JEE Hydrocarbons - Alkynes Questions

To determine the major product P, we analyze each step of the reaction sequence systematically.

The first two steps involve the conversion of 1,2-dibromocyclooctane into an alkyne. Treatment with aqueous $$KOH$$ initially causes dehydrohalogenation to form a vinyl bromide, and the subsequent treatment with the strong base $$NaNH_2$$ removes another molecule of $$HBr$$ to produce cyclooctyne through double dehydrohalogenation.

In the third step, cyclooctyne is treated with $$Hg^{2+}/H^+$$. Under these conditions, the alkyne undergoes acid-catalyzed hydration to form an enol, which rapidly tautomerizes to the corresponding ketone. Thus, cyclooctyne is converted into cyclooctanone.

In the final step, cyclooctanone is subjected to Clemmensen reduction using $$Zn\text{-}Hg/H^+$$. The carbonyl group is completely reduced to a methylene group $$(-CH_2-)$$, converting the ketone into the corresponding saturated hydrocarbon.

Therefore, the overall transformation is

$$\text{1,2-dibromocyclooctane} ;\longrightarrow; \text{cyclooctyne} ;\longrightarrow; \text{cyclooctanone} ;\longrightarrow; \text{cyclooctane}.$$

Hence, the final major product P is cyclooctane, corresponding to the eight-membered saturated ring shown in option B

We need to evaluate two statements about propyne ($$CH_3-C \equiv CH$$, molecular weight = 40 g/mol).

Statement I: One mole of propyne reacts with excess of sodium to liberate half a mole of H₂ gas.

Propyne has one terminal acidic hydrogen (on the $$\equiv CH$$). Only terminal alkynes react with sodium metal:

$$2 CH_3C \equiv CH + 2Na \rightarrow 2 CH_3C \equiv CNa + H_2$$

So 2 moles of propyne give 1 mole of H₂. Therefore, 1 mole of propyne gives $$\frac{1}{2}$$ mole of H₂.

Statement I is CORRECT.

Statement II: Four g of propyne reacts with NaNH₂ to liberate NH₃ gas which occupies 224 mL at STP.

4 g of propyne = $$\frac{4}{40} = 0.1$$ mol

$$CH_3C \equiv CH + NaNH_2 \rightarrow CH_3C \equiv CNa + NH_3$$

0.1 mol propyne produces 0.1 mol NH₃.

At STP, 0.1 mol of gas occupies $$0.1 \times 22400 = 2240$$ mL.

The statement says 224 mL, which would correspond to 0.01 mol. This is incorrect.

Statement II is INCORRECT.

The correct answer is Option 3: Statement I is correct but Statement II is incorrect.

To determine the least acidic compound among the given choices, we compare the stability of the conjugate bases formed after the loss of a proton ($$H^+$$). The greater the stability of the conjugate base, the stronger the corresponding acid.

Among the given compounds, $$m$$-ethoxyphenol is a phenol. Phenols are acidic with a typical $$pK_a$$ of about $$10$$ because the resulting phenoxide ion is stabilized by resonance over the benzene ring.

Benzenesulfonic acid is a sulfonic acid and is a very strong acid with $$pK_a < 0$$. Its conjugate base, the sulfonate ion, is highly stabilized by resonance, with the negative charge delocalized over three oxygen atoms.

Pentanoic acid is a carboxylic acid with a typical $$pK_a$$ of approximately $$4.8$$. The corresponding carboxylate ion is resonance-stabilized by delocalization of the negative charge over two equivalent oxygen atoms, making it considerably more acidic than phenols.

Ethyl propiolate ($$EtO_2C-C \equiv CH$$) contains a terminal alkyne hydrogen. Although terminal alkynes are acidic due to the $$sp$$-hybridized carbon atom, their acidity is much lower, with a typical $$pK_a$$ around $$25$$. The electron-withdrawing ester group increases its acidity slightly, but it is still far less acidic than phenols, carboxylic acids, or sulfonic acids.

Therefore, the order of acidity is

$$\text{Benzenesulfonic acid} > \text{Pentanoic acid} > m\text{-Ethoxyphenol} > \text{Ethyl propiolate}.$$

Hence, the least acidic compound is

$$\boxed{\text{(D) Ethyl propiolate}}.$$

For any hydrocarbon $$C_nH_m$$, the number of degrees of unsaturation (also called index of hydrogen deficiency, IHD) is calculated by the relation
$$\text{IHD}=\,\frac{(2n+2)-m}{2}\,.$$

Here $$n=6$$ and $$m=6$$, so
$$\text{IHD}=\,\frac{(2\times6+2)-6}{2}=\,\frac{14-6}{2}=4.$$ Each unit of IHD represents either one $$\pi$$-bond or one ring.

The statement says that the compound “takes up four moles of hydrogen per mole for complete hydrogenation”.
• One mole of $$H_2$$ adds across a double bond (or one of the two $$\pi$$ bonds in a triple bond).
• A ring does not consume $$H_2$$ during hydrogenation.

