Class 12Chemistry8-2.Carboxylic acids and Their derivativeMix Examples-Carboxylic acids and Their derivative
MediumWrite the chemical reactions for the following conversions:
$1.$ $1$-Bromopropane to Propan-$1$-ol and Propan-$2$-ol
$2.$ Ethanoic acid to Methanol
$1.$ $1$-Bromopropane to Propan-$1$-ol and Propan-$2$-ol
$2.$ Ethanoic acid to Methanol
(N/A) $1.$ Conversion of $1$-Bromopropane to Propan-$1$-ol and Propan-$2$-ol:
First,$1$-bromopropane $(CH_3CH_2CH_2Br)$ is treated with alcoholic $KOH$ and heated to undergo dehydrohalogenation to form propene $(CH_3CH=CH_2)$.
To obtain Propan-$1$-ol,propene undergoes hydroboration-oxidation: $(i) (BH_3)_2, (ii) H_2O_2, OH^-$.
To obtain Propan-$2$-ol,propene undergoes acid-catalyzed hydration: $H_2SO_4, H_2O$.
$2.$ Conversion of Ethanoic acid to Methanol:
First,ethanoic acid $(CH_3COOH)$ is esterified with methanol $(CH_3OH)$ in the presence of $H^+$ to form methyl ethanoate $(CH_3COOCH_3)$.
Methyl ethanoate is then reduced using $LiAlH_4$ or $H_2/Pd$ to yield ethanol $(CH_3CH_2OH)$ and methanol $(CH_3OH)$.
First,$1$-bromopropane $(CH_3CH_2CH_2Br)$ is treated with alcoholic $KOH$ and heated to undergo dehydrohalogenation to form propene $(CH_3CH=CH_2)$.
To obtain Propan-$1$-ol,propene undergoes hydroboration-oxidation: $(i) (BH_3)_2, (ii) H_2O_2, OH^-$.
To obtain Propan-$2$-ol,propene undergoes acid-catalyzed hydration: $H_2SO_4, H_2O$.
$2.$ Conversion of Ethanoic acid to Methanol:
First,ethanoic acid $(CH_3COOH)$ is esterified with methanol $(CH_3OH)$ in the presence of $H^+$ to form methyl ethanoate $(CH_3COOCH_3)$.
Methyl ethanoate is then reduced using $LiAlH_4$ or $H_2/Pd$ to yield ethanol $(CH_3CH_2OH)$ and methanol $(CH_3OH)$.
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Similar Questions
The reaction sequence is:
$2-\text{hydroxycyclohexylacetone}$ $\xrightarrow{I_2/NaOH}$ $\xrightarrow{\Delta}$ $\xrightarrow{H^+}$ $\xrightarrow{\Delta} A$
$A$ is:
$2-\text{hydroxycyclohexylacetone}$ $\xrightarrow{I_2/NaOH}$ $\xrightarrow{\Delta}$ $\xrightarrow{H^+}$ $\xrightarrow{\Delta} A$
$A$ is:
Difficult
View SolutionConsidering the reaction sequence given below,the correct statement$(s)$ is(are):
$A$. $P$ can be reduced to a primary alcohol using $NaBH_4$.
$B$. Treating $P$ with conc. $NH_4OH$ solution followed by acidification gives $Q$.
$C$. Treating $Q$ with a solution of $NaNO_2$ in aq. $HCl$ liberates $N_2$.
$D$. $P$ is more acidic than $CH_3CH_2COOH$.
$A$. $P$ can be reduced to a primary alcohol using $NaBH_4$.
$B$. Treating $P$ with conc. $NH_4OH$ solution followed by acidification gives $Q$.
$C$. Treating $Q$ with a solution of $NaNO_2$ in aq. $HCl$ liberates $N_2$.
$D$. $P$ is more acidic than $CH_3CH_2COOH$.
The products obtained in the given reactions are shown below. The number of possible products for $x$ and $y$ are:
Difficult
View SolutionThe end product of the above reaction is:
An organic acid $P$ $(C_{11}H_{12}O_2)$ can easily be oxidized to a dibasic acid which reacts with ethylene glycol to produce a polymer,dacron. Upon ozonolysis,$P$ gives an aliphatic ketone as one of the products. $P$ undergoes the following reaction sequences to furnish $R$ via $Q$. The compound $P$ also undergoes another set of reactions to produce $S$.
$(1)$ The compound $R$ is
$(2)$ The compound $S$ is
Give the answer for questions $(1)$ and $(2)$.
$(1)$ The compound $R$ is
$(2)$ The compound $S$ is
Give the answer for questions $(1)$ and $(2)$.
DifficultIIT 2018
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