Mix Examples-Hydrocarbon Questions in English

(B) Natalite is a mixture of $95\%$ ethanol and $5\%$ ether. It is used as a fuel or a substitute for petrol in internal combustion engines.

(A) The compound '$A$' must contain an alkene group to decolourise $Br_2 / CCl_4$ and a primary amine group $(-NH_2)$ to release $N_2$ gas upon reaction with nitrous acid $(HNO_2)$.
$1$. Reaction with $HNO_2$: Primary aliphatic amines react with $HNO_2$ to form unstable diazonium salts,which decompose to release $N_2$ gas and form alcohols.
$2$. Reaction with $Br_2 / CCl_4$: The presence of a carbon-carbon double bond $(C=C)$ leads to the addition of bromine,which causes the reddish-brown colour of $Br_2$ to disappear.
Among the given options,Cyclopent$-3-$en$-1-$amine contains both a primary amine group and a double bond,satisfying both conditions.

(B) Acidic strength $\propto \ \% \ s$-character.
The acidity of hydrogen atoms attached to carbon depends on the hybridization of the carbon atom. The order of electronegativity is $sp > sp^2 > sp^3$,which corresponds to $50\% \ s$,$33.3\% \ s$,and $25\% \ s$ character respectively.
$1$. $CH \equiv CH$ $(II)$: $sp$ hybridized carbon ($50\% \ s$-character),most acidic.
$2$. $CH_3-C \equiv CH$ $(III)$: Terminal alkyne,but the $+I$ effect of the $-CH_3$ group decreases the acidity compared to ethyne.
$3$. $H_2C=CH_2$ $(I)$: $sp^2$ hybridized carbon ($33.3\% \ s$-character).
$4$. $CH_3-CH_3$ $(IV)$: $sp^3$ hybridized carbon ($25\% \ s$-character),least acidic.
Thus,the correct order of acidic strength is $IV < I < III < II$.

(C) The reaction sequence is as follows:
$1$. Chlorination of propane $(CH_3CH_2CH_3)$ with $Cl_2$ in the presence of light $(hv)$ gives $1$-chloropropane as the major product: $CH_3CH_2CH_3 + Cl_2 \xrightarrow{hv} CH_3CH_2CH_2Cl + HCl$.
$2$. Dehydrohalogenation of $1$-chloropropane with alcoholic $KOH$ gives propene: $CH_3CH_2CH_2Cl \xrightarrow{\text{alc. } KOH} CH_3CH=CH_2 + KCl + H_2O$.
$3$. High-temperature chlorination of propene $(773 \ K)$ leads to allylic substitution,forming allyl chloride: $CH_3CH=CH_2 + Cl_2 \xrightarrow{773 \ K} ClCH_2CH=CH_2 + HCl$.
$4$. Nucleophilic substitution of the allylic chloride with aqueous $AgOH$ (or $NaOH$) replaces the $Cl$ atom with an $OH$ group to form allyl alcohol: $ClCH_2CH=CH_2 + AgOH_{(aq)} \longrightarrow HOCH_2CH=CH_2 + AgCl$.

(D) Step $1$: Dehydrohalogenation using $Alc. KOH/\Delta$ (E2 elimination). The starting material has two $-Br$ groups. Elimination of two moles of $HBr$ results in the formation of two double bonds: one terminal vinyl group $(-CH=CH_2)$ and one exocyclic double bond ($-C(CH_3)=CH_2$ is not formed,rather the double bond forms between the ring and the side chain). Wait,looking at the structure,the elimination creates a vinyl group and an exocyclic double bond. The alkyne remains unaffected.
Step $2$: Electrophilic addition of excess $Br_2/CCl_4$. The molecule now contains one terminal alkene,one exocyclic alkene,and one internal alkyne.
- The terminal alkene adds $1$ mole of $Br_2$ ($2$ $Br$ atoms).
- The exocyclic alkene adds $1$ mole of $Br_2$ ($2$ $Br$ atoms).
- The internal alkyne adds $2$ moles of $Br_2$ ($4$ $Br$ atoms).
Total $Br$ atoms added = $2 + 2 + 4 = 8$ atoms.
Including the $2$ $Br$ atoms already present in the starting material is incorrect as the question asks for the number of bromine atoms in the final product $P$. Counting the structure of $P$ in the solution image: there are $2$ $Br$ on the terminal side chain,$2$ $Br$ on the ring,$4$ $Br$ on the alkyne side chain. Total = $8$ atoms.

