Question 1

Which of the following reaction steps is known as propagation step

Accepted Answer

In a free radical substitution reaction, the mechanism involves three types of steps: 1. Initiation step: A neutral molecule breaks down to form free radicals. Cl-Cl Cl ^ + Cl ^ 2. Propagation step: A free radical reacts with a neutral molecule to generate a new free radical and a new neutral molecule. This step keeps the chain reaction going. CH _3^ + Cl _2 CH _3 -Cl + Cl ^ 3. Termination step: Two free radicals combine to form a neutral molecule, stopping the chain reaction. CH _3^ + Cl ^ CH _

Question 2

Which of the following alkenes on oxidation by KMnO _4 in dil H _2 SO _4 forms adipic acid ?

Accepted Answer

Oxidation of alkenes with acidic KMnO _4 leads to the cleavage of the carbon-carbon double bond, converting the vinylic carbons into carboxylic acids, ketones, or CO _2 depending on the substitution. Cyclohexene is a cyclic alkene with a six-membered ring. Upon oxidative cleavage, the ring opens to form a straight-chain dicarboxylic acid with six carbon atoms. The reaction is: Cyclohexene KMnO _4, H ^+ HOOC -( CH _2)_4- COOH (Adipic acid). Hex-1-ene gives pentanoic acid and CO _2. Hex-2-ene give

Question 3

Identify the major product of following reaction. Propene [ Peroxide ] HBr \,?

Accepted Answer

The reaction of propene with HBr in the presence of peroxide proceeds via a free radical mechanism, known as the peroxide effect or anti-Markovnikov addition. The peroxide undergoes homolytic cleavage to form alkoxy radicals, which then abstract a hydrogen atom from HBr to generate a bromine radical (Br^ ). The bromine radical adds to the less substituted carbon atom of the double bond in propene to generate a more stable secondary free radical: CH_3-CH=CH_2 + Br^ CH_3- C H-CH_2Br The secondary

Question 4

Which of the following hydrocarbons has stronger London dispersion forces ?

Accepted Answer

London dispersion forces increase with an increase in molecular mass and the surface area of the molecule. Methane (CH_4) and ethane (C_2H_6) have lower molecular masses compared to n-pentane and neo-pentane (C_5H_ 12 ), resulting in weaker dispersion forces. Between n-pentane and neo-pentane, n-pentane is a straight-chain alkane, whereas neo-pentane is highly branched and nearly spherical. The straight-chain structure of n-pentane provides a larger surface area for intermolecular contact compar

Question 5

The major product of the following reaction is

Accepted Answer

The given reactant is 3-methylhex-3-ene. The addition of HI to an alkene is an electrophilic addition reaction that proceeds via the formation of a carbocation intermediate. First, the electrophile H^+ attacks the double bond. This can result in two possible carbocations: 1. Protonation at C-4 yields a tertiary carbocation at C-3. 2. Protonation at C-3 yields a secondary carbocation at C-4. Since a tertiary carbocation is significantly more stable than a secondary carbocation due to greater hype

Question 6

Which of the following is NOT terminal alkyne ?

Accepted Answer

Terminal alkynes have a triple bond at the end of the carbon chain, meaning at least one hydrogen atom is directly attached to a triply bonded carbon atom (R-C C-H). Ethyne is H-C C-H, which is a terminal alkyne. Propyne is CH_3-C C-H, which is a terminal alkyne. But-1-yne is CH_3-CH_2-C C-H, which is a terminal alkyne. But-2-yne is CH_3-C C-CH_3, where the triple bond is between the second and third carbon atoms. It does not have a hydrogen atom attached to the triply bonded carbon atoms. Thus,

Question 7

Which of the following alkyl chlorides on Wurtz reaction gives 2, 2, 5, 5- tetramethylhexane ?

Accepted Answer

Wurtz reaction involves the coupling of two alkyl halide molecules in the presence of sodium and dry ether to form a symmetrical alkane. 2 R-Cl + 2 Na dry ether R-R + 2 NaCl The given product is 2,2,5,5-tetramethylhexane. Its structure is ( CH _3)_3 C - CH _2- CH _2- C ( CH _3)_3. Cleaving the molecule symmetrically at the central C - C bond gives two identical alkyl groups, which are neopentyl groups, ( CH _3)_3 C - CH _2-. Therefore, the required alkyl chloride is neopentyl chloride, ( CH _3)_

Question 8

Identify the 'A' in the following reaction. CaC _2 + H _2 O A + Ca(OH) _2

Accepted Answer

Calcium carbide reacts with water to form ethyne and calcium hydroxide. The balanced chemical equation is: CaC _2 + 2 H _2 O C _2 H _2 + Ca(OH) _2 Thus, the product 'A' is ethyne ( C _2 H _2). Answer: Ethyne

Question 9

Identify the product 'B' in the following reaction Toluene Chromyl chloride / CS _2 A H _3 O ^+ B

Accepted Answer

The given reaction is the Etard reaction. Toluene reacts with chromyl chloride ( CrO _2 Cl _2) in a non-polar solvent like CS _2 to form a brown chromium complex A. The chromium complex A on hydrolysis with H _3 O ^+ gives benzaldehyde B. Therefore, the product B is benzaldehyde. Answer: Benzaldehyde

Question 10

Alkenes on oxidation with KMnO _4 in dil H _2 SO _4 forms

Accepted Answer

Alkenes undergo oxidative cleavage when treated with a strong oxidizing agent like KMnO _4 in acidic medium (dilute H _2 SO _4). During this reaction, the carbon-carbon double bond is completely cleaved. A = CH - group is initially oxidized to an aldehyde (alkanal), which is highly susceptible to further oxidation under these conditions and immediately oxidizes to a carboxylic acid (alkanoic acid). For example: R - CH = CH - R ' KMnO _4 / H ^+ R - COOH + R '- COOH Thus, the final stable product

Question 11

When excess of chlorine is used in the chlorination of methane find the major product obtained.

Accepted Answer

Chlorination of methane is a free radical substitution reaction. When an excess of chlorine is used, all the hydrogen atoms of methane are successively replaced by chlorine atoms. The sequence of reactions is: CH_4 + Cl_2 CH_3Cl + HCl CH_3Cl + Cl_2 CH_2Cl_2 + HCl CH_2Cl_2 + Cl_2 CHCl_3 + HCl CHCl_3 + Cl_2 CCl_4 + HCl Thus, the final major product obtained is tetrachloromethane (CCl_4). Answer: Tetrachloromethane

Question 12

Identify the use of Baeyer's reagent by the following.

Accepted Answer

Baeyer's reagent is a cold, dilute, alkaline solution of potassium permanganate (KMnO_4). It is a strong oxidizing agent, but its primary application in qualitative organic analysis is to test for unsaturation. When Baeyer's reagent reacts with an organic compound containing a carbon-carbon double bond (alkene) or triple bond (alkyne), the purple colour of the permanganate solution disappears and a brown precipitate of manganese dioxide (MnO_2) is formed. Therefore, it is used for detecting the

Question 13

Identify the catalyst used in the Friedel crafts acylation reaction.

Accepted Answer

Friedel-Crafts acylation is an electrophilic aromatic substitution reaction in which an acyl group is introduced into an aromatic ring. The reaction requires a strong Lewis acid catalyst to facilitate the generation of the electrophilic acylium ion from an acyl halide or acid anhydride. Anhydrous AlCl _3 is the standard and most widely used Lewis acid catalyst for this purpose. The general reaction is: ArH + RCOCl Anhydrous AlCl _3 ArCOR + HCl Answer: Anhydrous AlCl _3

Question 14

Which of the following alkanes is NOT obtained when a mixture of ethyl iodide and n-propyl iodide is subjected to Wurtz reaction?

