Question 1

Why is ortho-nitrophenol less soluble in water than the meta- and para-nitrophenol isomers?

Accepted Answer

It forms intramolecular hydrogen bonding.. Ortho-nitrophenol has a hydroxyl group and a nitro group positioned close together, allowing it to form intramolecular hydrogen bonds, which stabilizes the molecule and reduces its ability to interact with water, making it less soluble compared to the meta and para isomers.

Question 2

Which of the following structures represents vinylcarbinol?

Accepted Answer

HO–CH2–CH=CH2. The correct structure for vinylcarbinol is HO–CH2–CH=CH2, which features a hydroxyl group (–OH) attached to a carbon that is part of a vinyl group (–CH=CH2), making it a vinyl alcohol.

Question 3

Which of the following is NOT a starting material used for the preparation of iodoform?

Accepted Answer

HCH2OH. The iodoform reaction needs a CH3-CO- or CH3-CH(OH)- group. HCH2OH (methanol, CH3OH has no such grouping... it is just CH3OH) cannot give iodoform, whereas isopropanol, ethanol, and acetone all can. So methanol is not a starting material.

Question 4

Iodoform is formed from all of the following except:

Accepted Answer

2-Methyl-1-propanol. Iodoform requires a CH3-CO- or CH3-CH(OH)- group. 2-Methyl-1-propanol, (CH3)2CHCH2OH, has no such group and cannot give iodoform. Ethyl methyl ketone, propan-2-ol, and 3-methyl-2-butanone all contain a CH3CO/CH3CHOH unit and do react.

Question 5

Which reagent can convert CH3CH2OH into CH3CHO?

Accepted Answer

Pyridinium chlorochromate (PCC) treatment. PCC is a mild oxidant that converts a primary alcohol to the aldehyde without over-oxidising to the acid: CH3CH2OH -> CH3CHO. LiAlH4 and catalytic hydrogenation are reducing agents, and KMnO4 would oxidise all the way to the carboxylic acid.

Question 6

Which reagent can be reacted with phenylmagnesium bromide to obtain 2-phenylethanol?

Accepted Answer

Ethylene oxide. Phenylmagnesium bromide opens the strained ring of ethylene oxide to give, after workup, PhCH2CH2OH (2-phenylethanol). Adding two carbons requires the epoxide; aldehydes/ketones would give different alcohols.

Question 7

When HBr is added to CH2=CH–OCH3 under dry conditions at room temperature, which product is formed?

Accepted Answer

H3C–CHBr–OCH3. In methyl vinyl ether (CH2=CH-OCH3) protonation occurs at the terminal CH2 so the positive charge sits on the carbon bearing OCH3, where it is stabilised as an oxocarbenium ion. Br- then adds there to give CH3-CHBr-OCH3.

Question 8

Which of the following alcohols undergoes the fastest reaction with concentrated HCl in the presence of anhydrous ZnCl2?

Accepted Answer

2-Methylpropan-2-ol. With conc. HCl/anhydrous ZnCl2 (Lucas reagent) the reaction goes via a carbocation (SN1). 2-Methylpropan-2-ol is tertiary and forms the most stable carbocation, so it reacts fastest.

Question 9

Which of the following ethers cannot be prepared by Williamson ether synthesis?

Accepted Answer

Di-tert-butyl ether. Williamson synthesis needs an alkoxide to do SN2 on an alkyl halide. For di-tert-butyl ether both partners are tertiary, so the tert-butyl halide undergoes E2 elimination instead of substitution; it cannot be made this way.

Question 10

Rectified spirit consists of which of the following mixtures?

Accepted Answer

95% ethyl alcohol and 5% water. Rectified spirit is the constant-boiling azeotrope of ethanol and water, about 95% ethanol and 5% water by volume (cannot be concentrated further by simple distillation).

Question 11

Absolute alcohol (100% ethanol) is obtained by distilling rectified spirit in the presence of which drying agent?

Accepted Answer

Magnesium ethoxide. Rectified spirit (~95% ethanol) is dried to absolute alcohol by treatment with magnesium, which forms magnesium ethoxide; the magnesium ethoxide reacts with the residual water [Mg(OC2H5)2 + H2O -> Mg(OH)2 + 2C2H5OH], removing it. CaCl2 cannot be used as it forms a complex with ethanol.

