Introduction Questions in English

(C) $Butanal$ $(CH_3-CH_2-CH_2-CHO)$ is an aliphatic aldehyde because it contains the $-CHO$ functional group attached to an alkyl chain.

(B) In the carbonyl group $(>C=O)$,the carbon atom is bonded to three atoms (one oxygen via a double bond and two other groups/atoms via single bonds) and has no lone pairs.
According to the valence bond theory,the carbon atom undergoes $sp^2$ hybridisation to form three $sp^2$ hybrid orbitals.
Therefore,the carbon atom of the carbonyl group is $sp^2$ hybridised.

(C) Glyoxal is the simplest dialdehyde with the chemical formula $CHO-CHO$. It is also known as ethanedial.

(A) Aldehydes are functional isomers of ketones.
Both aldehydes and ketones share the same general molecular formula,$C_nH_{2n}O$,for a given number of carbon atoms.
They exhibit functional isomerism because they contain different functional groups (aldehyde group $-CHO$ vs. ketone group $C=O$) despite having the same molecular formula.

(C) The functional groups of the given compounds are as follows:
$1$. Phenol: $Ar-OH$ (contains $-OH$ group)
$2$. Carboxylic acid: $R-COOH$ (contains $-OH$ group in the carboxyl moiety)
$3$. Aldehydes: $R-CHO$ (contains a carbonyl group,not an $-OH$ group)
$4$. Alcohols: $R-OH$ (contains $-OH$ group)
Therefore,aldehydes do not contain an $-OH$ group.

(B) carbonyl group $(>C=O)$ has two free valencies. When both of these valencies are satisfied by two alkyl groups $(R)$,the resulting compound is represented as $R-CO-R$. This class of organic compounds is known as a ketone.

(B) In the carbonyl group $(>C=O)$,the carbon atom is $sp^2$ hybridized.
It forms three sigma $(\sigma)$ bonds: two with the adjacent atoms (like $R$ and $R'$ or $H$) and one sigma bond with the oxygen atom.
Additionally,it forms one pi $(\pi)$ bond with the oxygen atom.
Therefore,the carbonyl carbon is joined to other atoms by three sigma and one pi bonds.

(B) Ketones are classified as simple (symmetrical) or mixed (unsymmetrical) based on the groups attached to the carbonyl carbon $(>C=O)$.
In simple ketones,both alkyl or aryl groups attached to the carbonyl carbon are the same.
In mixed ketones,the two groups attached to the carbonyl carbon are different (e.g.,one alkyl and one aryl group).
Acetophenone $(C_6H_5COCH_3)$ consists of a phenyl group and a methyl group attached to the carbonyl carbon,making it a mixed ketone.

(A) Chloral is the common name for $2,2,2-trichloroethanal$.
Its chemical formula is $CCl_3CHO$.

(B) Carbonyl compounds are defined as compounds containing the carbonyl group,which is a carbon atom double-bonded to an oxygen atom $(C=O)$.
While carboxylic acids and their derivatives also contain this group,the term 'carbonyl compounds' in general chemistry is most commonly used to refer specifically to aldehydes and ketones.

(B) Formalin is an aqueous solution containing approximately $40\% \, HCHO$ (formaldehyde),$8\% \, CH_3OH$ (methanol) as a stabilizer,and $52\%$ water.
It is widely used as a biological preservative for specimens.

(A) The given compound is $(CH_3)_3C-CHO$.
This compound consists of a $t$-butyl group attached to an aldehyde group $(-CHO)$.
The common name for this specific aldehyde is $Pivaldehyde$.
It is also known as $2,2$-dimethylpropanal in $IUPAC$ nomenclature.

(B) The molecular formula $C_5H_{10}O$ corresponds to a degree of unsaturation of $1$,which can be a ketone or an aldehyde. For ketones,the carbonyl group $(C=O)$ must be placed between two carbon atoms.
The possible ketone isomers are:
$1$. $CH_3-CO-CH_2-CH_2-CH_3$ (Pentan$-2-$one)
$2$. $CH_3-CH_2-CO-CH_2-CH_3$ (Pentan$-3-$one)
$3$. $CH_3-CO-CH(CH_3)_2$ ($3$-Methylbutan$-2-$one)
Thus,there are a total of $3$ ketone isomers.

