Versatile Nature of Carbon (Hydrocarbons, Functional Groups, Homologous Series, Nomenclature) Questions in English

(A) As the molecular mass of hydrocarbons increases,the intermolecular forces of attraction (van der Waals forces) between the molecules also increase.
Due to these stronger intermolecular forces,more energy is required to overcome them,which leads to an increase in both the boiling point and the melting point of the hydrocarbons.

(N/A) homologous series is a group of organic compounds having the same functional group and similar chemical properties,in which the successive members differ by a $CH_2$ unit.
Characteristics:
$(i)$ All members of a homologous series can be represented by the same general molecular formula.
(ii) Any two adjacent members of the series differ by one carbon atom and two hydrogen atoms,i.e.,a $CH_2$ unit in their molecular formula.

(C) The suffix '-one' in the name 'Pentanone' indicates the presence of a ketone functional group. $A$ ketone is characterized by a carbonyl group $(C=O)$ bonded to two carbon atoms.

(A) The general formula $C_{n}H_{2n-2}$ represents the homologous series of alkynes.
For the $1^{st}$ member,$n = 2$,which is Ethyne $(C_{2}H_{2})$.
For the $2^{nd}$ member,$n = 3$,which is Propyne.
The chemical formula for Propyne is $C_{3}H_{4}$.

(N/A) In a homologous series,as the molecular mass increases,the following characteristics change:
$(a)$ Melting and boiling points increase gradually due to the increase in intermolecular forces.
$(b)$ Physical properties such as density and solubility in a particular solvent show a regular gradation.
$(c)$ Molecular mass increases,which leads to a change in the physical state (e.g.,from gas to liquid to solid).
$(d)$ The chemical properties remain similar,but the intensity of reactions may vary slightly due to the change in the carbon chain length.

(A) $(i)$ The hydrocarbon is Ethene.
$(ii)$ It belongs to the Alkene family.
$(iii)$ The reaction of Ethene with Hydrogen in the presence of a catalyst (like Nickel or Palladium) results in the formation of Ethane:
$C_{2}H_{4} + H_{2} \xrightarrow{Ni/Pd} C_{2}H_{6}$

(N/A) For the molecular formula $C_{3}H_{6}O$,the two isomers are:
$1$. Propanal (Aldehyde): The structure is $CH_{3}-CH_{2}-CHO$.
$2$. Propanone (Ketone): The structure is $CH_{3}-CO-CH_{3}$.
Relation: These two compounds are functional isomers because they possess the same molecular formula but contain different functional groups (aldehyde group $-CHO$ and ketone group $>C=O$).

(N/A) Butanol.
$(b)$ Pentanoic acid.
$(c)$ Propan-$2$-one. An atom or group of atoms attached to the carbon chain by replacing one or more hydrogen atoms,which acts as a reactive site in the molecule and decides the chemical properties of the compound,is called a functional group.

Functional Group	Formula and Example
$1.$ Aldehyde	Formula: $-CHO$,Example: Ethanal $(CH_3CHO)$
$2.$ Ketone	Formula: $>C=O$,Example: Propanone $(CH_3COCH_3)$
$3.$ Carboxylic acid	Formula: $-COOH$,Example: Ethanoic acid $(CH_3COOH)$

(A) Butanone has the chemical formula $CH_3COCH_2CH_3$.
The suffix '-one' in the $IUPAC$ name indicates the presence of a ketone functional group $(>C=O)$.
In butanone,the carbonyl group is attached to two carbon atoms,which is the characteristic structure of a ketone.

(C) Alcohols are organic compounds containing a hydroxyl $(-OH)$ functional group attached to a saturated carbon atom.
The general formula for an alkane is $C_{n}H_{2n+2}$.
When one hydrogen atom is replaced by an $-OH$ group,the alkyl group becomes $C_{n}H_{2n+1}$.
Therefore,the general formula for a monohydric alcohol is $C_{n}H_{2n+1}-OH$.

(A) Alkynes are unsaturated hydrocarbons that contain at least one carbon-carbon triple bond.
The general formula for alkanes is $C_{n}H_{2n+2}$.
The general formula for alkenes is $C_{n}H_{2n}$.
The general formula for alkynes is $C_{n}H_{2n-2}$,where $n$ is the number of carbon atoms $(n \ge 2)$.

(B) Ethyne,also known as acetylene,has the chemical formula $C_{2}H_{2}$.
It consists of two carbon atoms connected by a triple bond,with each carbon atom also bonded to a single hydrogen atom.
The structural formula is $H - C \equiv C - H$.
Option $B$ correctly represents this structure.