Question 12

Which reagent is used to convert allyl alcohol into acrolein?

Accepted Answer

MnO2. Activated MnO2 selectively oxidises allylic (and benzylic) alcohols to the corresponding aldehydes without touching the C=C. Thus CH2=CH-CH2OH (allyl alcohol) -> CH2=CH-CHO (acrolein).

Question 13

In Williamson ether synthesis for preparing an unsymmetrical ether containing one primary and one tertiary alkyl group, what happens if the alkyl halide used is tertiary?

Accepted Answer

An alkene becomes the major product.. Using a tertiary alkyl halide in Williamson ether synthesis often leads to elimination reactions rather than substitution, resulting in the formation of an alkene as the major product due to the steric hindrance that prevents effective nucleophilic attack.

Question 14

Which of the following sequences represents the compounds in order of increasing acidity? o-cresol (a), salicylic acid (b), phenol (c)

Accepted Answer

a < c < b. o-Cresol is less acidic than phenol (methyl is electron-donating), while salicylic acid is a carboxylic acid (most acidic). So increasing acidity is a (o-cresol) < c (phenol) < b (salicylic acid).

Question 15

When wine is exposed to air, it turns sour because of

Accepted Answer

oxidation of C2H5OH to CH3COOH. On exposure to air, the ethanol in wine is oxidised (via acetic acid bacteria) to acetic acid: C2H5OH + O2 -> CH3COOH + H2O, which makes the wine sour.

Question 16

Williamson ether synthesis is employed in the preparation of which of the following compounds?

Accepted Answer

Diethyl ether. Williamson ether synthesis is a method used to create ethers by reacting an alkoxide ion with a primary alkyl halide, making it suitable for the preparation of diethyl ether, which is formed from ethanol and sodium metal.

Question 17

Which of the following compounds requires an oxidizing agent for its preparation from 1-butanol?

Accepted Answer

CH3CH2CH2CH = O. Converting 1-butanol to butanal (CH3CH2CH2CHO) is an oxidation and requires a mild oxidizing agent (e.g. PCC). The others (bromide, ether, alkene) are made by substitution/dehydration and need no oxidant.

Question 18

The Williamson ether synthesis is classified as which type of reaction?

Accepted Answer

Nucleophilic substitution. The Williamson ether synthesis involves a nucleophile attacking an electrophilic carbon in an alkyl halide, resulting in the formation of an ether. This process is characteristic of nucleophilic substitution reactions, where the nucleophile replaces a leaving group.

Question 19

Which one of the following diols undergoes periodate cleavage with HIO4 to give two separate fragments?

Accepted Answer

3,4-hexanediol. 3,4-hexanediol has hydroxyl groups on adjacent carbon atoms, allowing periodate to cleave the carbon-carbon bond between them, resulting in two separate fragments. In contrast, the other diols do not have the necessary structural arrangement for this type of cleavage.

Question 20

Which of the following can be oxidized to the corresponding carbonyl compound?

Accepted Answer

2-hydroxypropane. Only 2-hydroxypropane (isopropyl alcohol, a secondary alcohol) is oxidised to a carbonyl compound (acetone). tert-Butanol has no alpha-H and cannot give a carbonyl, and phenols/nitrophenols are not oxidised to simple carbonyls.

Question 21

Which of the following compounds is known as pinacolone?

Accepted Answer

3,3-Dimethyl-2-butanone. Pinacolone is the ketone (CH3)3C-CO-CH3, i.e. 3,3-dimethyl-2-butanone, formed by the pinacol-pinacolone rearrangement. The diol 2,3-dimethyl-2,3-butanediol is pinacol.

Question 22

Which reagent can be used to separate phenol from benzoic acid?

Accepted Answer

Sodium bicarbonate. Sodium bicarbonate can selectively deprotonate benzoic acid, forming its sodium salt, while phenol remains largely unreacted due to its weaker acidity. This difference allows for the separation of phenol from benzoic acid.

Question 23

ortho-Nitrophenol is less soluble in water than p- and m-nitrophenols because:

Accepted Answer

o-Nitrophenol shows intramolecular H-bonding. In o-nitrophenol the OH and NO2 groups are adjacent and form an intramolecular hydrogen bond (chelation), so the molecule cannot hydrogen-bond effectively with water and is less soluble than the m- and p-isomers.