(A-IV, B-II, C-III, D-I) . $1,6$-dibromohexane reacts with $Zn$ to undergo intramolecular cyclization to form cyclohexane. Thus,$A-iv$.
$B$. Ethanol $(C_2H_5OH)$ reacts with concentrated $H_2SO_4$ at $443 \ K$ to undergo dehydration to form ethene $(H_2C=CH_2)$. Thus,$B-ii$.
$C$. Propene reacts with $HBr$ in the presence of peroxide (anti-Markovnikov addition) to form $1$-bromopropane $(CH_3-CH_2-CH_2-Br)$. Thus,$C-iii$.
$D$. $1,1$-dibromopropane reacts with $NaNH_2$ (a strong base) to undergo dehydrohalogenation to form propyne $(H_3C-C \equiv CH)$. Thus,$D-i$.
Therefore,the correct matching is $A-iv, B-ii, C-iii, D-i$.

(A) The hydrocarbon has the molecular formula $C_4H_6$.
Cyclobutene $(C_4H_6)$ reacts with $Br_2$ to decolourise it,reacts with $HBr$ to form bromocyclobutane,and undergoes ozonolysis $(O_3 / Zn + H_2O)$ to produce butane$-1,4-$dial $(C_4H_6O_2)$.
Therefore,the correct structure is cyclobutene.

(A) . Hydration of propyne $(CH_3-C \equiv CH)$ in the presence of $Hg^{2+}/H^+$ follows Markovnikov's rule to form an enol,which tautomerizes to acetone $(CH_3-CO-CH_3)$,which is $II$.
$B$. Kolbe's electrolysis of sodium acetate $(CH_3COONa)$ yields ethane $(CH_3-CH_3)$,which is $IV$.
$C$. Acid-catalyzed hydration of propene $(CH_3-CH=CH_2)$ follows Markovnikov's rule to form propan$-2-$ol $(CH_3-CH(OH)-CH_3)$,which is $I$.
$D$. Hydroxylation of propene using cold dilute alkaline $KMnO_4$ (Baeyer's reagent) yields propane$-1,2-$diol $(CH_3-CH(OH)-CH_2OH)$,which is $III$.
Therefore,the correct match is $A-II, B-IV, C-I, D-III$.

(C) $1$. The reaction begins with propene $(C_3H_6)$.
$2$. The reagent $X$ is $HBr/ROOR$ (peroxide effect/anti-Markovnikov addition),which converts propene to $1$-bromopropane $(Y = CH_3CH_2CH_2Br)$.
$3$. In the presence of $AlCl_3$,$1$-bromopropane undergoes rearrangement to form a more stable isopropyl carbocation $(CH_3CH^+CH_3)$.
$4$. This isopropyl carbocation then undergoes Friedel-Crafts alkylation with benzene $(C_6H_6)$ to form isopropylbenzene (cumene) as the major product $(Z)$.
$5$. Thus,$X$ is $HBr/ROOR$ and $Z$ is isopropylbenzene.

(A) Step $1$: The reduction of but-$2$-yne with $Na/NH_3(liq)$ in the presence of ethanol is a Birch reduction,which yields $trans$-but-$2$-ene as the intermediate $X$.
Step $2$: The reaction of $trans$-but-$2$-ene with dilute alkaline $KMnO_4$ (Baeyer's reagent) proceeds via $syn$-dihydroxylation.
Step $3$: $Syn$-dihydroxylation of $trans$-alkenes results in the formation of a racemic mixture of enantiomers.
Step $4$: The product is a racemic mixture of $(2R, 3R)$-butane-$2,3$-diol and $(2S, 3S)$-butane-$2,3$-diol,which corresponds to the structures shown in option $A$.

(B) The reaction of $MeMgBr$ (a Grignard reagent) with hydrocarbons occurs if the hydrocarbon has an acidic hydrogen atom. $MeMgBr$ acts as a base and abstracts the most acidic proton to form methane $(CH_4)$.
Among the given options,$1,3$-cyclopentadiene contains a methylene group $(-CH_2-)$ between two double bonds. The hydrogen atoms on this carbon are significantly more acidic than those in other hydrocarbons because the resulting cyclopentadienyl anion is aromatic ($6\pi$ electrons,$H$ückel's rule),which provides high stability.
Therefore,$1,3$-cyclopentadiene reacts readily with $MeMgBr$ to form methane and the cyclopentadienyl Grignard reagent.

