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Multiple Choice Questions

1. When benzene sulfonic acid and p-nitrophenol are treated with $\text{NaHCO}_3$, the gases released respectively are: [JEE 2006]

(a) $\text{SO}_2, \text{NO}_2$

(b) $\text{SO}_2, \text{NO}$

(d) $\text{CO}_2, \text{CO}_2$

Answer: (d)

Explanation: Both benzene sulfonic acid ($\text{PhSO}_3\text{H}$) and p-nitrophenol are more acidic than carbonic acid ($\text{H}_2\text{CO}_3$). Thus, both react with $\text{NaHCO}_3$ to liberate $\text{CO}_2$ gas.

$$\text{PhSO}_3\text{H} + \text{NaHCO}_3 \longrightarrow \text{PhSO}_3\text{Na} + \text{H}_2\text{O} + \text{CO}_2 \uparrow$$ $$\text{O}_2\text{N}-\text{C}_6\text{H}_4-\text{OH} + \text{NaHCO}_3 \longrightarrow \text{O}_2\text{N}-\text{C}_6\text{H}_4-\text{ONa} + \text{H}_2\text{O} + \text{CO}_2 \uparrow$$

2. The increasing order of boiling points of above mentioned alcohols is: [JEE 2006]
(I) 1,2-dihydroxy benzene
(II) 1,3-dihydroxy benzene
(III) 1,4-dihydroxy benzene
(IV) Hydroxy benzene

(a) $\text{I} < \text{II} < \text{III} < \text{IV}$

(b) $\text{I} < \text{II} < \text{IV} < \text{III}$

(d) $\text{IV} < \text{II} < \text{I} < \text{III}$

Answer: (c)

Explanation: Hydroxybenzene (phenol) has only one $-\text{OH}$ group, so it has the lowest boiling point (IV). Among dihydroxy benzenes, 1,2-dihydroxybenzene (I) forms intramolecular hydrogen bonding, reducing intermolecular association and lowering its boiling point relative to 1,3- (II) and 1,4- (III) isomers. 1,4-dihydroxybenzene (III) has maximum symmetry and extensive intermolecular H-bonding, giving it the highest boiling point. Thus, the order is $\text{IV} < \text{I} < \text{II} < \text{III}$.

3. Cumene $\xrightarrow[\Delta]{\text{(i) }\text{O}_2\text{, (ii) }\text{H}_3\text{O}^+} P + Q$. The major products P and Q are: [JEE 2006]

(a) $\text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{CHO}$ and $\text{CH}_3\text{COCH}_3$

(b) $\text{C}_6\text{H}_5\text{COCH}_3$ and $\text{CH}_3\text{CH}_2\text{CHO}$

(d) $\text{C}_6\text{H}_5\text{CHO}$ and $\text{CH}_3\text{COCH}_3$

Answer: (c)

Explanation: It is the cumene hydroperoxide rearrangement reaction. Cumene on oxidation followed by acid hydrolysis yields Phenol ($\text{C}_6\text{H}_5\text{OH}$) and Acetone ($\text{CH}_3\text{COCH}_3$).

4. The correct acidity order of the following is: [JEE 2009 P-I]
(I) Phenol
(II) 4-Chlorophenol
(III) Benzoic acid
(IV) 4-Methylbenzoic acid

(a) $\text{III} > \text{IV} > \text{II} > \text{I}$

(b) $\text{IV} > \text{III} > \text{I} > \text{II}$

(d) $\text{II} > \text{III} > \text{IV} > \text{I}$

Answer: (a)

Explanation: Carboxylic acids are stronger acids than phenols ($pK_a$ of Benzoic acid (III) $= 4.17$, 4-Methylbenzoic acid (IV) $= 4.37$, 4-Chlorophenol (II) $= 9.38$, Phenol (I) $= 9.98$). Electron-withdrawing groups increase acidity while electron-donating groups ($-\text{CH}_3$) decrease acidity. Decreasing order of acidic strength: $\text{III} > \text{IV} > \text{II} > \text{I}$.

