Revisal Problems (Past 13 Year) JEE Advanced (Alkyl Halides)


31 Questions MCQ Test Chemistry for JEE Advanced | Revisal Problems (Past 13 Year) JEE Advanced (Alkyl Halides)


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This mock test of Revisal Problems (Past 13 Year) JEE Advanced (Alkyl Halides) for JEE helps you for every JEE entrance exam. This contains 31 Multiple Choice Questions for JEE Revisal Problems (Past 13 Year) JEE Advanced (Alkyl Halides) (mcq) to study with solutions a complete question bank. The solved questions answers in this Revisal Problems (Past 13 Year) JEE Advanced (Alkyl Halides) quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Revisal Problems (Past 13 Year) JEE Advanced (Alkyl Halides) exercise for a better result in the exam. You can find other Revisal Problems (Past 13 Year) JEE Advanced (Alkyl Halides) extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1

Only One Option Correct Type

Direction (Q. Nos. 1-12) This section contains 12 multiple choice questions. Each question has four choices (a), (b), (cj and (d), out of which ONLY ONE is correct.

Q. 

What is the major product of the reaction?

Solution:

H and Br from anti position eliminated.

QUESTION: 2

What is the major elimination product of the following reaction?

Solution:

Drawing Sawhorse projection of the given compound

QUESTION: 3

Which of the following is missing reagent in the following reaction sequence?

Solution:

E2 elimination first gives cyclopentene. In the next step, addition of bromine gives anti vicinal dibromide.

QUESTION: 4

Predict substitution product in the following reaction.

Solution:

SN2 reaction gives inversion of configuration at α-carbon.

QUESTION: 5

Propose product with a ppropriate stereochemistry in the following SN2 reaction.

Solution:

SN2 reaction gives inversion of configuration at α-carbon. If the compound also has chiral carbon other than α-carbon, configuration at those chiral carbons remains unchanged after tthe reaction.

QUESTION: 6

In SN1 reaction, racemisation occurs if the reaction occurs at a ste reogenic centre, however 50 : 50 mixture of enantiomers are rarely obtained, why?

Solution:

SN1 reaction involves planar carbocation, giving .racemic products predominantly. However same carbocation is not completely deteched from leaving group, exists as intimate ion pair. Such ion pairs prevent attacks from front side favouring backside attacks.

Due to above reason, in SN1 reaction, partial racemisation and net inversion is usually observed.

QUESTION: 7

Arrange the following halides in decreasing order of reactivity in SN1 reaction.




Solution:

SN1 reaction proceeds via carbocation intermediate, hence reactivity follow the order of stability of carbocation. In the present case, besides resonance stability of benzylic carbocations, hyperconjugation by alkyl group at para position is also affecting stability. Methyl group from para position will stabilise, the most by hyperconjugation effect due to three α - H followed by, ethyl group with two a-H and least by isopropyl group with only one α-H.

QUESTION: 8

Arrange the following compounds in decreasing order of reactivity for hydrolysis reaction.

I. PhCH2Br
 

Solution:

Hydrolysis will proceed by SN1 mechanism. Ill will be least reactive as it will involves the least stable, anti aromatic carbocation. IV will be most reactive as its carbocation will involve ring expansion of highly strained cyclopropyl ring.

QUESTION: 9

The in creasing order of reactivity of the following bromides in SN1 reaction is




Solution:

(III) is most reactive as it forms most stable, aromatic carbocation:

(II) is least reactive as it involves least stable, anti aromatic carbocation,

QUESTION: 10

Which is the least reactive in solvolysis reaction?

Solution:

It involves least stable, anti aromatic carbocation.

QUESTION: 11

Among the following given bromides, the correct decreasing order of reactivity in SN1 reaction is

Solution:

(III) is most reactive as it pro cee ds via m ost stable aromatic carbocation (II) is least reactive as it proceeds ! via least stable anti aromatic carbocation.

