Learn/ 12th Class Chemistry/ Unit 19 /Short Questions

Unit 19: Basics of Organic Chemistry — Short Questions

12th Class Chemistry · Unit 19: Basics of Organic Chemistry

Q2.a.Define enantiomers. Give enantiomers of 2-methylbutan-2-ol.
Q2.b.Alkanes have no functional group. Justify.

A functional group is an atom, group of atoms, or a double/triple bond that determines a compound's characteristic properties; its presence depends on bond polarity and bond order. Alkanes contain only C–C and C–H single bonds, which are essentially nonpolar, and they carry no charges, no lone-pair-bearing heteroatoms, and no double or triple bonds. Since they lack any such polar or multiple-bonded centre, alkanes (and cycloalkanes) possess no functional group.

Q2.c.a.Deduce the empirical and molecular formulae of the following compound: [a plain hexagon skeletal structure, i.e. a six-membered carbon ring with only single bonds shown, no substituents]

Each vertex of the hexagon is a carbon; since it is a saturated ring, each carbon completes its valency with 2 hydrogens (CH₂ units). Molecular formula: C₆H₁₂ (cyclohexane). Empirical formula: CH₂.

Q2.c.b.Deduce the empirical and molecular formulae of the following compound: [an open-chain zigzag skeletal structure with no branches, double bonds or substituents shown, i.e. a straight-chain saturated hydrocarbon]
Q2.c.c.Deduce the empirical and molecular formulae of the following compound: [a plain rectangle skeletal structure, i.e. a four-membered carbon ring with only single bonds shown, no substituents]

Each corner of the rectangle is a carbon; being a saturated ring, each carbon bears 2 hydrogens. Molecular formula: C₄H₈ (cyclobutane). Empirical formula: CH₂.

Q2.d.i.Give the systematic name of: CH₃-CH₂-CN

Propanenitrile.

Q2.d.ii.Give the systematic name of: CH₃COCH₂CH₂CH₃

Pentan-2-one.

Q2.d.iii.Give the systematic name of: CH₃CH₂COOCH₃

Methyl propanoate.

Q2.e.Define electrophile and nucleophile with two examples each.

Electrophile an atom or group of atoms that is able to accept a pair of electrons to form a new bond. Electrophiles are positive ions (including carbocations) or neutral electron-deficient molecules, e.g. H⁺ and BF₃. Nucleophile: an atom or group of atoms that is able to donate a pair of electrons to form a new bond. Nucleophiles are negative ions (including carbanions) or neutral molecules with lone pairs, e.g. OH⁻ and NH₃.

Q2.f.Pent-2-ene shows cis-trans isomerism but pent-1-ene does not. Give reason.

In pent-2-ene, CH₃-CH=CH-CH₂-CH₃, each double-bonded carbon carries two different groups (C-2: CH₃ and H; C-3: H and CH₂CH₃), satisfying the condition for cis-trans isomerism. In pent-1-ene, CH₂=CH-CH₂-CH₂-CH₃, the terminal double-bond carbon (C-1) bears two identical H atoms, so one of the two alkene carbons does not have two different groups — hence cis-trans isomerism is not possible.

Q2.g.i.Define: Constitutional isomerism

Constitutional (structural) isomerism occurs when compounds have the same molecular formula but the same groups or atoms are connected at different places, i.e. they differ in the bond connectivity of their substituents (e.g. chain, position, functional group isomerism).

Q2.g.ii.Define: Stereoisomerism

Stereoisomerism occurs when compounds have the same molecular formula and the same bond connectivity of substituents, but these substituents/groups have different spatial arrangement (configuration) in space, e.g. optical isomers and cis-trans (geometrical) isomers.

Q2.h.i.Define: racemic mixture

A racemic mixture is an equimolar (50:50) mixture of the (+) dextrorotatory and (-) levorotatory enantiomers of a chiral compound. Because the opposite rotations of the two enantiomers exactly cancel, the mixture as a whole is optically inactive. It is symbolised by placing (±) before the compound's name.

Q2.h.ii.Define: Optical activity

Optical activity is the property of a chiral (asymmetric) compound to rotate the plane of plane-polarised light when the light passes through its solution. A compound rotating the plane clockwise is dextrorotatory (+), and one rotating it anticlockwise is levorotatory (-); optical activity is measured using a polarimeter.