5. In the reaction, $\text{Ph}-\text{O}-\text{CH}_3 \xrightarrow{\text{HBr}} \text{Products}$, the products are: [JEE 2010 P-I]

(a) $\text{Ph}-\text{Br}$ and $\text{CH}_3\text{Br}$

(b) $\text{Ph}-\text{OCH}_3$ and $\text{H}_2\text{O}$

(d) $\text{Ph}-\text{OH}$ and $\text{CH}_3\text{Br}$

Answer: (d)

Explanation: Cleavage of anisole ($\text{Ph}-\text{O}-\text{CH}_3$) with $\text{HBr}$ gives phenol and methyl bromide because the $\text{C(phenyl)}-\text{O}$ bond has partial double bond character due to resonance and is stronger than the $\text{C(methyl)}-\text{O}$ bond.

$$\text{Ph}-\text{O}-\text{CH}_3 + \text{HBr} \longrightarrow \text{Ph}-\text{OH} + \text{CH}_3\text{Br}$$

6. In the reaction of Phenol with $\text{Br}_2/\text{NaOH(aq)}$, the intermediate(s) is (are): [JEE 2010 P-I]

(a) Phenoxide ion with negative charge delocalized at ortho position containing bromine substituent

(b) Phenoxide ion with negative charge delocalized at para position containing bromine substituent

(d) Bromobenzene intermediate

Answer: (a, b, c)

Explanation: Phenoxide ion is formed in alkaline medium ($\text{NaOH}$) which is strongly activated towards electrophilic aromatic substitution and directs to ortho and para positions via delocalized carbanion intermediates.

$$\text{Ph-O}^- \stackrel{\text{Br}_2}{\longrightarrow} \text{Resonance Intermediates (Ortho \& Para)} \longrightarrow \text{Brominated Products}$$

7. Arrange the following compounds in order of decreasing acidity: [JEE Main 2013]
(I) Phenol
(II) p-Chlorophenol
(III) p-Nitrophenol
(IV) p-Methoxyphenol

(a) $\text{II} > \text{IV} > \text{I} > \text{III}$

(b) $\text{I} > \text{II} > \text{III} > \text{IV}$

(d) $\text{IV} > \text{III} > \text{I} > \text{II}$

Answer: (c)

Explanation: Electron-withdrawing groups ($-\text{NO}_2, -\text{Cl}$) increase acidity while electron-releasing groups ($-\text{OCH}_3$) decrease acidity. Strongest $-\text{M}$ and $-\text{I}$ effect of $-\text{NO}_2$ makes p-nitrophenol (III) most acidic, followed by p-chlorophenol (II), phenol (I), and p-methoxyphenol (IV).

Acidic nature order: $\text{III} > \text{II} > \text{I} > \text{IV}$.

8. 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? [JEE Main 2013]

(a) secondary alcohol by $\text{S}_\text{N}1$

(b) tertiary alcohol by $\text{S}_\text{N}1$

(d) tertiary alcohol by $\text{S}_\text{N}2$

Answer: (b)

Explanation: Fastest reaction with Lucas reagent ($\text{conc. HCl} + \text{anc. ZnCl}_2$) is given by $3^\circ\text{ R}-\text{OH}$ (tertiary alcohol) instantaneously at room temperature, proceeding via carbocation intermediate following $\text{S}_\text{N}1$ mechanism.

9. The compound that does NOT liberate $\text{CO}_2$ on treatment with aqueous sodium bicarbonate solution, is: [JEE Adv. 2013 P-I]

(a) Benzoic acid

(b) Benzenesulphonic acid

(d) Carbolic acid (Phenol)

Answer: (d)

Explanation: $pK_a$ of Phenol (carbolic acid) is 9.98 and that of carbonic acid ($\text{H}_2\text{CO}_3$) is 6.63. Since phenol is a weaker acid than carbonic acid, it does not react with $\text{NaHCO}_3$ to liberate $\text{CO}_2$ gas.

10. The major product (s) of the following reaction is (are): [JEE Adv. 2013 P-II]

Phenol-4-sulfonic acid $\xrightarrow{\text{aqueous }\text{Br}_2\text{ (3.0 equivalents)}}$ ?

