Mix Examples-Hydrocarbon Questions in English

(C) The reactivity of hydrocarbons towards electrophilic addition reactions depends on the electron density of the carbon-carbon bond.
$C_2H_2$ (ethyne) contains a triple bond,$C_2H_4$ (ethene) contains a double bond,and $C_2H_6$ (ethane) contains only single bonds.
Triple bonds are more electron-rich and reactive than double bonds,which are in turn more reactive than single bonds.
Therefore,the order of reactivity is $C_2H_2 > C_2H_4 > C_2H_6$.

(B) The reaction involves the acid-catalyzed dehydration of cyclobutylmethanol.
$1$. Protonation of the hydroxyl group occurs to form a good leaving group $(-OH_2^+)$.
$2$. Loss of water generates a primary carbocation (cyclobutyl-methyl cation).
$3$. This primary carbocation undergoes a ring expansion rearrangement to form a more stable five-membered ring (cyclopentyl cation).
$4$. Loss of a proton from the cyclopentyl cation yields cyclopentene as the major product.
Note: The provided options do not include cyclopentene. Among the given structures,if we assume the question implies a rearrangement leading to a more stable alkene,the most stable product formed from such rearrangements is typically cyclopentene.

(A) The bond dissociation energy $(BDE)$ is inversely proportional to the stability of the free radical formed after the homolytic cleavage of the $C-H$ bond.
$I. \, CH_3-H \rightarrow \cdot CH_3$ (Methyl radical,least stable)
$II. \, CH_3CH_2-H \rightarrow CH_3\cdot CH_2$ (Primary radical,more stable than methyl)
$III. \, CH_2=CH-CH_2-H \rightarrow CH_2=CH-\cdot CH_2$ (Allyl radical,resonance stabilized,very stable)
$IV. \, C_6H_5-H \rightarrow \cdot C_6H_5$ (Phenyl radical,$sp^2$ hybridized carbon,least stable due to high $s$-character)
Comparing the stability of radicals: $Allyl > Ethyl > Methyl > Phenyl$.
Therefore,the order of $BDE$ is: $Phenyl > Methyl > Ethyl > Allyl$.
Thus,the correct order is $IV > I > II > III$.

(B) The rate of catalytic hydrogenation of alkenes is inversely proportional to the stability of the alkene.
More substituted alkenes are more stable due to the electron-donating inductive effect and hyperconjugation of the alkyl groups $(R)$.
Therefore,the stability order is: Tetrasubstituted > Trisubstituted > Disubstituted > Monosubstituted.
Since the rate of hydrogenation is inversely proportional to stability,the least substituted alkene will react the fastest.
Among the given options,the alkene with the fewest substituents (disubstituted) will react the fastest.
Comparing the options,the disubstituted alkenes ($A$ and $B$) are less stable than the trisubstituted $(C)$ and tetrasubstituted $(D)$ alkenes.
Between the two disubstituted alkenes,the cis-isomer $(B)$ is generally less stable than the trans-isomer $(A)$ due to steric hindrance.
Therefore,the cis-disubstituted alkene $(B)$ reacts the fastest.

(C) The acidity of hydrocarbons depends on the percentage of $s$-character in the $C-H$ bond.
$1$. Ethyne $(CH \equiv CH)$ has $sp$ hybridization ($50\% \ s$-character).
$2$. Propyne $(CH_3-C \equiv CH)$ has an acidic terminal hydrogen attached to an $sp$ hybridized carbon.
$3$. Ethene $(CH_2=CH_2)$ has $sp^2$ hybridization ($33.3\% \ s$-character).
$4$. Ethane $(CH_3-CH_3)$ has $sp^3$ hybridization ($25\% \ s$-character).
Comparing ethyne $(III)$ and propyne $(IV)$,the methyl group in propyne is electron-donating ($+I$ effect),which decreases the acidity of the terminal hydrogen compared to ethyne.
Thus,the decreasing order of acidity is: $III > IV > II > I$.

