Question 1

An optically active alkyl bromide C_4H_9Br, reacts with ethanolic KOH to form major compound [A] which reacts with bromine to give compound [B]. Compound [B] reacts with ethanolic KOH and sodamide to

Accepted Answer

The optically active alkyl bromide C _4 H _9 Br is 2-bromobutane, CH _3 CH ( Br ) CH _2 CH _3. Reaction with ethanolic KOH undergoes dehydrohalogenation to form but-2-ene as the major product [A] (Zaitsev's rule). But-2-ene [A] reacts with Br _2 to form 2,3-dibromobutane [B], CH _3 CH ( Br ) CH ( Br ) CH _3. Compound [B] undergoes double dehydrohalogenation with ethanolic KOH and NaNH _2 to give but-2-yne [C], CH _3 C CCH _3. Hydration of but-2-yne [C] with HgSO _4 and dilute H _2 SO _4 yields b

Question 2

An alkane (Y) requires 8 moles of oxygen for complete combustion and on chlorination with Cl_2/h , (Y) gives only one monochlorinated product (Z). The total number of primary carbon atoms in (Y) is __

Accepted Answer

The general formula of an alkane is C_n H_ 2n+2 . The balanced chemical equation for the complete combustion of an alkane is: C_n H_ 2n+2 + ( 3n+1 2 ) O_2 n CO_2 + (n+1) H_2O Given that 8 moles of oxygen are required for complete combustion: 3n+1 2 = 8 3n + 1 = 16 n = 5 The molecular formula of the alkane (Y) is C_5H_ 12 . Since (Y) yields only one monochlorinated product (Z) upon reaction with Cl_2/h , all the hydrogen atoms in the alkane must be equivalent. Among the isomers of C_5H_ 12 (n-pen

Question 3

R_f value for 2-methylpropene in a solvent system (Ethyl acetate + ether) is 0.42. 2-methylpropene is treated with dilute H _2 SO _4 to give major organic product (X). R_f value for (X) in the same so

Accepted Answer

2-methylpropene reacts with dilute H_2SO_4 to undergo hydration, forming tert-butyl alcohol as the major product (X). The reaction is: CH_2=C(CH_3)_2 dil. H_2SO_4 (CH_3)_3C-OH Product (X) is an alcohol, which is significantly more polar than the starting alkene due to the presence of the -OH group. In normal phase chromatography, the stationary phase (such as silica gel) is highly polar. The polar -OH group forms strong hydrogen bonds with the stationary phase, causing the alcohol to be adsorbed

Question 4

Given below are two statements: Statement I: Methane can be prepared by decarboxylation of sodium ethanoate, Kolbe's electrolysis of sodium acetate and reaction of CH _3 MgBr with water. Statement II:

Accepted Answer

Statement I is false because Kolbe's electrolysis of sodium acetate produces ethane, not methane. The reaction is: 2 CH _3 COONa + 2 H _2 O Electrolysis CH _3- CH _3 + 2 CO _2 + H _2 + 2 NaOH . Decarboxylation of sodium ethanoate and the reaction of CH _3 MgBr with water do produce methane. Statement II is true because the simplest unsaturated hydrocarbon is ethene ( C _2 H _4), which upon hydrogenation yields ethane, meaning methane cannot be prepared from unsaturated hydrocarbons. Additionally

Question 5

Given below are two statements: Statement I: 3-phenylpropene reacts with HBr and gives secondary alkyl bromide having a chiral carbon atom as the major product. Statement II: Aryl chlorides and aryl c

Accepted Answer

Statement I: When 3-phenylpropene reacts with HBr, protonation of the alkene occurs to form a secondary carbocation. C_6H_5-CH_2-CH=CH_2 + H^+ C_6H_5-CH_2-CH^+-CH_3 This secondary carbocation undergoes a 1,2-hydride shift to form a more stable, resonance-stabilized benzylic carbocation. C_6H_5-CH_2-CH^+-CH_3 C_6H_5-CH^+-CH_2-CH_3 The bromide ion then attacks this benzylic carbocation to form the major product, 1-bromo-1-phenylpropane. C_6H_5-CH^+-CH_2-CH_3 + Br^- C_6H_5-CH(Br)-CH_2-CH_3 The carb

Question 6

IUPAC name of the some alkenes are given below. Find out the correct stability order. A. 2-Methylbut-2-ene B. cis-But-2-ene C. 2,3-Dimethylbut-2-ene D. Prop-1-ene Choose the correct answer from the op

