Q.Propan-1-ol is treated with pyridinium chlorochromate (PCC). Identify the product and explain why PCC, rather than acidified $\text{KMnO}_4$ or $\text{K}_2\text{Cr}_2\text{O}_7$, is used when a primary alcohol must be stopped at the aldehyde stage.

Hg$^{2+}$/H$_2$SO$_4$ hydration of propyne (Markovnikov) gives propanone via the enol.

Terminal alkynes undergo Markovnikov (mercuric-ion catalysed) hydration with dilute $H_2SO_4$ in presence of 1% $HgSO_4$, adding water across the triple bond to give an enol which tautomerises to a ketone (following Markovnikov's rule).

$CH_3-C\equiv CH + H_2O \xrightarrow[40\%\ H_2SO_4]{1\%\ HgSO_4} CH_3-\underset{OH}{C}=CH_2 \ (\text{enol}) \longrightarrow CH_3-CO-CH_3$

The product is propanone (acetone).

2,3-Dimethylbut-2-ene, $(CH_3)_2C=C(CH_3)_2$, is symmetric, so ozonolysis cleaves it into two identical acetone molecules.

2,3-Dimethylbut-2-ene has the structure $(CH_3)_2C=C(CH_3)_2$ — both carbons of the double bond bear two identical methyl substituents. Ozonolysis (formation of the ozonide, followed by reductive cleavage with Zn dust/water) cleaves the $C=C$ bond, converting each doubly-bonded carbon into a carbonyl carbon:

$(CH_3)_2C=C(CH_3)_2 \xrightarrow[\text{ii) Zn/}H_2O]{\text{i) }O_3} 2\,(CH_3)_2C=O$

Since the molecule is symmetric, both fragments are identical: acetone (propan-2-one).

Hydration of acetylene over HgSO4/dil. H2SO4 gives acetaldehyde.

When acetylene (HC≡CH) is passed through dilute sulphuric acid in the presence of mercuric sulphate as catalyst, water adds across the triple bond (Markovnikov):

HC≡CH + H2O --(HgSO4/dil.H2SO4)--> [CH2=CH-OH] -> CH3-CHO

The unstable enol (vinyl alcohol) tautomerises to acetaldehyde. This is the Kucherov reaction. (Note: terminal alkynes give aldehydes only for acetylene; other alkynes give ketones.)

A nitrile can be reduced selectively to an aldehyde by the Stephen reduction (SnCl2/HCl) or by DIBAL-H, both followed by hydrolysis of the intermediate imine.

Identifying the reaction: Nitrile (-CN) converting to aldehyde (-CHO) is a partial (controlled) reduction — the nitrile must be stopped at the imine stage rather than going all the way to the primary amine.

Method 1 — Stephen reduction: Benzonitrile is treated with stannous chloride (SnCl2) and HCl gas, which reduces it to an imine-stannic chloride complex; this is then hydrolysed with water to release the aldehyde.

$C_6H_5CN \xrightarrow{SnCl_2/HCl} C_6H_5CH=NH\cdot HCl \xrightarrow{H_2O} C_6H_5CHO$

Method 2 (equally valid): DIBAL-H (diisobutylaluminium hydride) at low temperature reduces the nitrile to an imine-aluminium complex, which on hydrolysis gives the aldehyde.

Dry distillation of the calcium salt of formic acid (calcium formate) yields formaldehyde, HCHO.

Dry distillation of the calcium salt of a carboxylic acid is a classic route to aldehydes/ketones. Calcium formate, (HCOO)2Ca, on dry distillation gives formaldehyde:

(HCOO)2Ca ---> HCHO + CaCO3

Note the special case: only calcium formate gives an aldehyde (formaldehyde). A mixture of calcium formate with the calcium salt of another acid gives higher aldehydes, whereas two identical higher salts give ketones. Here pure calcium formate gives HCHO.

Dry distillation of the calcium salt of a carboxylic acid is a classic laboratory route to a ketone — for calcium acetate, this gives acetone.

When the calcium salt of acetic acid (calcium acetate) is heated strongly in the absence of air (dry distillation), it undergoes ketonic decarboxylation: two acetate groups combine, releasing calcium carbonate and forming a ketone — in this case, the simplest ketone, acetone (propan-2-one):

$Ca(CH_3COO)_2 \xrightarrow{\Delta,\,dry\ distillation} CaCO_3 + CH_3COCH_3\ (acetone)$

This reaction (and its general form: dry distillation of a calcium/barium salt of a carboxylic acid to give a ketone) is a standard preparative method for symmetrical ketones from carboxylic acids.