(C) The complete combustion (oxidation) of any hydrocarbon in the presence of excess oxygen results in the formation of carbon dioxide $(CO_2)$ and water $(H_2O)$ as the final products.
For a general hydrocarbon $C_xH_y$,the reaction is: $C_xH_y + (x + y/4) O_2 \rightarrow x CO_2 + (y/2) H_2O$.

(B) The general formula $C_nH_{2n}$ corresponds to either alkenes or cycloalkanes. Since $C_6H_{12}$ represents cyclohexane,it is an alicyclic compound.

(C) $1,3$-Butadiene requires $2 \ mol$ of $H_2$ for complete hydrogenation.
Theoretical heat of hydrogenation would be $2 \times 30 = 60 \ kcal$.
However,$1,3$-butadiene is stabilized by resonance energy.
Therefore,the actual heat of hydrogenation is less than the theoretical value.
Thus,the value is approximately $57 \ kcal$.

(C) The general formula for alkanals (aldehydes) is $C_nH_{2n}O$.
For example,for $n=1$ (methanal),the formula is $CH_2O$.
The given formula $C_nH_{2n+1}O$ is incorrect because it does not account for the correct number of hydrogen atoms required to satisfy the valency of the carbonyl group and the alkyl chain.

(A) Ethylene glycol $(HOCH_2-CH_2OH)$ undergoes oxidation when treated with strong oxidizing agents like acidified $KMnO_4$.
The reaction proceeds as follows:
$HOCH_2-CH_2OH + 2[O] \xrightarrow{H^+/KMnO_4} HOOC-COOH + 2H_2O$
The final product obtained is oxalic acid $(HOOC-COOH)$.

(A) The reaction sequence is as follows:
$1$. ${C_2H_5I} \xrightarrow{\text{alc. KOH}} CH_2=CH_2$ (Ethene,$X$)
$2$. $CH_2=CH_2 \xrightarrow[CCl_4]{Br_2} BrCH_2-CH_2Br$ ($1$,$2$-Dibromoethane,$Y$)
$3$. $BrCH_2-CH_2Br \xrightarrow{KCN} NC-CH_2-CH_2-CN$ (Butanedinitrile,$Z$)
$4$. $NC-CH_2-CH_2-CN \xrightarrow{H_3O^+} HOOC-CH_2-CH_2-COOH$ (Succinic acid,$A$)
Thus,the final product $(A)$ is Succinic acid.

(A) The rate of catalytic hydrogenation of an alkene is inversely proportional to its stability.
More substituted alkenes are more stable due to hyperconjugation and inductive effects,and therefore,they react more slowly with $H_2$.
The order of stability of alkenes is:
Tetrasubstituted > Trisubstituted > Disubstituted > Monosubstituted > Ethene.
Conversely,the order of reactivity towards catalytic hydrogenation is:
Monosubstituted > Disubstituted > Trisubstituted > Tetrasubstituted.
Among the given options,the alkene with the least number of substituents (monosubstituted or the least substituted one) will react fastest.
Based on the provided structures,the alkene with the fewest $R$ groups (most $H$ atoms) will be the most reactive.

(A)
$Cyclohexene$ on ozonolysis followed by oxidation gives adipic acid $(hexanedioic \ acid)$.
$C_6H_{10}$ $\xrightarrow{O_3} OHC-(CH_2)_4-CHO$ $\xrightarrow{[O]} HOOC-(CH_2)_4-COOH$ (Adipic acid).

(D) The reaction of benzene $(C_6H_6)$ with dichloromethane $(CH_2Cl_2)$ in the presence of anhydrous $AlCl_3$ is a Friedel-Crafts alkylation reaction. Since benzene is in excess,both chlorine atoms of $CH_2Cl_2$ are replaced by phenyl groups,resulting in the formation of diphenylmethane $(C_6H_5-CH_2-C_6H_5)$ and $2HCl$. The reaction is: $2C_6H_6 + CH_2Cl_2 \xrightarrow{AlCl_3} C_6H_5-CH_2-C_6H_5 + 2HCl$.

