Class 11Chemistry8-1.Organic Chemistry : Classification and Nomenclature of Organic compoundsBonding and hybridisation in organic compounds
MediumWhich of the following represents the given mode of hybridization $sp^2 - sp^2 - sp - sp$ from left to right?
- A$H_2C = CH - C \equiv CH$
- B$HC \equiv C - C \equiv CH$
- C$H_2C = C = C = CH_2$
- D$CH_2 = CH - CH = CH_2$
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Similar Questions
In the compound $HC \equiv C-CH=CH_2$, the hybridization of $C-2$ and $C-3$ carbons are respectively:
In which of the compounds given below is there more than one kind of hybridisation $(sp, sp^2, sp^3)$ for carbon?
$(i)$ $CH_3CH_2CH_2CH_3$
$(ii)$ $CH_3-CH=CH-CH_3$
$(iii)$ $CH_2=CH-CH=CH_2$
$(iv)$ $HC\equiv CH$
$(i)$ $CH_3CH_2CH_2CH_3$
$(ii)$ $CH_3-CH=CH-CH_3$
$(iii)$ $CH_2=CH-CH=CH_2$
$(iv)$ $HC\equiv CH$
DifficultAIPMT 1995
View SolutionFrom the following compounds,the ones which contain both $sp$ and $sp^2$ hybridized carbons are:
Amongst 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 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
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