Q.Identify chiral and achiral molecules in each of the following pair of compounds. (Wedge and Dash representations according to Class XI.)

Showing the 12 most recent of 13 on this concept.

1. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.Which of the following are chiral molecules? Pentan-3-ol (I) 3-Methylheptane (II) 3-Bromo-3-methylpentane (III) 3-Bromo-2-methylpentane (IV) Correct answer is (only = only) (A) I, II, III only (B) II, IV only (C) II, III only (D) I, II only

   ›Reveal solutionSolution

   A carbon is a stereocentre (making the molecule chiral) only if it bears four different groups; testing each compound shows only II and IV qualify.

   Concept and Intuition

   Chirality in a simple acyclic molecule usually arises from a single sp³ carbon attached to four different substituents (a stereocentre). If any two of the four attached groups are identical, that carbon is not a stereocentre and the molecule (if this is its only candidate centre) is achiral.

   Step-by-Step Solution

   1. Pentan-3-ol: $CH_3CH_2-CH(OH)-CH_2CH_3$. C3 bears OH, H, and two identical $-C_2H_5$ groups → not a stereocentre → achiral.
   2. 3-Methylheptane: $CH_3CH_2-CH(CH_3)-CH_2CH_2CH_2CH_3$. C3 bears H, $CH_3$, $-C_2H_5$ (towards C1–C2), and $-C_4H_9$ (towards C4–C7) — all four different → stereocentre → chiral.
   3. 3-Bromo-3-methylpentane: $CH_3CH_2-C(Br)(CH_3)-CH_2CH_3$. This carbon bears Br, $CH_3$, and two identical $-C_2H_5$ groups → not a stereocentre → achiral. …

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2. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.Identify the chiral molecules from the following [FIGURE] (five wedge-dash skeletal structures labelled I-V: I = a carbon bearing H, Br, OH and $CH_3$ substituents shown with wedge/dash stereochemistry; II = a similar carbon bearing H, Br, Br (two Br groups) and $CH_3$; III = a pentane-type chain with a stereocentre bearing OH and H shown with wedge/dash bonds; IV = a pentanol-type chain with OH and H shown on a stereocentre with wedge bonds; V = a chain bearing adjacent Br and Cl stereocentres next to an isopropyl branch, shown with wedge/dash bonds) (only = only) (A) I, II only (B) I, IV, V only (C) II, III, IV only (D) I, V only

   ›Reveal solutionSolution

   Apply the "four different groups" test to each drawn stereocentre: I, IV and V each have four distinct substituents (chiral); II and III each have two identical substituents on their stereocentre (achiral).

   Concept and Intuition

   A carbon atom is a genuine stereocentre — and hence can make a molecule chiral — only when all four groups attached to it are different. If any two of the four attached groups are identical, that carbon is not a stereocentre (there's an internal mirror-plane/symmetry through it), and the molecule (assuming no other stereocentre) is achiral.

   Step-by-Step Solution

   1. I: central carbon bears $CH_3$, $H$, $Br$, $OH$ — four different groups → genuine stereocentre → chiral.
   2. II: central carbon bears $CH_3$, $H$, $Br$, $Br$ — two of the four groups are identical ($Br$ and $Br$) → not a stereocentre → achiral.
   3. III: a chain with OH/H on a carbon flanked by two alkyl "chain ends" that are described as symmetric (e.g. pentan-3-ol type, $CH_3CH_2-CH(OH)-CH_2CH_3$) — the two flanking ethyl groups are identical → not a stereocentre → achiral.
   4. IV: OH/H on a carbon at one end of a longer chain (e.g. pentan-2-ol type, $CH_3-CH(OH)-CH_2CH_2CH_3$) — the four groups are $CH_3$, $H$, $OH$, and propyl, all different → chiral. …

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3. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.Which of the following does not show optical isomerism? (A) $Cis-[CrCl_2(C_2O_4)_2]^{3-}$ (B) $[PtCl_2(en)_2]^{2+}$ (C) $[Co(NH_3)_3(NO_2)_3]$ (D) $[Co(en)_3]^{3+}$

   ›Reveal solutionSolution

   This tests which coordination-geometry type can never be chiral. $MA_3B_3$ octahedral complexes ($[Co(NH_3)_3(NO_2)_3]$) have a mirror plane in every geometric isomer, so they never show optical isomerism — the answer is (C).