(a) P

(b) Q

(d) S

Answer: (d)

Explanation: Treatment of phenol-4-sulfonic acid with excess aqueous bromine (3.0 equivalents) results in ipso-bromination with displacement of the $-\text{SO}_3\text{H}$ group, yielding 2,4,6-tribromophenol as product S (or Q/R/S as depicted in options).

11. In the following reaction, the product (s) formed is (are): [JEE Adv. 2013 P-II]

(a) P (major)

(b) Q (minor)

(d) S (major)

Answer: (b, d)

Explanation: The reaction of 3-methylphenol with chloroform and sodium hydroxide (Reimer-Tiemann reaction) introduces a $-\text{CHO}$ group ortho to $-\text{OH}$, yielding ortho-formylated phenolic isomers (Q as minor and S as major product).

12. The acidic hydrolysis of ether (X) shown below is fastest when: [JEE Adv. 2014 P-II]

(a) one phenyl group is replaced by a methyl group

(b) one phenyl group is replaced by a para-methoxyphenyl group

(d) no structural change is made to X.

Answer: (c)

Explanation: Acidic hydrolysis of ethers involves the formation of a carbocation intermediate ($\text{S}_\text{N}1$ pathway). When two phenyl groups are replaced by two para-methoxyphenyl groups, the carbocation formed is exceptionally stabilized by resonance involving the strong $+M$ effect of the two $-\text{OCH}_3$ groups, making hydrolysis fastest.

13. The reactivity of compound Z with different halogens under appropriate conditions is given below: [JEE Adv. 2014 P-I]
Z $\xrightarrow{\text{I}_2}$ mono-halo substituted derivative
Z $\xrightarrow{\text{Br}_2}$ di-halo substituted derivative
Z $\xrightarrow{\text{Cl}_2}$ tri-halo substituted derivative
The observed pattern of electrophilic substitution can be explained by:

(a) the steric effect of the halogen

(b) the steric effect of the tert-butyl group

(d) the electronic effect of the tert-butyl group

Answer: (a)

Explanation: Iodine, being largest in size, experiences severe steric hindrance with the bulky tert-butyl group and phenolic $-\text{OH}$, leading to only mono-substitution. Smaller bromine permits di-substitution, while the smallest chlorine allows tri-substitution. Thus, the observed pattern is explained by the steric effect of the halogen.

14. The major product of the following reaction is: [JEE Adv. 2015 P-I]

(a) Product A

(b) Product B

(d) Product D

Answer: (a)

Explanation: Intramolecular nucleophilic attack of phenoxide ion on the epoxy ester/carbonyl followed by hydrolysis and elimination gives the bicyclic lactone/ether product corresponding to option (a).

15. The major product U in the following reactions is: [JEE Adv. 2015 P-II]

(a) Product A

(b) Product B

(d) Product D

Answer: (a)

Explanation: Stepwise reaction involving base-mediated intramolecular aldol/condensation followed by ring construction yields the stable cyclic derivative (a).

16. 2-chloro-2-methylpentane on reaction with sodium methoxide in methanol yields: [JEE Main 2016]
(a) $\text{CH}_3-\text{C}(\text{CH}_3)=\text{CH}-\text{CH}_2-\text{CH}_3$
(b) $\text{CH}_2=\text{C}(\text{CH}_3)-\text{CH}_2-\text{CH}_2-\text{CH}_3$
(c) $\text{CH}_3-\text{C}(\text{OCH}_3)(\text{CH}_3)-\text{CH}_2-\text{CH}_2-\text{CH}_3$

(a) All of these

(b) (a) and (c)

(d) (a) and (b)

Answer: (a)

Explanation: Being a tertiary alkyl halide reacting with a strong base/nucleophile ($\text{NaOCH}_3/\text{CH}_3\text{OH}$), both elimination (E2/E1 yielding Saytzeff alkene (a) and Hofmann alkene (b)) and substitution ($\text{S}_\text{N}1$ yielding ether (c)) take place. Therefore, all of these products (a, b, and c) are formed.

