Hydrocarbons

The complete reaction is as follows,

The final product is benzene, which is also formed by aromatisation of $n$-hexane.

The reaction is a Wurtz reaction, In the given case, three alkanes are formed

$ 2 \mathrm{C}_2 \mathrm{H}_5 \mathrm{Br}+2 \mathrm{Na} \xrightarrow[\text { Ether }]{\text { Dry }} \underset{\text { Butane }}{\mathrm{C}_4 \mathrm{H}_{10}}+2 \mathrm{NaBr} $

$ 2 \mathrm{C}_3 \mathrm{H}_7 \mathrm{Br}+2 \mathrm{Na} \xrightarrow[\text { Ether }]{\text { Dry }} \underset{\text { Hexane }}{\mathrm{C}_6 \mathrm{H}_{14}}+2 \mathrm{NaBr} $

$ \begin{aligned} \mathrm{C}_2 \mathrm{H}_5 \mathrm{Br}+\mathrm{C}_3 \mathrm{H}_7 \mathrm{Br}+2 \mathrm{Na} \longrightarrow & \underset{\text { Pentene }}{\mathrm{C}_5 \mathrm{H}_{12}} +2 \mathrm{NaBr} \end{aligned} $

The complete reaction sequence is as follows

Thus, statement given in option $(C)$ is incorrect $z$ on reaction with $\mathrm{CH}_3 \mathrm{MgBr}$ followed by hydrolysis will give $3^{\circ}$ alcohol.

The complete reaction is as follows,

$ \begin{aligned} & x=\mathrm{H}_2 \mathrm{O} / \mathrm{H}_2 \mathrm{SO}_4, \mathrm{Hg}^{2+} \\ & y=\mathrm{PCC} \end{aligned} $

To evaluate which statement regarding ethyne is incorrect, it's necessary to review the characteristics of the chemical bonds in ethyne (acetylene, $\mathrm{C_2H_2}$) compared to ethene (ethylene, $\mathrm{C_2H_4}$).

Option A: The statement that the carbon-carbon bonds in ethyne are weaker than that in ethene is incorrect. In ethyne, the carbon-carbon triple bond is composed of one sigma ($\sigma$) bond and two pi ($\pi$) bonds. This configuration makes the triple bond in ethyne much stronger and shorter compared to the double bond in ethene, which consists of one $\sigma$ bond and one $\pi$ bond. Thus, the carbon-carbon bond in ethyne is actually stronger than that in ethene.

Option B: The statement that both carbons are sp hybridised in ethyne is correct. In ethyne, each carbon atom is bonded to the other carbon atom and a single hydrogen atom, necessitating the sp hybridisation to form a linear molecule with $180^\circ$ angles between the bonds.

Option C: The statement that the $\mathrm{C-C}$ bonds in ethyne are shorter than that in ethene is correct. The presence of a triple bond between the carbon atoms in ethyne results in a shorter bond length compared to the double-bonded carbons in ethene. This is because the additional pi bonds in a triple bond pull the carbon atoms closer together.

Option D: The statement that ethyne is linear is correct. Due to the sp hybridisation of carbon atoms in ethyne, the molecule adopts a linear geometry with bond angles of $180^\circ$.

Therefore, the incorrect statement regarding ethyne is Option A: The carbon-carbon bonds in ethyne are weaker than that in ethene.

Statement (I) explains a characteristic of $\mathrm{S}_{\mathrm{N}}2$ (Substitution Nucleophilic Bimolecular) reactions. In $\mathrm{S}_{\mathrm{N}}2$ reactions, the attack of the nucleophile and the departure of the leaving group occur simultaneously in a single step. This reaction has a back-side attack mechanism, where the nucleophile attacks the electrophilic carbon from the opposite side of the leaving group. This back-side attack leads to the inversion of configuration at the carbon atom undergoing substitution, a phenomenon often referred to as 'Walden inversion'. Thus, $\mathrm{S}_{\mathrm{N}}2$ reactions are stereospecific, leading to the formation of a product with a specific configuration (either inversion or retention of configuration, but typically inversion in the case of $\mathrm{S}_{\mathrm{N}}2$). Therefore, Statement I is true.