(A) The reaction of an internal alkyne with $Hg^{2+}/H^{+}$ (Kuccherov reaction) involves the hydration of the alkyne to form a ketone.
For the unsymmetrical alkyne $Ph-C \equiv C-CH_3$,the hydration can occur at either carbon of the triple bond.
However,the formation of the ketone is governed by the stability of the intermediate carbocation or the electronic effects of the substituents.
In this case,the reaction proceeds to form $Ph-CO-CH_2-CH_3$ (propiophenone derivative) as the major product,which corresponds to the structure in image $287-$a474.

(D) $2(HCOO)_2Ca \xrightarrow{\text{Dry distillation}} 2HCHO + 2CaCO_3$
$6HCHO + 4NH_3 \to (CH_2)_6N_4 + 6H_2O$
The product $(CH_2)_6N_4$ is known as Urotropine.

(C) Let us analyze each option:
$(i)$ $CH_3-CH_2-CH_2Br$ undergoes elimination to form propene $(CH_3-CH=CH_2)$.
(ii) Cyclopropane undergoes ring-opening elimination to form propene $(CH_3-CH=CH_2)$.
(iii) $CH_3-CH_2-CH_2OH$ reacts with $HBr$ to form $CH_3-CH_2-CH_2Br$,which then undergoes elimination to form propene $(CH_3-CH=CH_2)$.
(iv) $CH_2=C=O$ (ketene) does not produce propene by reaction with $HBr$ or by direct elimination.
Therefore,the correct option is $C$.

(D) The correct option is $(d)$.
In the given hydrocarbon structure:
$1.$ $C_2$ is involved in a triple bond $(C \equiv C)$,so it is $sp$ hybridized.
$2.$ $C_3$ is bonded to four other atoms ($C_2, C_4, H$,and the $C$ of the $CH_3$ group) via single bonds,so it is $sp^3$ hybridized.
$3.$ $C_5$ is involved in a double bond $(C=C)$,so it is $sp^2$ hybridized.
$4.$ $C_6$ is bonded to four other atoms ($C_5, C_7$,and the $C$ atoms of two $CH_3$ groups) via single bonds,so it is $sp^3$ hybridized.
Thus,the sequence for $C_2, C_3, C_5$ and $C_6$ is $sp, sp^3, sp^2, sp^3$.

(B) To determine the hybridization of carbon atoms,we count the number of sigma bonds and lone pairs attached to each carbon atom.
$1$. Carbon $1$ $(CH_3)$: It is bonded to $3$ hydrogen atoms and $1$ carbon atom via single bonds. Total $4$ sigma bonds $\rightarrow sp^3$ hybridization.
$2$. Carbon $3$ $(-CH=)$: It is bonded to $1$ hydrogen atom,$1$ carbon atom via a single bond,and $1$ carbon atom via a double bond. Total $3$ sigma bonds $\rightarrow sp^2$ hybridization.
$3$. Carbon $5$ $(-C \equiv)$: It is bonded to $1$ carbon atom via a single bond and $1$ carbon atom via a triple bond. Total $2$ sigma bonds $\rightarrow sp$ hybridization.
Thus,the sequence of hybridization for carbons $1, 3,$ and $5$ is $sp^3, sp^2, sp$.

(A) Bunsen flame consists of several zones based on the extent of combustion.
Region $1$ is the innermost dark zone containing unburnt fuel.
Region $2$ is the luminous zone where incomplete combustion occurs.
Region $3$ is the non-luminous zone where complete combustion occurs,making it the hottest part of the flame.
Region $4$ is the tip of the flame.
Therefore,the hottest region is region $3$.

(C) The heat of hydrogenation $(\Delta H_{hyd})$ is inversely proportional to the stability of the alkene.
$1$. More substituted alkenes are more stable,hence they have lower $\Delta H_{hyd}$.
$2$. Trans-alkenes are more stable than cis-alkenes,hence cis-alkenes have higher $\Delta H_{hyd}$.
$3$. Increased ring strain leads to higher $\Delta H_{hyd}$.
In option $C$,the first molecule is a highly strained bicyclic alkene (bridgehead double bond in a small ring system),which makes it significantly less stable than the isomer in the second part of the option. Therefore,the first molecule has a higher heat of hydrogenation,making the order $C$ correct.

