An organic compound X with molecular formula | vedclass

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

Question

(A) The molecular formula $C_6H_{10}$ corresponds to the general formula $C_nH_{2n-2}$,indicating an alkyne. Treatment with $HBr$ to form a $gem$-dibr

Vedclass · Accepted Answer

hex$-1-$yne

Vedclass · Answer

(A) The molecular formula $C_6H_{10}$ corresponds to the general formula $C_nH_{2n-2}$,indicating an alkyne. Treatment with $HBr$ to form a $gem$-dibromide indicates that the alkyne is a terminal alkyne (Markownikoff addition). Acid-catalyzed hydration of a terminal alkyne using $HgSO_4$ and dil. $H_2SO_4$ produces a methyl ketone. $A$ methyl ketone $(CH_3CO-R)$ gives a positive iodoform test. Among the options,hex$-1-$yne $(CH_3CH_2CH_2CH_2C \equiv CH)$ is a terminal alkyne. Upon hydration,it forms hexan$-2-$one $(CH_3CH_2CH_2CH_2COCH_3)$,which is a methyl ketone and gives a positive iodoform test. Therefore,the compound $X$ is hex$-1-$yne.

An organic compound $X$ with molecular formula $C_6H_{10}$,when treated with $HBr$,forms a $gem$-dibromide. The compound $X$ upon warming with $HgSO_4$ and dil. $H_2SO_4$,produces a ketone,which gives a positive iodoform test. The compound $X$ is

Similar Questions

$Ph-CCl_2-CH_3 \xrightarrow{3NaNH_2} \text{Product } (A)$; Product $(A)$ is

Write the following reactions:
$1$. Complete combustion of ethyne in the presence of air.
$2$. Reaction of propyne with alkaline $KMnO_4$.

Which of the following compounds forms a $Na$-salt upon reaction with $NaNH_2$?

The reagent used for converting acetylene to oxalic acid is

In the following reaction sequence: $Ph-CH=CH-Ph$ $\xrightarrow{Br_2/CCl_4} (A)$ $\xrightarrow{2NaNH_2} (B)$ $\xrightarrow{H_2/Pd-CaCO_3} (C)$,the product $(C)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module