Accepted Answer

The stability of alkenes is directly proportional to the number of -hydrogen atoms due to hyperconjugation. Let us count the number of -hydrogen atoms in each given alkene: A. 2-Methylbut-2-ene: (CH_3)_2C=CHCH_3 Number of -hydrogen atoms = 3 + 3 + 3 = 9 B. cis-But-2-ene: CH_3CH=CHCH_3 Number of -hydrogen atoms = 3 + 3 = 6 C. 2,3-Dimethylbut-2-ene: (CH_3)_2C=C(CH_3)_2 Number of -hydrogen atoms = 3 + 3 + 3 + 3 = 12 D. Prop-1-ene: CH_3CH=CH_2 Number of -hydrogen atoms = 3 Since the stability of an

Question 7

"P" is a hydrocarbon of molecular formula: C_8H_ 14 . On ozonolysis, "P" forms "Q". "Q" on treatment with alkali under reflux condition produces "R", which on treatment with I_2/NaOH gives a yellow pr

Accepted Answer

The structure of "S" is 2-methylcyclopent-1-ene-1-carboxylic acid. Since "S" is obtained by the acidification of the product from the iodoform reaction of "R", "R" must contain a methyl ketone group (-COCH_3). The iodoform reaction converts the -COCH_3 group into a -COOH group alongside the formation of a yellow precipitate of iodoform (CHI_3). Therefore, the structure of "R" is 1-acetyl-2-methylcyclopentene. "R" is formed by the intramolecular aldol condensation of "Q". Performing a retro-aldol

Question 8

For the following Friedel Craft's alkylation reaction, which of the statements are correct? A. Major product is n-propyl benzene. B. iso-propyl carbocation intermediate is also generated. C. Multiple

Accepted Answer

In the given Friedel-Crafts alkylation reaction, 1-chloropropane reacts with anhydrous AlCl _3 to initially form a primary carbocation, the n-propyl carbocation ( CH _3- CH _2- CH _2^+). This primary carbocation is less stable and undergoes a 1,2-hydride shift to form a more stable secondary carbocation, the iso-propyl carbocation ( CH _3- CH ^+- CH _3). Therefore, statement B is correct. The iso-propyl carbocation then acts as an electrophile and attacks the benzene ring, yielding iso-propylben

Question 9

Compound (X) [styrene] is subjected to the sequence of reactions as shown above: (i) Br_2/CHCl_3 (ii) NaNH_2 excess (iii) CH_3I (iv) H_2, Na/NH_3 (l), giving Major Product (Y). Molar mass of the major

Accepted Answer

Addition of Br_2 to styrene (X) yields 1,2-dibromoethylbenzene. C_6H_5-CH=CH_2 Br_2/CHCl_3 C_6H_5-CH(Br)-CH_2(Br) Reaction with excess NaNH_2 causes double dehydrohalogenation followed by deprotonation of the terminal alkyne, forming sodium phenylacetylide. C_6H_5-CH(Br)-CH_2(Br) NaNH_2 (excess) C_6H_5-C C^- Na^+ Nucleophilic substitution with CH_3I yields 1-phenylpropyne. C_6H_5-C C^- Na^+ CH_3I C_6H_5-C C-CH_3 Reduction with Na/NH_3(l) reduces the internal alkyne to a trans-alkene without affe

Question 10

An alkene (X) on ozonolysis followed by reduction gives following products: H-CHO (2 moles), H-CO-CO-CHO, CH_3-CO-CO-CH_3. The alkene (X) is:

Accepted Answer

Ozonolysis (followed by reductive workup with Zn/H_2O) cleaves C=C bonds and converts each alkene carbon into a carbonyl. We can reverse-engineer the parent alkene from the fragments obtained. Analysing the products: HCHO (2 moles ): Two equivalents of formaldehyde indicate the presence of two terminal or exocyclic =CH_2 groups in the parent molecule. CH_3-CO-CO-CH_3 (biacetyl): To get this discrete fragment, the parent alkene must contain the structural unit -C(CH_3)=C(CH_3)-, i.e., an endocycl

Question 11

Consider the following sequence of reactions to give the major product (X) P g of the major product (X) formed is reacted with NaHCO_3 solution to liberate a gas which occupied 11.2 dm^3 at STP. P = _

Accepted Answer

Step 1: The first reaction is a Friedel-Crafts alkylation. Benzene reacts with CH_3Cl in the presence of anhydrous AlCl_3 to form toluene. Step 2: The second reaction is electrophilic aromatic substitution. Toluene reacts with Cl_2 in the presence of FeCl_3 to form a mixture of o-chlorotoluene and p-chlorotoluene. Due to steric hindrance at the ortho position, the major product is p-chlorotoluene. Step 3: The third reaction is the oxidation of the alkyl side chain. p-chlorotoluene is oxidized by

Question 12

The compound (X) on (i) on heating in the presence of anhydrous AlCl_3 and HCl gas gives 2,4-dimethyl pentane (ii) aromatization gives toluene and (iii) cyclisation gives methyl cyclohexane The correc

Accepted Answer

Compound (X) undergoes isomerization on heating with anhydrous AlCl_3 and HCl gas to give branched-chain alkanes. The formation of 2,4-dimethylpentane (a 7-carbon branched alkane) indicates that (X) is a 7-carbon straight-chain alkane. Aromatization of n-heptane at high temperature and pressure in the presence of oxides of V, Mo, or Cr supported over alumina gives toluene. Cyclisation of n-heptane under similar conditions initially yields methylcyclohexane, which can further undergo dehydrogenat

Question 13

Identify compounds A and E in the following reaction sequence.