Since four moles of $$H_2$$ are absorbed, all four units of IHD must be $$\pi$$ bonds; there are no rings contributing to IHD. Thus the molecule contains four distinct $$\pi$$ bonds.

Each $$\pi$$ bond is formed by a pair of electrons, therefore the total number of $$\pi$$ (unsaturated) electrons is
$$4 \times 2 = 8.$$

Hence, the compound possesses 8 electrons in its $$\pi$$ system (unsaturated electrons).

We consider Hex-1,3-dien-5-yne, whose structure is $$CH_2=CH-CH=CH-C\equiv CH$$, a six-carbon chain with double bonds at positions 1-2 and 3-4 and a triple bond at positions 5-6.

For $$\sigma$$ bonds, there are C-C single bonds between C2-C3 and C4-C5 (2 $$\sigma$$), one $$\sigma$$ each from the two C=C bonds at C1=C2 and C3=C4 (2 $$\sigma$$), and one $$\sigma$$ from the C≡C bond at C5-C6 (1 $$\sigma$$), giving a total of 5 C-C $$\sigma$$ bonds. The C-H bonds contribute $$\sigma$$ bonds as follows: C1 has 2H, C2 has 1H, C3 has 1H, C4 has 1H, C5 has 0H and C6 has 1H, totaling 6 $$\sigma$$ bonds. Hence the overall $$\sigma$$ count is 5 + 6 = 11.

Each double bond has one $$\pi$$ bond, so the two C=C bonds provide 2 $$\pi$$ bonds, while the C≡C bond provides 2 $$\pi$$ bonds, for a total of 4 $$\pi$$ bonds.

Therefore, the molecule contains $$\sigma + \pi = 11 + 4 = 15$$ bonds in total.

Identifying incorrect statements about primary standards:

(A) "It should be purely available in dry form" — Correct property of a primary standard.

(B) "It should not undergo chemical change in air" — Correct property.

(C) "It should be hygroscopic and should react with another chemical instantaneously and stoichiometrically" — Incorrect. A primary standard should be NON-hygroscopic (not absorb moisture from air).

(D) "It should be readily soluble in water" — Correct property.

(E) "$$KMnO_4$$ and NaOH can be used as primary standard" — Incorrect. Neither $$KMnO_4$$ nor NaOH are primary standards. $$KMnO_4$$ is not pure enough, and NaOH is hygroscopic and absorbs CO₂.

The incorrect statements are (C) and (E).

The correct answer is Option 2: (C) and (E) only.

Ethyne (acetylene, C₂H₂) has a C≡C triple bond.

(A) C-C bond in ethyne is shorter than ethene — TRUE (triple bond < double bond).

(B) Both carbons are sp hybridized — TRUE.

(C) Ethyne is linear — TRUE.

(D) C-C bond in ethyne is weaker than ethene — FALSE. Triple bond (837 kJ/mol) is stronger than double bond (614 kJ/mol).

The incorrect statement is Option 4.

Which halogens can undergo disproportionation reactions?

Disproportionation is a reaction where the same element is simultaneously oxidized and reduced. For halogens in their elemental form (oxidation state 0), disproportionation requires the element to form both positive and negative oxidation states.

The analysis of each halogen is as follows.

$$F_2$$: Fluorine is the most electronegative element and can only exist in oxidation states 0 and $$-1$$. It cannot achieve positive oxidation states, so it cannot disproportionate.

$$Cl_2$$: Can form $$-1$$ (in $$Cl^-$$) and positive states like $$+1$$ (in $$OCl^-$$). For example: $$Cl_2 + 2NaOH \rightarrow NaCl + NaOCl + H_2O$$. It can disproportionate.

$$Br_2$$: Similarly, can form $$-1$$ and positive oxidation states. It can disproportionate.

$$I_2$$: Can form $$-1$$ and positive states (e.g., $$+5$$ in $$IO_3^-$$). It can disproportionate.

The correct answer is Option (4): $$Cl_2$$, $$Br_2$$ and $$I_2$$.

Explain why thiosulphate reacts differently with iodine and bromine.

With $$I_2$$, thiosulphate reacts according to $$2S_2O_3^{2-} + I_2 \rightarrow S_4O_6^{2-} + 2I^-$$, in which sulfur is oxidized from +2 to +2.5, indicating mild oxidation; with $$Br_2$$, the reaction $$S_2O_3^{2-} + 5Br_2 + 5H_2O \rightarrow 2SO_4^{2-} + 4Br^- + 10H^+$$ shows sulfur oxidized from +2 to +6, indicating strong oxidation.

Bromine, being a stronger oxidizing agent than iodine, can oxidize thiosulphate all the way to sulfate ($$SO_4^{2-}$$, S in the +6 state), while iodine, being milder, oxidizes it only partially to tetrathionate ($$S_4O_6^{2-}$$).

The correct answer is Option (1): Bromine is a stronger oxidant than iodine.