QUESTION: 12

Arrange the following compounds in increasing order of their rate hydrolysis by SN1 mechanism.


Solution:

(IV) is most reactive as it involves the most stable carbocation.

(III) is least reactive as it is vinylic halide. (II) is more reactive than (I) because (II) proceeds via benzylic carbocation.

*Multiple options can be correct
QUESTION: 13

One or More than One Options Correct Type

Direction (Q. Nos. 13-22) This section contains 10 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct.

Q. 

What are the expected products of the following reaction?

Solution:

The reaction may proceed either by E2 or SN2 mechanism. The only elimination product possible here is 1-pentene. I 2-pentene is not possible because its formation will involve I rearrangement which is not possible in E2 reaction.
Also due to following equilibrium : 


Both tertiary butoxide and an ethoxide undergo SN2 reaction forming (b) and (d).

*Multiple options can be correct
QUESTION: 14

What is (are) true about E1cb reaction?

Solution:

E1cb is a modification of E2 mechanism that involve formation of a carbanion intermediate in the first, fast step followed by unimolecular elimination of halide ion in the second slow rate determining step. Overall order is two. Electron withdrawing group stabilises carbanion, favours the mechanism. Since first step is reversible, deuterium incorporation may take place as :

*Multiple options can be correct
QUESTION: 15

Which of the following is expected to give more than two products in an E2 elimination reaction?

Solution:




*Multiple options can be correct
QUESTION: 16

Consider the following elimination reaction,

Q. 

The statement that is (are) true regarding the above elimination reaction is

Solution:

Cis and trans 2-butenes are pair of diastereomers. C—H bond is broken in the rate determ ining slow step, stronger C—D bond will slow down the reaction. C—X bond is broken in slow, ate determining step, C—Cl bond will break slowly than C—Br bond.
Bulky base like tertiary butoxide results in less substituted Hofmann’s elimination product.

*Multiple options can be correct
QUESTION: 17

Consider the following reactions:

Q.

The correct statement regarding above substitution reaction is/are

Solution:

Reaction-I has involved rearrangement, must have proceeded by SN1 mechanism. Reaction-ll involves strong nucleophile and devoids of rearrangement, proceeded by SN2 mechanism.

*Multiple options can be correct
QUESTION: 18

Which of the following will result in racemic mixture as major solvolysis products?

Solution:

In option (a) and option (b) achiral products are formed via tertiary carbocations.

*Multiple options can be correct
QUESTION: 19

Consider the following solvolysis reaction.

Q. 

The expected product(s) is

Solution:


*Multiple options can be correct
QUESTION: 20

Select the correct statement.

Solution:

In SN2 reaction, order of reactivity is 1° > 2° > 3°. In option (c), although both are primary bromides, tertiary butyl group at a-carbon gives large steric hindrance, decreases reactivity in SN2 reaction.

*Multiple options can be correct
QUESTION: 21

Consider the follow ing SN2 reaction,

Solution:

Quarternary ammonium chloride brings CN- in organic phase by the following reaction:

If AgCN is used in place of KCN, isocyanide (linking from nitrogen) is preferred due to covalent character of Ag—C bond.

*Multiple options can be correct
QUESTION: 22

What is (are) true about the following free radical chlorination reaction ?

Solution:



In free radical chlorination reaction, the major product is determined by combining the stability of free radicals and probability of reaction at a particular carbon both. Hence, only the basis of stability of tertiary free radical alone, tertiary chloride can't be predicted as major product.

QUESTION: 23

Comprehension Type

Direction (Q. Nos. 23-25) This section contains a paragraph, describing theory, experiments, data, etc.
Three questions related to the paragraph have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d).

Passage

Compounds A, B and C are alkyl bromides with the molecular formulate C5H11Br. Treatment of A or B with potassium ferf-butoxide gives the same alkene D. If the base is changed to sodium ethoxide, A and C give the same alkene E. Treatment of C with potassium ferf-butoxide gives yet another alkene F.