(D) $1$. $CaC_2 + 2H_2O \rightarrow Ca(OH)_2 + C_2H_2$ (Acetylene). Acetylene is a terminal alkyne $(HC \equiv CH)$,which reacts with $AgNO_3 / NH_4OH$ (Tollens' reagent) to form a white precipitate of silver acetylide $(AgC \equiv CAg)$.
$2$. $2CHCl_3 + 6Ag \xrightarrow{\Delta} HC \equiv CH + 6AgCl$. Acetylene is formed,which gives a white precipitate with $AgNO_3 / NH_4OH$.
$3$. $Mg_2C_3 + 4H_2O \rightarrow 2Mg(OH)_2 + CH_3-C \equiv CH$ (Propyne). Propyne is a terminal alkyne,which reacts with $AgNO_3 / NH_4OH$ to form a white precipitate of silver propyne $(CH_3-C \equiv CAg)$.
$4$. $Al_4C_3 + 12H_2O \rightarrow 4Al(OH)_3 + 3CH_4$ (Methane). Methane is an alkane and does not react with $AgNO_3 / NH_4OH$ to form a precipitate.
Therefore,the reaction that does not give a white precipitate is $Al_4C_3 \xrightarrow{H_2O}$.

(B) The addition of $HCl$ to $2$-methylpropene follows Markovnikov's rule,where the $Cl^-$ attaches to the more substituted carbon atom.
$CH_3-C(CH_3)=CH_2 + HCl \rightarrow CH_3-CCl(CH_3)-CH_3$
The addition of $HCl$ to ethyne follows Markovnikov's rule in the second step,leading to the formation of a geminal dihalide.
$CH \equiv CH + HCl$ $\rightarrow CH_2=CHCl$ $\xrightarrow{HCl} CH_3-CHCl_2$
Thus,the products are $CH_3-CCl(CH_3)-CH_3$ and $CH_3-CHCl_2$.

(D) The hydrocarbon $C_5H_8$ consumes $2$ moles of $H_2$,which indicates the presence of two double bonds (a diene).
Ozonolysis of $2-$methyl$-1,3-$butadiene $(CH_2=C(CH_3)-CH=CH_2)$ breaks the double bonds to form $HCHO$ (methanal) and $CH_3-CO-CHO$ ($2-$oxopropanal).
The reaction is:
$CH_2=C(CH_3)-CH=CH_2 \xrightarrow{1. O_3, 2. Zn/H_2O} HCHO + CH_3-CO-CHO + HCHO$.

(B) The heat of hydrogenation $(HOH)$ is directly proportional to the number of $\pi-$bonds present in the molecule.
$1$. $1,4-$cyclohexadiene has $2$ $\pi-$bonds.
$2$. $1-$methylenecyclohex$-2-$ene has $3$ $\pi-$bonds.
$3$. $1,3-$cyclohexadiene has $2$ $\pi-$bonds.
$4$. Methylenecyclohexane has $1$ $\pi-$bond.
Since $1-$methylenecyclohex$-2-$ene has the highest number of $\pi-$bonds $(3)$,it will have the highest heat of hydrogenation.

(D) The reaction involves the acid-catalyzed dehydration of $1$-hydroxybicyclo[$4.4$.$0$]decane (decalol).
$H_2SO_4$ protonates the hydroxyl group, forming a good leaving group $(-OH_2^+)$.
Loss of water generates a carbocation at the bridgehead position.
According to Bredt's rule, a double bond at the bridgehead of a small bicyclic system is unstable.
However, in a decalin system (bicyclo[$4.4$.$0$]decane), the double bond can form at the bridgehead position (forming $\Delta^{1,6}$-octalin or bicyclo[$4.4$.$0$]dec-$1$$(6)$-ene) as it is the most substituted and stable alkene isomer (Zaitsev's rule).
Thus, the major product is bicyclo[$4.4$.$0$]dec-$1$$(6)$-ene.