Statement (II) concerns $\mathrm{S}_{\mathrm{N}}1$ (Substitution Nucleophilic Unimolecular) reactions, which proceed in two steps: first, the leaving group departs, forming a carbocation intermediate, and then the nucleophile attacks. The attack of the nucleophile can occur from either side of the planar carbocation, resulting in the formation of both enantiomers of the product. When the reactant is chiral, the product is usually a racemic mixture, consisting of equal amounts of both enantiomers, assuming that there is no steric or electronic preference for the attack on the carbocation. This makes $\mathrm{S}_{\mathrm{N}}1$ reactions non-stereospecific. Therefore, Statement II is true as well.

Based on the explanations provided, the correct answer is Option D: Both Statement I and Statement II are true.

$\text { Number of } 2^{\circ} \text { carbon atom }=1$

In aqueous medium the reaction follows $\mathrm{S_N 1}$ mechanism majorly

Due to Lewis base formation electrophile stops/greatly slowed. Hence statement 2 is false while statement 1 is true.

Hence correct Ans. (4).

$\mathrm{PhCH}=\mathrm{CH}_2 \xrightarrow{\mathrm{B}_2 \mathrm{H}_6 / \mathrm{H}_2 \mathrm{O}_2 \mathrm{OH}^{-}} \mathrm{PhCH}_2 \mathrm{CH}_2 \mathrm{OH}$

The complete reaction is as follows

respectively.

Next alkane to methane is ethane and it is prepared by reaction given in statement (I) and statement (IV). The reactions are

$ 2 \mathrm{CH}_{3} \mathrm{Br} \xrightarrow{2 \mathrm{Na}} \mathrm{CH}_{3}-\mathrm{CH}_{3}+2 \mathrm{NaBr} $

$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}+2[\mathrm{H}] \xrightarrow[\mathrm{H}^{+}]{\mathrm{Zn}} \mathrm{C}_{2} \mathrm{H}_{6}+\mathrm{HBr} $

When methane is heated with catalyst $ X $ (which is copper at 573 K ) it gives methanol on methyl alcohol. When methane is reacted with $ \mathrm{O}_{2} $ in presence of $ \mathrm{MO}_{2} \mathrm{O}_{3} $ then methanal (formaldehyde) is formed.

The complete reaction are as follows,

$ \begin{aligned} & \underset{\text { Methane }}{2 \mathrm{CH}_4}+\mathrm{O}_2 \xrightarrow[573 \mathrm{~K} / 100 \text { atm }]{\mathrm{Cu}} \underset{\text { Methanol }}{2 \mathrm{CH}_3 \mathrm{OH}} \\\\ & \underset{\text { Methane }}{\mathrm{CH}_4}+\mathrm{O}_2 \xrightarrow[\Delta]{\mathrm{Mo}_2 \mathrm{O}_3}\underset{\text { Methanal }}{\mathrm{HCHO}}+\mathrm{H}_2 \mathrm{O} \end{aligned} $

The complete reaction sequence is as follows

The compound $ A $ is sodium butanoate.

When sodium salt of butanoic acid is heated with soda lime, it readily undergoes decarboxylation. The reaction is as follows,

$\underset{\text { Sodium butanoate }}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{COONa}}+\underset{\text { Sodalime }}{(\mathrm{NaOH}+\mathrm{CaO})} \longrightarrow \underset{\text { Propane }}{\mathrm{Na}_2 \mathrm{CO}_3+\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_3}$

$\begin{gathered}\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCH}_3 \\ {\left[X: \mathrm{C}_4 \mathrm{H}_8\right]}\end{gathered}\xrightarrow[{\mathrm{[O]}}]{\mathrm{KMnO}_4 / \mathrm{H}^{+}} 2 \mathrm{CH}_3 \mathrm{COOH} \underset{(Y)}{ }$

$\Rightarrow \underset{(\text { Sodium salt of } Y)}{\mathrm{CH}_3 \mathrm{COONa}} \xrightarrow[\substack{\text { (Sodalime) } \\-\mathrm{Na}_2 \mathrm{CO}_3}]{(\mathrm{NaOH}+\mathrm{CaO}) / \Delta} \underset{(Z)}{\mathrm{CH}_4}$

Note: But-2-ene $\left(\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3\right)$ exhibits geometrical isomerism which undergoes oxidative cleavage of $\mathrm{C}=\mathrm{C}$ with $\mathrm{KMnO}_4 / \mathrm{H}^{+}$to give $Y$.