1. Identification of Reactant 'A'

• The Reaction: Alkyne reduction over palladium on carbon ($$\text{Pd/C}$$) usually reduces an alkyne completely to an alkane. 
• Analysis: The product has a 5-carbon continuous chain with a double bond at position 2. Therefore, the starting material A must be 2-Pentyne ($$\text{CH}_3-\text{C}\equiv\text{C}-\text{C}_2\text{H}_5$$).

2. Identification of Product 'B' (Birch Reduction)

• Mechanism: This is a classic Birch Reduction of an internal alkyne. It proceeds via a radical-anion mechanism that favors anti-addition of hydrogen atoms due to minimizing steric repulsion between the bulky alkyl groups.
• Product 'B': An internal alkyne reduced with $$\text{Na/liquid NH}_3$$ selectively yields a trans-alkene. Thus, B is trans-2-butene.

But-2-yne ($$CH_3-C \equiv C-CH_3$$) undergoes:

- With Pd/C + H₂ (catalytic hydrogenation): gives cis-but-2-ene (A) (syn addition)

- With Na/liq NH₃ (Birch reduction): gives trans-but-2-ene (B) (anti addition)

Let us check each statement:

Statement A: "A is more soluble than B" - cis-but-2-ene (A) has a higher dipole moment and is more polar, so it is slightly more soluble in polar solvents. This statement can be considered correct depending on context.

Statement B: "The boiling point and melting point of A are higher and lower than B respectively." - While cis isomers generally have higher boiling points due to higher polarity, and trans isomers have higher melting points due to better packing, the overall comparison needs careful analysis. This statement is INCORRECT as stated in the given context.

Statement C: "A is more polar than B because dipole moment of A is zero." - The first part is correct (cis is more polar), but the reasoning is completely wrong. The dipole moment of A (cis) is NOT zero; in fact, it is the trans isomer B that has zero (or near-zero) dipole moment due to symmetry. This statement is INCORRECT.

Statement D: "Br₂ adds easily to B than A." - Generally, cis-alkenes are more reactive toward electrophilic addition due to less steric strain in the transition state. So Br₂ should add more easily to A (cis) than B (trans). This statement is INCORRECT.

Therefore, the incorrect statements are B, C and D only.

We need to find the amount of 1,1,2,2-tetrabromopropane obtained from 1 g of bromine in the bromination of propyne, given that the yield is 27%.

The bromination of propyne ($$CH_3C \equiv CH$$) with two equivalents of $$Br_2$$ gives 1,1,2,2-tetrabromopropane:

$$CH_3C \equiv CH + 2Br_2 \rightarrow CH_3CBr_2CHBr_2$$

We have 1 g of bromine. The molar mass of $$Br_2$$ is $$2 \times 80 = 160$$ g/mol. So the moles of $$Br_2$$ available are:

$$n_{Br_2} = \frac{1}{160} \text{ mol}$$

Since 2 moles of $$Br_2$$ are needed per mole of product, the moles of 1,1,2,2-tetrabromopropane that can be formed (at 100% yield) are:

$$n_{product} = \frac{1}{160 \times 2} = \frac{1}{320} \text{ mol}$$

The molar mass of 1,1,2,2-tetrabromopropane ($$C_3H_6Br_4$$) is $$3(12) + 6(1) + 4(80) = 36 + 6 + 320 = 362$$ g/mol.

The theoretical mass of product is:

$$m_{theoretical} = \frac{362}{320} = 1.13125 \text{ g}$$

With a yield of 27%, the actual mass obtained is:

$$m_{actual} = 1.13125 \times 0.27 = 0.3054 \text{ g}$$

Expressing this as $$\times 10^{-1}$$ g: $$m_{actual} = 3.054 \times 10^{-1} \approx 3 \times 10^{-1}$$ g.

Hence, the correct answer is 3.

The reaction requires identifying a hydrocarbon whose ozonolysis product gives a positive Tollens’ test.

Tollens’ reagent gives a positive test for aldehydes.

Among carboxylic acids, only $$\mathrm{HCOOH}$$ can also reduce Tollens’ reagent because it contains an aldehydic hydrogen.

For option $$\mathrm{(A)}$$, ozonolysis of $$\mathrm{2,3\text{-}Dimethyl\text{-}2\text{-}butene}$$ forms acetone, which is a ketone and does not respond to Tollens’ reagent.

For option $$\mathrm{(B)}$$, ozonolysis of isopropylidenecyclopropane forms acetone and cyclopropanone, both ketones, so Tollens’ test is negative.

For option $$\mathrm{(D)}$$, ozonolysis of $$\mathrm{2\text{-}Butyne}$$ forms acetic acid $$\mathrm{(CH_3COOH)}$$, which does not respond to Tollens’ reagent.

For option $$\mathrm{(C)}$$, ozonolysis of $$\mathrm{1\text{-}Butyne\ (HC\equiv C-CH_2-CH_3)}$$ produces propanoic acid and:

$$\mathrm{HCOOH}$$

Formic acid gives a positive Tollens’ test and forms the silver mirror.

