Mix Examples-General Organic Chemistry Questions in English

(A) In assigning $R-S$ configuration,the priority is determined by the Cahn-Ingold-Prelog $(CIP)$ sequence rules.
According to these rules,the atom with the higher atomic number attached directly to the chiral center is given higher priority.
Comparing the atoms attached to the chiral center: $S$ (atomic number $16$) in $-SO_3H$ has a higher atomic number than $C$ (atomic number $6$) in $-COOH$,$-CHO$,and $-C_6H_5$.
Therefore,the $-SO_3H$ group has the highest priority.
The decreasing order of priority is $-SO_3H > -COOH > -CHO > -C_6H_5$.

(A) According to $Cahn-Ingold-Prelog$ sequence rules,the priority of groups is decided by the atomic number of the atoms directly attached to the chiral center.
$1$. The atom with the highest atomic number gets the highest priority. Here,$O$ (atomic number $8$) has higher priority than $C$ (atomic number $6$). Thus,$-OH$ is first.
$2$. For the remaining carbon-containing groups,we look at the atoms attached to the first carbon: $-COOH$ ($C$ bonded to $O, O, O$),$-CHO$ ($C$ bonded to $O, O, H$),and $-CH_2OH$ ($C$ bonded to $O, H, H$).
$3$. Comparing these,the priority order is $-COOH > -CHO > -CH_2OH$.
$4$. Therefore,the overall order is $-OH > -COOH > -CHO > -CH_2OH$.

(C) $I$. Torsional strain is higher in eclipsed ethane due to repulsion between hydrogen atoms on adjacent carbons. Thus,$\text{staggered ethane} < \text{eclipsed ethane}$. This statement is incorrect.
$II$. $2, 2-\text{Dimethylbutane}$ is more branched than $2-\text{methylpentane}$. Increased branching decreases the surface area,leading to weaker van der Waals forces and a lower boiling point. Thus,$2, 2-\text{Dimethylbutane} < 2-\text{methylpentane}$. This statement is incorrect.
$III$. $cis-\text{but}-2-\text{ene}$ has a net dipole moment due to the same side orientation of methyl groups,whereas $trans-\text{but}-2-\text{ene}$ has a zero net dipole moment due to symmetry. Thus,$cis-\text{but}-2-\text{ene} > trans-\text{but}-2-\text{ene}$. This statement is correct.

(B) The molecular weight of the dibromo derivative is $186 \ u$. Given $Br = 80 \ u$,the mass of the alkene part is $186 - (2 \times 80) = 186 - 160 = 26 \ u$. The general formula for an alkene is $C_nH_{2n}$,so $12n + 2n = 26$,which gives $14n = 26$. Since $n$ must be an integer,we re-evaluate: the dibromo derivative of ethene $(C_2H_4)$ is $C_2H_2Br_2$. Its molecular weight is $(2 \times 12) + (2 \times 1) + (2 \times 80) = 24 + 2 + 160 = 186 \ u$. Thus,the compound is $C_2H_2Br_2$. The possible isomers for $C_2H_2Br_2$ are:
$1$. $1,1-dibromoethene$ $(CH_2=CBr_2)$
$2$. $cis-1,2-dibromoethene$
$3$. $trans-1,2-dibromoethene$
Therefore,there are $3$ possible isomers.