17. The correct statement(s) about the following reaction sequence is(are): [JEE Adv. 2016 P-I]
Cumene ($\text{C}_9\text{H}_{12}$) $\xrightarrow[\text{(ii) }\text{H}_3\text{O}^+]{\text{(i) }\text{O}_2} P + Q$
$P \xrightarrow[\text{NaOH}]{\text{CHCl}_3/\text{NaOH}} R \text{ (minor)} + Q \text{ (major)}$
$Q + \text{PhCH}_2\text{Br} \xrightarrow{\text{NaOH}} S$

(a) R is steam volatile

(b) Q gives dark violet coloration with 1% aqueous $\text{FeCl}_3$ solution

(d) S gives dark violet coloration with 1% aqueous $\text{FeCl}_3$ solution

Answer: (a, b, c)

Explanation: $P$ is Phenol ($\text{C}_6\text{H}_5\text{OH}$). Reimer-Tiemann formylation of phenol gives salicylaldehyde ($R$, ortho-isomer, minor) and p-hydroxybenzaldehyde ($Q$, para-isomer, major). Salicylaldehyde ($R$) exhibits intramolecular hydrogen bonding and is steam volatile. $Q$ has free phenolic $-\text{OH}$ group, so it gives dark violet color with 1% $\text{FeCl}_3$. $S$ is an ether derivative with an aldehyde group, so it reacts with 2,4-DNP to yield a yellow precipitate. $S$ lacks free phenolic $-\text{OH}$ group and will not give violet color with $\text{FeCl}_3$. Hence, (a, b, c) are correct.

18. Phenol reacts with methyl chloroformate in the presence of $\text{NaOH}$ to form product A. A reacts with $\text{Br}_2$ to form product B. A and B are respectively: [JEE Main 2018]

(a) Methyl phenyl carbonate and 4-Bromophenyl methyl carbonate

(b) Methyl phenyl carbonate and 2-Bromophenyl methyl carbonate

(d) Phenyl acetate and 4-Bromophenyl acetate

Answer: (a)

Explanation: Phenol reacts with methyl chloroformate ($\text{ClCOOCH}_3$) in basic medium to form methyl phenyl carbonate ($A$). Subsequent bromination with $\text{Br}_2$ directs substitution to the para-position (due to ortho/para-directing nature of $-\text{OCOOCH}_3$ group), forming 4-bromophenyl methyl carbonate ($B$). Hence, option (a) is correct.

19. The major product formed in the reaction of 4-methylphenol with $\text{H}_2/\text{Pd-BaSO}_4$ or specific reagents is: [JEE Main 2018]

(a) Product A

(b) Product B

(d) Product D

Answer: (b, c)

Explanation: Catalytic hydrogenation selectively reduces or retains oxygenated/unsaturated functional groups based on conditions yielding corresponding cyclohexanol or substituted phenol isomers.

20. An organic compound 'X' showing the following solubility profile: insoluble in water, soluble in $\text{NaOH(aq)}$, insoluble in $\text{NaHCO}_3\text{(aq)}$, is: [JEE Main 2019]

(a) m-Cresol

(b) Oleic acid

(d) Benzamide

Answer: (a)

Explanation:
- Phenols like m-Cresol are weakly acidic: soluble in strong base ($\text{NaOH}$) but insoluble in weak base solution ($\text{NaHCO}_3$).
- Oleic acid is a carboxylic acid, so it is soluble in both $\text{NaOH}$ and $\text{NaHCO}_3$.
- o-Toluidine is basic, insoluble in $\text{NaOH}$ and $\text{NaHCO}_3$.
- Benzamide is neutral, insoluble in $\text{NaOH}$ and $\text{NaHCO}_3$.
Thus, compound 'X' is m-Cresol.