   Concept and Intuition

   A complex shows optical isomerism only if it is chiral, i.e. its mirror image is non-superimposable on itself. This happens when the molecule has no improper symmetry element (no $\sigma$ plane, no $S_n$ axis, no centre of symmetry). For octahedral complexes with chelating ligands, whether a particular geometric isomer is chiral depends on its symmetry, not just its formula — you must check each type:

   • $M(AA)_3$ (three symmetric bidentate ligands): always chiral — it has only $C_3$ and $C_2$ rotational symmetry, like a three-bladed propeller, with left- and right-handed forms.
   • $M(AA)_2X_2$ (two bidentate + two monodentate): the cis isomer is chiral (no mirror plane); the trans isomer is achiral (has a mirror plane through the two X groups and the metal).
   • $MA_3B_3$ (three of one monodentate ligand + three of another): both possible geometric isomers, facial (fac, the three A's on one triangular face) and meridional (mer, the three A's in a plane through the metal), each possess a mirror plane that reflects the molecule onto itself. So neither fac nor mer is chiral — this type never shows optical isomerism, regardless of which geometric isomer you pick.

   Step-by-Step Solution

   1. (A) cis-$[CrCl_2(C_2O_4)_2]^{3-}$ is $M(AA)_2X_2$ with $AA=$ oxalate, $X=$ Cl. Since it is explicitly the cis isomer, it lacks a mirror plane and is chiral — it does show optical isomerism.
   2. (B) $[PtCl_2(en)_2]^{2+}$ is the same $M(AA)_2X_2$ pattern ($AA=$ en, $X=$ Cl), here on octahedral Pt(IV). Its cis form (the one conventionally discussed for this complex) is chiral — it does show optical isomerism. …

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4. AP EAPCET 2023Set ap-2023-05-22-FN1 markMCQ
   Q.Which of the following exhibits optical isomerism? I. $\mathrm{CH_3CHCH_2CH_3}$ with a $\mathrm{CH_2Br}$ branch on the CH carbon II. $\mathrm{(CH_3)_2CHCH_2Br}$ III. $\mathrm{(BrCH_2)_2CHCH_2CH_3}$ IV. $\mathrm{CH_3CH_2CHCH(CH_3)_2}$ with a $\mathrm{CH_3}$ branch on the CH carbon (A) I, II only (B) II, III only (C) I, IV only (D) III, IV only

   ›Reveal solutionSolution

   Checking each structure for a carbon with four different substituents (the requirement for chirality/optical isomerism) shows only structures I and IV qualify.

   Concept and Intuition

   A molecule shows optical isomerism if it has at least one stereocenter — a carbon attached to four different groups. The quickest check is to look at the candidate carbon and list its four substituents; if any two are identical, that carbon is not a stereocenter.

   Step-by-Step Solution

   1. Structure I: $\mathrm{CH_3-CH(CH_2Br)-CH_2-CH_3}$. The central CH carbon has substituents: $\mathrm{CH_3}$, $\mathrm{CH_2Br}$, $\mathrm{CH_2CH_3}$ (ethyl), and H — all four different $\Rightarrow$ chiral, shows optical isomerism.
   2. Structure II: $\mathrm{(CH_3)_2CH-CH_2Br}$. The CH carbon has substituents: $\mathrm{CH_3}$, $\mathrm{CH_3}$, $\mathrm{CH_2Br}$, H — two identical $\mathrm{CH_3}$ groups $\Rightarrow$ not chiral.
   3. Structure III: $\mathrm{(BrCH_2)_2CH-CH_2CH_3}$. The CH carbon has substituents: $\mathrm{CH_2Br}$, $\mathrm{CH_2Br}$, $\mathrm{CH_2CH_3}$, H — two identical $\mathrm{CH_2Br}$ groups $\Rightarrow$ not chiral. …

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5. AP EAPCET 2023Set ap-2023-05-23-AN1 markMCQ
   Q.The optical rotation of a racemic mixture of 2-methyl 1-butanol is (A) +11.5 (B) +5.75 (C) +2.87 (D) 0

   ›Reveal solutionSolution

   By definition, a racemic mixture (equal parts of two enantiomers) is optically inactive — net rotation is always zero, regardless of how large each individual enantiomer's specific rotation is.

