Online Test: Organic Chemistry: Some Basic Principles and Techniques | Class 11 Chemistry

Question 1 of 18

The shape around an $sp^2$ hybridised carbon is:

tetrahedral

trigonal planar

linear

pyramidal

Explanation: $sp^2$ carbon (one double bond) is trigonal planar with $120^\circ$ bond angles.

Question 2 of 18

In a bond-line (skeletal) formula, each line-end and vertex represents:

a hydrogen

a carbon

an oxygen

a bond pair

Explanation: Vertices and ends are carbons; the attached hydrogens are understood.

Question 3 of 18

The IUPAC name of $(CH_3)_2CHCH_2CH_3$ is:

pentane

2-methylbutane

neopentane

2-methylpropane

Explanation: Longest chain is 4 C (butane) with a methyl on C-2: 2-methylbutane.

Question 4 of 18

Ethanol and dimethyl ether ($C_2H_6O$) are:

chain isomers

position isomers

functional isomers

optical isomers

Explanation: They have the same formula but different functional groups (alcohol vs ether) — functional isomerism.

Question 5 of 18

Which molecule shows geometrical (cis–trans) isomerism?

but-1-ene

but-2-ene

propene

2-methylpropene

Explanation: Each doubly bonded carbon of but-2-ene bears two different groups, giving cis and trans forms.

Question 6 of 18

Successive members of a homologous series differ in molar mass by:

12 u

14 u

16 u

28 u

Explanation: They differ by a $CH_2$ unit, which is $12+2=14$ u.

Question 7 of 18

Heterolytic fission of $CH_3−Br$ most readily gives:

$CH_3^{\bullet}$ and $Br^{\bullet}$

$CH_3^+$ and $Br^-$

$CH_3^-$ and $Br^+$

two free radicals

Explanation: Bromine, being more electronegative, takes both electrons, giving $CH_3^+$ and $Br^-$.

Question 8 of 18

Which of the following is a nucleophile?

$NO_2^+$

$BF_3$

$CN^-$

$H^+$

Explanation: $CN^-$ is electron-rich and donates a lone pair, so it is a nucleophile.

Question 9 of 18

The correct order of carbocation stability is:

$CH_3^+ > 1^\circ > 2^\circ > 3^\circ$

$3^\circ > 2^\circ > 1^\circ > CH_3^+$

$1^\circ > 2^\circ > 3^\circ$

all equal

Explanation: Tertiary cations are most stabilised by +I and hyperconjugation: $3^\circ > 2^\circ > 1^\circ > CH_3^+$.

Question 10 of 18

The $−NO_2$ group exerts which inductive effect?

$+I$

$−I$

no inductive effect

+M only

Explanation: Nitro is strongly electron-withdrawing, so it shows a $−I$ effect.

Question 11 of 18

Hyperconjugation involves delocalisation of electrons from:

a $\pi$ bond into a $\sigma$ bond

$\sigma(C−H)$ into an adjacent empty $p$/$\pi$ orbital

a lone pair into a $\sigma$ bond

a $d$ orbital into a $p$ orbital

Explanation: Hyperconjugation is 'no-bond resonance' — $\sigma(C−H)$ electrons delocalise into an adjacent empty $p$ or $\pi$ orbital.

Question 12 of 18

Dehydrohalogenation of an alkyl halide to an alkene is a reaction of type:

addition

substitution

elimination

rearrangement

Explanation: A small molecule (HX) is removed and a double bond forms — an elimination reaction.

Question 13 of 18

The carboxylate ion $RCOO^-$ is stabilised mainly by:

hyperconjugation

resonance over two oxygens

the electromeric effect

hydrogen bonding only

Explanation: Its negative charge is delocalised equally over two oxygens by resonance.

Question 14 of 18

Camphor is best purified by:

distillation

crystallisation

sublimation

steam distillation

Explanation: Camphor sublimes (solid → vapour) on heating, so sublimation separates it from non-sublimable impurities.

Question 15 of 18

In Lassaigne's test, halogens are confirmed by a precipitate with:

sodium nitroprusside

$FeSO_4$

$AgNO_3$

lime-water

Explanation: $X^-$ ions give silver-halide precipitates with $AgNO_3$ (AgCl white, AgBr pale yellow, AgI yellow).

Question 16 of 18

On combustion $0.25\ \text{g}$ of a compound gave $0.44\ \text{g}$ $CO_2$. The $\%C$ is:

$36\%$

$44\%$

$48\%$

$60\%$

Explanation: $\%C=\dfrac{12}{44}\times\dfrac{0.44}{0.25}\times100=\dfrac{12}{44}\times176=48\%$.

Question 17 of 18

Nitrogen present as a nitro ($−NO_2$) group is best estimated by the:

Kjeldahl method

Dumas method

Carius method

Liebig method

Explanation: Kjeldahl fails for nitro/azo/ring nitrogen; Dumas oxidises any N to $N_2$ and is general.

Question 18 of 18

Glycerol is purified by distillation under reduced pressure because it:

sublimes

decomposes at its normal boiling point

is immiscible with water

is a gas at room temperature

Explanation: Lowering the pressure lowers its boiling point so it distils without decomposing.

Online Test — Organic Chemistry: Some Basic Principles and Techniques