Q.Which of the following compounds will show cis-trans isomerism?

Concept and Intuition

Geometrical isomerism arises from restricted rotation about a C=C double bond. For two isomers (cis and trans) to be distinct, each carbon of the double bond must carry two different substituents — if either carbon has two identical groups, rotating that "end" 180° gives back the same structure, so there is no cis/trans distinction.

Step-by-Step Solution

1. I. $(CH_3)_2C=CH\,C_2H_5$: the left double-bond carbon carries two $\mathrm{CH_3}$ groups (identical) — fails the different-groups test, so no geometrical isomerism.
2. II. $C_6H_5\,CH=CH\,C_2H_5$: left carbon has $\mathrm{C_6H_5}$ and H (different); right carbon has $\mathrm{C_2H_5}$ and H (different) — geometrical isomerism exists.
3. III. $C_6H_5\,CH=CH_2$: the terminal carbon is $=\mathrm{CH_2}$, bearing two identical H atoms — fails the test, no geometrical isomerism (this is styrene, well known to lack cis/trans forms). …

Concept and Intuition

Among the $C_4H_8$ alkenes, only but-2-ene ($CH_3-CH=CH-CH_3$) has two different groups on each double-bond carbon arranged so that cis/trans isomers exist (1-butene and isobutylene each have two identical H's on one alkene carbon, ruling out cis-trans isomerism). Among $C_4H_6$ compounds, only a terminal alkyne has an acidic $\equiv C-H$ that reacts with the strong base $NaNH_2$ to form a sodium acetylide — identifying C as but-1-yne.

Step-by-Step Solution

1. A = but-2-ene ($CH_3-CH=CH-CH_3$), the only $C_4H_8$ alkene with cis-trans isomerism.
2. $Br_2/CCl_4$ adds across the double bond (anti addition) to give B = $CH_3-CHBr-CHBr-CH_3$ (2,3-dibromobutane).
3. C = but-1-yne ($HC\equiv C-CH_2-CH_3$), the $C_4H_6$ isomer with a terminal alkyne proton acidic enough to react with $NaNH_2$, forming the sodium acetylide (releasing $NH_3$).
4. Addition of the first HBr (Markovnikov) to but-1-yne: H adds to the terminal carbon, Br to the internal carbon, giving 2-bromobut-1-ene ($CH_2=CBr-CH_2CH_3$). …

Concept and Intuition

Markovnikov's rule: in HX addition to an unsymmetrical alkene, H goes to the carbon with more H's already, and X goes to the more substituted carbon (via the more stable carbocation). For $\mathrm{CH_3CH_2CH=CH_2}$, the major carbocation forms at C2 (secondary), so Br ends up there: Y = $\mathrm{CH_3CH_2CHBrCH_3}$ (2-bromobutane).

Dehydrohalogenation of a haloalkane with alcoholic KOH follows Zaitsev's rule: the more substituted (more stable) alkene is the major product. Eliminating HBr from 2-bromobutane can give but-1-ene (mono-substituted db) or but-2-ene (di-substituted, more stable) — Zaitsev favors but-2-ene, so Z = $\mathrm{CH_3-CH=CH-CH_3}$.

Hyperconjugation requires a C–H bond on a carbon directly attached (α) to an sp² (double-bond) carbon. In but-2-ene, C1 and C4 are the methyl carbons attached to the sp² C2 and C3 respectively, each contributing 3 α-hydrogens, for a total of 6 hyperconjugating C–H bonds — hence 6 no-bond resonance (hyperconjugative) structures.

Step-by-Step Solution

1. X = but-1-ene; HBr adds Markovnikov-wise → Y = 2-bromobutane.
2. Alc. KOH + Y → E2 elimination, Zaitsev product dominates → Z = but-2-ene.
3. Check each statement about Z (but-2-ene):
   • (A) Only one π bond (the single C=C) — statement false. …

Concept and Intuition

Two independent facts are being tested together:

1. Stereochemistry of alkyne reduction determines polarity of the alkene product. $H_2/Pd$-quinoline (Lindlar's catalyst, a poisoned catalyst) adds $H_2$ syn (same face), giving the cis-alkene. $Na$/liq. $NH_3$ reduces via a radical-anion mechanism giving anti addition, i.e. the trans-alkene. For but-2-ene specifically, the trans isomer has a centre of symmetry and its bond dipoles cancel (non-polar), while the cis isomer's dipoles do not cancel (net dipole, polar).
2. Strength/condition of oxidant determines whether a C=C survives as a diol or is cleaved to acids/ketones. Cold, dilute, neutral $KMnO_4$ (Baeyer's reagent, ~273 K) only syn-dihydroxylates the double bond (gives a diol). Hot, concentrated, acidic $KMnO_4$ (written $KMnO_4/H^+$) is a much stronger oxidant that cleaves the C=C bond completely; a fully-substituted carbon on each side (as in but-2-ene, $CH_3$–CH=CH–$CH_3$) gives two molecules of a carboxylic acid (here, $CH_3COOH$, acetic acid).