Thus, compound $$\mathrm{A}$$ is:

$$\mathrm{HCOOH}$$

Thus, the correct option is C.

Metallic sodium is a strong reducing agent and reacts with acidic hydrogen atoms in compounds. The key is to identify which compound among the given options does not have a sufficiently acidic hydrogen for sodium to react with under normal conditions.

Gaseous ammonia ($$\text{NH}_3$$) has N-H bonds that react with sodium: $$2\text{Na} + 2\text{NH}_3 \rightarrow 2\text{NaNH}_2 + \text{H}_2$$, though this reaction requires heating.

Ethyne ($$\text{HC \equiv CH}$$) has a terminal C-H bond that is acidic ($$pK_a \approx 25$$) because the carbon is $$sp$$ hybridised: $$2\text{Na} + 2\text{HC \equiv CH} \rightarrow 2\text{NaC \equiv CH} + \text{H}_2$$.

tert-Butyl alcohol has an O-H bond, which readily reacts with sodium: $$2\text{Na} + 2(CH_3)_3\text{COH} \rightarrow 2(CH_3)_3\text{CONa} + \text{H}_2$$.

But-2-yne ($$\text{CH}_3\text{C \equiv CCH}_3$$) is an internal alkyne. It has no terminal C-H bond on the triple-bond carbon. Both carbons of the triple bond bear methyl groups, so there is no acidic terminal alkynyl hydrogen. Sodium reacts with terminal alkynes by abstracting the acidic acetylenic proton, but since but-2-yne has no such proton, it does not react with metallic sodium under normal conditions.

Therefore, the correct answer is option (2): But-2-yne.

Lindlar catalyst is a heterogeneous catalyst that consists of palladium deposited on calcium carbonate (or barium sulfate), which is then treated (poisoned) with various forms of lead or quinoline to reduce its activity. It is commonly described as partially deactivated palladised charcoal.

Let us examine each option. Option (1), zinc chloride and HCl, is Lucas reagent used to distinguish between primary, secondary, and tertiary alcohols. Option (2), cold dilute $$KMnO_4$$ solution, is Baeyer's reagent used for the test of unsaturation. Option (3), sodium in liquid $$NH_3$$, is Birch reduction conditions used to reduce alkynes to trans-alkenes. Option (4), partially deactivated palladised charcoal, is the correct description of Lindlar catalyst, which is used to selectively reduce alkynes to cis-alkenes.

Hence, the correct answer is option (4).

The reaction sequence can be understood by analyzing each step individually.

In the first step, propyne,

$$\mathrm{CH_3-C\equiv C-H},$$

is treated with sodium amide,

$$\mathrm{NaNH_2}.$$

Since propyne is a terminal alkyne, the hydrogen attached to the triple-bonded carbon is acidic. The strong base $$\mathrm{NH_2^-}$$ abstracts this proton to form the corresponding acetylide ion.

Thus, intermediate A is sodium propynide,

$$\boxed{\mathrm{CH_3-C\equiv C^-Na^+.}}$$

In the second step, the acetylide ion reacts with 4-bromobutan-2-ol,

$$\mathrm{Br-CH_2-CH_2-CH(OH)-CH_3}.$$

The acetylide ion acts as a nucleophile and attacks the primary carbon bearing the bromine atom through an $$S_N2$$ mechanism, displacing $$\mathrm{Br^-}$$ and forming a new carbon-carbon bond.

The resulting intermediate B is

$$\boxed{\mathrm{CH_3-C\equiv C-CH_2-CH_2-CH(OH)-CH_3}.}$$

In the third step, intermediate B is treated with

$$\mathrm{H_2/Pd-C}.$$

Catalytic hydrogenation completely reduces the carbon-carbon triple bond to a single bond without affecting the alcohol group.

Hence, intermediate C becomes

$$\boxed{\mathrm{CH_3-CH_2-CH_2-CH_2-CH_2-CH(OH)-CH_3},}$$

which is heptan-2-ol.

In the final step, intermediate C is oxidized with

$$\mathrm{CrO_3}.$$

Chromium trioxide converts a secondary alcohol into the corresponding ketone.

Therefore, the hydroxyl-bearing carbon is oxidized to a carbonyl group, giving

$$\boxed{\mathrm{CH_3-CH_2-CH_2-CH_2-CH_2-C(=O)-CH_3}.}$$

This compound is heptan-2-one.