Q. 

The correct statement regarding A, B and C is

Solution:


B being 1° halide, undergoes faster SN2 reaction than 2° halideC. Both B and C are chiral while A is achiral.
Grignard's reagent (R- Mg X+) with most stable carbanion (1°) is formed with B.

QUESTION: 24

Compounds A, B and C are alkyl bromides with the molecular formulate C5H11Br. Treatment of A or B with potassium ferf-butoxide gives the same alkene D. If the base is changed to sodium ethoxide, A and C give the same alkene E. Treatment of C with potassium ferf-butoxide gives yet another alkene F.

Q. 

The correct statement concerning A, B and C is

Solution:


B being 1° halide, undergoes faster SN2 reaction than 2° halideC. Both B and C are chiral while A is achiral.
Grignard's reagent (R- Mg X+) with most stable carbanion (1°) is formed with B.

QUESTION: 25

Compounds A, B and C are alkyl bromides with the molecular formulate C5H11Br. Treatment of A or B with potassium ferf-butoxide gives the same alkene D. If the base is changed to sodium ethoxide, A and C give the same alkene E. Treatment of C with potassium ferf-butoxide gives yet another alkene F.

Q. 

Which forms most stable Grignard’s reagent when treated with Mg in THF?

Solution:


B being 1° halide, undergoes faster SN2 reaction than 2° halideC. Both B and C are chiral while A is achiral.
Grignard's reagent (R- Mg X+) with most stable carbanion (1°) is formed with B.

*Answer can only contain numeric values
QUESTION: 26

One Integer Value Correct Type

Direction (Q. Nos. 26-29) This section contains 4 questions. When worked out will result in an integer from 0 to 9 (both inclusive).

Q. 

Methyl cyclopentane on treatment with chlorine gas in sunlight undergo free radical chlorination reaction. How many of the above monochlorination products upon treatment with ethanolic KOH solution can give 1-methyl cyclopentane as one of the product?


Solution:

Only the following chlorination products, upon E2 reaction can give 1-methyl cyclopentene as one of the product.

*Answer can only contain numeric values
QUESTION: 27

In principle, how many different alkyl chlorides would be formed on treatment of 3-methyl-2-pentanol with concentrated HCI in the presence of ZnCI2?


Solution:

*Answer can only contain numeric values
QUESTION: 28

How many different isomers of alkyl bromide on treatment with ethanolic KOH solution can result in 3-methyl-3-hexene?


Solution:

*Answer can only contain numeric values
QUESTION: 29

If Br2CH—CHBr2 is treated with excess of NaCN solution such that all bromides are substituted by nucleophiles, how many different substitution products would be formed in principle which are chiral?


Solution:

QUESTION: 30

Matching List Type

Direction (Q. Nos. 30 and 31) Choices for the correct combination of elements from Column I and Column II are given as options (a), (b), (c) and (d), out of which one is correct.

Q. 

Match the reactions from Column I and their corresponding properties from Column II and mark the correct option from codes given below.

Solution:

(i) With SOCI2, an intramolecular nucleophilic substitution (SN1) reaction takes place. In absence of base like pyridine, complete retention of configuration takes place.

(ii) With SOCI2, in the presence of pyridine, SN2 reaction takes place with inversion of configuration at chiral α-carbon

(iii) SN1 reaction takes place giving racemic mixture of product.


QUESTION: 31

Match the reactions from Column I and their corresponding properties from Column II and mark the correct option from codes given below.

Solution:

(i) Deuterium incorporation indicates E1cb mechanism which follow 2nd order reaction rate law.

(ii) E2 reaction giving major product, which follow 2nd order rate law

(iii) Loss of fluoride indicates that carbanion intermediate is involved

(iv) Rearrangement of carbocation intermediate gives this product, indicating E1 mechanism