Basic Organic Chemistry and Hydrocarbons — Practice

Question 1

The most stable carbocation among the following is:
(a) (CH₃)₃C⁺ (b) CH₃CH₂⁺ (c) CH₃⁺ (d) (CH₃)₂CH⁺

Accepted Answer

(a) (CH₃)₃C⁺.

The tertiary carbocation is stabilised by three electron-donating alkyl groups (+I) and the most hyperconjugation, so stability runs 3° > 2° > 1° > methyl.

Question 2

Addition of HBr to propene (in the absence of peroxide) gives mainly:
(a) 2-bromopropane (b) 1-bromopropane (c) 1,2-dibromopropane (d) propane

Accepted Answer

(a) 2-bromopropane.

By Markovnikov's rule, H adds to the terminal CH₂ (more hydrogens) and Br to the middle carbon, via the more stable secondary carbocation → 2-bromopropane.

Question 3

Addition of HBr to propene in the presence of peroxide gives:
(a) 1-bromopropane (b) 2-bromopropane (c) propene oxide (d) no reaction

Accepted Answer

(a) 1-bromopropane.

The peroxide (Kharasch) effect reverses the regiochemistry by a free-radical mechanism, putting Br on the terminal carbon — anti-Markovnikov. This works only with HBr.

Question 4

Which of the following is the most acidic C–H?
(a) HC≡CH (b) H₂C=CH₂ (c) CH₃–CH₃ (d) CH₄

Accepted Answer

(a) HC≡CH.

The sp carbon in ethyne has 50% s-character, holding the C–H electrons closest to carbon, so its hydrogen is the most acidic: sp > sp² > sp³.

Question 5

Which reagent is decolourised by an alkene, serving as a test for unsaturation?
(a) bromine water (b) sodium chloride (c) dilute HCl (d) sodium bicarbonate

Accepted Answer

(a) bromine water.

An alkene (or alkyne) adds bromine across the multiple bond, decolourising the orange-brown bromine water — a classic test for unsaturation.

Question 6

According to Hückel's rule, a compound is aromatic if it is cyclic, planar, fully conjugated and has:
(a) (4n+2) π electrons (b) 4n π electrons (c) any even number of electrons (d) 6 sigma bonds

Accepted Answer

(a) (4n+2) π electrons.

Hückel's rule requires (4n+2) delocalised π electrons (n = 0, 1, 2…). Benzene, with 6 π electrons (n = 1), is the classic aromatic system.

Question 7

The nitration of nitrobenzene gives mainly:
(a) 1,3-dinitrobenzene (meta) (b) 1,2-dinitrobenzene (ortho) (c) 1,4-dinitrobenzene (para) (d) no reaction

Accepted Answer

(a) 1,3-dinitrobenzene (meta).

The –NO₂ group is electron-withdrawing and a meta director, so the second nitro group enters the meta position (and the ring reacts more slowly, being deactivated).

Question 8

Phenol undergoes electrophilic substitution faster than benzene because the –OH group is:
(a) activating and ortho/para directing (b) deactivating and meta directing (c) inert (d) meta directing

Accepted Answer

(a) activating and ortho/para directing.

The –OH group donates electron density into the ring (+M), activating it and directing incoming electrophiles to the ortho and para positions.

Question 9

Chlorobenzene, on nitration, gives mainly ortho and para products, yet chlorine is:
(a) deactivating (b) strongly activating (c) meta directing (d) inert

Accepted Answer

(a) deactivating.

Halogens are the exception: they are ortho/para directing (through +M donation of lone pairs) but overall deactivating (through strong −I withdrawal), so the ring reacts slower than benzene but still at o/p positions.

Question 10

The chlorination of methane to chloromethane proceeds by a:
(a) free-radical mechanism (b) electrophilic addition (c) nucleophilic substitution (d) elimination

Accepted Answer

(a) free-radical mechanism.

Methane's chlorination in UV light goes through initiation (Cl₂ → 2Cl•), propagation and termination steps — a free-radical chain reaction.

Question 11

Which compound shows optical isomerism?
(a) butan-2-ol (b) propan-1-ol (c) propan-2-ol (d) ethanol

Accepted Answer

(a) butan-2-ol.

C-2 of butan-2-ol carries four different groups (H, OH, CH₃, C₂H₅), making it a chiral centre — so the molecule exists as two optically active enantiomers.

Question 12

Which of the following shows cis–trans (geometrical) isomerism?
(a) 2-butene (b) 1-butene (c) propene (d) ethene

Accepted Answer

(a) 2-butene.

Each double-bond carbon of 2-butene carries a hydrogen and a methyl group (two different groups), so cis and trans forms exist. Terminal alkenes like 1-butene and propene cannot.

Question 13

In IUPAC naming, which functional group has the highest priority for the suffix?
(a) –COOH (b) –OH (c) –CHO (d) –NH₂

Accepted Answer

(a) –COOH.

Carboxylic acid outranks aldehyde, ketone, alcohol and amine, so it takes the principal suffix and the lowest locant.

Question 14

The greater stability of a tertiary carbocation compared with a primary one is best explained by:
(a) more hyperconjugation and +I effect (b) resonance in the ring (c) higher electronegativity (d) hydrogen bonding

Accepted Answer

(a) more hyperconjugation and +I effect.

Three alkyl groups on a tertiary carbocation donate electron density (+I) and provide many α C–H bonds for hyperconjugation, dispersing the positive charge and stabilising it.

Question 15

Ozonolysis of an alkene (O₃ followed by Zn/H₂O) cleaves the C=C to give:
(a) two carbonyl compounds (b) an alcohol (c) a dihalide (d) a carboxylic acid only

Accepted Answer

(a) two carbonyl compounds.

Ozonolysis breaks the double bond and, with reductive work-up (Zn/H₂O), gives aldehydes and/or ketones — used to locate the position of the double bond.

Practice — Basic Organic Chemistry and Hydrocarbons

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