Hence, the final product D is

$$\boxed{\mathrm{CH_3-CH_2-CH_2-CH_2-CH_2-C(=O)-CH_3}.}$$

Therefore, the correct answer is

$$\boxed{\text{Option (A).}}$$

• Step 1: Kucherov Reaction (Formation of X)

  When 3-methylbut-1-yne ($$\text{CH}_3\text{--CH(CH}_3\text{)--C}\equiv\text{CH}$$) undergoes hydration in the presence of $$\text{HgSO}_4$$ and $$\text{H}_2\text{SO}_4$$, Markovnikov addition of water occurs to form an enol intermediate, which rapidly tautomerizes to a stable methyl ketone:

  $$\text{Product X} = \text{3-methylbutan-2-one } \left[\text{CH}_3\text{--CH(CH}_3\text{)--C}(=\text{O})\text{--CH}_3\right]$$



• Step 2: Grignard Reaction

  The carbonyl carbon of ketone X is attacked by the ethyl nucleophile from ethylmagnesium bromide ($$\text{C}_2\text{H}_5\text{MgBr}$$). Subsequent acid hydrolysis ($$\text{H}_2\text{O}$$) gives a tertiary alcohol:

  $$\text{Intermediate Alcohol} = \text{3,4-dimethylhexan-3-ol } \left[\text{CH}_3\text{--CH(CH}_3\text{)--C(OH)(CH}_3\text{)--CH}_2\text{CH}_3\right]$$



• Step 3: Acid-Catalyzed Dehydration (Formation of Y)

  Heating the tertiary alcohol with concentrated $$\text{H}_2\text{SO}_4$$ drives a dehydration reaction via a stable carbocation intermediate. Following Saytzeff's rule, elimination yields the most highly substituted and thermodynamically stable alkene as the major product:

  $$\text{Product Y} = \text{2,3-dimethylpent-2-ene } \left[(\text{CH}_3)_2\text{C}=\text{C(CH}_3\text{)CH}_2\text{CH}_3\right]$$

Conclusion:

The sequence proceeds via oxymercuration-hydration of the alkyne, Grignard addition to form a tertiary alcohol, and subsequent elimination to deliver the most substituted alkene.

Answer: Option D — 2,3-dimethylpent-2-ene

We first look carefully at the given substrate. $$CH_3CH_2CH(Br)CH_2Br$$ Numbering the carbon chain from the left it is $$C_1 = CH_3,\;C_2 = CH_2,\;C_3 = CH(Br),\;C_4 = CH_2Br$$ - that is, a vicinal (adjacent) 1,2-dibromide.

Step (i) : treatment with alcoholic $$KOH$$ - E2 dehydro-halogenation. The E2 rule tells us that one bromine atom and a β-hydrogen situated anti-periplanar to it are lost simultaneously as $$KBr$$ and $$H_2O$$ to give an alkene. On $$C_3$$ there is the leaving group $$Br$$; the β-hydrogens that can be removed are found on $$C_4$$. Removing the hydrogen from $$C_4$$ together with the bromine on $$C_3$$ produces the following alkene:

$$\displaystyle CH_3CH_2C(Br)=CH_2 $$

This product is called 3-bromobut-1-ene (or 1-bromo-1-butene). Note that only ONE molecule of $$HBr$$ has been removed, so the molecule still contains one bromine atom, now attached directly to the vinylic carbon $$C_3$$.

Step (ii) : treatment with $$NaNH_2$$ in liquid $$NH_3$$ - very strong base. A vinylic bromide such as $$CH_3CH_2C(Br)=CH_2$$ undergoes another elimination of $$HBr$$ when it is exposed to a base as strong as sodamide. The general rule (which should be stated) is: “A geminal or vinylic halide, when treated with excess $$NaNH_2$$ in liquid $$NH_3$$, loses a second molecule of $$HBr$$ to give an alkyne.”

Applying the rule, the base abstracts the sole vinylic hydrogen on $$C_4$$, the bromide ion departs from $$C_3$$, and a second π-bond is introduced between $$C_3$$ and $$C_4$$. Because there is already one π-bond there, a second π-bond converts it into a triple bond:

$$ \begin{aligned} CH_3CH_2C(Br)=CH_2 &\xrightarrow{NaNH_2/NH_3} CH_3CH_2C\!\equiv\!CH \;+\; NH_3Br^- \;+\; NH_2^-H^+\\ &\qquad\qquad(\text{1-butyne}) \end{aligned} $$

Thus after the two successive eliminations (one normal E2 and one very strong-base vinylic E2) the vicinal dibromide is transformed into the terminal alkyne $$CH_3CH_2C\!\equiv\!CH$$, i.e. 1-butyne.

Among the options supplied, this structure corresponds to Option C.

Hence, the correct answer is Option C.

First we recall that the acid strength of an organic molecule depends upon the stability of the conjugate base that is produced after loss of a proton. The more stable the conjugate base, the larger the acid dissociation constant $$K_a$$ and the smaller the $$pK_a\;(\;pK_a = -\log K_a\;),$$ hence the stronger the acid.

All the three compounds in the question lose a proton from a carbon atom, so we have to compare the stabilities of the corresponding carbanions:

$$CH \equiv CH \longrightarrow HC \equiv C^- + H^+$$

$$CH_3 - C \equiv CH \longrightarrow CH_3 - C \equiv C^- + H^+$$

$$CH_2 = CH_2 \longrightarrow CH_2 = CH^- + H^+$$

Now we analyse the factors that stabilise or destabilise each carbanion.

