A compound with formula C_4H_{10}O on oxidati | vedclass

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

Question

(B) The compound $C_4H_{10}O$ is $2-butanol$ $(CH_3CH(OH)CH_2CH_3)$.On oxidation,$2-butanol$ gives $2-butanone$ $(CH_3COCH_2CH_3)$,which is a ketone.$

Vedclass · Accepted Answer

$CH_3CH(OH)CH_2CH_3$

Vedclass · Answer

(B) The compound $C_4H_{10}O$ is $2-butanol$ $(CH_3CH(OH)CH_2CH_3)$.On oxidation,$2-butanol$ gives $2-butanone$ $(CH_3COCH_2CH_3)$,which is a ketone.$2-butanone$ contains a $CH_3CO-$ group,so it gives a positive iodoform test and forms an oxime with $NH_2OH$.On dehydration with concentrated $H_2SO_4$,$2-butanol$ gives $2-butene$ $(CH_3CH=CHCH_3)$ as the major product.Thus,the correct structure is $CH_3CH(OH)CH_2CH_3$.

$A$ compound with formula $C_4H_{10}O$ on oxidation gives $C_4H_8O$,which forms an oxime and gives a positive iodoform test. The original compound on reaction with concentrated $H_2SO_4$ gives $C_4H_8$. What is the structure of the compound?

Similar Questions

Which of the following compounds on oxidation gives ethyl methyl ketone?

Lucas reagent is

$(CH_3)_2CHOH$ $\xrightarrow{dil. \, [O]} X$ $\xrightarrow[(ii) \, H_2O]{(i) \, CH_3MgBr} Y$. What is $Y$ here?

$HO-CH_2CH_2CH_2CH_2-Br \xrightarrow{KOH} \text{Product}$. The product is:

Which is the best reagent to convert isopropyl alcohol to isopropyl bromide?
$CH_3-CH(OH)-CH_3 \xrightarrow{?} CH_3-CH(Br)-CH_3$

Vedclass Test Series

Exam Paper Generator

Online Exam Module