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NEET Exam  >  NEET Tests  >  Test: Distinction of Alcohols (February 8) - NEET MCQ

Test: Distinction of Alcohols (February 8) - NEET MCQ


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10 Questions MCQ Test - Test: Distinction of Alcohols (February 8)

Test: Distinction of Alcohols (February 8) for NEET 2024 is part of NEET preparation. The Test: Distinction of Alcohols (February 8) questions and answers have been prepared according to the NEET exam syllabus.The Test: Distinction of Alcohols (February 8) MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Distinction of Alcohols (February 8) below.
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Test: Distinction of Alcohols (February 8) - Question 1
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Which of the following alcohols produce turbidity immediately when reacted with conc. HCl and ZnCl2?

Detailed Solution for Test: Distinction of Alcohols (February 8) - Question 1

Tert-butyl alcohol is tertiary alcohol and therefore reacts with conc.HCl and ZnCl2 (Lucas reagent) form alkyl halide. To distinguish between the reactivities of the three classes of alcohols, the Lucas test is carried out.

Alcohols react with Lucas reagent (Conc. HCl and ZnCl2) which results in turbidity in the solution due to the formation of alkyl halides.

  • Primary alcohols do not produce any turbidity.
  • Secondary alcohols turn the solution cloudy after gentle heating for a few minutes.
  • Tertiary alcohols produce turbidity immediately after reaction since they form halides easily.
Test: Distinction of Alcohols (February 8) - Question 2
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An organic compound X (C7H16O) on treatm ent with concentrated HCI solution forms immediate turbidity even in the absence of ZnCl2. Also, X is resolvable into enantiomers. The correct statement concerning X is

Detailed Solution for Test: Distinction of Alcohols (February 8) - Question 2

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Test: Distinction of Alcohols (February 8) - Question 3
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Which of the following is least likely to form turbidity with HCI in the presence of ZnCI2 at room temperature?

Detailed Solution for Test: Distinction of Alcohols (February 8) - Question 3

Although it is secondary alcohol, it forms a very unstable carbocation (Cl is electron withdrawing group), hence it does not give turbidity with Lucas reagent in cold condition.

*Multiple options can be correct
Test: Distinction of Alcohols (February 8) - Question 4
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One or More than One Options Correct Type

Direction (Q. Nos. 9-12) This section contains 4 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct.

Q. 

Which reagent(s) can be used to differentiate between 2-pentanol and 1-pentanol?
Detailed Solution for Test: Distinction of Alcohols (February 8) - Question 4

2-pentanol forms yellow precipitate of CHI3 with I2/NaOH but 1-pentanol does not. 2-pentanol give immediate turbidity with Lucas reagent but 1 -pentanoi does not gives turbidity with Lucas reagent at room temperature. Both alcohols (1° and 2°) change colour of chromic acid solution from orange to blue green and both give effervescence of NH3(g) with NaNH2, hence these reagents cannot be used for distinction between 1° and 2° alcohols.

*Multiple options can be correct
Test: Distinction of Alcohols (February 8) - Question 5
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Alcohols given below that behaves like 1°-aliphatic alcohol in Lucas test is/are
Detailed Solution for Test: Distinction of Alcohols (February 8) - Question 5

Both have electron withdrawing groups, destabilises carbocation, do not form turbidity with Lucas reagent at room temperature like primary alcohols. Option (b) and option (d) have electron donating groups, stabilises benzylic carbocation, forms immediate turbidity with Lucas reagent like 2° and 3° alcohols.

Test: Distinction of Alcohols (February 8) - Question 6
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Comprehension Type

Direction (Q. Nos. 13-15) This section contains a passage describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer out of the given 4 options (a), (b), (c) and (d).

Passage

An organic compound X (C9H12O) gives the following reactions :
i. Na - Slow gas bubble formation
ii. Acetic anhydride - Pleasent smelling liquid
iii. CrO3-H2SO4 - Blue-green solution
iv. Hot KMnO4 - Benzoic acid
v. Br2-CCI4 - No decolouration
vi. I2 + NaOH - Yellow solid is formed
vii. X rotates the plane polarised light

Q. 

The structure of X is
Detailed Solution for Test: Distinction of Alcohols (February 8) - Question 6


Also X, has CH3—CH(OH)—, gives iodoform test and it is chiral.

Test: Distinction of Alcohols (February 8) - Question 7
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An organic compound X (C9H12O) gives the following reactions :
i. Na - Slow gas bubble formation
ii. Acetic anhydride - Pleasent smelling liquid
iii. CrO3-H2SO4 - Blue-green solution
iv. Hot KMnO4 - Benzoic acid
v. Br2-CCI4 - No decolouration
vi. I2 + NaOH - Yellow solid is formed
vii. X rotates the plane polarised light

Q. 

If X is treated with HCI in the presence of ZnCI2, the major product would be
Detailed Solution for Test: Distinction of Alcohols (February 8) - Question 7


Also X, has CH3— CH(OH)—, gives iodoform test and it is chiral.
 

*Answer can only contain numeric values
Test: Distinction of Alcohols (February 8) - Question 8
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An alcohol X (C7H16O) gives immediate turbidity with HCI/ZnCI2 and can be resolved into enantiomers. Also X has a tertiary hydrogen. How many CH3— groups are present in X?

Detailed Solution for Test: Distinction of Alcohols (February 8) - Question 8

Compound X satisfying the given criteria is

*Answer can only contain numeric values
Test: Distinction of Alcohols (February 8) - Question 9
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An organic compound X (C7H16O) gives effervescence with Na. X has both enantiomers and diastereomers. X gives yellow precipitate with alkaline iodine solution. With CrO3 - H2SO4 , X is converted into C7H14O (Y) which has enantiom ers but not diastereom ers. X on refluxing with H2SO4 isomerises to Z which neither gives yellow precipitate with alkaline iodine nor changes colour of CrO3 - H2SO4 . What is the minim um number of carbons that can be present in the parent chain of X ?

Detailed Solution for Test: Distinction of Alcohols (February 8) - Question 9

From the give condition, structure of X is derived to be

Test: Distinction of Alcohols (February 8) - Question 10
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An organic compound C4H10O (X) on reaction with I2/red-P give s C4H9I which on further reaction with AgNO2 gives C4H9NO2 (Y). Y on treatment with HNO2 forms a blue solution which turns to red on making solution slightly alkaline. The possible identity of X is
Detailed Solution for Test: Distinction of Alcohols (February 8) - Question 10

X must be a primary alcohol as indicated by Victor Meyer's test. Both (a) and (b) are primary satisfy the condition.

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