1. Hybridisation (s-character)
The electronegativity of a carbon atom increases with the percentage of s-character in its hybrid orbital. The order of s-character is

$$sp\;(50\%\,s) > sp^2\;(33\%\,s) > sp^3\;(25\%\,s).$$

Therefore a negative charge on an $$sp$$ carbon is held more tightly (more stabilised) than on an $$sp^2$$ carbon. Both alkynes have the negative charge on an $$sp$$ carbon, while the alkene places the charge on an $$sp^2$$ carbon. So immediately we get

$$\text{any terminal alkyne} \;\; > \;\; CH_2 = CH_2$$

in acid strength.

2. Inductive ( +I ) effect of the $$CH_3$$ group
Between the two terminal alkynes, $$CH_3 - C \equiv C^-$$ carries an additional $$CH_3$$ group next to the charged carbon. A methyl group exerts a +I (electron-releasing) effect which pushes electron density towards the negatively charged carbon, destabilising it.

In contrast, the carbanion $$HC \equiv C^-$$ has no such electron-releasing group; hence it is intrinsically more stable. Therefore

$$HC \equiv C^- \; >\; CH_3 - C \equiv C^-$$

in terms of stability, and consequently

$$HC \equiv CH \; >\; CH_3 - C \equiv CH$$

in acid strength.

3. Consolidating both factors
Combining the hybridisation argument (which places both alkynes ahead of the alkene) with the inductive argument (which distinguishes between the two alkynes), we obtain the complete order:

$$HC \equiv CH \; > \; CH_3 - C \equiv CH \; > \; CH_2 = CH_2.$$

Numerically this order is supported by the experimental $$pK_a$$ values:

$$pK_a(HC \equiv CH)\;\approx\;25,\qquad pK_a(CH_3 - C \equiv CH)\;\approx\;26,\qquad pK_a(CH_2 = CH_2)\;\approx\;44,$$

and the smaller the $$pK_a,$$ the stronger the acid.

Thus the correct sequence exactly matches Option B in the list.

Hence, the correct answer is Option B.

Step 1: Reaction with DCl (1 equivalent)

The starting material is propyne (CH3-C≡CH). It undergoes electrophilic addition with deuterium chloride (DCl) according to Markovnikov's rule.

1. The electrophile (D+) attacks the terminal carbon to form the more stable secondary carbocation: CH3-C+=CH-D
2. The nucleophile (Cl-) then attacks the carbocation center.

Product after Step 1: CH3-C(Cl)=CHD (2-chloro-1-deuteropropene)

Step 2: Reaction with DI

The alkene product from Step 1 now reacts with deuterium iodide (DI), again following Markovnikov's rule.

1. The electrophile (D+) attacks the terminal carbon because the resulting carbocation at C-2 is highly stabilized by the resonance/lone pair donation of the chlorine atom (carbocation backbonding). CH3-C+(Cl)-CHD2
2. The nucleophile (I-) attacks the C-2 carbocation.

Final Product: CH3-C(Cl)(I)-CHD2

Final Answer

The final major product of the reaction sequence is:

CH3-C(Cl)(I)-CHD2 (2-chloro-2-iodo-1,1-diduteropropane)

Reaction with Ag2O / Heat (Tollens'-like test for Alkynes)

When compound A is treated with silver oxide (Ag2O) and heat, it forms a precipitate (ppt).

• Deduction: This is a characteristic test for terminal alkynes (alkynes with a -C triple bond CH group). The acidic terminal hydrogen is replaced by a silver ion to form an insoluble silver acetylide precipitate.
• Therefore, A must be a terminal alkyne.

2. Hydration of Alkyne A to form B

When compound A (a terminal alkyne) undergoes hydration in the presence of Hg2+ / H+:

• Markovnikov addition of water occurs across the triple bond to form an enol, which quickly tautomerizes into a stable methyl ketone.
• Reaction path: R-C triple bond CH --> [ R-C(OH)=CH2 ] --> R-CO-CH3 (Compound B)

3. Reduction of B to C

Compound B (the methyl ketone) is reduced using sodium borohydride (NaBH4):

• NaBH4 selectively reduces ketones to secondary (2 degree) alcohols.
• Reaction path: R-CO-CH3 --> R-CH(OH)-CH3 (Compound C)

4. Reaction of C with Lucas Reagent (ZnCl2 / conc. HCl)

Compound C is treated with Lucas reagent, giving turbidity within 5 minutes.

• Lucas Test Rules:
  • 3 degree alcohols give turbidity immediately.
  • 2 degree alcohols give turbidity within 5 minutes.
  • 1 degree alcohols do not give turbidity at room temperature.
• Since C gives turbidity within 5 minutes, it confirms that C is a secondary alcohol, which perfectly matches the deduction from Step 3.