Question 24

Among the following the one that gives positive iodoform test upon reaction with I2 and NaOH is

Accepted Answer

CH3CH2OH. A positive iodoform test requires a CH3-CO- or CH3-CH(OH)- group. Among the choices only CH3CH2OH (ethanol) qualifies; n-propanol, isobutanol and the ether do not.

Question 25

Consider the reaction sequence: CH3CH2OH is treated with P and I2 to give A, A is then reacted with Mg in dry ether to form B, B is allowed to react with HCHO to produce C, and C on hydrolysis with water gives D. What is D?

Accepted Answer

n-propyl alcohol. The reaction sequence starts with ethanol (CH3CH2OH) which, when treated with phosphorus and iodine, undergoes a halogenation to form an alkyl iodide. This alkyl iodide then reacts with magnesium to form a Grignard reagent, which upon reaction with formaldehyde (HCHO) produces a primary alcohol. Hydrolysis of this product yields n-propyl alcohol, confirming that D is n-propyl alcohol.

Question 26

When phenol is treated first with concentrated sulphuric acid and then with concentrated nitric acid, the product formed is

Accepted Answer

2,4,6-trinitrobenzene. Phenol with conc. H2SO4 then conc. HNO3 gives 2,4,6-trinitrophenol (picric acid), so the trinitro product is the answer rather than mono-nitrophenol.

Question 27

When phenol is treated with sodium hydroxide followed by carbon dioxide, which major product is formed?

Accepted Answer

Salicylic acid. The reaction of phenol with sodium hydroxide generates the phenoxide ion, which can then undergo a carboxylation reaction with carbon dioxide to form salicylic acid, a compound characterized by the presence of both a hydroxyl and a carboxylic acid group.

Question 28

Which of the following alcohols undergoes the quickest reaction with concentrated HCl in the presence of anhydrous ZnCl2?

Accepted Answer

2-Methylpropan-2-ol. 2-Methylpropan-2-ol is a tertiary alcohol, which means it has three alkyl groups attached to the carbon bearing the hydroxyl group. This structure allows for greater stability of the carbocation intermediate formed during the reaction with concentrated HCl and ZnCl2, leading to a faster reaction compared to primary or secondary alcohols.

Question 29

What is the major product formed when C6H5CH2CH(OH)CH(CH3)2 is treated with concentrated H2SO4?

Accepted Answer

C6H5CH=CHCH(CH3)2. The correct option is formed through dehydration of the alcohol group in the presence of concentrated sulfuric acid, leading to the formation of a double bond between the carbon atoms, resulting in an alkene. This reaction favors the formation of the most stable alkene, which in this case is the product with the phenyl group and the double bond adjacent to it.

Question 30

Consider the following reaction:
C2H5OH + H2SO4 → Product
Among the following, which one cannot be formed as a product under any conditions ?

Accepted Answer

Acetylene. With conc. H2SO4 ethanol gives ethyl hydrogen sulphate (cold), diethyl ether (~140 C) and ethylene (~170 C). Acetylene (C2H2) cannot be produced from ethanol with sulphuric acid under any condition.

Question 31

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 ?

Accepted Answer

tertiary alcohol by S_N1. Lucas reagent (conc. HCl + ZnCl2) reacts fastest with tertiary alcohols because they form the most stable carbocation, proceeding by an SN1 mechanism (immediate turbidity).

Question 32

The most suitable reagent for the conversion of R – CH2 – OH → R – CHO is:

Accepted Answer

PCC (Pyridinium chlorochromate). Pyridinium chlorochromate (PCC) is a mild oxidant that converts a primary alcohol R-CH2-OH to the aldehyde R-CHO without over-oxidising to the carboxylic acid, unlike KMnO4, K2Cr2O7 or CrO3. So PCC is the most suitable reagent.

Question 33

Iodoform can be prepared from all except:

Accepted Answer

Isobutyl alcohol. Isobutyl alcohol cannot be converted to iodoform because it lacks the necessary structural features, specifically a methyl ketone group, which is required for the iodoform reaction to occur.

Question 34

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?