Multiple Choice Questions

21. An organic compound neither reacts with neutral ferric chloride solution nor with Fehling solution. It however, reacts with Grignard reagent and gives positive iodoform test. The compound is - [JEE Main 2019]

(a) $\text{C}_2\text{H}_5-\text{CH}(\text{OH})-\text{C}(=\text{O})-\text{CH}_3$

(b) $\text{C}_2\text{H}_5-\text{C}(\text{CH}_3)_2-\text{O}-\text{CH}_3$

(d) $\text{C}_2\text{H}_5-\text{CH}(\text{CH}_3)-\text{CH}(\text{OH})-\text{CH}_3$

Answer: (a)

Explanation: The compound contains a $-\text{CH}(\text{OH})-\text{CH}_3$ group attached to a keto/alkyl structure which gives a positive iodoform test ($\text{NaOH} + \text{I}_2$). It has an acidic $\text{H}$ atom in the $-\text{OH}$ group which reacts with Grignard reagent ($\text{RMgX}$). It shows a negative neutral $\text{FeCl}_3$ test because it has no phenolic group, and a negative Fehling's test as it is not an aldehyde.

22. The compounds A and B in the following reaction are, respectively : [JEE Main 2019]

$$\text{HCHO} \xrightarrow{\text{HCl}} \text{Benzyl chloride} \xrightarrow{\text{AgCN}} \text{B}$$

(a) A = Benzyl alcohol, B = Benzyl isocyanide

(b) A = Benzyl alcohol, B = Benzyl cyanide

(d) A = Benzyl chloride, B = Benzyl isocyanide

Answer: (d)

Explanation:

$$\text{Benzene} + \text{HCHO} \xrightarrow{\text{HCl}} \text{Ph}-\text{CH}_2-\text{Cl} \text{ (A = Benzyl chloride)}$$

$$\text{Ph}-\text{CH}_2-\text{Cl} \xrightarrow{\text{AgCN (covalent)}} \text{Ph}-\text{CH}_2-\text{NC} \text{ (B = Benzyl isocyanide)}$$

23. The product formed in the reaction of cumene with $\text{O}_2$, followed by treatment with dil. $\text{HCl}$ are: [JEE Main 2019]

(a) $\text{C}_6\text{H}_5\text{OH}$ and $\text{CH}_3\text{OH}$

(b) $\text{C}_6\text{H}_5\text{COCH}_3$ and $\text{CH}_3\text{OH}$

(d) $\text{C}_6\text{H}_5\text{CH}_2\text{OH}$ and $\text{CH}_3\text{CHO}$

Answer: (c)

Explanation: Oxidation of cumene yields cumene hydroperoxide, which upon acid hydrolysis ($\text{H}^+, \text{H}_2\text{O}/\Delta$) cleaves into phenol ($\text{C}_6\text{H}_5\text{OH}$) and acetone ($\text{CH}_3\text{COCH}_3$).

$$\text{Ph}-\text{CH}(\text{CH}_3)_2 \xrightarrow{\text{O}_2} \text{Ph}-\text{C}(\text{CH}_3)_2-\text{O}-\text{O}-\text{H} \xrightarrow{\text{H}^+ / \text{H}_2\text{O}} \text{Ph-OH} + \text{CH}_3\text{COCH}_3$$

24. The major product of the following reaction is: [JEE Main 2019]

$$\text{PhCH}_2\text{CH}(\text{Br})\text{CH}_2\text{Br} \xrightarrow{\text{alc. KOH (Excess)}, \Delta} ?$$

(a) $\text{PhCH}=\text{C}=\text{CH}_2$

(b) $\text{PhC}\equiv\text{CCH}_3$

(d) $\text{PhCH}_2\text{C}\equiv\text{CH}$

Answer: (c)

Explanation: Example of $E_2$ elimination where a conjugated diene is formed with the phenyl ring in conjugation, which makes it very stable.

25. The increasing order of the $p K_a$ values of the following compounds is: [JEE Main 2019]

(A) 4-nitrophenol
(B) 2-nitrophenol
(C) phenol
(D) 3-methylphenol

(a) D < A < C < B

(b) B < C < D < A

(d) B < C < A < D

Answer: (d)

Explanation: Acidic strength is inversely proportional to $p K_a$. The order of acidic strength is:

$$\text{4-nitrophenol} > \text{2-nitrophenol} > \text{phenol} > \text{3-methylphenol}$$

Therefore, the order of $p K_a$ values is: $\text{B} < \text{C} < \text{A} < \text{D}$.

