Class 11Chemistry8-1.Organic Chemistry : Classification and Nomenclature of Organic compoundsBonding and hybridisation in organic compounds
MediumWhat is the hybridization state of carbon atoms $1, 2, 3,$ and $4$ in the given reaction?
- A$1$ and $2$ are $sp^2$; $3$ and $4$ are $sp^3$
- B$1$ and $2$ are $sp^2$; $3$ and $4$ are $sp$
- C$1, 2, 3,$ and $4$ are $sp$
- D$1, 2$ are $sp^3$; $3, 4$ are $sp^2$
Explore More
Similar Questions
Arrange the following compounds in decreasing order of acidity: $H_{2}O, ROH, HC \equiv CH$
Medium
View SolutionThe compounds containing $sp$-hybridised carbon atom are:
$(I)$ $H_3C-N(CH_3)-CHO$
$(II)$ Pyridine
$(III)$ $H_3C-CN$
$(IV)$ $H_2C=C=CH-CH_3$
$(I)$ $H_3C-N(CH_3)-CHO$
$(II)$ Pyridine
$(III)$ $H_3C-CN$
$(IV)$ $H_2C=C=CH-CH_3$
MediumKVPY 2016
View SolutionNumber of pure $p-$ orbitals in the given molecule $\text{hexa-}1, 3-\text{dien-}5-\text{yne}$.
Medium
View SolutionAmongst the given options,the compound$(s)$ in which all the atoms are in one plane in all the possible conformations (if any),is (are):
$(A)$ $CH_2=CH-CH=CH_2$
$(B)$ $HC\equiv C-CH=CH_2$
$(C)$ $H_2C=C=O$
$(D)$ $H_2C=C=CH_2$
$(A)$ $CH_2=CH-CH=CH_2$
$(B)$ $HC\equiv C-CH=CH_2$
$(C)$ $H_2C=C=O$
$(D)$ $H_2C=C=CH_2$
DifficultIIT 2011
View SolutionThe ascending acidity order of the following $H$ atoms is:
$A: HC \equiv C-H$
$B: H_2C=CH-H$
$C: (CH_3)_3C-H$
$D: CH_3-CH_2-H$
$A: HC \equiv C-H$
$B: H_2C=CH-H$
$C: (CH_3)_3C-H$
$D: CH_3-CH_2-H$
DifficultJEE MAIN 2024
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo