JEE Chemistry Hydrocarbons MCQs Set 21

Answer: (b) Oxalic acid

Explanation:
1. Both carbons of the triple bond are oxidised to carboxyl groups
2. Ethyne gives oxalic acid, \( \ce{HOOC-COOH} \)

Teacher's Note:
1. Cold dilute alkaline KMnO4 gives a diol from an alkene, while hot acidic KMnO4 cleaves the double bond.
2. Terminal alkynes are separated from other hydrocarbons by their acidic hydrogen.

Answer: (c) Ammoniacal cuprous chloride

Explanation:
1. Both gases decolourise permanganate and bromine
2. Only acetylene gives the red copper(I) acetylide precipitate

Teacher's Note:
1. Draw the possible intermediates and pick the most stable one (tertiary, then secondary, then primary).
2. Only HBr shows the peroxide effect, not HCl or HI.

Answer: (c) \( \ce{Ag(NH3)2^+} \)

Explanation:
1. Tollens' reagent gives a white precipitate of the silver alkynide
2. The other tests detect unsaturation in general

Teacher's Note:
1. Decide which functional groups can react with the reagent and which cannot.
2. A test that works for both compounds cannot distinguish them.

Answer: (a) HCN

Explanation:
1. At the high temperature of the arc \( \ce{C2H2} \) and \( \ce{N2} \) combine
2. \( \ce{C2H2 + N2 -> 2 HCN} \)

Teacher's Note:
1. Terminal alkynes have an acidic sp C-H bond (50% s-character) and give precipitates with ammoniacal AgNO3 and Cu2Cl2.
2. Mercuric hydration of an alkyne gives a ketone, except ethyne which gives ethanal.

Answer: (b) Glyoxal

Explanation:
1. Ozone adds to the triple bond and the ozonide is cleaved
2. This gives ethanedial (glyoxal), \( \ce{OHC-CHO} \), as the product from the two-carbon alkyne

Teacher's Note:
1. Ozonolysis splits C=C into two carbonyl groups, so join the carbonyl carbons to find the alkene.
2. Hydroboration-oxidation gives the anti-Markovnikov alcohol, while oxymercuration gives the Markovnikov alcohol.

Answer: (a) Propyne

Explanation:
1. The sp carbon is only present in propyne
2. The sp-sp3 single bond in propyne is shorter than other C-C bonds in the list

Teacher's Note:
1. The more s-character, the shorter and stronger the bond and the more acidic the C-H.
2. Li or Na in liquid NH3 gives the trans alkene, while Lindlar's catalyst gives cis.

Answer: (b) Silver acetylide

Explanation:
1. The acidic hydrogens are replaced by silver
2. The white precipitate is silver acetylide, \( \ce{Ag-C#C-Ag} \)

Teacher's Note:
1. Terminal alkynes have an acidic sp C-H bond (50% s-character) and give precipitates with ammoniacal AgNO3 and Cu2Cl2.
2. Mercuric hydration of an alkyne gives a ketone, except ethyne which gives ethanal.

Answer: (b) \( \ce{C2H2} \)

Explanation:
1. Ethyne adds two molecules of HCl in a Markovnikov way
2. \( \ce{HC#CH + 2 HCl -> CH3-CHCl2} \)

Teacher's Note:
1. Draw the possible intermediates and pick the most stable one (tertiary, then secondary, then primary).
2. Only HBr shows the peroxide effect, not HCl or HI.

Answer: (b) \( \ce{CuCl/NH4Cl} \) and \( \ce{H2/Pd(BaSO4)} \)

Explanation:
1. \( \ce{CuCl/NH4Cl} \) dimerises ethyne to vinylacetylene
2. Partial hydrogenation of the triple bond over poisoned palladium gives \( \ce{CH2=CH-CH=CH2} \)

Teacher's Note:
1. Identify the functional group that is removed or formed and count the carbon atoms of the product.
2. Wurtz and Kolbe couple two alkyl groups, soda lime removes one carbon, and Grignard reagents pick up an acidic proton.

Answer: (a) Ethyne

Explanation:
1. Three ethyne molecules combine in a red hot tube
2. The cyclic trimer is benzene

Teacher's Note:
1. Use Huckel's rule (4n + 2 pi electrons in a planar cyclic conjugated ring) to judge aromaticity.
2. Side chain reactions in light go through radicals, while ring reactions with a Lewis acid go through cations.

Answer: (a) \( \ce{CH3-CH(CH3COO)2} \)

Explanation:
1. Acetic acid adds twice across the triple bond
2. The product is ethylidene diacetate, \( \ce{CH3CH(OCOCH3)2} \)

Teacher's Note:
1. Ozonolysis splits C=C into two carbonyl groups, so join the carbonyl carbons to find the alkene.
2. Hydroboration-oxidation gives the anti-Markovnikov alcohol, while oxymercuration gives the Markovnikov alcohol.

Answer: (c) Acetaldehyde

Explanation:
1. Hydration via the enol \( \ce{CH2=CHOH} \) gives ethanal
2. The product is acetaldehyde

Teacher's Note:
1. Terminal alkynes have an acidic sp C-H bond (50% s-character) and give precipitates with ammoniacal AgNO3 and Cu2Cl2.
2. Mercuric hydration of an alkyne gives a ketone, except ethyne which gives ethanal.