The complete reaction is as follows,

$ \mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2 $

$ \left(\mathrm{C}_3 \mathrm{H}_6\right) \xrightarrow[(X)]{\mathrm{HBr} / R \mathrm{OOR}, \text { Kharasch affect }} $

$ \underset{(Y)}{\mathrm{C}_3 \mathrm{H}_7 \mathrm{Br}} \underset{\mathrm{AlCl}_3}{\mathrm{C}_6 \mathrm{H}_6} $

$\underset{\substack{\text { 3-hexene }}}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}}=\mathrm{CHCH}_2 \mathrm{CH}_3 \xrightarrow[\text { (ii) } \mathrm{Zn}, \mathrm{H}_2 \mathrm{O}]{\text { (i) } \mathrm{O}_3}$

$2 \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHO} \xrightarrow{\mathrm{Zn}-\mathrm{Hg}, \mathrm{HCl}} \underset{\text { Propane }}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_3}$

So, the correct sequence of reagent is (i) $\mathrm{O}_3$ (ii) $\mathrm{Zn}, \mathrm{H}_2 \mathrm{O} ; \mathrm{Zn}-\mathrm{Hg}, \mathrm{HCl}$.

The complete reaction is as follows

The given reaction is Wurtz Fifting reaction.

It is a chemical process that produces substituted aromatic compounds from aryl halides and alkyl halides in presence of sodium and dry ether. The complete reactions are as follow,

In Kolbe's electrolysis, the sodium or potassium salt of a carboxylic acid undergoes decarboxylation to form the corresponding alkanes. The overall reaction is outlined below:

$ \begin{aligned} & 2 \, \text{CH}_3\text{CH}_2\text{COONa} + 2 \, \text{H}_2\text{O} \xrightarrow[\text{electrolysis}]{\text{Kolbe's}} \\ & \underset{\text{(At anode)}}{\text{C}_4\text{H}_{10}} + 2 \, \text{CO}_2 + 2 \, \text{NaOH} + \underset{\text{(At cathode)}}{\text{H}_2} \end{aligned} $

Thus, the products at the anode and cathode are:

Anode: $\text{C}_4\text{H}_{10}$ (butane)

Cathode: $\text{H}_2$ (hydrogen gas)

Wurtz-Fittig reaction Aryl halide and alkyl halide react in presence of sodium metal/dry ether to form alkyl benzene.

The complete reaciton is as follows,

$ \text { This reaction is called as Wurtz reaction. } $

In the Wurtz reaction, alkanes are synthesized by reducing alkyl halides using sodium in dry ether. This process combines two alkyl groups to form higher alkanes.

$ R-X + 2 \mathrm{Na} + X-R_1 \xrightarrow{\text{Dry ether}} R-R_1 + 2 \mathrm{NaI} $

When a mixture of ethyl iodide and $n$-propyl iodide undergoes the reaction, the following higher alkanes are formed:

(i) Butane:

$ \begin{aligned} &\mathrm{CH}_3\mathrm{CH}_2-\mathrm{I} + 2\mathrm{Na} + \mathrm{I}-\mathrm{CH}_2\mathrm{CH}_3 \\ &\longrightarrow \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_3 + 2 \mathrm{NaI} \end{aligned} $

(ii) Hexane:

$ \begin{aligned} &\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2-\mathrm{I} + 2\mathrm{Na} + \mathrm{I}-\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_3 \\ &\longrightarrow \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_3 + 2 \mathrm{NaI} \end{aligned} $

(iii) Pentane:

$ \begin{aligned} &\mathrm{CH}_3\mathrm{CH}_2\mathrm{I} + 2\mathrm{Na} + \mathrm{I}-\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_3 \\ &\longrightarrow \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_3 + 2 \mathrm{NaI} \end{aligned} $

Thus, propane is the only hydrocarbon not formed in this reaction.

$S_{\mathrm{N}} 1$ mechanism are generally carried out in polar protic solvents and generally occurs in two steps.

Among the given reactions products of $B$ and $C$ will undergo hydrolysis by $S_N 1$ mechanism, as the products are $2^{\circ} / 3^{\circ}$. benzy bromides which will furnish very stable carbocations.

Thus, the compound $X$ and $Z$ are $\mathrm{HBr} /\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CO}\right)_2 \mathrm{O}_2$ and $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{CHO}$.

$ \text { The complete reaction is as follows } $

But-1-ene does not show cis-frans isomerism (geometrical).

The complete reaction is as follows,