Summary of Compounds

• A: A terminal alkyne (R-C triple bond CH)
• B: A methyl ketone (R-CO-CH3)
• C: A secondary alcohol (R-CH(OH)-CH3)

We have the starting alkyne $$CH_{3}-C\equiv C-CH_{3}$$, whose IUPAC name is 2-butyne.

First we consider the reaction with $$H_{2}$$ in the presence of Lindlar’s catalyst. The Lindlar system is a poisoned palladium catalyst that brings about syn addition of one molecule of hydrogen across the triple bond. The syn mode of addition places the two newly added hydrogens on the same side of the former $$\pi$$-bond. As a result the product is the cis (Z) alkene:

$$CH_{3}-C\equiv C-CH_{3} \overset{H_2,\;\text{Lindlar}}{\longrightarrow} \text{CH}_3\text{-CH}=\text{CH-CH}_3 \;(\text{cis})$$

Hence, the compound we call $$X$$ is $$\text{cis-2-butene}$$.

Now we examine the reaction with metallic sodium in liquid ammonia, $$Na/liq.\;NH_{3}$$. This Birch-type reduction donates electrons in anti fashion, giving trans addition of hydrogen atoms across the alkyne. Consequently the product is the trans (E) alkene:

$$CH_{3}-C\equiv C-CH_{3} \overset{Na,\;\text{liq. }NH_3}{\longrightarrow} \text{CH}_3\text{-CH}=\text{CH-CH}_3 \;(\text{trans})$$

Therefore, compound $$Y$$ is $$\text{trans-2-butene}$$.

Next we compare the physical properties of the two alkenes.

For dipole moment we recall that the molecular dipole $$\mu$$ is the vector sum of all individual bond dipoles. In $$\text{cis-2-butene}$$ both $$C-CH_{3}$$ bond dipoles are oriented toward the same side of the double bond, so they reinforce each other and give a net dipole moment different from zero. In $$\text{trans-2-butene}$$ the molecule is planar and the two $$C-CH_{3}$$ bond dipoles lie in opposite directions, so they effectively cancel:

$$\mu_{\text{cis}} \gt 0,\qquad \mu_{\text{trans}}\approx 0$$

Thus, $$X$$ (cis) has the higher dipole moment, while $$Y$$ (trans) has the lower dipole moment.

Boiling point is influenced both by molecular mass and by intermolecular attractions. The masses are identical, so the decisive factor is the strength of the attractive forces. Because $$X$$ possesses a non-zero dipole, it exhibits dipole-dipole interactions in addition to London dispersion forces. These interactions raise its boiling point. In contrast, $$Y$$ has almost no permanent dipole and relies mainly on dispersion forces, giving it a lower boiling point. Experimental data corroborate this reasoning:

$$\text{b.p. of cis-2-butene}\;(X)\approx 3.7^{\circ}\text{C}$$
$$\text{b.p. of trans-2-butene}\;(Y)\approx 1.0^{\circ}\text{C}$$

So we conclude that $$X$$ possesses both the higher dipole moment and the higher boiling point relative to $$Y$$.

Among the given statements, only Option D matches this conclusion.

Hence, the correct answer is Option D.

A: $$Ph-C \equiv C-Ph + 2H_2 \xrightarrow{Pd/C} Ph-CH_2-CH_2-Ph$$

B: $$Ph-C \equiv C-Ph + H_2 \xrightarrow{\text{Lindlar's}} \text{cis-}Ph-CH=CH-Ph$$

C: $$Ph-C \equiv C-Ph \xrightarrow{Li/liq. NH_3} \text{trans-}Ph-CH=CH-Ph$$

D: $$Ph-C \equiv C-Ph \xrightarrow{LiAlH_4} \text{No Reaction}$$

An alkyne can be converted to an alkene by partial hydrogenation. The stereochemistry (cis / trans) of the product depends on the reagent used.

Case 1: Catalytic hydrogenation over Pd, Pt, Ni or Lindlar’s catalyst
  • Reagents such as $$Pt/H_2$$ or $$Pd/BaSO_4\;( \text{Lindlar’s catalyst})$$ add molecular hydrogen to the alkyne by a syn mechanism.
  • Both hydrogen atoms approach the triple bond from the same side, giving the cis (Z) or syn-alkene.
  • Example: $$RC \equiv CR \xrightarrow[\text{H}_2]{\text{Lindlar}} cis\text{-}RCH\!=\!CHR$$.

Case 2: Dissolving-metal reduction in liquid ammonia
  • Metals such as Na or Li in liquid $$NH_3$$ (−78 °C) generate solvated electrons. These electrons reduce the alkyne by an anti mechanism.
  • Hydrogen (from $$NH_3$$) adds to opposite faces of the π-system, producing the trans (E) alkene.
  • Example: $$RC \equiv CR \xrightarrow[\text{NH}_3]{\text{Na or Li}} trans\text{-}RCH\!=\!CHR$$.