(C) The degree of unsaturation $(DU)$ for $C_3H_6O$ is calculated as: $DU = C + 1 - \frac{H}{2} = 3 + 1 - \frac{6}{2} = 1$.
This indicates the presence of either one double bond or one ring.
Possible acyclic isomers:
$1$. Propanal $(CH_3CH_2CHO)$
$2$. Propanone $(CH_3COCH_3)$
$3$. Allyl alcohol $(CH_2=CH-CH_2OH)$
$4$. $1$-propen$-1-$ol $(CH_3-CH=CH-OH)$
$5$. $2$-propen$-1-$ol $(CH_2=C(OH)-CH_3)$
$6$. Methoxyethene $(CH_3-O-CH=CH_2)$
Possible cyclic isomers:
$7$. Cyclopropanol
$8$. Methoxycyclopropane (Oxetane is not possible as it has $4$ atoms in ring,but $C_3H_6O$ can form $2$-methyloxirane or oxetane)
$9$. $2$-methyloxirane (propylene oxide)
$10$. Oxetane
Wait,let us list them carefully:
$(1)$ Propanal,$(2)$ Propanone,$(3)$ Allyl alcohol $(CH_2=CHCH_2OH)$,$(4)$ $1$-propen$-1-$ol $(CH_3CH=CHOH)$,$(5)$ $2$-propen$-1-$ol $(CH_2=C(OH)CH_3)$,$(6)$ Methoxyethene $(CH_3OCH=CH_2)$,$(7)$ Cyclopropanol,$(8)$ $2$-methyloxirane,$(9)$ Oxetane.
Total isomers = $9$.

(C) Structures $A$ and $B$ are mirror images of each other and are non-superimposable,therefore they are enantiomers.
Structures $C$ and $D$ have the same configuration at one chiral center but opposite configurations at the other,making them diastereomers.

(B) The molecular formula $C_5H_9Cl$ indicates a degree of unsaturation of $2$.
$X$ is $3$-chloropent-$1$-ene $(CH_3CH_2CH(Cl)CH=CH_2)$,which is optically active due to the chiral center at $C_3$.
Upon hydrogenation with $1 \ mole$ of $H_2$,$X$ forms $3$-chloropentane $(CH_3CH_2CH(Cl)CH_2CH_3)$,which is optically inactive $(Z)$ due to the presence of a plane of symmetry.
$Y$ is $4$-chloropent-$1$-ene $(CH_2=CHCH_2CH(Cl)CH_3)$,which is also optically active due to the chiral center at $C_4$.
Upon hydrogenation with $1 \ mole$ of $H_2$,$Y$ forms $2$-chloropentane $(CH_3CH_2CH_2CH(Cl)CH_3)$,which is still optically active $(P)$ because the chiral center at $C_2$ remains intact.
Thus,$X$ is $3$-chloropent-$1$-ene and $Y$ is $4$-chloropent-$1$-ene.

(B) The given reaction is the propagation step of a free radical substitution reaction.
Here,the trigonal planar free radical $(CH_3-\dot{C}H-C_2H_5)$ is attacked by $Cl^{\bullet}$ from either side with equal probability,leading to both retention and inversion of configuration.
This results in an equimolar mixture of $d$- and $l$-enantiomers,which is known as a racemic mixture.

(A) To exhibit geometrical isomerism,the molecule must have a double bond with different groups attached to each carbon atom of the double bond.
To exhibit optical isomerism,the molecule must contain at least one chiral center (a carbon atom bonded to four different groups).
In $4-$Bromopent$-2-$ene $(CH_3-CH(Br)-CH=CH-CH_3)$:
$1$. The double bond between $C_2$ and $C_3$ allows for geometrical isomerism ($cis$ and $trans$ forms).
$2$. The $C_4$ atom is a chiral center because it is bonded to a hydrogen atom,a methyl group,a bromine atom,and a propenyl group.
Therefore,$4-$Bromopent$-2-$ene exhibits both geometrical and optical isomerism.

(B) The correct matches are as follows:
$A$. Acetaldehyde $(CH_3CHO)$ and vinyl alcohol $(CH_2=CH-OH)$ are tautomers.
$B$. Eclipsed and staggered ethane are two conformations of ethane,hence they are conformational isomers.
$C$. $(+)2$-butanol and $(-)2$-butanol are non-superimposable mirror images of each other,hence they are enantiomers.
$D$. Methyl-$n$-propylamine $(CH_3-NH-CH_2CH_2CH_3)$ and diethylamine $((C_2H_5)_2NH)$ have different alkyl groups attached to the same functional group $(-NH-)$,hence they are metamers.
Therefore,the correct matching is $A-2, B-4, C-1, D-5$.

