NEET Exam  >  NEET Tests  >  Chemistry Class 12  >  Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - NEET MCQ

Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - NEET MCQ


Test Description

25 Questions MCQ Test Chemistry Class 12 - Test: Preparation of Aldehydes & Ketones (Morrison & Boyd)

Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) for NEET 2024 is part of Chemistry Class 12 preparation. The Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) questions and answers have been prepared according to the NEET exam syllabus.The Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) below.
Solutions of Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) questions in English are available as part of our Chemistry Class 12 for NEET & Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) solutions in Hindi for Chemistry Class 12 course. Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free. Attempt Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) | 25 questions in 35 minutes | Mock test for NEET preparation | Free important questions MCQ to study Chemistry Class 12 for NEET Exam | Download free PDF with solutions
Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - 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), (c) and (d), out of which ONLY ONE is correct.

Q. 

An optically active organic compound has molecular formula C5H12O(X). X on oxidation with CrO3/H2SO4 gives an achiral C5H10O. Hence, X could be

Detailed Solution for Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 1

Oxidation of X giving ketone as well as X is chiral, it must be a secondary alcohol with a-chiral carbon.

Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 2

Which of the following reaction will not produce an aldehyde?

Detailed Solution for Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 2

Anti vicinal-diols do not undergo oxidative cleavage with HIO4.

1 Crore+ students have signed up on EduRev. Have you? Download the App
Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 3

Which reagent below cannot reduce an acid chloride to an aldehyde?

Detailed Solution for Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 3

Na /C2 H5OH further reduces aldehydes to alcohols.

Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 4

The incorrect statement regarding oxo process for synthesis of an aldehyde is

Detailed Solution for Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 4

In an oxo process, formylation (addition of H and CHO) of double bond takes place, hence ketones cannot be synthesised in direct oxo process.

Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 5

All of the following reaction gives atleast one ketone as a significant organic product except

Detailed Solution for Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 5

It gives aldehydes as major product.

Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 6

All of the following results in the formation of an aldehyde except

Detailed Solution for Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 6

It gives a ketone.

Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 7

Consider the following reaction,

All of the following reagents can bring about the above transformation except

Detailed Solution for Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 7

Alkaline permanganate (Baeyer’s reagent) also oxidises olefinic bonds to syn vicinal-diols.

Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 8

A hydrocarbon X(C7H12) on ozonolysis followed by the treatment with (CH3)2S gives C7H12O2 which gives positive Tollen’s test as well as positive iodoform test. The compoppd below satisfying the criteria of X is

Detailed Solution for Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 8

The ozonolysis product of X is a dicarbonyl which contains both aldehyde group and CH3CO— group.

Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 9

A hydrocarbon X has molecular formula C5H10 X on treatment with B2H6 in H2O2 /NaOH gives an optically active C5H12O which on treatment with CrO3/HCI / pyridine gives C5H10O which is still chiral. Which of the following can be a product of reductive ozonolysis of X?

Detailed Solution for Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 9

Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 10

What is the final major product of the following reaction

Detailed Solution for Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 10

Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 11

Consider the following reaction,

(A pure enantiomer)

Q. 

The incorrect statement regarding X is

Detailed Solution for Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 11

 X will be a pure enantiomer of aldehyde.

Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 12

Consider the following reaction sequence,

Q. 

The correct statement regarding X is

Detailed Solution for Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 12

*Multiple options can be correct
Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 13

One or More than One Options Correct Type

Direction (Q. Nos. 13-17) This section contains 5 multiple choice questions. Each question has four choices (a), (b), (c) and (dj, out of which ONE or MORE THAN ONE are correct.

Q. 

In which of the following reactions, an aldehyde is formed as major product?

Detailed Solution for Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 13

(b) In option (b), aldehyde is not formed. When a terminal alkyne is oxidised with KMnO4, a carboxylic acid is always formed.
All options are preparation of aldehydes.

*Multiple options can be correct
Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 14

In which of the following reactions, ketone is formed as the major organic product?

Detailed Solution for Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 14

If acid derivatives like nitrile, acid chlorid e or ester is taken in excess in Grignard synthesis, second addition of Grignard’s reagent on carbonyl product does not succeed and carbonyls are obtained as major products.
In option (b), carboxylic acids and in option (d), an aldehyde is formed.

*Multiple options can be correct
Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 15

Consider the following reaction,

Q. 

The correct statement(s) regarding the above reaction is/are

Detailed Solution for Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 15


Trans diastereomer of the above diol does not react due to its inability to form cyclic intermediate.