(D) The molecular formulas for the given compounds are as follows:
$(i)$ Propanone is a ketone with three carbon atoms: $CH_3COCH_3$.
$(ii)$ Propanal is an aldehyde with three carbon atoms: $CH_3CH_2CHO$.
$(iii)$ $n$-Propanol is a primary alcohol with three carbon atoms: $CH_3CH_2CH_2OH$.
Therefore,the correct sequence is $(i) = CH_3COCH_3, (ii) = CH_3CH_2CHO, (iii) = CH_3CH_2CH_2OH$.

(B) The given chemical structure is $CH_2=CH-CHO$.
This compound is an unsaturated aldehyde containing three carbon atoms.
The $IUPAC$ name for this compound is $Prop-2-enal$.
The common name for $CH_2=CH-CHO$ is $Acrolein$.

(B) For the molecular formula $C_4H_8O$,we look for carbonyl isomers (aldehydes and ketones).
$1$. Aldehydes (containing $-CHO$ group):
- Butanal: $CH_3-CH_2-CH_2-CHO$
- $2-$Methylpropanal: $(CH_3)_2CH-CHO$
Total aldehydes = $2$.
$2$. Ketones (containing $C=O$ group):
- Butanone: $CH_3-CO-CH_2-CH_3$
Total ketones = $1$.
Total isomers = $2 \text{ (aldehydes)} + 1 \text{ (ketone)} = 3$.

(A) The $IUPAC$ name $3-$methyl$-2-$pentanone indicates a ketone functional group $(C=O)$ at the $C-2$ position of a $5-$carbon chain (pentane).
At the $C-3$ position,there is a methyl group $(-CH_3)$.
Writing the structure:
$C^1H_3 - C^2(=O) - C^3H(CH_3) - C^4H_2 - C^5H_3$.
This matches the structure in option $A$.

(A) The general formula for aliphatic aldehydes is $C_nH_{2n}O$ (where $n \geq 1$).
For $n=1$,the first member is $HCHO$ (methanal).
For $n=2$,the second member is $CH_3CHO$ (ethanal).
For $n=3$,the third member is $CH_3CH_2CHO$ (propanal).
Therefore,the correct structure for the third member is $CH_3CH_2CHO$.

(A) Acrolein is the common name for $CH_2=CH-CHO$.
It contains a carbon-carbon double bond (unsaturation) and an aldehyde functional group $(-CHO)$.
Therefore,it is classified as an unsaturated aldehyde.

(C) Formalin is a $40\%$ aqueous solution of formaldehyde $(HCHO)$.

(A) As an exception,formaldehyde exists as a gas at room temperature. In contrast,other low molecular weight members of aldehydes and ketones are liquids at room temperature.

(C) The correct formula for Acetaldehyde is $CH_3CHO$. The option $CH_3CH_2CHO$ represents Propanal. Therefore,the pair Acetaldehyde - $CH_3CH_2CHO$ is incorrectly matched.

(A) The structure provided is $C_6H_5COCH_3$.
According to $IUPAC$ nomenclature,ketones are named by replacing the suffix $-e$ of the corresponding alkane with $-one$.
For the compound $C_6H_5COCH_3$,the parent chain is an ethane chain (two carbons) attached to a phenyl group.
Thus,the $IUPAC$ name is $1$-phenylethanone (often simply called phenylethanone).
Acetophenone is the common name,and methyl phenyl ketone is another common name.

(C) Let us analyze the structure of the given molecule: $CH_3 - C(=O) - O - CH_2 - C(=O) - CH_2 - NH_2$.
$1$. The group $-C(=O)-O-$ is an $Ester$ group.
$2$. The group $-C(=O)-$ (where the carbonyl carbon is bonded to two carbon atoms) is a $Ketone$ group.
$3$. The group $-NH_2$ is a primary $Amine$ group.
$4$. An $Amide$ group requires a nitrogen atom bonded directly to a carbonyl carbon ($-C(=O)-NH_2$ or similar).
In the given structure,there is no nitrogen atom bonded to a carbonyl carbon. Therefore,the $Amide$ functional group is not present.