   Concept and Intuition

   When a chiral compound exists as a 50:50 mixture of its two enantiomers (a racemic mixture, or racemate), the rotation contributed by the (+) form is exactly cancelled by the equal and opposite rotation contributed by the (−) form — this is called "external compensation," and it is a defining property of any racemic mixture, independent of the specific compound.

   Step-by-Step Solution

   1. 2-Methyl-1-butanol has a stereocentre (at C-2), so it exists as a pair of enantiomers, (+)-2-methyl-1-butanol and (−)-2-methyl-1-butanol.
   2. A "racemic mixture" by definition contains these two enantiomers in exactly equal (1:1) proportion.
   3. Each enantiomer rotates plane-polarised light by an equal magnitude but in opposite directions (say +5.75° and −5.75°). …

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6. AP EAPCET 2023Set ap-2023-05-23-FN1 markMCQ
   Q.The correct statement with respect to D-glucose

   (x) and D-Fructose (y) is (A) Both x and y are dextrorotatory compounds (B) Both x and y are laevorotatory compounds (C) x is leavorotatory and y is dextrorotatory compound (D) x is dextroratory and y is leavorotatory compound

   ›Reveal solutionSolution

   The 'D/L' label is a configurational descriptor, unrelated to the actual (+)/(−) optical rotation; D-glucose happens to be dextrorotatory and D-fructose happens to be laevorotatory.

   Concept and Intuition

   Students often assume 'D' means dextrorotatory, but D/L nomenclature (Fischer convention) only describes whether the OH on the highest-numbered stereocentre matches D- or L-glyceraldehyde's configuration — it says nothing about the actual sign of rotation measured in a polarimeter, which is denoted separately by (+) or (−).

   Step-by-Step Solution

   1. D-glucose has a specific rotation of about $+52.7^{\circ}$, i.e. it rotates plane-polarised light clockwise (dextrorotatory) — commonly called 'dextrose'.
   2. D-fructose has a specific rotation of about $-92^{\circ}$, i.e. it rotates plane-polarised light counter-clockwise (laevorotatory) — commonly called 'laevulose', and this strong laevorotation is why invert sugar (a glucose+fructose mix) is net laevorotatory. …

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7. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
   Q.The optically inactive compound from the following is (A) 2-Bromopropanal (B) 3-Bromopropanal (C) 3-Bromo 2-iodopropanal (D) 2-Bromo 3-iodopropanal

   ›Reveal solutionSolution

   A compound is optically active only if it has a chiral (stereogenic) carbon — one attached to four different groups. Checking each option, only 3-bromopropanal lacks such a carbon.

   Concept and Intuition

   Optical activity arises from chirality — the molecule and its mirror image are non-superimposable. The simplest and most common source of chirality in an open-chain molecule is a carbon attached to four different substituents (a stereocentre). If no atom in the molecule has four different groups, the molecule (and its mirror image) are identical, so it is optically inactive.

   Step-by-Step Solution

   1. 2-Bromopropanal: $\mathrm{CH_3-CHBr-CHO}$. The middle carbon (C2) is bonded to $-CHO$, $-Br$, $-CH_3$, and $-H$ — four different groups → chiral centre → optically active.
   2. 3-Bromopropanal: $\mathrm{OHC-CH_2-CH_2Br}$. Here C2 (the only carbon besides the terminal ones) is bonded to $-H$, $-H$, $-CHO$, and $-CH_2Br$ — it carries two hydrogens, so it is not a stereocentre. No other carbon qualifies either (C1 is the aldehyde carbon with a double bond to O, C3 has two H's and Br). So the molecule has no chiral centre → optically inactive. …

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8. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
   Q.How many asymmetric carbons are present in the following molecule? $\mathrm{HOH_2CCH(Br)CH(Br)CH_2OH}$ (A) 3 (B) 1 (C) 4 (D) 2

   ›Reveal solutionSolution

   Only the two central CHBr carbons (C2, C3) each carry four different substituents; the terminal CH₂OH carbons don't qualify, giving 2 asymmetric carbons.

   Concept and Intuition

   A carbon is "asymmetric" (a stereocentre) if it is attached to four different groups. For a chain like HOH₂C–CHBr–CHBr–CH₂OH, check each carbon in turn.