Answer: (a) Ethyne

Explanation:
1. The sp carbon of ethyne is the most electronegative among the choices
2. So the C-H bond is the most acidic

Teacher's Note:
1. The more s-character, the shorter and stronger the bond and the more acidic the C-H.
2. Li or Na in liquid NH3 gives the trans alkene, while Lindlar's catalyst gives cis.

Answer: (d) All of these

Explanation:
1. Each methyl on the ring is oxidised to \( \ce{-COOH} \)
2. para-Xylene gives terephthalic, ortho gives phthalic and meta gives isophthalic acid

Teacher's Note:
1. Cold dilute alkaline KMnO4 gives a diol from an alkene, while hot acidic KMnO4 cleaves the double bond.
2. Terminal alkynes are separated from other hydrocarbons by their acidic hydrogen.

Answer: (c)

Explanation:
1. Acetylide adds to acetone to give \( \ce{HC#C-C(CH3)2-OH} \) (X)
2. Mercuric hydration is Markovnikov, giving a methyl ketone, \( \ce{CH3-CO-C(CH3)2-OH} \)

Teacher's Note:
1. Electrophilic addition goes through the more stable carbocation, so H adds to the carbon with more hydrogens.
2. With HBr and a peroxide the radical path reverses the orientation.

Answer: (c) Ethyne

Explanation:
1. Bond length decreases from single to double to triple bond
2. Ethyne (120 pm) is shortest

Teacher's Note:
1. Decide which functional groups can react with the reagent and which cannot.
2. A test that works for both compounds cannot distinguish them.

Answer: (a) Ammoniacal silver nitrate

Explanation:
1. Bromine water and Baeyer's reagent only show unsaturation
2. A white precipitate with ammoniacal \( \ce{AgNO3} \) specifically shows a terminal alkyne

Teacher's Note:
1. Draw the possible intermediates and pick the most stable one (tertiary, then secondary, then primary).
2. Only HBr shows the peroxide effect, not HCl or HI.

Answer: (c) NaOH

Explanation:
1. Acetylene is a very weak acid and is not deprotonated by NaOH
2. It reacts with Na, ammoniacal silver nitrate and HCl

Teacher's Note:
1. Ozonolysis splits C=C into two carbonyl groups, so join the carbonyl carbons to find the alkene.
2. Hydroboration-oxidation gives the anti-Markovnikov alcohol, while oxymercuration gives the Markovnikov alcohol.

Answer: (d) \( \ce{Cl2CHCHO} \)

Explanation:
1. \( \ce{HOCl} \) adds twice across the triple bond
2. The geminal chlorohydrin loses water to give dichloroacetaldehyde, \( \ce{Cl2CH-CHO} \)

Teacher's Note:
1. Electrophilic addition goes through the more stable carbocation, so H adds to the carbon with more hydrogens.
2. With HBr and a peroxide the radical path reverses the orientation.

Answer: (a) 1

Explanation:
1. Only the terminal sp C-H of \( \ce{CH3CH2C#CH} \) is acidic
2. So one mole of protons is given

Teacher's Note:
1. The more s-character, the shorter and stronger the bond and the more acidic the C-H.
2. Li or Na in liquid NH3 gives the trans alkene, while Lindlar's catalyst gives cis.

Answer: (a) \( \ce{CH3-CH=CH-C#CH} \)

Explanation:
1. Only a compound with a terminal alkyne reacts to form the silver derivative
2. \( \ce{CH3-CH=CH-C#CH} \) (pent-3-en-1-yne) is the only one

Teacher's Note:
1. Terminal alkynes have an acidic sp C-H bond (50% s-character) and give precipitates with ammoniacal AgNO3 and Cu2Cl2.
2. Mercuric hydration of an alkyne gives a ketone, except ethyne which gives ethanal.

Answer: (b) \( \ce{HgSO4} \)

Explanation:
1. Mercuric sulphate is the catalyst for hydration of alkynes
2. The product is acetaldehyde via the enol

Teacher's Note:
1. The more s-character, the shorter and stronger the bond and the more acidic the C-H.
2. Li or Na in liquid NH3 gives the trans alkene, while Lindlar's catalyst gives cis.

Answer: (b) \( \ce{CH3-C#C-CH3} \)

Explanation:
1. A white precipitate needs an acidic terminal C-H of an alkyne
2. But-2-yne is an internal alkyne and has none

Teacher's Note:
1. Terminal alkynes have an acidic sp C-H bond (50% s-character) and give precipitates with ammoniacal AgNO3 and Cu2Cl2.
2. Mercuric hydration of an alkyne gives a ketone, except ethyne which gives ethanal.

Answer: (b) Alkyne

Explanation:
1. Alkanes are \( \ce{CnH2n+2} \) and alkenes are \( \ce{CnH2n} \)
2. Alkynes have \( \ce{CnH2n-2} \)

Teacher's Note:
1. The more s-character, the shorter and stronger the bond and the more acidic the C-H.
2. Li or Na in liquid NH3 gives the trans alkene, while Lindlar's catalyst gives cis.

Answer: (a) (E)-2,3-Dichloropent-2-ene

Explanation:
1. One mole of chlorine adds to the triple bond to give a dichloroalkene
2. The addition is anti, so the two Cl atoms are on opposite sides, which is the E isomer
3. Tetrachloride needs two moles

Teacher's Note:
1. Ozonolysis splits C=C into two carbonyl groups, so join the carbonyl carbons to find the alkene.
2. Hydroboration-oxidation gives the anti-Markovnikov alcohol, while oxymercuration gives the Markovnikov alcohol.