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Products of the following reaction :
$CH_3C\equiv CCH_2CH_3 \xrightarrow[(ii)\ \text{Hydrolysis}]{(i)\ O_3}$
are -
$CH_3C\equiv CCH_2CH_3 \xrightarrow[(ii)\ \text{Hydrolysis}]{(i)\ O_3}$
are -
A
$CH_3COOH + CH_3COCH_3$
B
$CH_3COOH + HOOC-CH_2CH_3$
C
$CH_3CHO + CH_3CH_2CHO$
D
$CH_3COOH + CO_2$
Detailed Solution
Ozonolysis of an alkyne cleaves the carbon-carbon triple bond, and on hydrolysis each carbon of the triple bond becomes the carbon of a carboxylic acid group (via a 1,2-diketone that is cleaved by the $H_2O_2$ formed).
The given alkyne is pent-2-yne, $CH_3-C\equiv C-CH_2CH_3$.
Cleavage between C2 and C3 gives a two-carbon fragment and a three-carbon fragment.
$CH_3-C\equiv C-CH_2CH_3 \xrightarrow{(i)\ O_3,\ (ii)\ H_2O} CH_3COOH + CH_3CH_2COOH$
The products are ethanoic (acetic) acid and propanoic acid, $HOOC-CH_2CH_3$.
So the products are $CH_3COOH + HOOC-CH_2CH_3$.
The given alkyne is pent-2-yne, $CH_3-C\equiv C-CH_2CH_3$.
Cleavage between C2 and C3 gives a two-carbon fragment and a three-carbon fragment.
$CH_3-C\equiv C-CH_2CH_3 \xrightarrow{(i)\ O_3,\ (ii)\ H_2O} CH_3COOH + CH_3CH_2COOH$
The products are ethanoic (acetic) acid and propanoic acid, $HOOC-CH_2CH_3$.
So the products are $CH_3COOH + HOOC-CH_2CH_3$.