(A) straight chain hydrocarbon is one where carbon atoms are linked in a single continuous chain without any branching.
Option $A$ represents a branched structure ($2$-methylpentane).
Option $B$ represents a branched structure ($2$-methylpentane).
Option $C$ represents a branched structure ($3$-methylhexane).
Option $D$ represents a straight chain hydrocarbon (n-pentane).
Since the question asks for the structure that is $NOT$ a straight chain,and options $A, B,$ and $C$ are all branched,the question implies identifying the branched ones. However,based on standard chemistry nomenclature,$D$ is the only straight chain. If the question asks which is $NOT$ straight,$A, B,$ and $C$ are correct. Given the typical format of such questions,$A$ is a clear example of a branched chain.

(C) Unsaturated hydrocarbons are those that contain at least one carbon-carbon double bond $(C=C)$ or triple bond $(C\equiv C)$.
$(i)$ $H_3C-CH_2-CH_2-CH_3$ is Butane,which is a saturated hydrocarbon (alkane).
$(ii)$ $H_3C-C\equiv C-CH_3$ is But$-2-$yne,which contains a triple bond,so it is unsaturated.
$(iii)$ $H_3C-CH(CH_3)-CH_3$ is Isobutane ($2$-methylpropane),which is a saturated hydrocarbon (alkane).
$(iv)$ $H_3C-C(CH_3)=CH_2$ is $2-$methylpropene,which contains a double bond,so it is unsaturated.
Therefore,$(ii)$ and $(iv)$ are unsaturated hydrocarbons.

(C) homologous series is a group of organic compounds having the same functional group and similar chemical properties,where each successive member differs by a $CH_{2}$ unit.
For alkanes,the general formula is $C_{n}H_{2n+2}$.
$CH_{4}$ $(n=1)$ follows $C_{n}H_{2n+2}$.
$C_{2}H_{6}$ $(n=2)$ follows $C_{n}H_{2n+2}$.
$C_{3}H_{8}$ $(n=3)$ follows $C_{n}H_{2n+2}$.
$C_{4}H_{8}$ follows the general formula $C_{n}H_{2n}$,which represents an alkene.
Therefore,$C_{4}H_{8}$ belongs to a different homologous series.

(D) The given compound is $CH_{3}-CH_{2}-CHO$.
This compound contains a chain of $3$ carbon atoms.
The functional group present is the aldehyde group $(-CHO)$.
For a $3$-carbon chain,the prefix is 'prop-'.
Since it is an aldehyde,the suffix '-al' is added to the alkane name.
Therefore,the $IUPAC$ name is Propanal.

(A) In organic chemistry,a heteroatom is any atom other than carbon or hydrogen present in an organic molecule.
The given molecule is $CH_{3}-CH_{2}-O-CH_{2}-CH_{2}Cl$.
$1$. Carbon $(C)$ and Hydrogen $(H)$ are the basic elements of the organic backbone.
$2$. Oxygen $(O)$ is a heteroatom because it is not carbon or hydrogen.
$3$. Chlorine $(Cl)$ is a heteroatom because it is not carbon or hydrogen.
Therefore,the heteroatoms present are oxygen $(i)$ and chlorine $(iv)$.

(B) The general formula for the alkyne homologous series is $C_nH_{2n-2}$,where $n$ represents the number of carbon atoms.
For the first member,we set $n = 2$ because a triple bond requires at least two carbon atoms.
Substituting $n = 2$ into the formula: $C_2H_{2(2)-2} = C_2H_{4-2} = C_2H_2$.
This compound is known as ethyne (also called acetylene).
Therefore,the first member of the alkyne series is ethyne.

(B) The functional group $(-CHO)$ is known as the aldehyde group.
Compounds containing this group are classified as aldehydes.
Amides contain the $(-CONH_2)$ group,ketones contain the $(>C=O)$ group,and alcohols contain the $(-OH)$ group.

(B) The functional group present in carboxylic acid is the carboxyl group,which is represented by the formula $-COOH$.
In this group,a carbon atom is double-bonded to an oxygen atom and single-bonded to a hydroxyl group $(-OH)$.