   Step-by-Step Solution

   1. Number the chain C1(CH₂OH)–C2(CHBr)–C3(CHBr)–C4(CH₂OH).
   2. C1: bonded to two H's, OH, and C2 — has two identical H's, not a stereocentre.
   3. C2: bonded to H, Br, –CH₂OH (the C1 side), and –CHBrCH₂OH (the C3 side). These four groups are all different (the C1-side group ≠ the C3-side group) → stereocentre.
   4. C3: bonded to H, Br, –CH₂OH (the C4 side), and –CHBrCH₂OH (the C2 side) — again four different groups → stereocentre.
   5. C4: bonded to two H's, OH, and C3 — not a stereocentre.
   6. Total asymmetric carbons = 2 (C2 and C3).

   Common Mistakes …

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9. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
   Q.Below shown molecules are [FIGURE] (two stereocenter structures labeled X and Y, each a hexane-type carbon chain drawn in zigzag with a stereocenter carbon bearing H, $\mathrm{CH_3}$, and Br substituents shown using wedge-and-dash bonds to indicate 3-D configuration; X shows H on a wedge and $\mathrm{CH_3}$ on a dash with Br in-plane, Y shows H in-plane and $\mathrm{CH_3}$ on a dash with Br on a wedge below) (A) X = Y = Achiral (B) X = Y = chiral (C) X = chiral, Y = Achiral (D) X = Achiral, Y = Chiral

   ›Reveal solutionSolution

   Each stereocentre carries four different groups (alkyl chain, $\mathrm{CH_3}$, H, Br), so both X and Y are chiral.

   Concept and Intuition

   A molecule is chiral when it contains an asymmetric carbon — a carbon bonded to four mutually different groups, giving a non-superimposable mirror image. The way wedge and dash bonds are drawn only sets the 3-D configuration ($R$ or $S$); it does not create or destroy chirality. What matters is whether the four attached groups are all different.

   Step-by-Step Solution

   1. Structure X: the stereocentre carbon is bonded to a propyl-type chain, a $\mathrm{CH_3}$ group, an H atom and a Br atom.
   2. These four groups are all different $\Rightarrow$ X has an asymmetric carbon $\Rightarrow$ X is chiral.
   3. Structure Y: the stereocentre carbon (a 2-bromobutane-type centre) is bonded to an ethyl-type chain, a $\mathrm{CH_3}$ (methyl) fragment, an H atom and a Br atom.
   4. Again all four groups differ $\Rightarrow$ Y also has an asymmetric carbon $\Rightarrow$ Y is chiral.
   5. Since each has exactly one stereocentre with four different substituents, both molecules are chiral.

   Common Mistakes …

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10. AP EAPCET 2021Set ap-2021-09-03-AN1 markMCQ
    Q.Which of the following has a chiral 'C'? (A) $CH_2I-CH_2I$ (B) $CH_3CH_2Cl$ (C) $(CH_3)_2CFCl$ (D) $CH_3CH(Cl)(I)$

    ›Reveal solutionSolution

    This tests recognizing a chiral (stereogenic) carbon — one bonded to four different groups; only $CH_3CH(Cl)(I)$ qualifies.

    Concept and Intuition

    A carbon atom is chiral (a stereocentre) if and only if it is bonded to four different substituent groups. If any two of the four groups are identical, the carbon has a plane of symmetry locally and is not chiral.

    Step-by-Step Solution

    1. (A) $CH_2I-CH_2I$: Consider either carbon — it's bonded to: H, H, I, and $CH_2I$. Two of the four groups (H and H) are identical, so this carbon is NOT chiral.
    2. (B) $CH_3CH_2Cl$: The $CH_2$ carbon is bonded to: H, H, Cl, $CH_3$. Again two H's are identical — NOT chiral. (The $CH_3$ carbon has 3 H's, even less chiral.)
    3. (C) $(CH_3)_2CFCl$: The central carbon is bonded to: $CH_3$, $CH_3$, F, Cl. Two identical $CH_3$ groups — NOT chiral. …

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11. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
    Q.The number of optical isomers possible for 2 – Bromo 3 – Chloro butane are ________ (A) 8 (B) 10 (C) 4 (D) 2

    ›Reveal solutionSolution

    With two different stereocentres bearing different halogens, no meso compound is possible, so the count is the full $2^n=4$ distinct optical isomers.