Case 3: Other reagents in the list
  • $$LiAlH_4$$ is a hydride donor; it reduces alkyl halides, carbonyls, etc., but it does not reduce an isolated carbon-carbon triple bond.
  • Therefore it leaves $$2$$-hexyne unchanged.

Applying these facts to $$2$$-hexyne, the desired product is trans-$$2$$-hexene. Only the dissolving-metal reduction (Li in liquid $$NH_3$$) provides this anti addition.

Hence, the correct reagent is

Option B which is: $$Li/NH_3$$

When an unsymmetrical alkyne reacts with $$\text{H}_2\text{SO}_4$$ and $$\text{HgSO}_4$$, water adds across the triple bond following Markovnikov's rule. The electrophile ($$\text{H}^+$$) attacks the carbon that yields the more stable carbocation intermediate.

The triple bond is between the carbon attached to the phenyl ring ($$\text{C}_1$$) and the carbon attached to the methyl group ($$\text{C}_2$$):

$$\text{Ph}-\text{C}\equiv\text{C}-\text{CH}_3$$

Because the benzylic carbocation is significantly more stable, the nucleophile ($$-\text{OH}$$ from water) will attack $$\text{C}_1$$.

The addition of $$-\text{H}$$ and $$-\text{OH}$$ results in an unstable enol intermediate:

$$\text{Ph}-\overset{\text{OH}}{\text{C}}=\text{CH}-\text{CH}_3$$

Enols are generally unstable and rapidly undergo tautomerization to form the more thermodynamically stable carbonyl compound (keto form).

$$\text{Ph}-\overset{\text{OH}}{\text{C}}=\text{CH}-\text{CH}_3 \rightleftharpoons \text{Ph}-\overset{\text{O}}{\overset{\parallel}{\text{C}}}-\text{CH}_2-\text{CH}_3$$

The hydration of an alkyne is normally carried out with dilute $$H_2SO_4$$ in the presence of $$HgSO_4$$ (Kucherov reaction). Under these conditions, water adds across the triple bond according to Markovnikov’s rule, giving an enol that rapidly rearranges (tautomerises) to a carbonyl compound.

Propyne is $$CH_3-C\equiv CH$$. Add the elements of water, $$HO-H$$, so that $$OH$$ goes to the more substituted carbon of the triple bond (Markovnikov addition) and $$H$$ goes to the terminal carbon.

Step-1 (enol formation):
$$CH_3-C\equiv CH \xrightarrow[{HgSO_4}]{H_2O/H_2SO_4} CH_3-C(OH)=CH_2$$ (prop-1-en-2-ol)

Step-2 (keto-enol tautomerism): the enol rearranges by shifting the double bond and relocating the proton to give the more stable carbonyl (keto) form.
$$CH_3-C(OH)=CH_2 \longrightarrow CH_3-CO-CH_3$$

The carbonyl compound formed is $$CH_3-CO-CH_3$$, which is acetone (propan-2-one).

Therefore the major product obtained from hydration of propyne is acetone.

Option A which is: acetone

The reaction required is an electrophilic addition of two moles of $$HBr$$ to an unsymmetrical alkyne or alkene. For an alkyne, the first mole of $$HBr$$ gives a vinyl bromide; the second mole adds to the double bond formed in the first step. In the absence of peroxides, each addition follows Markovnikov’s rule: the hydrogen attaches to the carbon that already has more hydrogens and the bromine attaches to the other carbon.

Case A: $$CH_3\!-\!C\equiv CH$$ (propyne) + $$2\,HBr$$
  Step 1 (first mole of $$HBr$$):
  H attaches to the terminal carbon, Br to the internal carbon, giving $$CH_3\!-\!C(Br)=CH_2$$ (2-bromoprop-1-ene).
  Step 2 (second mole of $$HBr$$):
  H again attaches to the terminal carbon of the new double bond, Br to the same internal carbon, yielding $$CH_3\!-\!CBr_2\!-\!CH_3$$, i.e. $$2,2$$-dibromopropane.

Thus Option A furnishes the required geminal dihalide.

Case B: $$CH_3CH=CHBr$$ + $$HBr$$ → Markovnikov addition places H on $$CH_3$$ carbon and Br on the vinylic carbon that already bears Br, giving $$CH_3CH_2CHBr_2$$ (1,1-dibromopropane), not the desired product.

Case C: $$CH\equiv CH$$ + $$2\,HBr$$ → successively gives $$CH_2=CHBr$$ and then $$CH_3CHBr_2$$ (1,1-dibromoethane), which has only two carbon atoms.

Case D: $$CH_3CH=CH_2$$ + $$HBr$$ → gives the mono-bromide $$CH_3CHBrCH_3$$ (2-bromopropane), containing only one bromine atom.

Only Option A produces $$2,2$$-dibromopropane.

Answer: Option A which is: $$CH_3\!-\!C\equiv CH + 2\,HBr \longrightarrow CH_3\!-\!CBr_2\!-\!CH_3\; (2,2\text{-dibromopropane})$$