26. The major product of the following reaction is :- [JEE Main 2019]

$$\text{Ph-CH}_2-\text{CH}=\text{CH}_2 \xrightarrow{(1) \text{ HCl}} \xrightarrow{(2) \text{ Anhyd. AlCl}_3} ?$$

(a) 1-chloro-1-phenylpropane

(b) 1-methylindane

(d) tetralin

Answer: (b)

Explanation: Electrophilic addition of $\text{HCl}$ according to Markovnikov's rule yields $\text{Ph-CH}_2-\text{CH}(\text{Cl})-\text{CH}_3$. Intramolecular Friedel-Crafts alkylation in the presence of $\text{AlCl}_3$ yields 1-methylindane.

27. $\text{CH}_3\text{CH}_2-\text{C}(\text{OH})(\text{Ph})-\text{CH}_3$ cannot be prepared by: [JEE Main 2019]

(a) $\text{HCHO} + \text{PhCH}(\text{CH}_3)\text{CH}_2\text{MgX}$

(b) $\text{PhCOCH}_2\text{CH}_3 + \text{CH}_3\text{MgX}$

(d) $\text{CH}_3\text{CH}_2\text{COCH}_3 + \text{PhMgX}$

Answer: (a)

Explanation: Reaction of $\text{HCHO}$ with Grignard reagent $\text{PhCH}(\text{CH}_3)\text{CH}_2\text{MgX}$ produces a primary alcohol $\text{PhCH}(\text{CH}_3)\text{CH}_2\text{CH}_2\text{OH}$, whereas the target compound is a tertiary alcohol $\text{CH}_3\text{CH}_2-\text{C}(\text{OH})(\text{Ph})-\text{CH}_3$.

28. The major product in the following conversion is: [JEE Main 2019]

$$\text{CH}_3\text{O}-\text{C}_6\text{H}_4-\text{CH}_2-\text{CH}=\text{CH}_2 \xrightarrow{\text{HBr (excess)}, \text{Heat}} ?$$

(a) $\text{CH}_3\text{O}-\text{C}_6\text{H}_4-\text{CH}_2-\text{CH}(\text{Br})-\text{CH}_3$

(b) $\text{HO}-\text{C}_6\text{H}_4-\text{CH}_2-\text{CH}(\text{Br})-\text{CH}_3$

(d) $\text{CH}_3\text{O}-\text{C}_6\text{H}_4-\text{CH}_2-\text{CH}_2-\text{CH}_2\text{Br}$

Answer: (b)

Explanation: Excess $\text{HBr}$ with heat performs two reactions: (1) Hydrolysis/cleavage of the aryl methyl ether to form phenol ($\text{HO}-\text{C}_6\text{H}_4-$), and (2) Markovnikov addition of $\text{HBr}$ across the double bond to form $-\text{CH}_2-\text{CH}(\text{Br})-\text{CH}_3$.

Matrix Match/Column Match

29. Match the compounds in Column I with their characteristic test(s)/reaction(s) given in Column II. [JEE 2008 P-II]

Column I:

(A) Glycine

(B) $\text{C}_6\text{H}_5\text{NH}_3^+\text{Cl}^-$

(D) $\text{C}_6\text{H}_5\text{CONHNH}_2$

Column II:

1. Sodium fusion extract of the compound gives Prussian blue colour with $\text{FeSO}_4$

2. Gives white precipitate with $\text{AgNO}_3$

3. Reacts with aldehydes to form the corresponding hydrazone derivative

4. Reacts with $\text{NaOH}$

(a) A $\rightarrow$ 3, 4; B $\rightarrow$ 1, 2; C $\rightarrow$ 1, 2, 3; D $\rightarrow$ 1, 4

(b) A $\rightarrow$ 2, 4; B $\rightarrow$ 1, 2; C $\rightarrow$ 1, 2, 4; D $\rightarrow$ 1, 3

(d) A $\rightarrow$ 1, 2; B $\rightarrow$ 2, 3; C $\rightarrow$ 1, 2, 3; D $\rightarrow$ 1, 4

Answer: (a)

Explanation: Matching characteristics of organic nitrogen compounds.

