Properties Questions in English

(B) $1$. Nucleophilic addition reactions in carbonyl compounds depend on the electrophilicity of the carbonyl carbon and steric hindrance.
$2$. Aldehydes have one alkyl group and one hydrogen atom attached to the carbonyl carbon, whereas ketones have two alkyl groups.
$3$. Alkyl groups are electron-donating ($+I$ effect), which reduces the positive charge (electrophilicity) on the carbonyl carbon.
$4$. Ketones have two such groups, making them less electrophilic than aldehydes.
$5$. Additionally, two alkyl groups in ketones create more steric hindrance compared to one alkyl group and one small hydrogen atom in aldehydes, making the approach of the nucleophile more difficult in ketones.

(A) $1$. The Cannizzaro reaction is a disproportionation reaction involving aldehydes that do not have an $\alpha$-hydrogen atom.
$2$. In the presence of a concentrated base, one molecule of the aldehyde is reduced to the corresponding alcohol, while another molecule is oxidized to the salt of the corresponding carboxylic acid.
$3$. Therefore, the correct statement is that some aldehydes are converted to the respective alcohols and the salt of the respective carboxylic acid.

(D) Step $1$: Reaction in container $A$ with $H_2 / Ni$. $H_2 / Ni$ is a strong reducing agent that reduces both the alkene double bond and the aldehyde group. Thus, $CH_3-CH=CH-CH_2-CHO + 2H_2 \xrightarrow{Ni} CH_3-CH_2-CH_2-CH_2-CH_2OH$ (or $CH_3(CH_2)_3CH_2OH$). So, $P = CH_3(CH_2)_3CH_2OH$.
Step $2$: Reaction in container $B$ with $LiAlH_4$. $LiAlH_4$ is a selective reducing agent that reduces the aldehyde group to a primary alcohol but does not reduce the isolated alkene double bond. Thus, $CH_3-CH=CH-CH_2-CHO \xrightarrow{LiAlH_4 / H_3O^+} CH_3-CH=CH-CH_2-CH_2OH$. So, $Q = CH_3-CH=CH-CH_2-CH_2OH$.
Step $3$: Comparing with options, $P = CH_3(CH_2)_3CH_2OH$ and $Q = CH_3-CH=CH-CH_2-CH_2OH$ matches option $D$.

(A) Step $1$: Acetone $(CH_3COCH_3)$ is a ketone.
Step $2$: $CrO_3$ is a strong oxidizing agent.
Step $3$: Oxidation of ketones with strong oxidizing agents like $CrO_3$ leads to the cleavage of $C-C$ bonds, resulting in a mixture of carboxylic acids with fewer carbon atoms than the original ketone.
Step $4$: The oxidation of acetone $(CH_3COCH_3)$ yields acetic acid $(CH_3COOH)$ and formic acid $(HCOOH)$, where acetic acid is the major product.
Therefore, the correct option is $A$.

(A) $1$. The reduction of alkanones (ketones) to alkanes can be achieved using Clemmensen reduction.
$2$. In Clemmensen reduction, the carbonyl group $(>C=O)$ is reduced to a methylene group $(-CH_2-)$ using zinc amalgam $(Zn-Hg)$ and concentrated hydrochloric acid $(HCl)$.
$3$. The reaction is: $R-CO-R' + 4[H] \xrightarrow{Zn-Hg/HCl} R-CH_2-R' + H_2O$.
$4$. Etard reaction is used for the oxidation of toluene to benzaldehyde. Rosenmund reduction is used for the reduction of acid chlorides to aldehydes. Haloform reaction is used for the detection of methyl ketones.

(B) $1$. The general reaction of an aldehyde with a Grignard reagent $(R'MgX)$ is: $RCHO + R'MgX \rightarrow RCH(OMgX)R'$.
$2$. Upon subsequent hydrolysis, the intermediate forms a secondary alcohol: $RCH(OMgX)R' + H_2O \rightarrow RCH(OH)R' + Mg(OH)X$.
$3$. Since formaldehyde $(HCHO)$ reacts with Grignard reagents to form primary alcohols, all other aldehydes ($RCHO$ where $R \neq H$) yield secondary alcohols.

(B) Step $1$: Ethyl methyl ketone is $CH_3-CO-C_2H_5$.
Step $2$: Grignard reagent $CH_3MgBr$ acts as a nucleophile, where $CH_3^-$ attacks the carbonyl carbon.
Step $3$: The reaction is: $CH_3-CO-C_2H_5 + CH_3MgBr \rightarrow CH_3-C(OMgBr)(CH_3)-C_2H_5$.
Step $4$: Upon hydrolysis, the intermediate forms $CH_3-C(OH)(CH_3)-C_2H_5$, which is $2-methylbutan-2-ol$ or $C_2H_5-C(CH_3)_2-OH$.

(B) $1$. Hydroboration-oxidation of propene $(CH_3-CH=CH_2)$ follows anti-Markovnikov addition to yield propan$-1-$ol $(P = CH_3-CH_2-CH_2-OH)$.
$2$. Oxidation of primary alcohol $(P)$ with $CrO_3$ in anhydrous medium stops at the aldehyde stage, yielding propanal $(Q = CH_3-CH_2-CHO)$.
$3$. Reaction of propanal $(Q)$ with methylmagnesium bromide $(CH_3MgBr)$ followed by acid hydrolysis yields butan$-2-$ol $(R = CH_3-CH_2-CH(OH)-CH_3)$.

(A) $1$. Reaction of carbonyl compounds with hydroxylamine $(NH_2OH)$ yields an oxime: $(a) \rightarrow (ii)$.
$2$. Reaction of carbonyl compounds with primary amines $(R-NH_2)$ yields a Schiff base: $(b) \rightarrow (iii)$.
$3$. Reaction of carbonyl compounds with alcohols $(R-OH)$ in the presence of dry $HCl$ yields an acetal: $(c) \rightarrow (iv)$.
$4$. Reaction of carbonyl compounds with hydrogen cyanide $(HCN)$ yields a cyanohydrin: $(d) \rightarrow (i)$.
Therefore, the correct matching is $a-ii, b-iii, c-iv, d-i$.

(D) The reaction between benzaldehyde and acetophenone in the presence of a dilute base $(OH^-)$ is a Claisen-Schmidt condensation reaction.
$1$. Acetophenone acts as the nucleophile because it has $\alpha$-hydrogens, which are removed by the base to form an enolate ion.
$2$. The enolate ion attacks the carbonyl carbon of benzaldehyde to form a $\beta$-hydroxy ketone.
$3$. The $\beta$-hydroxy ketone undergoes dehydration (loss of $H_2O$) to form an $\alpha, \beta$-unsaturated ketone, which is the stable major product.
$4$. The product is $1,3-\text{diphenylprop-2-en-1-one}$ (also known as chalcone), represented by option $D$.