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All Physics Chemistry Botany Zoology

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p-nitrophenol is stronger acid than phenol because nitro group is:

electron withdrawing

electron donating

basic

acidic

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Explanation

(a) It is a reason for the fact.

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Which of the following has highest boiling point?

Benzene

Phenol

Toluene

Ethyl benzene

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Explanation

(b)

Due to intermolecular H-bonding in phenol

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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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