Test: Reaction Mechanism- 1


30 Questions MCQ Test Organic Chemistry | Test: Reaction Mechanism- 1


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QUESTION: 1

Which of the following statements regarding the E2 mechanism is wrong?

Solution:

Answer: D

Solution: --

E2 mechanism

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).

QUESTION: 2

The major product P formed in the given:

Solution:
QUESTION: 3

 above reaction is

Solution:

It is nucleophilic substitution reaction which is in first order.

QUESTION: 4

The major product formed in the following reaction is:

Solution:
QUESTION: 5

Predict the major product P in the following reaction:

Solution:

correct ans is 'A'

QUESTION: 6

The major product formed in the following reaction is:

Solution:

Answer: A

Solution: --

QUESTION: 7

The major product obtained treatment of compound X with H2SO4 at 80°C is:

Solution:
QUESTION: 8

In the following compound, the hydroxy group that is most readily Methylated with C2H2N2 is

Solution:

Answer: B

Solution: -- The conjugate base obtained by the elimination of protons of q -OH gr 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 gr has stability ,so the H corresponding to this OH gr is highly acidic. Consequently the H is captured readily by CH2N2 and readily gets methylated.

QUESTION: 9

In the reaction, the major product X is:

Solution:
QUESTION: 10

In the reaction, the major product X is:

Solution:
QUESTION: 11

The major product formed in the following reaction is :

Solution:

Answer: D

Solution: --

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.

QUESTION: 12

The major product formed in the following reaction is:

Solution:

Answer: B

Solution: --

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.

QUESTION: 13

In the cyclisation reaction given below, the most probable product formed is:

Solution:

Answer: D

Solution: --

QUESTION: 14

The major product formed in the following reaction is:

Solution:
QUESTION: 15

The body of car is repaired with help of a filler which is hardened by addition of

Solution:

Answer: B

Solution: -- Benzoyl peroxide induces chain growth polymerization.

QUESTION: 16

The correct order of reactivity of p-halonitrobenzens in the following reaction is 

Solution:

Answer: B

Solution: -- 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

QUESTION: 17

Solvolysis of the optically active compound X gives, mainly:

Solution:

Answer: C

Solution: -- -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 and 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.

QUESTION: 18

The major product obtained in the following reaction, is 

Solution:
QUESTION: 19

The major product formed in the following reaction is:

Solution:

Answer: C

Solution: -- 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.

QUESTION: 20

Increasing order of stability of following carbocations (give least stable first)?

(I) Tropylium 

(III) (C6H5)2C+
(IV) CH3+

Solution:

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

QUESTION: 21

What is the nucleophilicity order for SN2 reaction:

(I) C6H5S
(II) C2H5O
(III) NO3
(IV) CN
(V) I

Solution:
QUESTION: 22

Select order of effectiveness of Lewis acid catalyst in Friedel-Crafts react ion:

Solution:

Answer: D

Solution: -- 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.

QUESTION: 23

For the reaction between alkyl halide and OH- increase in solvent polarity generally

Solution:

Answer: B

Solution: --

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

1.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.

2.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.

QUESTION: 24

In an SN2 reaction there is:

Solution:

Answer: B

Solution: --

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

QUESTION: 25

Reactive intermediate formed in the following reaction is:

Solution:

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.

QUESTION: 26

Cannizzaro reaction is not given by:

Solution:

Answer: B

Solution: --

Cannizzaro reaction is given by aldehydes that do not have a α hydrogen atom. Acetaldehyde (CH3​CHO) has 3 α hydrogen atoms and thus does not undergo Cannizzaro reaction.

QUESTION: 27

An SN2 reaction at an asymmetric carbon atom of a dextro alkyl halide always gives a:

Solution:

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.

QUESTION: 28

Reaction of ethyne with HCN in presence of Ba (CN)2 is an example of:

Solution:

Answer: B

Solution: --

QUESTION: 29

Consider the following carbocations is most stable:

Solution:

Answer: D

Solution: --

(a) The stability of C6H5CH2+ is due to resonance.

(b) The stability of C6H5CH2CH2+ is due to only inductive effect.

(c) The stability of C6H5CH+CH3 is due to resonance and +I effect of 1 methyl group.

(d) 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.

QUESTION: 30

The major product formed in the reaction given below is:

Solution:

Answer: D

Solution: --

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