(C) The hydrogenation of benzene $(C_6H_6)$ requires $3 \text{ equivalents}$ of $H_2$ to fully reduce it to cyclohexane $(C_6H_{12})$.
$C_6H_6 + 3H_2 \rightarrow C_6H_{12}$
In this reaction,we are given $1 \text{ equivalent}$ of benzene and only $1 \text{ equivalent}$ of $H_2$.
Since the stoichiometry requires $3 \text{ equivalents}$ of $H_2$ for $1 \text{ equivalent}$ of benzene,$1 \text{ equivalent}$ of $H_2$ will react with $1/3$ $(0.33)$ of the benzene molecules to fully reduce them to cyclohexane,while $2/3$ $(0.66)$ of the benzene molecules will remain unreacted.
Therefore,the final mixture will contain $0.66 \text{ equivalents}$ of unreacted benzene and $0.33 \text{ equivalents}$ of cyclohexane.

(B) $1$. Compound $'X'$ $(C_3H_6Cl_2)$ reacts with excess $NaNH_2$ to form propyne $(CH_3-C \equiv CH)$ as compound $'Y'$.
$2$. Hydration of propyne $(Y)$ with $dil. H_2SO_4/Hg^{2+}$ gives acetone $(CH_3-CO-CH_3)$ as compound $'Z'$. Acetone contains a methyl ketone group and gives a yellow precipitate of iodoform $(CHI_3)$ with $NaOI$. Thus,Statement $I$ is true.
$3$. Cyclic trimerization of propyne $(Y)$ using a red hot iron tube gives $1,3,5$-trimethylbenzene (mesitylene) as compound $'Q'$ $(C_9H_{12})$.
$4$. In $1,3,5$-trimethylbenzene,there are $3$ aromatic $H$ atoms (on the ring) and $9$ aliphatic $H$ atoms (in the three methyl groups). The ratio of aromatic to aliphatic $H$ atoms is $3:9 = 1:3$. Thus,Statement $II$ is true.

(C) The reaction sequence is as follows:
$1$. The compound $[P]$ is $1,1-dibromo-1-(4-methylphenyl)methane$ (also known as $4-methylbenzal$ bromide,but with the formula $C_9H_{10}Br_2$,it is $1-(1,1-dibromoethyl)-4-methylbenzene$ or similar gem-dibromide structure).
$2$. Treatment of $[P]$ with excess $NaNH_2$ followed by $HCl$ leads to dehydrohalogenation,forming the terminal alkyne $[Q]$,which is $1-ethynyl-4-methylbenzene$ $(p-methylphenylacetylene)$.
$3$. Hydration of the alkyne $[Q]$ with $HgSO_4$ and $dil. H_2SO_4$ (Kucherov reaction) yields the ketone $[R]$,which is $4-methylacetophenone$.
$4$. $4-methylacetophenone$ contains a $CH_3CO-$ group,thus it gives a positive Iodoform test. It is a ketone,so it gives a negative Tollen's test.
$5$. Therefore,$[P]$ is $1,1-dibromoethyl-4-methylbenzene$ (represented by option $C$).

(B) The bromination of a cycloalkane $(X)$ consuming one mole of $Br_2$ implies an addition reaction across a double bond or ring opening,resulting in a dibromo derivative $(Y)$.
Let the formula of $(Y)$ be $C_n H_m Br_2$.
Given the $C:Br$ ratio is $3:1$,we have $\frac{n}{2} = \frac{3}{1}$,which gives $n = 6$.
The product $(Y)$ is $C_6 H_{10} Br_2$.
The molar mass of $C_6 H_{10} Br_2 = (6 \times 12) + (10 \times 1) + (2 \times 80) = 72 + 10 + 160 = 242 \ g \ mol^{-1}$.
The percentage of bromine in $(Y) = \frac{\text{Mass of } Br}{\text{Molar mass of } Y} \times 100 = \frac{160}{242} \times 100 \approx 66.11 \%$.
Rounding to the nearest integer,we get $66 \%$.

(A) $1$. The formation of a yellow precipitate with $I_2/NaOH$ (iodoform test) indicates the presence of a methyl ketone group $(CH_3CO-)$ or a secondary alcohol that can be oxidized to a methyl ketone.
$2$. "$P$" $(C_8H_{14})$ undergoes ozonolysis to form "$Q$".
$3$. "$Q$" undergoes an intramolecular aldol condensation (alkali under reflux) to form "$R$",which contains a methyl ketone moiety.
$4$. The structure "$S$" is a cyclic carboxylic acid derivative formed after the haloform reaction and acidification.
$5$. Analyzing the degree of unsaturation ($C_8H_{14}$ has $8 - 14/2 + 1 = 2$ degrees of unsaturation),and the reaction sequence,$1$,$2$-dimethylcyclohexene (option $A$) is the most suitable precursor that fits the molecular formula and the described chemical transformations.