Accepted Answer

tertiary alcohol by S_N1. Tertiary alcohols react fastest with Lucas reagent because they undergo the S_N1 mechanism, which involves the formation of a stable carbocation. The tertiary carbocation is more stable than secondary or primary ones, allowing for a quicker reaction.

Question 35

The most suitable reagent for the conversion of R–CH2OH → R–CHO is-

Accepted Answer

PCC (Pyridinium Chlorochromate). PCC is a mild oxidizing agent that selectively oxidizes primary alcohols to aldehydes without further oxidizing them to carboxylic acids, making it the most suitable choice for this conversion.

Question 36

Which of the following compounds will most readily be dehydrated to give alkene under acidic condition ? [JEE-Main On line-2018]

Accepted Answer

4-Hydroxypentan-2-one. 4-Hydroxypentan-2-one has a hydroxyl group adjacent to a carbonyl group, which facilitates the formation of a stable carbocation upon dehydration under acidic conditions, making it more reactive compared to the other options.

Question 37

Which is the most suitable reagent for the following transformation?

CH3–CH=CH–CH2–CH(OH)–CH3 → CH3–CH=CH–CH2–COOH

Accepted Answer

I2/NaOH. The fragment CH3-CH(OH)- is a methyl carbinol that gives a positive iodoform (haloform) reaction. I2/NaOH cleaves it to a carboxylate (losing CHI3), converting CH3-CH(OH)- into -COOH while leaving the C=C double bond intact. Alkaline KMnO4 would instead cleave the double bond.

Question 38

CH3-CH2-C(OH)(Ph)-CH3 cannot be prepared by -

Accepted Answer

HCHO + PhCH(CH3)CH2MgX. The target 2-phenylbutan-2-ol is tertiary (C bearing OH, methyl, ethyl, phenyl). A tertiary alcohol needs a ketone + RMgX; HCHO + a Grignard yields only a primary alcohol RCH2OH, so the HCHO route cannot give it. The ketone routes (e.g. ethyl methyl ketone + PhMgX) all work.

Question 39

Two compounds A and B with same molecular formula (C3H6O) undergo Grignard's reaction with methylmagnesium bromide to give products C and D. products C and D show following chemical tests.

Accepted Answer

C = H3C–CH2–CH(OH)–CH3; D = H3C–C(CH3)2–OH. Propanal + CH3MgBr -> 2-butanol (CH3CH2CH(OH)CH3): secondary, Lucas in ~5 min, positive iodoform = C. Acetone + CH3MgBr -> tert-butanol (CH3C(CH3)2OH): tertiary, immediate Lucas, negative iodoform = D.

Question 40

The major product [B] in the following reactions is: CH3–CH2–CH(CH3)–CH2–OCH2–CH3
HI, Heat → [A] alcohol
H2SO4, Δ → [B]

Accepted Answer

CH3–CH=C(CH3)–CH3. The correct option is CH3–CH=C(CH3)–CH3 because the dehydration of the alcohol under acidic conditions leads to the formation of an alkene, and the structure indicates that a double bond is formed between the second and third carbon atoms, with a methyl group attached to the third carbon, which is consistent with the elimination reaction.

Question 41

Which of the following derivatives of alcohols is unstable in an aqueous base ?
(1) RO – CMe3
(2) RO – C(=O)Me
(3) RO – CH2 – C6H5
(4) a six-membered cyclic ether derivative with RO substituent and one oxygen atom in the ring

Accepted Answer

RO – C(=O)Me. The derivative RO – C(=O)Me is unstable in an aqueous base because the carbonyl group can undergo hydrolysis, leading to the formation of a carboxylic acid and making it more susceptible to nucleophilic attack, which destabilizes the compound.

Question 42

Consider the following reactions:
(a) (CH3)3CCH(OH)CH3 —conc. H2SO4→
(b) (CH3)2CHCH(Br)CH3 —conc. KOH→
(c) (CH3)2CHCH(Br)CH3 —(CH3)3O−K+→
(d) (CH3)2C(OH)CH2CHO —Δ→
Which of these reactions(s) will not produce Saytzeff product ?

Accepted Answer

(2) (c) only. With bulky (CH3)3CO-K+ (potassium tert-butoxide) on 2-bromo-3-methylbutane, the less-substituted Hofmann alkene dominates, so reaction (c) does NOT give the Saytzeff product; the others give the more-substituted/stable alkene.