(B) $CH_3-CH(CH_3)-CHO$ is Isobutyraldehyde.
$CH_3-CN$ is Acetonitrile or Methyl cyanide.
$CH_2=CH-CHO$ is Acrolein or Acrylaldehyde.
$CH_3-C(CH_3)(CH_3)-OH$ is tert-Butyl alcohol.

(C) ketone is classified as unsymmetrical (or mixed) if the two alkyl or aryl groups attached to the carbonyl carbon atom are different.
$a.$ $CH_3-CH(CH_3)-O-CH_2-CH_3$ is an ether.
$b.$ $CH_3-CH_2-CO-CH_2-CH_3$ (Pentan-$3$-one) has two ethyl groups (symmetrical).
$c.$ $CH_3-CH(CH_3)-CO-CH_2-CH_2-CH_3$ ($2$-methylhexan-$3$-one) has an isopropyl group and a propyl group (unsymmetrical).
$d.$ $CH_3-CO-CH_3$ (Propanone) has two methyl groups (symmetrical).

(A) The given compound is $C_6H_5COCH_3$.
In the $IUPAC$ system,ketones are named by replacing the suffix $-e$ of the corresponding alkane with $-one$.
The parent chain is the two-carbon chain (ethane),and the phenyl group is attached to the first carbon.
Therefore,the $IUPAC$ name is $1-$phenylethanone (commonly known as acetophenone).

(B) The given structure is an aldehyde with a $6-$carbon chain (hexane) and a methyl group attached to the second carbon atom from the aldehyde group.
According to $IUPAC$ nomenclature rules,the carbon atom of the aldehyde group $(-CHO)$ is assigned position $1$.
Counting from the aldehyde carbon,the methyl group is at the $C-2$ position.
Therefore,the correct $IUPAC$ name is $2-$methylhexanal.

(N/A) The structures of the given compounds are as follows:

$(i)$ $\alpha$-Methoxypropionaldehyde	$CH_3-CH(OCH_3)-CHO$
$(ii)$ $3$-Hydroxybutanal	$CH_3-CH(OH)-CH_2-CHO$
$(iii)$ $2$-Hydroxycyclopentanecarbaldehyde	$C_5H_8(OH)CHO$
$(iv)$ $4$-Oxopentanal	$CH_3-CO-CH_2-CH_2-CHO$
$(v)$ Di-sec-butyl ketone	$CH_3-CH_2-CH(CH_3)-CO-CH(CH_3)-CH_2-CH_3$
$(vi)$ $4$-Fluoroacetophenone	$F-C_6H_4-COCH_3$

(N/A) $(i)$ $CH_{3}CO(CH_{2})_{6}CH_{3}$
$IUPAC$ name: Octan$-2-$one
Common name: Methyl $n-$heptyl ketone
$(ii)$ $CH_{3}CH_{2}CHBrCH_{2}CH(CH_{3})CHO$
$IUPAC$ name: $4-$Bromo$-2-$methylhexanal
$(iii)$ $CH_{3}(CH_{2})_{5}CHO$
$IUPAC$ name: Heptanal
$(iv)$ $Ph-CH=CH-CHO$
$IUPAC$ name: $3-$phenylprop$-2-$enal
Common name: $\beta-$Phenylacrolein
$(v)$ Cyclopentanecarbaldehyde
$(vi)$ $PhCOPh$
$IUPAC$ name: Diphenylmethanone
Common name: Benzophenone

