Alcohol & Ether – JEE Main Previous Year Questions with Solutions

Class 9-10, JEE & NEET

JEE Main Previous Year Papers Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Mains chapter wise questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas. Simulator Previous Years AIEEE/JEE Mains Questions
Q. The number of stereoisomers possible for a compound of the molecular formula (1) 4        (2) 6         (3) 3           (4) 2 AIEEE-2009

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Sol. (1)

Q. A liquid was mixed with ethanol and a drop of concentrated $\mathrm{H}_{2} \mathrm{SO}_{4}$ was added. A compound with a fruity smell was formed. The liquid was :- (1) $\mathrm{CH}_{3} \mathrm{COCH}_{3}$ (2) $\mathrm{CH}_{3} \mathrm{COOH}$ (3) $\mathrm{CH}_{3} \mathrm{OH}$ (4) HCHO AIEEE-2009

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Sol. (2)

Q. From amongst the following alcohols the one that would react fastest with conc. HCl and anhydrous $\mathrm{ZnCl}_{2}$, is :- (1) 1–Butanol (2) 2–Butanol (3) 2–Methylpropan–2–ol (4) 2–Methylpropanol AIEEE-2010

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Sol. (3)

Q. Consider the following reaction : $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow$ Produce Among the following, which one cannot be formed as a product under any conditions ? (1) Ethyl-hydrogen sulphate (2) Ethylene (3) Acetylene (3) Diethyl ethe AIEEE-2011

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Sol. (3)

Q. Iodoform can be prepared from all except :- (1) Isobutyl alcohol (2) Ethyl methyl ketone (3) Isopropyl alcohol (4) 3-Methyl–2–butanone AIEEE-2012

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Sol. (1)

Q. An unknown alcohol is treated with the “Lucas reagent’ to determine whether the alcohol is primary, secondary or tertiary. Which alcohol reacts fastest and by what mechanism :- (1) secondary alcoholby SN $^{1}$ (2) tertiary alcohol by SN $^{1}$ (3) secondary alcoholby SN $^{2}$ (4) tertiary alcohol by SN $^{2}$ AIEEE-2013

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Sol. (2)

Q. The most suitable reagent for the conversion of $\mathrm{R}-\mathrm{CH}_{2}-\mathrm{OH} \rightarrow \mathrm{R}-\mathrm{CHO}$ is : (1) $\mathrm{CiO}_{3}$ (2) PCC (Pyridinium chlorochromate) (3) KMNO $_{4}$ (4) $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ Jee-Main-2014

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Sol. (2)

Q. Allyl phenyl ether can be prepared by heating: (1) $\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{Br}+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{ONa}$ (2) $\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CH}=\mathrm{CH}-\mathrm{Br}+\mathrm{CH}_{3}-\mathrm{ONa}$ (3) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br}+\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{ONa}$ (4) $\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{Br}+\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CH}_{2}-\mathrm{ONa}$ Jee-Main-2014

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Sol. (1)

Q. In the Victor-Meyer’s test, the colour given by $1^{\circ}, 2^{\circ}$ and $3^{\circ}$ alcohols are respectively :- (1) Red, blue, colourless (2) Colourless, red, blue (3) Red, blue, violet (4) Red, colourless, blue Jee-Main-2014

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Sol. (1)

Q. Williamson synthesis of ether is an example of (1) Nucleophilic addition (2) Electrophilic substitution (3) Nucleophilic substitution (4) Electrophilic addition Jee-Main-2014

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Sol. (3) Nucleophilic substitution

Q. Phenol on treatment with $\mathrm{CO}_{2}$ in the presence of NaOH followed by acidification produces compound X as the major product. X on treatment with $\left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{O}$ in the presence of catalytic amount of $\mathrm{H}_{2} \mathrm{SO}_{4}$ produces : Jee-Main-2018

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Sol. (4)

Q. Phenol reacts with methyl chloroformate in the presence of NaOH to form product A. A reacts with $\mathrm{Br}_{2}$ to form product B. A and B are respectively : Jee-Main-2018

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Sol. (2)

• April 8, 2021 at 5:00 pm

THIS COMMENT HAS 20 LIKES

21
• April 10, 2021 at 4:42 pm

lol

2
• January 23, 2021 at 10:14 pm

abe ye log kam questions dete h aur bacho ko bewakoof banate h

239
• December 22, 2020 at 8:22 pm

tqsm…..

1
• December 22, 2020 at 8:21 pm

tqsm…..😊

0
• September 3, 2020 at 5:28 pm

bht bhala

0
• August 26, 2020 at 8:06 am

Very useful.
Thank you very much

0
• August 20, 2020 at 8:00 pm

higher level questions……

0
• August 18, 2020 at 11:44 am

Thanks very useful

0
• August 7, 2020 at 10:37 am

Helpful but if you give answer with explanation….then it would be great help

0
• July 15, 2020 at 10:55 am

its nice

12
• June 7, 2020 at 12:37 pm