JEE Chemistry Hydrocarbons MCQs Set 22

Answer: (b) 3-Methylbut-1-yne

Explanation:
1. A white precipitate shows a terminal alkyne
2. Oxidation of \( \ce{(CH3)2CH-C#CH} \) cleaves the triple bond to give \( \ce{(CH3)2CHCOOH} \)
3. Pent-1-yne would give butanoic 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: (d) 2-Butyne

Explanation:
1. Alkenes follow \( \ce{CnH2n} \)
2. 2-Butyne is an alkyne, \( \ce{C4H6} \)

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: (c) trans-2-Butene

Explanation:
1. Dissolving metal reduction (Li or Na in liquid ammonia) adds H anti
2. The trans alkene is formed

Teacher's Note:
1. Write the balanced equation first and track every carbon.
2. Check whether the reagent gives elimination (alcoholic KOH) or substitution (aqueous KOH).

Answer: (b) \( \ce{[Ag(NH3)2]+} \)

Explanation:
1. Pent-1-yne gives a white silver alkynide
2. Pent-1-ene gives no precipitate, while the other reagents react with both

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) Ethane and ethene

Explanation:
1. Only ethyne forms the silver salt and is removed
2. Ethane and ethene pass through

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) Acidic property

Explanation:
1. Acetylene gives up its hydrogen as \( \ce{H+} \) and forms silver acetylide
2. This shows acidic property

Teacher's Note:
1. Aromatic rings keep their stability by electrophilic substitution, not addition.
2. Activating groups direct ortho and para, while strongly withdrawing groups direct meta.

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

Explanation:
1. The first member of the alkyne series is ethyne
2. Its formula is \( \ce{C2H2} \)

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) \( \ce{sp^3} \), \( \ce{sp^2} \) and sp

Explanation:
1. Methane has four sigma bonds, ethene three sigma bonds and ethyne two sigma bonds per carbon
2. These correspond to sp3, sp2 and sp

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: (c) 1

Explanation:
1. Only the hydrogen on the sp carbon is acidic
2. \( \ce{CH3CH2C#CH} \) has one such hydrogen

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: (c) Acetylene

Explanation:
1. Each carbon in ethyne is sp hybridised with bond angle 180 degrees
2. So the molecule is linear

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: (c) Acetylene

Explanation:
1. \( \ce{CaC2 + 2 H2O -> Ca(OH)2 + C2H2} \)
2. Acetylene is formed

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: (c) Vinyl cyanide

Explanation:
1. \( \ce{HC#CH + HCN -> CH2=CH-CN} \)
2. Vinyl cyanide (acrylonitrile) is the product

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) \( \ce{C2H2} \)

Explanation:
1. Ethyne has acidic hydrogen
2. \( \ce{2 HC#CH + 2 Na -> 2 HC#CNa + H2} \)

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: (c) Both \( \ce{CH3COOH} \) and \( \ce{HCOOH} \)

Explanation:
1. Each carbon of the triple bond becomes a carboxylic acid group
2. The methyl end gives acetic acid and the terminal CH end gives formic acid

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) One

Explanation:
1. The product is ethanal, \( \ce{CH3CHO} \)
2. It has only the pi bond of the C=O group

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) \( \ce{CH3-C#CH} \)

Explanation:
1. Propyne has an acidic hydrogen on the sp carbon
2. The others have no sp C-H bonds and are far less acidic

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: (d) All of these

Explanation:
1. Alkynes add reagents across the triple bond and polymerise
2. Terminal alkynes also undergo substitution of the acidic hydrogen

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) \( \ce{CHI3 + 6 Ag} \) heated

Explanation:
1. Heating iodoform with silver powder removes iodine: \( \ce{2 CHI3 + 6 Ag -> HC#CH + 6 AgI} \)
2. Potassium formate gives \( \ce{H2} \), ethanol gives ethene and \( \ce{Be2C} \) gives methane

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) \( \ce{Li-NH3/C2H5OH} \)

Explanation:
1. Dissolving metal reduction gives anti addition of \( \ce{H2} \)
2. This produces the trans alkene
3. Lindlar-type catalysts give cis, and \( \ce{PtO2} \) gives the alkane

Teacher's Note:
1. Link each named reaction to its reagent and the change in carbon number.
2. Elimination gives alkenes, reduction gives alkanes, and coupling doubles the alkyl chain.

Answer: (b) Ethylene is formed

Explanation:
1. Lindlar's catalyst is poisoned palladium
2. It stops the hydrogenation at the alkene stage, so ethylene is formed

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) One sigma bond and two pi bonds

Explanation:
1. The C#C triple bond has one sp-sp sigma bond
2. and two perpendicular pi bonds

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) Propyne

Explanation:
1. Only a terminal alkyne has an acidic C-H
2. Propyne gives a white precipitate of silver propynide

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: (d) \( \ce{CH3CH(Br)CH3} \)

Explanation:
1. Hydration of ethyne gives ethanal, and the Grignard reagent adds to it to give propan-2-ol
2. \( \ce{P/Br2} \) replaces OH by Br, giving 2-bromopropane

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) Produce an electrophile

Explanation:
1. \( \ce{AlCl3} \) is a Lewis acid
2. It removes halide from the alkyl or acyl halide, generating the electrophile \( \ce{R+} \) or \( \ce{RCO+} \)

Teacher's Note:
1. Work out the electrophile first (NO2+, SO3, R+, RCO+, Cl+) and then where it attacks.
2. Alkyl groups on the ring raise reactivity, while nitro and carbonyl groups lower it.

Answer: (c) \( \ce{C6H5COCH3} \)

Explanation:
1. This is Friedel-Crafts acylation
2. The acylium ion \( \ce{CH3CO+} \) substitutes a ring hydrogen, giving acetophenone

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.