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Alcohol, phenol & ether MCQs for NEET — Chemistry Questions with Answers

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n-propyl alcohol and iso-propyl alcohol can be chemically distinguished by which reagent

PCl₅

reduction

oxidation with potassium dichromate

ozonolysis

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Explanation

n-propyl alcohol and iso-propyl alcohol can be distinguished by oxidation with potassium dichromate. n-propyl alcohol is oxidized to propanoic acid, while iso-propyl alcohol is oxidized to acetone.

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Ethanol and dimethyl ether form a pair of functional isomers. The boiling point of ethanol is higher than that of dimethyl ether due to the presence of

H-bonding in ethanol

H-bonding in dimethyl ether

-CH₃ group in ethanol

-CH₃ group in dimethyl ether

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Explanation

Ethanol has a higher boiling point than dimethyl ether due to the presence of hydrogen bonding in ethanol. The hydroxyl group in ethanol can form intermolecular hydrogen bonds, increasing its boiling point.

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The compound B formed in the following sequence of reactions,

CH₃CH₂CH₂OH $\overset{{PCl}_{5}}{\to}$ A $\overset{Alc . NaOH}{\to}$ B will be:

propyne

propene

propanal

propane

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Explanation

(b) A is CH₃CH₂CH₂CH₂Cl; B is CH₃.CH=CH₂

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Ethyl alcohol is denatured by:

methanol and formic acid

KCN

CH₃OH and C₆H₆

CH₃OH and pyridine

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Explanation

(d) Denaturation is made by addition of pyridine, CH₃OH or naphtha.

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The reaction products of C₆H₅-O-CH₃+ HI $\overset{A}{\to}$

C₆H₅OH + CH₃l

C₆H₅l + CH₃OH

C₆H₅CH₃ +HOI

C₆H₆+ CH₃Ol

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Explanation

C₆H₅-O-CH₃+ HI $\overset{∆}{\to}$ C₆H₅OH + CH₃l

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The Strongest acid among the following aromatic compunds is

p-chlorophenol

p-nitrophenol

m-nitrophenol

o- nitrophenol

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Explanation

Nitro group ( $- {NO}_{2}$ ) is strong electron withdrawing group and because of resonance ( $- {NO}_{2}$ group) withdraw electron effectively from para position.

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(CH₃)₃C-O-CH₃ + HI $\to$ Products, the product are

(CH₃)₃C-OH + CH₃l

(CH₃)₃C-l + CH₃OH

(CH₃)₃C-l + CH₃l

(CH₃)₃C-OH + CH₃OH

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Explanation

(CH₃)₃C-O-CH₃ + HI+ HI $\to$ (CH₃)₃C-l + CH₃OH

Here, $S_{N} 1$ mechanism takes place.

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Compounds I and II can be distinguished by using reagent

(I)	(II)
4–Amino–2–methlbut–3–en–2–ol	4–Amino–2, 2–dimethylbut–3–yn–1–ol

NaNO₂/HCI

Br₂/H₂O

HCI/ZnCI₂ (anhydrous)

Cu2CI₂ + NH₄OH

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Which of the following statements is true?

CH₃CH₂S^- is both a stronger base and more nucleophilic than CH₃CH₂O^-

CH₃CH₂S^- is a stronger base but is less nucleophilic than CH₃CH₂O^-

CH₃CH₂S^- is a weaker base but is more nucleophilic than CH₃CH₂O^-

CH₃CH₂S^- is both a weaker base and less nucleophilic than CH₃CH₂O^-

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Explanation

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$\begin{matrix} \underset{|}{{CO}_{2} H} \\ {CO}_{2} H \end{matrix} + \begin{matrix} \underset{|}{{CH}_{2}} & - & \overset{18}{OH} \\ {CH}_{2} & - & \overset{18}{OH} \end{matrix} \to_{∆}^{H^{+}}$ (A); Product (A) is:

$\begin{matrix} \overset{O}{\overset{∥}{\underset{|}{C}}} & - & O & - & {CH}_{2} \\ \underset{O}{\underset{∥}{C}} & - & O & - & {CH}_{2} \end{matrix}$

$\begin{matrix} \overset{O}{\overset{∥}{\underset{|}{C}}} & - & \overset{18}{O} & - & \underset{|}{{CH}_{2}} \\ \underset{O}{\underset{∥}{C}} & - & \underset{18}{O} & - & {CH}_{2} \end{matrix}$

$\begin{matrix} \overset{^{18} O}{\overset{∥}{\underset{|}{C}}} & - & O & - & \underset{|}{{CH}_{2}} \\ \underset{^{18} O}{\underset{∥}{C}} & - & O & - & {CH}_{2} \end{matrix}$

$\begin{matrix} H & - & O & - & \overset{O}{\overset{∥}{C}} & - & \overset{O}{\overset{∥}{C}} & - O & - & {CH}_{2} & - & {CH}_{2} & - & OH \end{matrix}$

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Explanation

The reaction involves the electrophilic addition of H+ to the carbonyl group of the carboxylic acid, followed by the nucleophilic addition of H2O18 (labeled water) to the resulting carbocation. The product will have the 18O label incorporated into the carboxyl group.

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