Step-by-Step Solution

1. But-2-yne + A → X (polar). Since the polar product must be cis-but-2-ene, A must give syn addition → A = $H_2/Pd$-quinoline. …

Concept and Intuition

Reduction of an internal alkyne can be steered to give either the cis- or trans-alkene depending on the reagent: Lindlar's catalyst (Pd/BaSO$_4$ poisoned with quinoline) gives syn addition of H$_2$ → the cis-alkene, while dissolving-metal reduction with Na (or Li) in liquid NH$_3$ proceeds through a radical-anion mechanism that favours anti addition → the trans-alkene. Since trans-2-butene is symmetric (the two methyl groups are on opposite sides), its bond dipoles cancel, making it essentially nonpolar — unlike cis-2-butene, which retains a small net dipole.

Step-by-Step Solution

1. X (a dibromide, $\mathrm{C_4H_8Br_2}$) undergoes dehydrohalogenation (elimination of HBr, here effectively twice) to give an alkyne Y.
2. For Y to be 2-butyne ($\mathrm{CH_3-C\equiv C-CH_3}$), X must have both bromines on the internal carbons: $\mathrm{CH_3-CHBr-CHBr-CH_3}$ (2,3-dibromobutane). [If X were $\mathrm{CH_3-CH_2-CHBr-CH_2Br}$, elimination would give 1-butyne, a terminal alkyne, which cannot give a symmetric nonpolar butene on reduction — ruling out options C/D.]
3. Reduction of 2-butyne with Z gives the nonpolar isomer of $\mathrm{C_4H_8}$, i.e. trans-2-butene (symmetric, cancelling dipoles). …

Concept and Intuition

$\mathrm{Br_2}$ reacts with alkenes in two very different ways depending on conditions:

• In the dark (ionic mechanism), $\mathrm{Br_2}$ adds across the C=C double bond (electrophilic addition) — this is the classic bromine-water/$\mathrm{Br_2}$-in-$\mathrm{CCl_4}$ decolourisation test for unsaturation.
• Under UV light (photochemical, free-radical mechanism), at low effective bromine concentration the dominant pathway is abstraction of an allylic hydrogen atom, giving allylic substitution instead of addition (analogous to NBS bromination).

Step-by-Step Solution

1. Identify $X$: among $\mathrm{C_4H_8}$ alkenes — but-1-ene ($\mathrm{CH_2=CHCH_2CH_3}$), cis/trans-but-2-ene, and isobutylene ($\mathrm{(CH_3)_2C=CH_2}$) — the one with no cis-trans isomerism and a straight-chain skeleton matching the reaction products offered is but-1-ene (2-butene would show cis-trans isomerism, so it's excluded by the given condition).
2. But-1-ene: $\mathrm{CH_2=CH-CH_2-CH_3}$ (C1=C2, allylic carbon = C3). …

Concept and Intuition

When a dibromo derivative of an alkene keeps the C=C double bond (substitution product, not an addition product), its molecular formula is $C_nH_{2n-2}Br_2$. Once the molecular formula is pinned down from the given molar mass, we must count BOTH constitutional isomers (where the two Br atoms sit) AND, for any isomer that still has a C=C double bond with two different groups on each carbon, the cis/trans geometrical isomers (since rotation about a double bond is restricted).

Step-by-Step Solution

1. Let the alkene be $C_nH_{2n}$; a dibromo derivative that keeps the double bond replaces 2 H's with 2 Br's: formula $C_nH_{2n-2}Br_2$, molar mass $=12n + (2n-2)(1) + 2(80) = 14n + 158$.
2. Set $14n+158 = 186 \Rightarrow 14n = 28 \Rightarrow n = 2$. (Equivalently, direct substitution check: $C_2H_2Br_2$ mass $=24+2+160=186$ ✓.)
3. So the compound is $C_2H_2Br_2$ — dibromoethylene, derived from ethene by replacing 2 H's with 2 Br's while keeping the double bond.
4. Enumerate isomers of $C_2H_2Br_2$:
   • 1,1-dibromoethene: $CH_2=CBr_2$ — only one possible structure (no cis/trans, since one carbon bears two identical H's). …

Concept and Intuition

Cis/trans (E/Z) isomerism arises from restricted rotation about a C=C double bond, but it is only observable when each doubly-bonded carbon carries two DIFFERENT substituents. If either carbon has two identical groups (including two H's), the "cis" and "trans" arrangements become identical, so no geometrical isomerism exists.

