R-CO-OR' xrightarrow{LiAlH_4} B + C. Both B a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The reduction of an ester $R-CO-OR'$ with $LiAlH_4$ yields two alcohols: $R-CH_2OH$ and $R'-OH$. For an alcohol to give a positive iodoform test,i

Vedclass · Accepted Answer

$-CH_3, -CH(CH_3)_2$

Vedclass · Answer

(C) The reduction of an ester $R-CO-OR'$ with $LiAlH_4$ yields two alcohols: $R-CH_2OH$ and $R'-OH$. For an alcohol to give a positive iodoform test,it must be ethanol $(CH_3CH_2OH)$ or a secondary alcohol with a methyl group attached to the carbinol carbon $(CH_3CH(OH)R)$. $1.$ If $R = -CH_3$,then $B = CH_3CH_2OH$ (ethanol),which gives a positive iodoform test. $2.$ If $R' = -CH(CH_3)_2$ (isopropyl group),then $C = (CH_3)_2CHOH$ (propan$-2-$ol),which also gives a positive iodoform test. Therefore,the correct combination is $R = -CH_3$ and $R' = -CH(CH_3)_2$.

$R-CO-OR' \xrightarrow{LiAlH_4} B + C$. Both $B$ and $C$ give a positive iodoform test. $R$ and $R'$ can be:

Similar Questions

The reaction $R-CH_2-CH_2-COOH \xrightarrow[Br_2]{Red \ P} R-CH_2-CH(Br)-COOH$ is called as:

The reducing agent used in the following reaction is:

$G$ in the above sequence of reactions is

$A$ colourless water-soluble organic liquid decomposes sodium carbonate and liberates carbon dioxide. It produces a black precipitate with Tollen's reagent. The liquid is

Vedclass Test Series

Exam Paper Generator

Online Exam Module