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, $4x+y=20$

$x=3$

$y=8$

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

The product S is

In this reaction, NH₃ chooses to attack at aliphatic aldehyde group than a less reactive aromatic aldehyde group. Now,

The reaction is

1, 4-addition is the major product as it is the result of the formation of a stable allylic carbocation.

The reactions involved are

The sequence of reactions for formation of Y are

As it has a CH₃CO$-$group, it gives positive iodoform test and it is a functional isomer of X as only the functional groups are different.

The reactions involved are

(A)

(B)

(C) ${(N{H_4})_2}C{O_3}\buildrel \Delta \over \longrightarrow 2N{H_3} + C{O_2} + {H_2}O$

(D)

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.

The structure of allene is CH₂ = C = CH₂ . Here both the terminal carbons are sp² hybridised while the central carbon is sp hybridised.

In presence of mineral acid $\mathrm{H}^{+}$(with no water) positively charged centre is generated at second position on the ring.

Since, oxygen is electronegative, it shifts pi electron towards itself generating a carbocation which attack one lone pair of electron on oxygen of alcohol.

Loss of proton $\left(\mathrm{H}^{+}\right)$from oxygen generates acetal.

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 ).

To get appropriate bromoalkane an alkyne that were involved in the synthesis, we have to cleave the molecule and we know that sodium amide is involved, so the synthesis of 3-octyne using the alkyne and bromoalkane was $\mathrm{S}_{\mathrm{N}}2$ reaction. And as the sodium amide is involved, the alkyne involved must be terminal then only it could abstract the acidic hydrogen 3-octyne is a molecule that will have eight carbon atoms in their parent chain since the prefix-oct represents 8 and as in the name it is given that 3 octyne, ' 3 ' represents the position of triple bond. Thus suffix-one represents triple bond.

So, in the molecule triple bond will be at the third carbon. Structure of 3-octyne :

Now, cleave the molecule between $\mathrm{C}_2-\mathrm{C}_3$ or $\mathrm{C}_4$ - $C_5$.

Now, see the products formed if cleavage takes place between C₂-C₃, the reactants will be

If cleavage takes place between C₄-C₅, the reactants will be :

Then correct option is D.

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.

Concentrated HNO$_3$ and concentrated H$_2$SO$_4$ will cause nitration of the ring to introduce $-$NO$_2$ in one of the rings.

The ring which is adjacent to N atom is more electron rich than the other. This is due to +R effect of NH group.

Further, it will attack on more activated ortho or para positions. Since, ortho attack is not possible due to bulky ring present adjacent to it, there will be an attack on para position.

Alcoholic KOH helps in beta elimination in 1, 2-dibromoethane to form ethene.

The proton of ethene cannot be abstracted by a second molecule of ethene; hence, a much stronger base such as sodium amide is used to carry out another elimination to form ethyne.

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

The propene on reaction with nitrosyl chloride, NOCl produces an adduct as shown below.

The propylene undergoes an electrophilic addition reaction, following the Markownikoff's rule.

Step 1:

Decomposition of nitrosyl chloride

$ \mathrm{NOCl} \rightleftharpoons \mathrm{NO}^{+}+\mathrm{Cl}^{-} $

Step 2:

Addition of positive part of nitrosyl chloride on carbon atom of double bond containing highest number (2 degree carbocation) of hydrogen atoms. A stable intermediate is produced, during this slow step.

Step 3:

Addition of chloride ion on carbon atom of double bond containing least number of hydrogen atoms. The adduct is obtained during a very fast reaction.

$ \text { The structure of } \mathrm{C}_6 \mathrm{H}_5 \mathrm{COCl} \text { is shown below: } $

The IUPAC name of the compound is benzene carbonyl chloride.

The compound is derived from benzoic acid, $\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}$ by replacing - OH group by Cl group.

The $\mathrm{C}_6 \mathrm{H}_5$ group stands for benzene, CO group stands for carbonyl and Cl stands for chloride group. The functional group in the compound is carbonyl.

The complete name of the compound $\mathrm{C}_6 \mathrm{H}_5 \mathrm{COCl}$ is benzene carbonyl chloride.

$ \text { The chemical reaction is given below. } $

$ \text { It is a two step reaction: } $

Step 1: Friedel Craft Alkylation

The product is P is formed by reaction between benzene and 1-chloropropane in presence of anhydrous $\mathrm{AlCl}_3$.

Hence, product $P$ is cumene.

Step 2: Cumene Hydroperoxide Rearrangement

The second step is oxidation of cumene to form Cumene hydroperoxide.

The cumene hydroperoxide on acidic hydrolysis produces phenol and acetone.

On hydrolysis cumene hydroperoxide rearranges to give phenol and acetone as final products.

Hence, the product $P$ is isopropyl benzene and Q is acetone with chemical formula $\mathrm{CH}_3 \mathrm{COCH}_3$.

$2,3$-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.