Hydrocarbons

We know, conbustion of Hydrocarbon

C_x H_y + (x + ${y \over 4}$)O₂  $ \to $  x CO₂ + ${y \over 2}$ H₂O

So, to consume 1 mole of C_x H_y we need

(x + ${y \over 4}$) mole of O₂ gas.

As both C_x H_y and O₂ are gas,

So avogadro's law is applicable on them.

At constant temperature and pressure according to avogadro's law, volume $ \propto $ mole

So, for 5L of C_x H_y the

volume of O₂ = 5(x + ${y \over 4}$)

According to the question,

5(x + ${y \over 4}$) = 25

$\therefore\,\,\,$ (x + ${y \over 4}$) = 5 . . . . . (1)

For Ethane (C₂ H₆), x = 2 and y = 6

$\therefore\,\,\,$ x + ${y \over 4}$ = 2 + ${6 \over 4}$ $ \ne $ 5

$\therefore\,\,\,$ C₂ H₆ can't be the required alkane.

For Propane (C₃ H₈), x = 3. and y = 8

$\therefore\,\,\,$ x + ${y \over 4}$ = 3 + ${8 \over 4}$ = 5

$\therefore\,\,\,$ Propane (C₃ H₈) is the right alkane.

Similarly for Butane (C₄ H₁₀) and Isobutane (C₄ H₁₀) you can check
x + ${y \over 4}$ $ \ne $ 5.

${C_x}{H_{y\left( g \right)}} + \left( {{{4x + y} \over 4}} \right){O_{2\left( g \right)}}$ $ \to xC{O_{2\left( g \right)}} + {y \over 2}{H_2}O\left( l \right)$

Volume of ${O_2}$ used

$ = 375 \times {{20} \over {100}} = 75ml$

$\therefore$ From the reaction of combination

$1\,\,ml\,{C_x}{H_y}\,\,$ requires $ = {{4x + y} \over 4}ml\,{O_2}$

$15\left( {{{4x + y} \over 4}} \right) = 75$

So, $4x+y=20$

$x=3$

$y=8$

$\therefore$ ${C_3}{H_8}$

Anti addition of hydrogen atoms to the the triple bond occurs when alkynes are reduced with sodium (or lithium) metal in ammonia, ethylamine, or alcohol at low temperatures. This reaction called, a dissolving metal reduction, produces an $(E)$- or $trans$-alkene.

Sodium in liq. $N{H_3}$ is used as a source of electrons in the reduction of an alkyne to a $trans$ alkene.

Whenever dehydration can produce two different alkenes, major product is formed according to Saytzeff rule i.e. more substituted alkene (alkene having lesser number of hydrogen atoms on the two doubly bonded carbon atoms ) is the major product.

Such reactions which can produce two or more structural isomers but one of them in greater amounts than the other are called regioselective; in case a reaction is $100$% regioselective, it is termed as regiospecific.

In addition to being regioselective, alcohol dehydrations are stereoselective (a reaction in which a single starting material can yield two or more stereoismeric products, but gives one of them in greater amount than any other ).

on reduction with $Sn/HCl$ they will form corresponding anilines in which $ - N{O_2}$ group changes to $ - N{H_2}.$



These anilines when diazotized and then treated with $CuBr$ forms $o-, p-$ bromotoluenes.

Writing the reaction we get



So we find that $C{H_4}$ is produced in this reaction.

Alkynes having terminal $ - C \equiv H$ react with Na in liquid ammonia to yield ${H_2}$ gas of the given compounds $C{H_3}C{H_2}C \equiv CH$ can react with Na in liquid $N{H_3}$ so the correct answer is $(b).$

Completing the sequence of given reactions,



Thus $'B'\,\,$ is $\,\,C{H_3}CHO$

Hence $(d)$ is correct answer.

The reaction follows Markownikoff rule which states that when unsymmetrical reagent adds across unsymmetrical double or triple bond the negative part adds to carbon atom having lesser number of hydrogen atoms.

$FeC{l_3}\,\,$ is Lewis acid. In presence of $FeC{l_3}\,\,$ side chain hydrogen atoms of toluene are substituted.