(N/A) Aldehydes and ketones are often referred to by their common names.
$(a)$ Common names of Aldehydes:
$(i)$ The common names of aldehydes are derived from the common names of the corresponding carboxylic acids by replacing the terminal '-ic acid' with '-aldehyde'.
$(ii)$ These names often reflect the Latin or Greek origin of the source of the acid or aldehyde.
$(iii)$ The position of substituents in the carbon chain is indicated by Greek letters $\alpha, \beta, \gamma, \delta$,etc. The $\alpha$-carbon is the one directly linked to the $-CHO$ group,the $\beta$-carbon is the next,and so on.
$(iv)$ The carbon atom next to the $-CHO$ group is designated as $\alpha$: $\stackrel{\delta}{C}-\stackrel{\gamma}{C}-\stackrel{\beta}{C}-\stackrel{\alpha}{C}-CHO$.
Examples include $HCHO$ (Formaldehyde),$CH_3CHO$ (Acetaldehyde),and $\beta$-Bromobutyraldehyde.
$(b)$ Common names of Ketones:
$(i)$ Common names are obtained by naming the two alkyl or aryl groups attached to the carbonyl group $(>C=O)$ as two separate words in alphabetical order,followed by the word 'ketone'.
$(ii)$ For example,$CH_3COCH_3$ is dimethyl ketone (acetone) and $CH_3COC_2H_5$ is ethyl methyl ketone.

(A) $IUPAC$ nomenclature of aldehydes and ketones:
$1$. $IUPAC$ nomenclature of Aldehydes:
- The $IUPAC$ names of open-chain aliphatic aldehydes are derived from the names of the corresponding alkanes by replacing the terminal $'e'$ with the suffix $'al'$.
- In substituted aldehydes,the carbon chain containing the aldehyde group is selected as the parent chain. The numbering starts from the aldehyde carbon,which is always assigned position $1$.
- Example: $CH_3CH_2CHO$ is Propanal.
$2$. $IUPAC$ nomenclature of Ketones:
- The $IUPAC$ names of open-chain aliphatic ketones are derived from the names of the corresponding alkanes by replacing the terminal $'e'$ with the suffix $'one'$.
- The chain is numbered in such a way that the carbonyl carbon gets the lowest possible number.
- Example: $CH_3COCH_2CH_3$ is Butan$-2-$one.

(A) The structures for the given $IUPAC$ names are as follows:
$(i)$ Methanal: $HCHO$
$(ii)$ Ethanal: $CH_3CHO$
$(iii)$ $2$-Methylpropanal: $(CH_3)_2CHCHO$
$(iv)$ $3$-Methylcyclohexanecarbaldehyde: $C_6H_{10}(CH_3)CHO$ (with the aldehyde group attached to the ring)
$(v)$ $2$-Methoxypropanal: $CH_3CH(OCH_3)CHO$
$(vi)$ Pentanal: $CH_3CH_2CH_2CH_2CHO$
$(vii)$ Prop-$2$-enal: $CH_2=CHCHO$
$(viii)$ Benzene-$1,2$-dicarbaldehyde: $C_6H_4(CHO)_2$ (ortho-substituted)
$(ix)$ $3$-Bromobenzaldehyde: $C_6H_4(Br)CHO$ (meta-substituted)

(A) The structures for the requested compounds are as follows:
$(i)$ Propanone: $CH_3COCH_3$
$(ii)$ Butanone: $CH_3COCH_2CH_3$
$(iii)$ Hexan$-3-$one: $CH_3CH_2COCH_2CH_2CH_3$
$(iv)$ $2,4-$dimethylpentan$-3-$one: $(CH_3)_2CHCOCH(CH_3)_2$
$(v)$ Pentan$-3-$one: $CH_3CH_2COCH_2CH_3$
$(vi)$ $2-$methylcyclohexanone: $\text{A}$ cyclohexanone ring with a methyl group at position $2$.
$(vii)$ $4-$methylpent$-3-$en$-2-$one: $CH_3COCH=C(CH_3)_2$
$(viii)$ Benzophenone: $C_6H_5COC_6H_5$
$(ix)$ Acetophenone: $C_6H_5COCH_3$
$(x)$ $3-$methylbutan$-2-$one: $CH_3COCH(CH_3)_2$
$(xi)$ $5-$hydroxy$-5-$methylhexan$-3-$one: $CH_3CH_2COCH_2C(OH)(CH_3)_2$
$(xii)$ $3-$oxobutanoic acid: $CH_3COCH_2COOH$
$(xiii)$ $3,3-$dimethylheptan$-2,6-$dione: $CH_3COCH_2C(CH_3)_2CH_2CH_2COCH_3$