Accepted Answer

In the Wurtz reaction, alkyl halides react with sodium in dry ether to form alkanes through the coupling of alkyl radicals. The given mixture contains ethyl iodide (C_2H_5I) and n-propyl iodide (C_3H_7I). The free radicals generated are the ethyl radical (C_2H_5^ ) and the n-propyl radical (C_3H_7^ ). The possible coupling products are: 1. Self-coupling of ethyl radicals: C_2H_5^ + C_2H_5^ C_4H_ 10 (n-Butane) 2. Cross-coupling of ethyl and n-propyl radicals: C_2H_5^ + C_3H_7^ C_5H_ 12 (n-Pentane

Question 15

Which from following alkenes is highly stabilized ?

Accepted Answer

The stability of an alkene increases with the number of alkyl groups attached to the double-bonded carbon atoms due to hyperconjugation and the +I effect. The order of stability of alkenes based on substitution is: Tetrasubstituted > Trisubstituted > Disubstituted > Monosubstituted. Among the given options, R _2 C = CR _2 is a tetrasubstituted alkene, having the maximum number of alkyl groups attached to the double bond. Thus, it has the maximum number of hyperconjugative structures and is the m

Question 16

What are the reactants involved in Friedal Crafts alkylation?

Accepted Answer

Friedel-Crafts alkylation is an electrophilic aromatic substitution reaction in which an alkyl group is introduced into an aromatic ring. The typical reactants are an aromatic compound, such as benzene ( C _6 H _6), and an alkyl halide, such as methyl chloride ( CH _3 Cl ). The reaction is carried out in the presence of a Lewis acid catalyst like anhydrous AlCl _3. C _6 H _6 + CH _3 Cl Anhydrous AlCl _3 C _6 H _5 CH _3 + HCl The reaction of benzene with acetyl chloride ( CH _3 COCl ) corresponds

Question 17

Identify the product of reforming reaction of n-hexane in the presence of V _2 O _5 over alumina.

Accepted Answer

Alkanes containing six or more carbon atoms on heating to 773 K at 10-20 atm pressure in the presence of oxides of vanadium, molybdenum or chromium supported over alumina get dehydrogenated and cyclised to benzene and its homologues. This reaction is known as aromatization or reforming. When n-hexane is passed over V _2 O _5 supported on alumina, it undergoes aromatization to form benzene. CH _3-( CH _2)_4- CH _3 V _2 O _5 / Al _2 O _3, 773 K , 10-20 atm C _6 H _6 + 4 H _2 Answer: Benzene

Question 18

When a mixture of two different alkyl halides reacts with metallic sodium in dry ether, the formation of a possible number of alkanes are-

Accepted Answer

Let the two different alkyl halides be R-X and R'-X. When this mixture reacts with metallic sodium in dry ether (Wurtz reaction), three types of coupling reactions take place. R-X + 2Na + X-R R-R + 2NaX R'-X + 2Na + X-R' R'-R' + 2NaX R-X + 2Na + X-R' R-R' + 2NaX Thus, a total of three different alkanes are formed. Answer: Three

Question 19

Which one of the following compound will show addition of HBr to alkene according to markownikoff's rule?

Accepted Answer

Markovnikov's rule is applicable for the addition of unsymmetrical reagents to unsymmetrical alkenes. But-2-ene (CH_3-CH=CH-CH_3) is a symmetrical alkene. Hex-3-ene (CH_3-CH_2-CH=CH-CH_2-CH_3) is a symmetrical alkene. Ethene (CH_2=CH_2) is a symmetrical alkene. 2-methylpropene (CH_3-C(CH_3)=CH_2) is an unsymmetrical alkene. Therefore, 2-methylpropene will undergo addition of HBr according to Markovnikov's rule.

Question 20

Identify the compound formed when But-2-ene is treated with KMnO _4 in dilute H _2 SO _4.

Accepted Answer

But-2-ene has the structure CH _3- CH = CH - CH _3. When treated with acidic KMnO _4 ( KMnO _4 in dilute H _2 SO _4), it undergoes oxidative cleavage. The strong oxidizing agent completely breaks the carbon-carbon double bond, oxidizing each of the resulting fragments into a carboxylic acid. CH _3- CH = CH - CH _3 KMnO _4, H ^+ 2 CH _3 COOH The product formed is acetic acid (ethanoic acid). Answer: Acetic acid

Hydrocarbons — Previous Year Questions & Solutions