Hofmann's rule : When theoretically more than one type of alkenes are possible in eliminations reaction, the alkene containing least alkylated double bond is formed as major product. Hence



NOTE : It is less stearically $\beta $-hydrogen is removed

$2,3$-dichloro butane will exhibit optical isomerism due to the presence of two asymmetric carbon atom.

Ease of replacement of $H$-atom ${3^ \circ } > {2^ \circ } > {1^ \circ }.$

Since it contains only two types of $H$-atoms hence it will give only two mono chlorinated compounds viz.



and

NOTE : Among isomeric alkanes, the straight chain isomer has higher boiling point than the branched chain isomer. The greater the branching of the chain, the lower is the boiling point. Further due to the presence of $\pi $ electrons, these molecules are slightly polar and hence have higher boiling points than the corresponding alkanes.

Thus $B.pt.$ follows the order

alkynes $>$ alkene $>$ alkanes (straight chain) > branched chain alkanes.

In neopentane all the $H$ atoms are same $\left( {{1^ \circ }} \right).$

Alkenes combine with hydrogen under pressure and in presence of a catalyst $(Ni, Pt$ or $Pd)$ and form alkanes.

Acetylene reacts with the other three as :

1. Acetylene reacts with sodium :

$\mathrm{2CH} \equiv \mathrm{CH} + 2\mathrm{Na} \longrightarrow 2\mathrm{CH} \equiv \mathrm{C}^{-}\mathrm{Na}^{+} + \mathrm{H}_2$

Here, acetylene reacts with sodium to produce sodium acetylide and hydrogen gas.

2. Acetylene reacts with ammoniacal silver nitrate :

$\mathrm{CH} \equiv \mathrm{CH} + 2\mathrm{AgNO}_3 \longrightarrow \mathrm{Ag}_2\mathrm{C}_2 + 2\mathrm{HNO}_3$

In this reaction, acetylene reacts with ammoniacal silver nitrate to form a white precipitate of silver acetylide and nitric acid.

3. Acetylene reaction with hydrochloric acid in the presence of a $\mathrm{Hg}^{2+}$ catalyst (oxymercuration-demercuration) :

$\mathrm{CH} \equiv \mathrm{CH} + \mathrm{Hg(OAc)_2} + \mathrm{H}_2\mathrm{O} \longrightarrow \mathrm{CH}_2=\mathrm{CH}-\mathrm{OH} + \mathrm{HgOAc}$

$\mathrm{CH}_2=\mathrm{CH}-\mathrm{OH} + \mathrm{NaBH}_4 \longrightarrow \mathrm{CH}_2=\mathrm{CH}-\mathrm{H} + \mathrm{B(OH)_3} + \mathrm{NaOAc}$

Acetylene does not readily react with $\mathrm{HCl}$. However, in the presence of a catalyst like $\mathrm{Hg}^{2+}$, a specific type of electrophilic addition known as oxymercuration can occur. The result is vinyl chloride. Note that $\mathrm{NaBH}_4$ is used to reduce the mercurinium ion intermediate and to replace the $\mathrm{Hg}$ group with a hydrogen atom.

4. Acetylene does not react with sodium hydroxide :

$\mathrm{CH} \equiv \mathrm{CH} + \mathrm{NaOH} \longrightarrow \text { no reaction }$

According to the principles of the Hard and Soft Acids and Bases (HSAB) theory, soft acids prefer to bind with soft bases and hard acids prefer to bind with hard bases. In this case, $\mathrm{OH}^-$ is a hard base, while the $\mathrm{C}$ in acetylene is a soft acid, so they are not particularly reactive with each other.

5. Acetylene reacts with sodium amide :

$\mathrm{CH} \equiv \mathrm{CH} + \mathrm{NaNH}_2 \longrightarrow \mathrm{CH} \equiv \mathrm{C}^{-}\mathrm{Na}^{+} + \mathrm{NH}_3$

Here, sodium amide acts as a strong base and nucleophile, deprotonating the acetylene to form sodium acetylide and ammonia.