(B) The dehydration of malonic acid $(CH_2(COOH)_2)$ using phosphorus pentoxide $(P_4O_{10})$ at $150^\circ C$ results in the formation of carbon suboxide $(C_3O_2)$.
The chemical reaction is:
$3 CH_2(COOH)_2 \xrightarrow{P_4O_{10}, \, 150^\circ C} 3 C_3O_2 + 4 H_3PO_4$
Therefore,compound $(X)$ is carbon suboxide.

(C) The heat of hydrogenation is inversely proportional to the stability of the alkene.
For option $A$,cyclobutene is more strained than cyclopentene,so it has a higher heat of hydrogenation.
For option $B$,$cis-2-$butene is less stable than $trans-2-$butene due to steric hindrance,so it has a higher heat of hydrogenation.
For option $C$,the second compound (bicyclo[$4.2$.$0$]oct$-2-$ene) has a double bond in a smaller,more strained ring (cyclobutene ring) compared to the first compound (bicyclo[$4.2$.$0$]oct$-1-$ene),where the double bond is shared between the rings. Therefore,the second compound has higher ring strain and higher heat of hydrogenation.
For option $D$,$1-$butene is less stable than $2-$butene,so it has a higher heat of hydrogenation.
Thus,the correct pair where the second compound has a higher heat of hydrogenation is $C$.

(C) When glycerol $(CH_2OH-CHOH-CH_2OH)$ is heated with a dehydrating agent like potassium bisulfate $(KHSO_4)$,it undergoes dehydration.
Two molecules of water are removed from the glycerol molecule to form acrolein $(CH_2=CH-CHO)$.

(A) When an electrophile attacks a substituted benzene ring,the orientation is determined by the existing groups.
In $m-$Nitrotoluene,the $-CH_3$ group is $o/p-$directing and the $-NO_2$ group is $m-$directing.
The positions available for electrophilic substitution are the $2, 4,$ and $6$ positions relative to the methyl group.
Due to the directing effects of both groups,substitution at these positions leads to three distinct isomers in significant amounts.
In contrast,$o-$Nitrotoluene and $p-$Nitrotoluene have symmetry that reduces the number of distinct isomers formed,and $2, 4, 6-$Trinitrotoluene is highly deactivated towards electrophilic substitution.

(C) The four compounds are toluene $(C_6H_5CH_3)$,ethylbenzene $(C_6H_5C_2H_5)$,benzyl alcohol $(C_6H_5CH_2OH)$,and benzaldehyde $(C_6H_5CHO)$.
Strong oxidizing agents (like $KMnO_4/H^+$ or $K_2Cr_2O_7/H^+$) oxidize alkyl groups attached to a benzene ring to a carboxylic acid group,provided there is at least one benzylic hydrogen.
$1$. Toluene $(C_6H_5CH_3)$ is oxidized to benzoic acid $(C_6H_5COOH)$.
$2$. Ethylbenzene $(C_6H_5CH_2CH_3)$ is oxidized to benzoic acid $(C_6H_5COOH)$.
$3$. Benzyl alcohol $(C_6H_5CH_2OH)$ is oxidized to benzoic acid $(C_6H_5COOH)$.
$4$. Benzaldehyde $(C_6H_5CHO)$ is oxidized to benzoic acid $(C_6H_5COOH)$.
Since all four compounds are converted into benzoic acid,the final product is only $C_6H_5COOH$.

(D) The reaction of styrene $(C_6H_5CH=CH_2)$ with $Br_2$ is an electrophilic addition reaction across the double bond.
This results in the formation of styrene dibromide $(C_6H_5CH(Br)CH_2Br)$,which is compound $A$.
When styrene dibromide is treated with $KMnO_4$ (a strong oxidizing agent),the side chain is oxidized.
Specifically,the alkyl group attached to the benzene ring is oxidized to a carboxylic acid group,resulting in the formation of benzoic acid $(C_6H_5COOH)$,which is compound $B$.
Therefore,$A$ is styrene dibromide and $B$ is benzoic acid.