(N/A) Electronic configuration of carbon and oxygen is as follows:
$C^{*} [He] 2s^{1} 2p_{x}^{1} 2p_{y}^{1} 2p_{z}^{1}$
$O [He] 2s^{2} 2p_{x}^{2} 2p_{y}^{1} 2p_{z}^{1}$
$(b)$ $(i)$ The $C=O$ bond of the carbonyl group in aldehydes and ketones is made up of one $\sigma$-bond and one $\pi$-bond. The carbonyl carbon atom is $sp^{2}$-hybridised and forms three sigma $(\sigma)$ bonds.
$(ii)$ The fourth valence electron of carbon remains in its $p$-orbital and forms a $\pi$-bond with oxygen by overlapping with the $p$-orbital of oxygen.
$(iii)$ In addition,the oxygen atom also has two non-bonding electron pairs. Thus,the carbonyl carbon and the three atoms attached to it lie in the same plane,and the $p$-electron cloud is above and below this plane $\left( C \frac{\pi}{\sigma} \ddot{O} \right)$.
$(iv)$ In carbonyl group compounds,the carbonyl carbon atom is $sp^{2}$ hybridised and forms three $sp^{2}$ hybrid orbitals.
$(v)$ One of the $sp^{2}$ hybridised orbitals of carbon overlaps with the $p$-orbital of the oxygen atom,while the $\pi$ bond is formed by the lateral overlap of the $p$-orbitals of carbon and oxygen.
$(vi)$ The remaining two hybridised $sp^{2}$ orbitals of the carbon atom form two additional $\sigma$-bonds by overlapping with the $1s$-orbital of two hydrogen atoms in formaldehyde.
$(vii)$ For any other aldehyde,two hybridised $sp^{2}$ orbitals of the carbon atom form two $\sigma$-bonds by overlapping with the $1s$-orbital of one hydrogen atom and one $sp^{3}$ hybrid orbital of an alkyl group carbon.
$(viii)$ For any ketone,two hybridised $sp^{2}$ orbitals of the carbon atom form two $\sigma$-bonds by overlapping with two $sp^{3}$ hybrid orbitals of two alkyl group carbons.
$(ix)$ All three $\sigma$-bonds lie in the same plane inclined to one another by $120^{\circ}$ as shown in the figure.

(N/A) Electronic configuration of carbon and oxygen is as follows:
$C^{*}[He] 2s^{1} 2p_{x}^{1} 2p_{y}^{1} 2p_{z}^{1}$
$O [He] 2s^{2} 2p_{x}^{2} 2p_{y}^{1} 2p_{z}^{1}$
$(b)$ $(i)$ The $C=O$ bond of the carbonyl group in aldehydes and ketones is made up of one $\sigma$-bond and one $\pi$-bond. The carbonyl carbon atom is $sp^{2}$-hybridised and forms three sigma $(\sigma)$ bonds.
$(ii)$ The fourth valence electron of carbon remains in its $p$-orbital and forms a $\pi$-bond with oxygen by overlap with the $p$-orbital of an oxygen atom.
$(iii)$ In addition,the oxygen atom also has two non-bonding electron pairs. Thus,the carbonyl carbon and the three atoms attached to it lie in the same plane,and the $p$-electron cloud is above and below this plane $\left(C \frac{\pi}{\sigma} \ddot{O}\right)$.
$(iv)$ In carbonyl group compounds,the carbonyl carbon atom is $sp^{2}$ hybridised and forms three $sp^{2}$ hybrid orbitals.
$(v)$ One of the $sp^{2}$ hybridised orbitals of carbon overlaps with the $p$-orbital of the oxygen atom forming a $\sigma$ bond,while the $p$-orbital overlap forms the $\pi$ bond.
$(vi)$ The remaining two hybridised $sp^{2}$ orbitals of the carbon atom form two additional $\sigma$-bonds by overlapping with $1s$-orbitals of two hydrogen atoms in formaldehyde.
$(vii)$ For any other aldehyde,two hybridised $sp^{2}$ orbitals of the carbon atom form two $\sigma$-bonds by overlapping with the $1s$-orbital of one hydrogen atom and one $sp^{3}$ hybrid orbital of an alkyl group carbon.
$(viii)$ For any ketone,two hybridised $sp^{2}$ orbitals of the carbon atom form two $\sigma$-bonds by overlapping with two $sp^{3}$ hybrid orbitals of two alkyl group carbons.
$(ix)$ All three $\sigma$-bonds lie in the same plane,inclined to one another by $120^{\circ}$ as shown in the figure.