(C) In $IUPAC$ nomenclature,functional groups are identified by specific suffixes.
$1$. The suffix $-ol$ is used for the alcohol functional group,which is represented by the hydroxyl group $-OH$.
$2$. The suffix $-al$ is used for aldehydes $(-CHO)$.
$3$. The suffix $-one$ is used for ketones $(>C=O)$.
$4$. Halogens $(-X)$ are indicated by prefixes like chloro-,bromo-,etc.
Therefore,the correct option is $C$.

(D) Methyl ethanoate is an organic compound with the chemical formula $CH_3COOCH_3$.
It belongs to the class of organic compounds known as esters.
The general formula for an ester is $R-COO-R'$,where $R$ and $R'$ are alkyl groups.
In methyl ethanoate,the functional group present is the ester group $(-COO-)$.
Therefore,the correct option is $D$.

(C) The functional group $>C=O$ represents a ketone.
For a ketone to exist,the carbonyl carbon $(C=O)$ must be bonded to two other carbon atoms (one on each side).
Therefore,the simplest ketone is propanone $(CH_3-CO-CH_3)$,which contains $3$ carbon atoms.
In contrast,$-COOH$ (carboxylic acid) and $-CHO$ (aldehyde) can exist with only $1$ carbon atom (e.g.,methanoic acid $HCOOH$ and methanal $HCHO$).
$-C-O-$ (ether) can exist with $2$ carbon atoms (e.g.,dimethyl ether $CH_3-O-CH_3$).

(B) The functional group present in alcohols is the hydroxyl group,which is represented as $-OH$.
- Option $A$ $(-CHO)$ represents the aldehyde group.
- Option $B$ $(-OH)$ represents the hydroxyl group,characteristic of alcohols.
- Option $C$ $(-COOH)$ represents the carboxylic acid group.
- Option $D$ $(>C=O)$ represents the carbonyl group found in ketones.

(C) When one hydrogen atom of an alkane is replaced by a hydroxyl group $[-OH]$,the resulting compound belongs to the class of organic compounds known as alcohols.
The general formula for an alkane is $C_nH_{2n+2}$.
Replacing one hydrogen atom with an $-OH$ group gives the general formula $C_nH_{2n+1}OH$,which represents an alcohol.
For example,replacing one hydrogen in methane $(CH_4)$ with an $-OH$ group results in methanol $(CH_3OH)$.

(C) The given molecular formula is $CH_{3}COOCH_{3}$.
This compound belongs to the ester functional group,which has the general formula $R-COO-R'$.
In $CH_{3}COOCH_{3}$,the acyl group is derived from acetic acid (ethanoic acid),which contains two carbon atoms $(CH_{3}CO-)$.
The alkyl group attached to the oxygen is a methyl group $(-CH_{3})$.
Therefore,the $IUPAC$ name is methyl ethanoate.

(B) The functional group present in aldehyde compounds is the formyl group,which is represented as $-CHO$.
- Option $A$ $(-Cl)$ represents a chloro group (haloalkane).
- Option $B$ $(-CHO)$ represents an aldehyde group.
- Option $C$ $(-COOH)$ represents a carboxylic acid group.
- Option $D$ $(-OH)$ represents a hydroxyl group (alcohol).

(D) ketone is an organic compound characterized by the presence of a carbonyl group $(>C=O)$ bonded to two carbon atoms.
In a ketone,the carbonyl carbon is attached to two alkyl or aryl groups.
Option $A$ $(-CHO)$ represents an aldehyde group.
Option $B$ $(-COOH)$ represents a carboxylic acid group.
Option $C$ $(-OH)$ represents a hydroxyl group (alcohol).
Option $D$ $(>C=O)$ represents the carbonyl group found in ketones.

(C) The general formula for an alkane is $C_nH_{2n+2}$. For methane $(n=1)$,the formula is $CH_4$.
An aldehyde is a functional group characterized by a carbonyl group $(C=O)$ bonded to at least one hydrogen atom. The general formula for an aldehyde is $R-CHO$.
For the simplest aldehyde,where $R=H$,the compound is $HCHO$,which is known as methanal (formaldehyde).
Since methane has one carbon atom,its corresponding aldehyde must also contain one carbon atom,which is methanal.

(B) An aldehyde is represented by the general formula $R-CHO$,where $R$ is an alkyl group or a hydrogen atom.
In $Methanal$ $(HCHO)$,the $R$ group is replaced by a hydrogen atom $(H)$.
Therefore,$Methanal$ is the only aldehyde that does not contain an alkyl group attached to the carbonyl carbon.
$Methanol$ is an alcohol,not an aldehyde.
$Ethanal$ $(CH_3CHO)$ contains a methyl group $(CH_3)$,and $Propanal$ $(CH_3CH_2CHO)$ contains an ethyl group $(CH_3CH_2)$.