(C) $1$. Cyclohexene reacts with $NBS$ followed by $Mg/ether$ to form cyclohex$-2-$en$-1-$ylmagnesium bromide.
$2$. This Grignard reagent reacts with isobutyraldehyde to form an alkoxide,which is then alkylated with ethyl bromide $(CH_3CH_2Br)$ to form an ether derivative of cyclohexenol.
$3$. Finally,the reaction with $OsO_4/H_2O_2$ performs syn-dihydroxylation of the double bond,resulting in the product shown in the solution image.

(C) The reaction of $Na$ metal with compounds containing acidic hydrogen atoms (like $-OH$ or $-COOH$ groups) or compounds that can form stable aromatic anions (like cyclopentadiene) results in the liberation of $H_2$ gas.
$1$. Cyclopentadiene $(C_5H_6)$ reacts with $Na$ to form the aromatic cyclopentadienyl anion $(C_5H_5^-)$,releasing $H_2$ gas.
$2$. tert-Butyl alcohol $((CH_3)_3COH)$ reacts with $Na$ to form sodium tert-butoxide $((CH_3)_3CONa)$,releasing $H_2$ gas.
$3$. Acetic acid $(CH_3COOH)$ reacts with $Na$ to form sodium acetate $(CH_3COONa)$,releasing $H_2$ gas.
$4$. Cyclooctatetraene $(C_8H_8)$ is a non-aromatic,tub-shaped molecule. While it can react with $Na$ to form the aromatic cyclooctatetraenide dianion $(C_8H_8^{2-})$,this reaction involves the addition of electrons (reduction) rather than the liberation of $H_2$ gas.
Therefore,cyclooctatetraene does not liberate $H_2$ gas.

(A) The question is incomplete as it does not specify the reactants or the reaction conditions. However,based on the options provided,if the reaction involves the hydroboration-oxidation of methylenecyclohexane,the product would be cyclohexylmethanol (Option $A$). If it involves the hydration of $1-$methylcyclohexene,the product would be $1-$methylcyclohexanol (Option $B$). Without the specific reaction,a definitive answer cannot be determined.

(C) The molecular formula $C_8H_{12}$ indicates a degree of unsaturation of $2$.
Ozonolysis of the compound yields $2 \ moles$ of $3-$oxobutanal $(CH_3COCH_2CHO)$.
This implies the original compound must be a cyclic diene that,upon cleavage of its two double bonds,produces two identical fragments of $3-$oxobutanal.
$1$,$4$-dimethylcyclohexa$-1,4-$diene (Option $A$) on ozonolysis gives $2 \ moles$ of $3-$oxobutanal.
$1$,$5$-dimethylcyclohexa$-1,4-$diene (Option $B$) is the same as $1,4-$dimethylcyclohexa-$1,4-$diene.
$1$,$2$-dimethylcyclohexa$-1,4-$diene (Option $C$) on ozonolysis gives $1 \ mole$ of $2,3-$dioxobutane and $1 \ mole$ of $1,4-$butanedial.
$2$,$3$-dimethylcyclohexa$-1,3-$diene (Option $D$) on ozonolysis gives $1 \ mole$ of $2,3-$dioxobutane and $1 \ mole$ of $1,4-$butanedial.
Since the question asks for the structure that $CAN$ $NOT$ be the compound,and both $C$ and $D$ do not yield $3-$oxobutanal,there is an ambiguity in the provided options. However,based on standard chemistry problems of this type,$1,4-$dimethylcyclohexa-$1,4-$diene is the intended structure that $DOES$ produce $3-$oxobutanal. Given the options,$C$ and $D$ are incorrect structures.

(C) Let us analyze each reaction:
$(A)$ Propane $(CH_3CH_2CH_3)$ reacts with $Br_2/h\nu$ to form $2-$bromopropane $(CH_3CH(Br)CH_3)$. Further reaction with $Na/ether$ (Wurtz reaction) gives $2,3-$dimethylbutane.
$(B)$ Isobutyric acid $((CH_3)_2CHCOOH)$ reacts with $NaOH$ to form sodium isobutyrate. Electrolysis (Kolbe's electrolysis) of sodium isobutyrate gives $2,3-$dimethylbutane.
$(C)$ $2-$bromo$-2,3-$dimethylbutane $((CH_3)_2CH-C(Br)(CH_3)_2)$ reacts with $Mg/ether$ to form a Grignard reagent,which upon hydrolysis with $H_2O$ gives $2,3-$dimethylbutane.
Thus,all three reactions $(A, B, C)$ produce the same product,$2$,$3$-dimethylbutane.