30. The only correct combination in which the reaction proceeds through radical mechanism is [JEE 2008 P-II]

(a) (II) (iii) (C)

(b) (III) (ii) (A)

(d) (I) (ii) (C)

Answer: (d)

Explanation: Toluene (I) + $\text{Br}_2 / h\nu$ (ii) $\longrightarrow$ Free Radical Substitution (C) to form benzyl bromide.

31. For the synthesis of benzoic acid, the only correct combination is [JEE 2008 P-II]

(a) (III) (iv) (C)

(b) (IV) (ii) (A)

(d) (I) (iv) (B)

Answer: (c)

Explanation: Acetophenone (II) + $\text{NaOH} / \text{Br}_2$ (i) $\longrightarrow$ Haloform reaction (D) to give sodium benzoate, which yields benzoic acid upon acidification.

32. The only correct combination that gives two different carboxylic acids is [JEE 2008 P-II]

(a) (IV) (iii) (B)

(b) (I) (i) (D)

(d) (II) (iv) (C)

Answer: (c)

Explanation: Benzaldehyde (III) + $(\text{CH}_3\text{CO})_2\text{O} / \text{CH}_3\text{COOK}$ (iii) $\longrightarrow$ Perkin condensation (A) yielding cinnamic acid and acetic acid.

33. The correct match between Column I and Column II is: [JEE Main 2019]

Column - I (compound):
(A) Lysine
(B) Furfural
(C) Benzyl alcohol
(D) Styrene

Column – II (reagent):
1. 1-naphthol
2. ninhydrin
3. $\text{KMnO}_4$
4. Ceric ammonium nitrate

(a) A $\rightarrow$ 2; B $\rightarrow$ 1; C $\rightarrow$ 4; D $\rightarrow$ 3

(b) A $\rightarrow$ 2; B $\rightarrow$ 3; C $\rightarrow$ 4; D $\rightarrow$ 1

(d) A $\rightarrow$ 3; B $\rightarrow$ 1; C $\rightarrow$ 2; D $\rightarrow$ 4

Answer: (a)

Explanation: Lysine gives positive Ninhydrin test (2). Furfural gives Molisch test with 1-naphthol (1). Benzyl alcohol reacts with Ceric ammonium nitrate (4). Styrene decolorizes alkaline $\text{KMnO}_4$ (3).

Paragraph

34. Which one of the following reagents is used in the Reimer-Tiemann reaction? [IIT JEE 2007 P-II]

(a) $\text{aq. NaOH} + \text{CH}_2\text{Cl}_2$

(b) $\text{aq. NaOH} + \text{CHCl}_3$

(d) $\text{aq. NaOH} + \text{CCl}_4$

Answer: (c)

Explanation: Reimer-Tiemann reaction utilizes aqueous $\text{NaOH}$ and chloroform ($\text{CHCl}_3$).

35. The electrophile in this reaction is [IIT JEE 2007 P-II]

(a) $:\text{CHCl}$

(b) $\text{CHCl}_2^+$

(d) $\cdot \text{CCl}_3$

Answer: (c)

Explanation: $\text{OH}^- + \text{CHCl}_3 \rightleftharpoons \text{H}_2\text{O} + :\text{CCl}_3^- \longrightarrow :\text{CCl}_2 + \text{Cl}^-$. Dichlorocarbene ($:\text{CCl}_2$) acts as the active electrophile.

36. The structure of the intermediate I is [IIT JEE 2007 P-II]

(a) $\text{C}_6\text{H}_4(\text{O}^-)(\text{CH}_3)(\text{CH}_2\text{Cl})$

(b) $\text{C}_6\text{H}_3(\text{O}^-)(\text{CH}_3)(\text{CHCl}_2)$

(d) $\text{C}_6\text{H}_4(\text{ONa})(\text{CH}_3)(\text{CCl}_2^-)$

Answer: (b)

Explanation: Electrophilic addition of $:\text{CCl}_2$ on the phenoxide ion forms the phenolate salt containing $-\text{CHCl}_2$ intermediate at the ortho position.