(C) The degree of unsaturation for $C_4H_6$ is $4 - (6/2) + 1 = 2$. The possible isomers are:
$(i)$ $CH_3-CH_2-C \equiv CH$ (but-$1$-yne)
$(ii)$ $CH_3-C \equiv C-CH_3$ (but-$2$-yne)
$(iii)$ $CH_2=CH-CH=CH_2$ (buta-$1,3$-diene)
$(iv)$ $CH_3-CH=C=CH_2$ (buta-$1,2$-diene)
$(v)$ cyclobutene
$(vi)$ $1$-methylcyclopropene
$(vii)$ $3$-methylcyclopropene
$(viii)$ methylenecyclopropane
$(ix)$ bicyclo[$1.1.0$]butane
Thus,there are $9$ possible isomers. Hence,option $(C)$ is correct.

$(A)$. Acetaldehyde $(CH_3CHO)$ and vinyl alcohol $(CH_2=CH-OH)$ are tautomers,as they exist in dynamic equilibrium involving the migration of an $\alpha$-hydrogen atom.
$B$. Eclipsed and staggered ethane are different spatial arrangements of the same molecule,known as conformational isomers.
$C$. $(+)$$2$-butanol and $(-)$$2$-butanol are non-superimposable mirror images of each other,hence they are enantiomers.
$D$. Methyl-$n$-propylamine $(CH_3-NH-CH_2CH_2CH_3)$ and diethylamine $(CH_3CH_2-NH-CH_2CH_3)$ have different alkyl groups attached to the same nitrogen atom,which defines them as metamers.

(B) To determine the relationship between the two molecules,we analyze their stereochemistry and connectivity.
Both molecules have the same connectivity and the same configuration at the chiral center (both are $S$-configuration).
By rotating the molecule $II$ in space,it can be superimposed on molecule $I$.
Since they are superimposable,they are identical molecules,which are known as homomers.

(D) Ethyl $3$-oxobutanoate (also known as ethyl acetoacetate) exists in keto-enol tautomerism. The enol form is stabilized by intramolecular hydrogen bonding between the hydroxyl group and the carbonyl oxygen of the ester group. The keto form is $CH_3COCH_2COOC_2H_5$. The enol form is $CH_3C(OH)=CHCOOC_2H_5$,which forms a stable six-membered ring due to hydrogen bonding.

(C) Pair $1$: Both structures represent the same molecule. By rotating the second structure by $180^{\circ}$ in the plane,it superimposes on the first. Thus,they are homomers (identical).
Pair $2$: Both structures are $m$-nitrotoluene. They are identical (homomers).
Pair $3$: The first structure is $1,1$-dichloro-$2$-methylprop-$1$-ene and the second is $1,2$-dichloro-$2$-methylprop-$1$-ene. These are constitutional isomers (specifically,positional isomers).

(D) Heating $\beta$-keto acids or substituted malonic acids leads to decarboxylation (loss of $CO_2$).
For a compound to become achiral after decarboxylation,the resulting product must have a plane of symmetry,a center of symmetry,or be a molecule where the chiral center is destroyed (i.e.,the carbon atom becomes bonded to two identical groups).
In option $(D)$,the starting material is $2$-ethyl-$3$-oxobutanoic acid derivative. Upon heating,it undergoes decarboxylation to form $2$-butanone (or a similar derivative depending on the specific structure). Specifically,the structure in $(D)$ is $2$-ethyl-$3$-oxobutanoic acid. Upon decarboxylation,the chiral center at the $\alpha$-carbon (which has $H$,$Et$,$CO_2H$,and $COCH_3$ groups) loses the $CO_2H$ group and gains an $H$ atom. If the resulting product has two identical groups attached to the central carbon (e.g.,two $H$ atoms or two $Et$ groups),it becomes achiral.
Looking at the provided solution image,the transformation in $(d)$ results in a molecule where the central carbon is bonded to $H$,$Et$,$Me$,and $H$ (if the $CO_2H$ is replaced by $H$). Wait,the image shows the product of $(d)$ as having $H$,$Et$,$Me$,and $H$ attached to the central carbon,making it achiral because it has two identical $H$ atoms.