(N/A) Polar bond: In the carbonyl group,the carbon-oxygen double bond is polarized due to the higher electronegativity of oxygen $(3.5)$ compared to that of the carbon $(2.5)$ atom.
As a result,the oxygen atom tends to attract the electron cloud of the $\pi$-bond towards itself,as shown by the polarization: $C=O \leftrightarrow C^{+\delta}=O^{-\delta}$.
Carbonyl compounds have substantial dipole moments and are more polar than ethers.
$(b)$ Resonance structure: The high polarity of the carbonyl group $(C^{+\delta}=O^{-\delta})$ is explained on the basis of resonance involving a neutral structure and a dipolar structure,as shown in the figure.
$(c)$ Electrophilic (Lewis acid) and Nucleophilic (Lewis base) centers: The carbon-oxygen double bond is polarized due to the higher electronegativity of oxygen relative to carbon. Hence,the carbonyl carbon acts as an electrophilic (Lewis acid) center,and the carbonyl oxygen acts as a nucleophilic (Lewis base) center.

(N/A) $(i)$ $4-$Nitropropiophenone: The structure consists of a benzene ring with a nitro group $(-NO_2)$ at the $4-$position and a propiophenone group $(-CO-CH_2-CH_3)$ at the $1-$position.
$(ii)$ $2-$Hydroxycyclopentanecarbaldehyde: The structure consists of a cyclopentane ring with a carbaldehyde group $(-CHO)$ at position $1$ and a hydroxyl group $(-OH)$ at position $2$.
$(iii)$ Phenyl acetaldehyde: The structure consists of a phenyl group $(-C_6H_5)$ attached to an acetaldehyde group $(-CH_2-CHO)$.

(A-4, B-5, C-1, D-2, E-3) The correct matches are as follows:
$A$. Cinnamaldehyde is $3$-Phenylprop-$2$-en-$al$ $(4)$.
$B$. Acetophenone is $1$-Phenylethanone $(5)$.
$C$. Valeraldehyde is Pentanal $(1)$.
$D$. Acrolein is Prop-$2$-en-$al$ $(2)$.
$E$. Mesityl oxide is $4$-Methylpent-$3$-en-$2$-one $(3)$.
Therefore,the correct sequence is $A-4, B-5, C-1, D-2, E-3$.

(A) $(i) \rightarrow (b), (c), (d), (e)$
$(ii) \rightarrow (a), (f)$
$(iii) \rightarrow (g)$

(A) $(i)$ False: Acetone is propanone,but it contains a ketone group,not an amide group.
$(ii)$ True: Acetone is dimethyl ketone because two methyl groups are attached to the carbonyl carbon.
$(iii)$ True: The chemical formula of acetone is $CH_3COCH_3$.
$(iv)$ True: Acetone contains a carbonyl group $(>C=O)$.

(A) The correct matching is as follows:
$(A)$ Resonance structures of the carbonyl group represent the polarity of the bond,where carbon acts as an electrophilic centre and oxygen as a nucleophilic centre. Thus,$A \rightarrow iv$.
$(B)$ The geometry shows the $sp^2$ hybridized carbon with a planar triangular structure and bond angles of approximately $120^{\circ}$. Thus,$B \rightarrow iii$.
$(C)$ The $\sigma$-bonding involves the overlap of orbitals forming the skeleton of the carbonyl group. Thus,$C \rightarrow ii$ (referring to the non-bonding $2p$ orbitals perpendicular to the plane).
$(D)$ The $\pi$-bonding involves the overlap of $p$-orbitals above and below the plane. Thus,$D \rightarrow i$.
Therefore,the correct match is $A$ $\rightarrow iv, B$ $\rightarrow iii, C$ $\rightarrow ii, D$ $\rightarrow i$.