(B) $1$. The reaction starts with cyclopentane. Treatment with $Br_2$ in the presence of $hv$ (free radical substitution) gives bromocyclopentane as product $(A)$.
$2$. Bromocyclopentane $(A)$ reacts with alcoholic $KOH$ (dehydrohalogenation) to form cyclopentene as product $(B)$.
$3$. Cyclopentene $(B)$ reacts with $NBS$ ($N$-Bromosuccinimide),which is a specific reagent for allylic bromination,to form $3-$bromocyclopent$-1-$ene as product $(C)$.

(C) The starting material is $1$-hydroxycyclopentane-$1,1$-dicarboxylic acid.
When this compound is heated with an oxidizing agent like $K_2Cr_2O_7$,it undergoes oxidation of the secondary alcohol group to a ketone,followed by decarboxylation of the gem-dicarboxylic acid group.
Specifically,the $1$-hydroxy group is oxidized to a ketone,forming $1$-oxocyclopentane-$1,1$-dicarboxylic acid.
This $\beta$-keto acid (or specifically,a gem-dicarboxylic acid derivative) is unstable upon heating and undergoes decarboxylation (loss of $CO_2$) to yield cyclopentanone.

(C) The heat of combustion is directly proportional to the number of carbon atoms in the molecule.
Cyclopropane $(C_3H_6)$ has $3$ carbon atoms.
Cyclobutane $(C_4H_8)$ has $4$ carbon atoms.
Cyclopentane $(C_5H_{10})$ has $5$ carbon atoms.
Therefore,the order of heat of combustion is $III > II > I$.

(A) The stability of a conjugated diene is determined by the extent of conjugation and the number of hyperconjugative structures (alpha-hydrogens).
In $1,3$-cyclohexadiene derivatives,conjugation is present in all options.
Comparing the number of alpha-hydrogens:
$A$) $1,3$-dimethyl$-1,3-$cyclohexadiene has $10$ alpha-hydrogens.
$B$) $2,4$-dimethyl$-1,3-$cyclohexadiene has $8$ alpha-hydrogens.
$C$) $1,5$-dimethyl$-1,3-$cyclohexadiene has $8$ alpha-hydrogens.
$D$) $2,6$-dimethyl$-1,3-$cyclohexadiene has $8$ alpha-hydrogens.
Since $1,3$-dimethyl$-1,3-$cyclohexadiene has the highest number of alpha-hydrogens $(10)$,it is the most stable due to greater hyperconjugation.

(D) Organometallic compounds like Grignard reagents $(R-MgX)$ and organolithium compounds $(R-Li)$ are strong bases.
They react rapidly with any compound containing an acidic hydrogen $(H)$ or deuterium $(D)$ atom (such as water,alcohols,or acids) to form hydrocarbons.
These reactions are classified as acid-base reactions because the organometallic species acts as a Brønsted-Lowry base,abstracting a proton or deuteron from the substrate.
Therefore,all the given reactions involve an acid-base mechanism.

(D) An elimination reaction involves the removal of atoms or groups from adjacent carbon atoms to form a $\pi$-bond (double or triple bond).
In reaction $(a)$,cyclohexanol undergoes dehydration in the presence of $H_2SO_4$ to form cyclohexene,which is an elimination reaction.
In reaction $(b)$,trans$-1,2-$dibromocyclohexane reacts with $KI$ to undergo debromination,forming cyclohexene,which is also an elimination reaction.
Reaction $(c)$ is a substitution reaction where the methoxy group replaces the bromine atom.
Therefore,both $(a)$ and $(b)$ are elimination reactions.

