b or b<a<c<d.