(B, D) Option $A$: The given structures are $cis$ and $trans$ isomers of $but-2-ene$,which are diastereomers,not enantiomers. Thus,this is false.
Option $B$: The reaction of $CH_3CHO$ with $HCN$ produces a cyanohydrin. Since the carbonyl carbon is $sp^2$ hybridized (planar),the cyanide ion can attack from either side with equal probability,resulting in a racemic mixture of the cyanohydrin. Thus,this is true.
Option $C$: By assigning $R/S$ configurations to the two Fischer projections of $butan-2-ol$,both are found to have the same configuration (e.g.,$R$). Therefore,they represent the same molecule and are not enantiomers. Thus,this is false.
Option $D$: $CH_3-CH=NOH$ (acetaldoxime) exhibits geometrical isomerism due to the restricted rotation around the $C=N$ bond and the presence of a lone pair on the nitrogen atom,leading to $syn$ and $anti$ (or $cis$ and $trans$) isomers. Thus,this is true.

(B) meso-compound is an achiral molecule that contains stereocenters but is superimposable on its mirror image due to an internal plane of symmetry.
$1$. $trans-hex-3-ene + Br_2$ undergoes anti-addition to form a racemic mixture of $(3R, 4R)$ and $(3S, 4S)-3,4-dibromohexane$.
$2$. $cis-hex-3-ene + H_2 \xrightarrow{Pd-C}$ undergoes syn-addition to form a meso-compound,$(3R, 4S)-hexane-3,4-diol$ (if it were dihydroxylation) or in this case,the product of hydrogenation is $hexane$,which is achiral but not a meso-compound in the stereochemical sense. However,looking at the options provided,the question asks for the reaction that yields a meso-compound.
$3$. $hex-3-yne + H_2 \xrightarrow{Lindlar's \ catalyst}$ gives $cis-hex-3-ene$,which is not a meso-compound.
$4$. $cyclohexene + Br_2 \xrightarrow{CCl_4}$ gives $trans-1,2-dibromocyclohexane$,which is a racemic mixture,not a meso-compound.
Re-evaluating the options based on standard textbook examples: The hydrogenation of $cis-3,4-dimethylhex-3-ene$ (if that were the substrate) would yield a meso-compound. Given the provided image for option $B$ shows $cis-3,4-dimethylhex-3-ene$,the product of syn-addition is indeed a meso-compound.

(D) Statement $A$ is true: Stability increases with more covalent bonds and less charge separation.
Statement $B$ is false: In electromeric effect,$+E$ effect occurs with electrophiles and $-E$ effect occurs with nucleophiles.
Statement $C$ is true: Inductive effect causes polarity,affecting physical properties like melting and boiling points.
Statement $D$ is false: The heat of hydrogenation decreases as the number of alkyl groups attached to the double bond increases (due to increased stability of the alkene).
Statement $E$ is true: Stability of carbanion increases with $s$-character (e.g.,$sp > sp^2 > sp^3$).
Therefore,statements $B$ and $D$ are not true.

(A) . $2-$Methylpropene $(C_4H_8)$ and but$-1-$ene $(C_4H_8)$ differ in the carbon skeleton,so they are chain isomers $(III)$.
$B$. $Cis-$but$-2-$ene and $trans-$but$-2-$ene are geometric isomers,which are a type of stereoisomers $(I)$.
$C$. $2-$Butanol (alcohol) and diethyl ether (ether) have different functional groups,so they are functional group isomers $(IV)$.
$D$. But$-1-$ene and but$-2-$ene differ in the position of the double bond,so they are position isomers $(II)$.
Therefore,the correct matching is $A-III, B-I, C-IV, D-II$.