(C) The reaction involves the treatment of acetylacetone $(CH_3COCH_2COCH_3)$ with $2$ equivalents of $KNH_2$ in liquid ammonia $(NH_3(l))$.
$KNH_2$ is a strong base that removes both acidic protons from the central methylene group,forming a dianion: $[CH_3COCHCOCH_3]^{2-}$.
This dianion is highly nucleophilic.
When treated with $n-C_4H_9Br$,the more nucleophilic carbon (the terminal carbon of the enolate,which is less sterically hindered) attacks the alkyl halide via an $S_N2$ mechanism.
After acidic workup with $H_3O^+$,the product formed is $CH_3COCH_2COCH_2CH_2CH_2CH_2CH_3$ (octane$-2,4-$dione substituted at the terminal position).
Looking at the options,the structure corresponding to the alkylation at the terminal methyl group is $C$.

(B) In an $E2$ elimination reaction,the base $(NaOEt)$ abstracts a $\beta$-hydrogen that is anti-periplanar to the leaving group $(Cl^-)$.
For the given molecule,there are two possible $\beta$-hydrogens available for elimination.
Elimination of the hydrogen at $C_2$ leads to a less substituted alkene,while elimination of the hydrogen at $C_6$ leads to a more substituted,and thus more stable,alkene.
According to Zaitsev's rule,the more substituted alkene is the major product.
Therefore,the product with the double bond formed towards the more substituted side is the major product.

(D) The transformation involves converting $2,2-dimethylbutane$ to $3,3-dimethylbutan-1-ol$.
Step $1$: Radical bromination using $Br_2, hv$ selectively replaces the hydrogen at the $C-3$ position to form $3-bromo-2,2-dimethylbutane$.
Step $2$: Elimination using a bulky base like $t-BuO^-$ ($E_2$ mechanism) yields the less substituted alkene,$3,3-dimethylbut-1-ene$.
Step $3$: Hydroboration-oxidation ($BH_3, THF$ followed by $H_2O_2, OH^-$) performs an anti-Markovnikov hydration of the terminal alkene to give the primary alcohol,$3,3-dimethylbutan-1-ol$.
Thus,the correct sequence is $1$. $Br_2, hv$ $2$. $t-BuO^-$ $3$. $BH_3, THF$ $4$. $H_2O_2, OH^-$.

(B) The molecular formula of cubane is $C_8H_8$.
For a hydrocarbon with formula $C_xH_y$,the double bond equivalent $(DBE)$ is calculated using the formula: $DBE = x - \frac{y}{2} + 1$.
Substituting the values $x = 8$ and $y = 8$:
$DBE = 8 - \frac{8}{2} + 1 = 8 - 4 + 1 = 5$.
Alternatively,the $DBE$ is equal to the number of rings plus the number of pi bonds. Cubane has $5$ rings and $0$ pi bonds,so $DBE = 5 + 0 = 5$.

(D) Cubane is a polycyclic hydrocarbon with the formula $C_8H_8$,consisting of eight carbon atoms arranged in a cube.
To convert a cyclic structure into a non-cyclic (acyclic) skeleton,we must break enough bonds to eliminate all rings.
For a polycyclic system,the number of bonds to be broken is equal to the number of rings present.
Cubane has $5$ faces (rings) that are fused together.
As shown in the process:
$1$. Breaking $2$ bonds opens the first set of rings.
$2$. Breaking another $2$ bonds opens further rings.
$3$. Breaking $1$ final bond results in a non-cyclic structure.
Total bond cleavages required = $2 + 2 + 1 = 5$.

(B) The reaction involves the catalytic hydrogenation of $3$-methylmethylenecyclopentane using $Pd$ catalyst.
Hydrogenation is a $syn$-addition process,meaning both hydrogen atoms add to the same face of the double bond.
Since the starting material has a chiral center at the $C-3$ position,the addition of $H_2$ to the exocyclic double bond creates a new chiral center at the $C-1$ position.
Because the $H_2$ can add from either the same side as the methyl group or the opposite side,two stereoisomeric products are formed.
These two products are non-mirror image stereoisomers,which are classified as diastereomers.