Question 43

Arrange the following compounds in increasing order of C – OH bond length: methanol, phenol, p-ethoxyphenol

Accepted Answer

phenol < p-ethoxyphenol < methanol. The C – OH bond length is influenced by the electron-donating or withdrawing effects of substituents. In this case, phenol has a shorter bond due to resonance stabilization, while p-ethoxyphenol has a longer bond than phenol but shorter than methanol due to the electron-donating ethoxy group, making the order phenol < p-ethoxyphenol < methanol.

Question 44

Given below are two statements:
Statement I: o-Nitrophenol is steam volatile due to intramolecular hydrogen bonding.
Statement II: o-Nitrophenol has high melting point due to hydrogen bonding.
In the light of the above statements, choose the most appropriate answer from the options given b

Accepted Answer

Statement I is true but statement II is false. o-Nitrophenol exhibits intramolecular hydrogen bonding, which allows it to be steam volatile, making Statement I true. However, its high melting point is primarily due to intermolecular hydrogen bonding, not intramolecular, rendering Statement II false.

Question 45

Ceric ammonium nitrate and CHCl3/alc. KOH are used for the identification of functional groups present in ____ and ____, respectively.

Accepted Answer

Alcohol, amine. Ceric ammonium nitrate gives a red color with alcohols, confirming the -OH group. CHCl3 with alcoholic KOH is the carbylamine (isocyanide) test, specific for primary amines. So the groups identified are alcohol and amine, respectively.

Question 46

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Synthesis of ethyl phenyl ether may be achieved by Williamson's synthesis.
Reason (R): Reaction of bromobenzene with sodium ethoxide yields ethyl phenyl ether. In the li

Accepted Answer

(A) is correct but (R) is not correct. The assertion is true because ethyl phenyl ether can indeed be synthesized using Williamson's synthesis. However, the reason is incorrect as bromobenzene does not react with sodium ethoxide to produce ethyl phenyl ether; instead, it requires a different alkyl halide.

Question 47

Hex-4-ene-2-ol on treatment with PCC gives 'A' on reaction with sodium hypoiodite gives 'B', which on further heating with soda lime gives 'C'. The compound 'C' is

Accepted Answer

2-butene. The compound 'C' is 2-butene because the reaction sequence involves the oxidation of hex-4-ene-2-ol to form a ketone, which upon treatment with sodium hypoiodite undergoes a rearrangement to yield a compound that, when heated with soda lime, eliminates a small molecule to form 2-butene.

Question 48

Given below are two statements:
Statement I: On heating with KHSO4, glycerol is dehydrated and acrolein is formed.
Statement II: Acrolein has fruity odour and can be used to test glycerol’s presence.
Choose the correct option.
(1) Both Statement I and Statement II are correct.
(2) Both Sta

Accepted Answer

(3) Statement I is correct but Statement II is incorrect.. Statement I is correct: glycerol heated with KHSO4 is dehydrated to acrolein (CH2=CH-CHO). Statement II is incorrect: acrolein has a sharp, pungent, irritating odour, not a fruity one. So Statement I is correct but Statement II is incorrect.

Question 49

The difference in the reaction of phenol with bromine in chloroform and bromine in water medium is due to:

Accepted Answer

Polarity of solvent. The reaction of phenol with bromine varies in different solvents due to the polarity of the solvent, which affects the solubility and reactivity of the phenol and bromine, leading to different reaction pathways and products.

Question 50

Choose the correct option for the following reactions.
B —(BH3, H2O2/OH−)→ [product]
[alkene] —(Hg(OAc)2, H2O; NaBH4)→ A

Accepted Answer

‘A’ is Markovnikov product and ‘B’ is anti-Markovnikov product. Reaction A uses Hg(OAc)2/H2O then NaBH4 (oxymercuration-demercuration), which gives the Markovnikov alcohol with no rearrangement. Reaction B uses BH3 then H2O2/OH- (hydroboration-oxidation), giving the anti-Markovnikov alcohol. So A is Markovnikov and B is anti-Markovnikov.

JEE Main Chemistry: Alcohols, Phenols and Ethers questions with solutions

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