The enolic form of acetone contains

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

Question

(A) The enolic form of acetone is propen$-2-$ol,which has the structure $CH_2=C(OH)-CH_3$.Counting the bonds: there are $3$ $C-H$ bonds in the methyl

Vedclass · Accepted Answer

$9$ sigma bonds,$1$ pi bond and $2$ lone pairs

Vedclass · Answer

(A) The enolic form of acetone is propen$-2-$ol,which has the structure $CH_2=C(OH)-CH_3$.Counting the bonds: there are $3$ $C-H$ bonds in the methyl group,$2$ $C-H$ bonds in the methylene group,$1$ $O-H$ bond,$1$ $C-O$ bond,$1$ $C-C$ bond,and $1$ $C=C$ bond (which consists of $1$ $\sigma$ and $1$ $\pi$ bond).Total $\sigma$ bonds = $3 + 2 + 1 + 1 + 1 + 1 = 9$.Total $\pi$ bonds = $1$.Oxygen atom in the hydroxyl group has $2$ lone pairs.Thus,the structure contains $9$ $\sigma$ bonds,$1$ $\pi$ bond,and $2$ lone pairs.

Similar Questions

What is $Z$ in the following reaction sequence ?
$C_3H_6$ $\xrightarrow[CCl_4]{Br_2} \underset{\text{(colourless)}}{X}$ $\xrightarrow[\text{(ii) } NaNH_2, \Delta]{\text{(i) } KOH / \text{alcohol}} Y$ $\xrightarrow[\substack{Hg^{2+}, H^{+} \\ 333 \ K}]{H_2O} Z$

The end product $(P)$ of the following reaction is:

Identify product $(P)$

Which type of reactions are easily performed by alkane and alkene?

The unknown $(P)$ of the reaction is

Vedclass Test Series

Exam Paper Generator

Online Exam Module