(A) The functional groups are analyzed as follows:
$1$. The hydroxyl group $(-OH)$ consists of an oxygen atom bonded to a hydrogen atom via a single covalent bond. It does not contain any double bonds.
$2$. The carbonyl group $(>C=O)$ contains a carbon atom double-bonded to an oxygen atom.
$3$. The carboxyl group $(-COOH)$ contains a carbonyl group $(C=O)$ and a hydroxyl group $(-OH)$,thus it possesses a double bond.
$4$. The aldehyde group $(-CHO)$ contains a carbon atom double-bonded to an oxygen atom $(C=O)$.
Therefore,the hydroxyl group $(-OH)$ is the only one among the options that does not possess a double bond.

(B) $Methanal$ is the $IUPAC$ name for the simplest aldehyde,which has the chemical formula $HCHO$.
Its common name is $Formaldehyde$.
$Acetaldehyde$ is the common name for $Ethanal$ $(CH_3CHO)$.
$Methyl$ alcohol is $Methanol$ $(CH_3OH)$.
$Formic$ acid is $Methanoic$ acid $(HCOOH)$.

(C) Methanal is the simplest aldehyde,containing one carbon atom.
Its $IUPAC$ name is methanal,and its common name is formaldehyde.
The general formula for an aldehyde is $R-CHO$.
For methanal,the $R$ group is a hydrogen atom $(H)$,resulting in the formula $HCHO$.

(D) The chemical formula of ethanal is $CH_3CHO$.
It belongs to the aldehyde functional group with two carbon atoms.
The $IUPAC$ name is ethanal,and its common name is acetaldehyde.

(D) The molecular formula $CH_{3}CHO$ represents an aldehyde with two carbon atoms.
In the $IUPAC$ nomenclature, an aldehyde with two carbon atoms is named as $Ethanal$.
Its common name is $Acetaldehyde$.
Therefore, the correct option is $D$.

(D) Propanone is a ketone with the chemical formula $CH_3COCH_3$.
Its $IUPAC$ name is propanone,while its widely used common name is acetone.
Therefore,the correct option is $D$.

(B) Propanone is a ketone with the chemical formula $CH_3COCH_3$.
The functional group present in ketones is the carbonyl group,represented as $>C=O$.
Option $A$ $(-CHO)$ represents an aldehyde.
Option $C$ $(-COOH)$ represents a carboxylic acid.
Option $D$ $(-COOR)$ represents an ester.
Therefore,the correct functional group for propanone is $>C=O$.

(C) ketone functional group requires a carbonyl group $(C=O)$ to be bonded to two other carbon atoms.
Therefore,the simplest ketone is propanone,which contains $3$ carbon atoms.
It is impossible to form a ketone with only $2$ carbon atoms because the carbonyl carbon must be flanked by two alkyl groups.

(C) The $-COOH$ functional group is known as the carboxylic acid group.
Among the given options:
$A$. Methanol $(CH_3OH)$ is an alcohol.
$B$. Methanal $(HCHO)$ is an aldehyde.
$C$. Formic acid $(HCOOH)$ contains the $-COOH$ functional group.
$D$. Acetone $(CH_3COCH_3)$ is a ketone.
Therefore,the correct answer is Formic acid.

(B) Formic acid has the chemical formula $HCOOH$.
According to $IUPAC$ nomenclature,a carboxylic acid with one carbon atom is named as methanoic acid.
The prefix 'meth-' indicates the presence of one carbon atom,and the suffix '-oic acid' denotes the carboxylic acid functional group.

(B) homopolymer is a polymer formed from a single type of monomer unit.
Polythene is formed by the polymerization of a single type of monomer,which is ethene $(CH_2=CH_2)$.
Styrene-butadiene rubber is a copolymer formed from two different monomers: styrene and butadiene.
Nylon and polyester are also copolymers formed by the condensation of different monomer units.
Therefore,polythene is the correct example of a homopolymer.

(C) The functional groups and their chemical representations are as follows:
$1$. Hydroxyl group is represented as $-OH$ (matches $b$).
$2$. Aldehyde group is represented as $-CHO$ (matches $c$).
$3$. Carboxylic acid group is represented as $-COOH$ (matches $d$).
$4$. Ketone group is represented as $>C=O$ (matches $a$).
Therefore, the correct matching is $1-b, 2-c, 3-d, 4-a$.