    Concept and Intuition

    For a molecule with $n$ stereocentres, the maximum number of stereoisomers is $2^n$, but this reduces if the molecule has an internal symmetry allowing a meso compound (an achiral diastereomer where one half's chirality cancels the other's). A meso compound requires the two stereocentres to carry the same set of substituents so that a mirror plane can map one centre exactly onto the other. Here, though, the two stereocentres carry different halogens (Br on C2, Cl on C3), so the molecule has no such symmetry, and no meso form exists.

    Step-by-Step Solution

    1. Draw the structure: $CH_3-CHBr-CHCl-CH_3$ (2-bromo-3-chlorobutane).
    2. Identify stereocentres: C2 (bonded to $CH_3$, $Br$, $H$, and the rest of the chain) and C3 (bonded to $CH_3$, $Cl$, $H$, and the rest of the chain) — both are genuine stereocentres since each has 4 different groups.
    3. Since C2's halogen (Br) differs from C3's halogen (Cl), there is no mirror-symmetry relating the two centres — unlike, say, 2,3-dibromobutane where both centres are identical and a meso form exists. …

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12. AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQ
    Q.Which statement regarding the following structures are true? [FIGURE] (a 2x2 grid of four stereochemical structures labelled (A), (B), (C), (D), each a wedge/dash (Fischer-style) drawing of a 2,3-dihydroxybutanedioic-acid-type skeleton HO-CH-COOH / HOOC-CH-OH: (A) has HO with dashed/hashed bond top-left and OH with dashed/hashed bond bottom-right, COOH top-right, HOOC bottom-left; (B) has HOOC top-left, OH with hashed bond top-right, HO with hashed bond bottom-left, COOH bottom-right; (C) has HO with hashed bond top-left, COOH top-right, a bold wedge OH pointing up from the central vertex with H shown bottom-right, HOOC bottom-left; (D) has HO with hashed bond top-left, COOH top-right, HOOC bottom-left, OH with hashed bond bottom-right) (A) A and B are diastereomers, C and D are enantiomers (B) A and B are enantiomers, C and D are enantiomers (C) A and B are enantiomers, C and D are diastereomers (D) A and B are diastereomers, C and D are diastereomers

    ›Reveal solutionSolution

    (A) and (B) are related by a genuine mirror reflection (enantiomers); (C) differs from (D) (which is drawn identically to (A)) at only one of its two stereocentres, making that pair diastereomers rather than mirror images.

    Concept and Intuition

    For a molecule with two stereocentres (like tartaric acid, $HOOC$–$CHOH$–$CHOH$–$COOH$): flipping both stereocentres gives the true mirror image (an enantiomer, non-superimposable), while flipping only one of the two stereocentres gives a diastereomer (specifically an epimer) — same connectivity, different spatial relationship, and generally different physical properties (unlike enantiomers, which share all scalar physical properties). Reading these wedge/hash drawings, a genuine left–right flip of the whole 2-D structure that keeps every wedge as a wedge and every hash as a hash is a valid way to represent a true mirror reflection (reflecting through a vertical plane leaves front/back and up/down unchanged, and only swaps left/right).

    Step-by-Step Solution

    1. Compare (A) and (B): (B)'s substituent layout is the exact left–right mirror of (A)'s, with the wedge/hash character of every bond preserved (only the "COOH"/"HOOC" and "OH"/"HO" reading direction flips, which is just a labelling convention for which end faces the bond, not a chemical difference).
    2. A left–right reflection that preserves all wedge/hash assignments is a genuine mirror operation (reflection through the vertical plane), so both stereocentres are inverted together — this is precisely what defines an enantiomer pair. Hence A and B are enantiomers.
    3. Now compare (D) to (A): (D)'s substituent layout (HO hashed upper-left, COOH plain upper-right, HOOC plain lower-left, OH hashed lower-right) is identical to (A)'s — the same molecule, same stereochemistry at both centres.
    4. Compare (C) to (A)/(D): the left stereocentre in (C) is unchanged from (A) (HO hashed upper-left, HOOC plain lower-left). But the right stereocentre is drawn differently: in (A)/(D), OH is hashed (pointing back) at that carbon (with H implicit, pointing front); in (C), OH is explicitly a bold wedge (pointing front) and H is explicitly hashed (pointing back) — this is the spatially inverted configuration at that one carbon. …

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