Q2.i.Explain why meso tartaric acid is optically inactive although it has two chiral carbon atoms.

Meso tartaric acid has two similar chiral carbon atoms and also possesses an internal plane of symmetry that divides the molecule into two equal halves, each half being the mirror image of the other. Each half rotates the plane-polarised light by an equal amount but in opposite directions. The rotation produced by one half is therefore exactly cancelled (internally compensated) by the rotation of the other half, so the molecule as a whole does not rotate plane-polarised light and is optically inactive, despite having two chiral carbons.

Q3.b.i.Draw the mirror image of the following compound. Label the molecule as chiral or achiral: 2-bromo-2-chloropropane

2-bromo-2-chloropropane is (CH₃)₂CBrCl. Its central carbon is bonded to Br, Cl, and two identical CH₃ groups, so it is not bonded to four different groups. [Mirror image: reflecting (CH₃)₂CBrCl through a plane gives a structure identical to, and superimposable on, the original, since two of the substituents (the two CH₃ groups) are the same.] The molecule is therefore achiral.

Q3.b.ii.Draw the mirror image of the following compound. Label the molecule as chiral or achiral: 1-bromo-1-chloroethane

1-bromo-1-chloroethane is CH₃CHBrCl. C-1 is bonded to four different groups: CH₃, H, Br and Cl. [Mirror image: with Br and Cl (and H) arranged oppositely in 3-D space about C-1, the reflected structure cannot be superimposed on the original by any rotation.] The molecule is therefore chiral, existing as a pair of non-superimposable enantiomers.

Q3.b.iii.Draw the mirror image of the following compound. Label the molecule as chiral or achiral: ethoxyethane

Ethoxyethane (diethyl ether) is CH₃CH₂-O-CH₂CH₃. It has no carbon bonded to four different groups (the molecule is symmetric about the oxygen). [Its mirror image is identical to, and superimposable on, the original structure.] The molecule is therefore achiral.

Q3.b.iv.Draw the mirror image of the following compound. Label the molecule as chiral or achiral: 2-aminopropanoic acid

2-aminopropanoic acid (alanine) is CH₃CH(NH₂)COOH. C-2 is bonded to four different groups: CH₃, H, NH₂ and COOH. [Mirror image: the two spatial arrangements of NH₂, H, CH₃ and COOH about C-2 are non-superimposable.] The molecule is therefore chiral, existing as a pair of enantiomers.

Q3.c.What is the alkane of lowest relative molecular mass that contains a chiral centre? Give its systematic name and molecular formula. Also draw mirror image of this compound.

3-methylhexane, CH₃CH₂CH(CH₃)CH₂CH₂CH₃, molecular formula C₇H₁₆, is the smallest (lowest molar mass) alkane with a chiral centre. Its C-3 is bonded to four different groups: H, CH₃ (branch), C₂H₅ (ethyl, towards C-1–C-2) and C₃H₇ (propyl, towards C-4–C-5–C-6). [Its mirror image, with these four groups arranged oppositely in space about C-3, is non-superimposable on the original — the molecule is chiral, existing as a pair of enantiomers.] No C₆H₁₄ (hexane) isomer has a true chiral centre, because in every branched hexane isomer the substituted carbon always ends up bonded to two identical groups (e.g. in 3-methylpentane, C-3 carries two identical ethyl groups).

Q3.d.3-chlorobutan-2-ol is an optically active compound. Draw all its possible optical isomers. Is there any meso form of this compound? Give reason.

3-chlorobutan-2-ol, CH₃-CH(OH)-CH(Cl)-CH₃, has two chiral carbons: C-2 (bonded to OH, H, CH₃, CHClCH₃) and C-3 (bonded to Cl, H, CH₃, CH(OH)CH₃). Since these two chiral centres are not similar (C-2 bears OH while C-3 bears Cl, so the two halves of the molecule are different), the full 2ⁿ = 2² = 4 stereoisomers are possible: (2R,3R) and (2S,3S) — a pair of enantiomers — and (2R,3S) and (2S,3R) — another pair of enantiomers (these two pairs are diastereomers of each other). There is no meso form, because a meso form requires two similar chiral carbons related by an internal plane of symmetry; here the two chiral carbons carry different substituents (OH vs Cl), so no such symmetry plane exists and the rotation contributed by the two stereocentres cannot internally cancel.