Accepted Answer

The starting compound is 1-ethyl-4-nitrobenzene. Reaction with Br _2/ AlBr _3 is an electrophilic aromatic substitution. The - C _2 H _5 group is activating and ortho/para-directing, while the - NO _2 group is deactivating and meta-directing. Both groups direct the incoming electrophile ( Br ^+) to the position ortho to the - C _2 H _5 group. Thus, compound A is 2-bromo-1-ethyl-4-nitrobenzene. Reaction with Sn / HCl reduces the - NO _2 group to an - NH _2 group, forming compound B (4-amino-2-bro

Question 14

For the given molecule, "x", the preferred site for the attack of the electrophile is :

Accepted Answer

The given molecule, benzanilide, contains two phenyl rings with different reactivities towards electrophilic aromatic substitution. The left phenyl ring is attached to the carbonyl carbon of the amide group (-C(=O)NHPh). The carbonyl group is electron-withdrawing due to its -M and -I effects, which deactivates this ring towards electrophilic attack. The right phenyl ring is attached to the nitrogen atom of the amide group (-NHC(=O)Ph). The nitrogen atom possesses a lone pair of electrons that ca

Question 15

Consider the isomers of hydrocarbon with molecular formula C_5H_ 10 . These isomers do not decolourise KMnO_4 solution. These isomers are subjected to chlorination with chlorine in presence of light t

Accepted Answer

The hydrocarbon with molecular formula C_5H_ 10 has a degree of unsaturation of 1. Since its isomers do not decolourise KMnO_4 solution, they do not contain any carbon-carbon double bonds. Thus, they must be cycloalkanes. The possible structural isomers of cyclic C_5H_ 10 and the number of their monochloro structural isomers are as follows: 1. Cyclopentane: All carbon atoms are equivalent. Chlorination gives 1 monochloro product (chlorocyclopentane). 2. Methylcyclobutane: There are 4 different t

Question 16

One mole of an alkane (x) requires 8 mole oxygen for complete combustion. Sum of number of carbon and hydrogen atoms in the alkane (x) is _______.

Accepted Answer

The general formula of an alkane is C_n H_ 2n+2 . The balanced chemical equation for the complete combustion of an alkane is: C_n H_ 2n+2 + ( 3n+1 2 ) O_2 n CO_2 + (n+1) H_2O Given that 1 mole of the alkane requires 8 moles of oxygen for complete combustion, we equate the stoichiometric coefficient of O_2 to 8: 3n+1 2 = 8 3n + 1 = 16 3n = 15 n = 5 The molecular formula of the alkane is C_5H_ 12 . The number of carbon atoms is 5 and the number of hydrogen atoms is 12. Sum of the number of carbon

Question 17

Identify the correct statements : The presence of - NO _ 2 group in benzene ring A. activates the ring towards electrophilic substitutions. B. deactivates the ring towards electrophilic substitutions.

Accepted Answer

Presence of NO_2 group in Benzene ring deactivate ring towards electrophilic substitution reaction due to -M effective & activate ring towards nucleophilic substitution. So, B & C are correct.

Question 18

Ph - CH = CH _ 2 [ HBr ] ( PhCOO )_ 2 Product Consider the above reaction A. The reaction proceeds through a more stable radical intermediate. B. The role of peroxide is to generate H ^ (Hydrogen radi

Accepted Answer

Ph-CH=CH_2 [HBr] (PhCOO)_2 Ph-CH_2-CH_2-Br Anti Markovnikov addition • Reaction follow radical addition in presence of peroxide • In absence of peroxide follow carbocation mechanism • Benzene also formed

Question 19

Method used for separation of mixture of products (B and C) obtained in the following reaction is

Accepted Answer

Benzene reacts with Br _2 in the presence of FeBr _3 (Friedel-Crafts halogenation) to give bromobenzene. Bromobenzene on nitration with HNO _3 / H _2 SO _4 gives a mixture of ortho-bromonitrobenzene and para-bromonitrobenzene (since - Br is an ortho/para director). The two products (B and C) are separated by fractional distillation method due to their different boiling points.

Question 20

Given below are two statements for the following reaction sequence. Compound ' Z ' will give yellow precipitate with NaOI. Statement II: Compound ' Q ' has two different types of ' H ' atoms (aromatic

Accepted Answer

Both Statement I and Statement II are true.

Hydrocarbons — Previous Year Questions & Solutions