37. Compound H is formed by the reaction of [JEE 2008 Paper -II]

(a) $\text{PhCOCH}_3 + \text{PhCH}_2\text{COCH}_3$

(b) $\text{PhCHO} + \text{PhCH}_2\text{MgBr}$

(d) $\text{PhCOCH}_3 + \text{PhMgBr}$

Answer: (b)

Explanation: Reaction of benzaldehyde ($\text{PhCHO}$) with benzyl magnesium bromide ($\text{PhCH}_2\text{MgBr}$) yields the secondary/tertiary alcohol precursor $\text{PhCH}(\text{OH})\text{CH}_2\text{Ph}$.

38. The structure of compound I is [JEE 2008 Paper -II]

(a) $\text{PhCH}=\text{CHPh}$

(b) $\text{Ph(CH}_3)\text{C}=\text{CHPh}$

(d) $\text{PhCH}=\text{C}(\text{CH}_3)_2$

Answer: (a)

Explanation: Acid-catalyzed dehydration of alcohol H yields 1,2-diphenylethane derivative / stilbene ($\text{PhCH}=\text{CHPh}$).

39. The structures of compounds J, K and L, respectively, are [JEE 2008 Paper -II]

(a) $\text{PhCHO}, \text{PhCH}_2\text{CHO}$ and $\text{PhCH}_2\text{COOK}$

(b) $\text{PhCOCH}_3, \text{PhCH}_2\text{CHO}$ and $\text{CH}_3\text{COOK}$

(d) $\text{PhCHO}, \text{PhCOCH}_3$ and $\text{PhCOOK}$

Answer: (d)

Explanation: Ozonolysis of compound I produces benzaldehyde ($\text{J}$) and acetophenone ($\text{K}$). Benzaldehyde under Cannizzaro reaction with $\text{KOH}$ gives benzyl alcohol and potassium benzoate ($\text{L} = \text{PhCOOK}$).

40. The product X is [JEE Adv. 2014 P-II]

(a) $(2E, 4E)$-hexadiene derivative / alkyne addition product

(b) $(2Z, 4E)$-isomer

(d) terminal alkyne

Answer: (a)

Explanation: Sequential treatment of alkyne M with excess $\text{NaNH}_2$, followed by alkylation with $\text{CH}_3\text{CH}_2\text{I}$ and $\text{CH}_3\text{I}$, then hydrogenation over Lindlar's catalyst yields product X.

41. The correct statement with respect to product Y is [JEE Adv. 2014 P-II]

(a) It gives a positive Tollens test and is a functional isomer of X

(b) It gives a positive Tollens test and is a geometrical isomer of X

(d) It gives a positive iodoform test and is a geometrical isomer of X

Answer: (c)

Explanation: X and Y are functional isomers of each other, and Y contains a methyl ketone group ($\text{CH}_3\text{CO}-$), thus giving a positive iodoform test.

Integer and Subjective

42. The number of resonance structures for N is [JEE Adv. 2015 P-I]

Answer: 9

Explanation: Phenoxide derivative / substituted aromatic anion N exhibits 9 distinct resonating canonical forms involving the aromatic ring and substituent groups.

43. Calculate the total number of cyclic ether (including stereo) having formula $\text{C}_4\text{H}_8\text{O}$. [JEE 2019 P-II]

Answer: 10

Explanation: Total cyclic ethers including all stereoisomers (R, S pairs) with molecular formula $\text{C}_4\text{H}_8\text{O}$ is equal to 10.

44. $\text{Alkene} \xrightarrow{\text{dil. KMnO}_4, 273\text{ K}} \text{'Y'}$. Number of hydroxyl group in compound 'Y' is: [JEE 2019 P-II]

Answer: 6

Explanation: Dilute cold $\text{KMnO}_4$ (Baeyer's reagent) converts each double bond into a vicinal diol (2 $-\text{OH}$ groups). Since the reactant contains 3 double bonds, syn-hydroxylation adds $3 \times 2 = 6$ hydroxyl groups in compound Y.