Step-by-Step Solution

1. List the five acyclic structural isomers of pentene ($C_5H_{10}$):
   • Pent-1-ene: $CH_2=CH-CH_2-CH_2-CH_3$ — terminal carbon has two H's → no isomerism.
   • Pent-2-ene: $CH_3-CH=CH-CH_2CH_3$ — left carbon has $CH_3$/H, right carbon has $C_2H_5$/H, all four substituents distinct pairwise on each carbon → cis/trans isomerism EXISTS.
   • 2-Methylbut-1-ene: $CH_2=C(CH_3)-CH_2CH_3$ — terminal carbon has two H's → no isomerism.
   • 3-Methylbut-1-ene: $CH_2=CH-CH(CH_3)_2$ — terminal carbon has two H's → no isomerism.
   • 2-Methylbut-2-ene: $CH_3-C(CH_3)=CH-CH_3$ — the left double-bond carbon carries two identical methyl groups → no isomerism. …

Concept and Intuition

Geometrical isomerism about a C=C double bond requires restricted rotation (present in any double bond) AND each carbon of the double bond must carry two DIFFERENT groups. If either carbon has two identical groups, cis/trans forms become indistinguishable (no isomerism).

Step-by-Step Solution

1. 2-Butene: $\mathrm{CH_3-CH=CH-CH_3}$ — each double bond carbon has $\mathrm{CH_3}$ and H, both different $\Rightarrow$ shows cis/trans isomerism.
2. 3-Hexene: $\mathrm{CH_3CH_2-CH=CH-CH_2CH_3}$ — each double bond carbon has ethyl and H, different $\Rightarrow$ shows geometrical isomerism.
3. But-2-enal: $\mathrm{CH_3-CH=CH-CHO}$ — each double bond carbon has a different group (CH3/H and CHO/H) $\Rightarrow$ shows geometrical isomerism. …

Concept and Intuition

Cis isomers have their similar groups on the same side, giving a net molecular dipole (more polar), while trans isomers are more symmetric with dipoles that largely cancel (less polar, often near zero). Polarity affects boiling point (stronger intermolecular dipole-dipole attraction raises boiling point), while melting point depends more on how well molecules pack into a crystal lattice — symmetric trans isomers usually pack better and so have HIGHER melting points than the less symmetric cis isomers.

Step-by-Step Solution

1. Statement I: cis-2-butene has both methyl groups on the same side, giving a net dipole; trans-2-butene's methyl groups are on opposite sides, largely cancelling the dipole. So cis is more polar than trans — TRUE.
2. Statement II: because trans-2-butene is more symmetric, it packs more efficiently into a solid lattice, giving it a HIGHER melting point than cis-2-butene — so "cis melting point > trans" is FALSE. …

Concept and Intuition

With alcoholic KOH the tertiary halide loses HX by E2, removing a $\beta$-hydrogen from any adjacent carbon. Carbon-3 bears three distinct kinds of $\beta$-carbon: the ethyl side (C-2), the propyl side (C-4) and the branch methyl. Each gives a different alkene, and where both doubly-bonded carbons carry two different groups, cis-trans (E/Z) isomers are separate compounds.

Step-by-Step Solution

1. Structure: $\mathrm{CH_3CH_2\!-\!C(Br)(CH_3)\!-\!CH_2CH_2CH_3}$ (Br on C-3, with a methyl branch on C-3).
2. Eliminate toward C-2 $\Rightarrow$ 3-methylhex-2-ene; both alkene carbons carry different groups $\Rightarrow$ E and Z (2 alkenes).
3. Eliminate toward C-4 $\Rightarrow$ 3-methylhex-3-ene; again both carbons differ $\Rightarrow$ E and Z (2 alkenes).
4. Eliminate the branch-methyl H $\Rightarrow$ 3-methylenehexane (terminal $=\mathrm{CH_2}$); no geometrical isomerism (1 alkene).
5. Total distinct alkenes $= 2+2+1 = 5$.

Common Mistakes …