(A) The correct matches are:
$A. 4$-oxopentanal corresponds to structure $ii$ $(CH_3COCH_2CH_2CHO)$ and common name $r$ ($\beta$-acetyl propionaldehyde).
$B. \text{Ethane-}1,2\text{-dial}$ corresponds to structure $iii$ $(CHOCHO)$ and common name $s$ (Glyoxal).
$C. \text{Pentane-}2,4\text{-dione}$ corresponds to structure $iv$ $(CH_3COCH_2COCH_3)$ and common name $p$ (Acetyl acetone).
$D. \text{Pentanal}$ corresponds to structure $i$ $(CH_3CH_2CH_2CH_2CHO)$ and common name $q$ (Valeraldehyde).
Thus,the correct sequence is $A-ii-r, B-iii-s, C-iv-p, D-i-q$.

(D) Mesityl oxide is an $\alpha, \beta-$unsaturated ketone formed by the aldol condensation of acetone.
Its chemical structure is $(CH_3)_2C=CHCOCH_3$.
According to $IUPAC$ nomenclature,the longest carbon chain containing the ketone group has $5$ carbons.
The double bond is at position $3$ and the ketone group is at position $2$.
There is a methyl group at position $4$.
Therefore,the $IUPAC$ name is $4-$Methylpent$-3-$en$-2-$one.

(B)
$A$ formyl group consists of a carbonyl group attached to a hydrogen atom,represented as the $-CHO$ group.
The structures of the given organic compounds are as follows:
$(i)$ Acetone: $CH_3-CO-CH_3$
$(ii)$ Acetaldehyde: $CH_3-CHO$
$(iii)$ Acetic acid: $CH_3-COOH$
$(iv)$ Acetic anhydride: $(CH_3CO)_2O$
Since acetaldehyde contains the $-CHO$ group,it is the correct answer.
Thus,the correct option is $(b)$.

(C) simple or symmetrical ketone is one in which both alkyl or aryl groups attached to the carbonyl carbon are identical.
Benzophenone,with the structure $(C_6H_5)_2C=O$,has two phenyl groups attached to the carbonyl carbon,making it a symmetrical ketone.
Acetophenone $(C_6H_5COCH_3)$,Butanone $(CH_3COCH_2CH_3)$,and Pentan-$2$-one $(CH_3COCH_2CH_2CH_3)$ are all mixed or unsymmetrical ketones.

(C) Butyraldehyde $(CH_3CH_2CH_2CHO)$ is known for its characteristic buttery or rancid butter-like odour.

(D) The structure of mesityl oxide is $(CH_3)_2C=CHCOCH_3$.
To determine the $IUPAC$ name,we identify the longest carbon chain containing the principal functional group (ketone) and the double bond.
The chain has $5$ carbon atoms,so the parent alkane is pentane.
The ketone group is at position $2$,and the double bond starts at position $3$.
There is a methyl substituent at position $4$.
Thus,the $IUPAC$ name is $4-$methylpent$-3-$en$-2-$one.

(C) The structure of mesityl oxide is $(CH_3)_2C=CHCOCH_3$.
To name this compound according to $IUPAC$ rules:
$1$. Identify the longest carbon chain containing the functional group (ketone) and the double bond. The chain has $5$ carbon atoms,so the parent alkane is pentane.
$2$. Number the chain starting from the end closer to the ketone group: $C_1$ is the methyl carbon,$C_2$ is the carbonyl carbon,$C_3$ is the $CH$ group,$C_4$ is the $C$ group,and $C_5$ is the methyl group attached to $C_4$.
$3$. The ketone is at position $2$,and the double bond starts at position $3$. There is a methyl group at position $4$.
$4$. Combining these,the name is $4-$methylpent$-3-$en$-2-$one.

(C) The chemical formula of salicylaldehyde is $C_7H_6O_2$.
It consists of a benzene ring substituted with an aldehyde group $(-CHO)$ and a hydroxyl group $(-OH)$ at the ortho position.
Looking at the structure,there is one oxygen atom in the aldehyde group and one oxygen atom in the hydroxyl group.
Therefore,the total number of oxygen atoms present in salicylaldehyde is $1 + 1 = 2$.