*Multiple options can be correct
Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 16

Consider the reaction mentioned below,

Q. 

The expected organic product(s) is/are

Detailed Solution for Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 16

Formylation of double bond (oxo process) occurs. However, the reaction is not regioselective, hence both isomers of aldehyde are formed.

*Multiple options can be correct
Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 17

Consider the following reaction, 

Q. 

Reagent(s) that can bring about the above reaction successfully is/are

Detailed Solution for Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 17

Both Rosenmund reduction (Pd/BaSO4)and reduction with Li[{(CH3)CO}3AIH] at - 80°C are effective in reducing acid chlorides into aldehydes.

Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 18

Comprehension Type

Direction (Q. Nos. 18-20) 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

A hydrocarbon A (C10H18) is capable of showing both enantiomerism as well as diastereomerism. Treatment of A either with HgSO4 / H2SO4 or B2H/ H2O2 -NaOH results in the same carbonyl compound B. Also,

  C can also be obtained as one of the product in the following reaction.

Q. 

What is the most likely structure of B?

Detailed Solution for Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 18

Compound A has a triple bond and it is symmetrical because its partially reduced product B gives single ozonolysis product C, Also, A shows both enantiomerism and diastereomerism, it must be

As shown above, A has two chiral carbons but simultaneously, it is symmetrical. Hence, it has both meso and a pair of enantiomers as stereoisomers.

Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 19

A hydrocarbon A (C10H18) is capable of showing both enantiomerism as well as diastereomerism. Treatment of A either with HgSO4 / H2SO4 or B2H/ H2O2 -NaOH results in the same carbonyl compound B. Also,

  C can also be obtained as one of the product in the following reaction.

Q. 

What is the structure of compound C?

Detailed Solution for Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 19

Compound A has a triple bond and it is symmetrical because its partially reduced product B gives single ozonolysis product C, Also, A shows both enantiomerism and diastereomerism, it must be

As shown above, A has two chiral carbons but simultaneously, it is symmetrical. Hence, it has both meso and a pair of enantiomers as stereoisomers.

Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 20

A hydrocarbon A (C10H18) is capable of showing both enantiomerism as well as diastereomerism. Treatment of A either with HgSO4 / H2SO4 or B2H/ H2O2 -NaOH results in the same carbonyl compound B. Also,

 C can also be obtained as one of the product in the following reaction.

Consider the reaction given below,

Q. 

How many different alcohols are expected?

Detailed Solution for Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 20

Compound A has a triple bond and it is symmetrical because its partially reduced product B gives single ozonolysis product C, Also, A shows both enantiomerism and diastereomerism, it must be

As shown above, A has two chiral carbons but simultaneously, it is symmetrical. Hence, it has both meso and a pair of enantiomers as stereoisomers.

*Answer can only contain numeric values
Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 21

One Integer Value Correct Type

Direction (Q. Nos. 21-25) This section contains 5 questions. When worked out will result in an integer from 0 to 9 (both inclusive).

Q. 

How many different aldehyde isomers exist for C5H10O ?


Detailed Solution for Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 21

*Answer can only contain numeric values
Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 22

How many different ketones isomer exist for C6H12O ?


Detailed Solution for Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 22

*Answer can only contain numeric values
Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 23

How many different alcohol isomers with molecular formula C5H12O can be oxidised to ketones using K2Cr2O7 - H2SO4?


Detailed Solution for Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 23

All secondary alcohol isomers can be oxidised to ketones.

*Answer can only contain numeric values
Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 24

If all the ketone isomers of C6H10O are reduced independently with NaBH4 , how many of them will produce racemic mixture of alcohols?


Detailed Solution for Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 24


All the above shown ketones on reduction with NaBH4 will produce racemic mixtures of alcohols. The isomer below already has a chiral carbon, show enantiomerism. A pure enantiomer of this ketone, on reduction with NaBH4, will produce pair of diastereomers.

*Answer can only contain numeric values
Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 25

Consider the following reaction,


Q. 

How many different diols are expected at the end of the above reaction?


Detailed Solution for Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) - Question 25


108 videos|286 docs|123 tests
Information about Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) Page
In this test you can find the Exam questions for Test: Preparation of Aldehydes & Ketones (Morrison & Boyd) solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Preparation of Aldehydes & Ketones (Morrison & Boyd), EduRev gives you an ample number of Online tests for practice

Top Courses for NEET

108 videos|286 docs|123 tests
Download as PDF

Top Courses for NEET