Hydrocarbons

Among the given sets, only III is correct i.e.

cis-but-2-ene > trans-but-2-ene (dipole moment) while I and II are incorrect. Their correct form is (I) torsional strain is more in eclipsed ethane than in staggered ethane. It is because in eclipsed form hydrogen atoms attached to two carbons are as closed together as possible while in staggered form hydrogen are as far apart as possible. 2-dimethyl butane < 2-methyl pentane (B.P)

The complete reaction sequence is as follows

So, the molecular formula is $\mathrm{C}_6 \mathrm{H}_6 \mathrm{Cl}$ and its empirical formula is CHCl . Thus, the empirical formula weight of $Z$ is 48.5

$t$-butyl benzene does not give benzoic acid when treated with alkaline $\mathrm{KMnO}_4$ because it lack the necessary benzylic hydrogen required for oxidation to form benzoic acid, since the tertiary carbon in $t$-butylbenzene has no hydrogen attached directly to it, the oxidation cannot proceeds to form carboxylic acid.

Statement I is wrong, but Statement II is right.

Statement I should say: In Kolbe's electrolysis of sodium propionate, the product is butane, not n-hexane.

The chemical reaction is:

$ \begin{aligned} & 2 \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COONa} + 2 \mathrm{H}_2 \mathrm{O} \longrightarrow \\ & \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3 + 2 \mathrm{CO}_2 + 2 \mathrm{NaOH} + \mathrm{H}_2 \uparrow \\ & \text { Butane } \end{aligned} $

In Statement II, it is correct that in Kolbe's process, carbon dioxide ($\mathrm{CO}_2$) is released at the anode and hydrogen gas ($\mathrm{H}_2$) is released at the cathode.

The primary catalyst used for the isomerisation of $n$-alkanes to branch

chain alkanes are acid catalyst, particularly zeolites and sulphated zirconia.

Anhydrous $\mathrm{AlCl}_3$ or HCl can be also used.

The complete reaction is as follows

This is Wurtz-fittig reaction

$ \text { This is Friedel-Craft alkylation. } $

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} $

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

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

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,

Here for unknown formula given $\mathrm{C}_6 \mathrm{H}_{10}$, we can predict molecular structure of given formula by using double bond equivalent.

$\begin{aligned} & \mathrm{DBE}=\mathrm{C}-\frac{\mathrm{H}}{2}+\frac{\mathrm{N}}{2}-\frac{X}{2}+1 \\ & \Rightarrow \quad=6-\frac{10}{2}+0-0+1 \\ & \Rightarrow \quad=6-5+1=2 \end{aligned}$

Hence, the hydrocarbon X is 3-methylpent-1-yne.

When chlorobenzene is heated with 6% solution of NaOH at 623 K and under pressure 300 atm to form sodium phenoxide.

Further on acidified with dilute HCl, it will form phenol.