Select the correct order of acidic strength.
(This question has multiple correct options.)
A) ${ BF }_{ 3 }{ >Me }_{ 3 }{ B }$
B) ${ CH }_{ 3 }{ \equiv C-H}$$ < $${CH }_{ 2 }{ =CH-H>CH }_{ 3 }{ -CH }_{ 2 }{ -H }$
C) ${ BF }_{ 3 }$$ < $${Me }_{ 3 }{ B }$
D) ${ CH }_{ 3 }{ \equiv C-H > CH }_{ 2 }{ =CH-H>CH }_{ 3 }{ -CH }_{ 2 }{ -H }$
Answer
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Hint: Acidic strength refers to its ability to lose a proton ${ H }^{ + }$.
Back bonding is an additional pi bond formed between two atoms of which one has an empty orbital and another must-have lone pair.
Complete step-by-step answer:
The order of acidic strength is alkane < alkene < alkyne. and this is because of their hybridization and hence alkanes are the least electronegative whereas, an alkyne is the most electronegative.
${ H-C\equiv HC\rightarrow }{ H-C\equiv C }^{ \_ }$
${ R-CH=CH }_{ 2 }{ \rightarrow R-CH=C^{ \_ }H^{ + } }$
${ H-CH }_{ 2 }{ -CH }_{ 3 }{ \rightarrow H-CH }_{ 2 }{ C }^{ - }{ H }_{ 2 }$
The ${ C }^{ - }$ is sp hybridized, so it can hold negative.
The carbon with a triple bond is more electronegative and hence, can hold a negative charge and is most stable. Hence, an alkyne is more stable than alkene and alkene is more than alkane.
So, the correct order of acidic strength will be
${ CH }_{ 3 }{ \equiv C-H>CH }_{ 2 }{ =CH-H>CH }_{ 3 }{ -CH }_{ 2 }{ -H }$
Now, among boron trihalides, maximum ${ p\Pi -p\Pi }$, back bonding exists in ${ BF }_{ 3 }$. This is because as the size of the halogen atom increases down the group, the overlapping of the vacant ${ 2p }$ orbital of boron cannot take place easily and efficiently with the p-orbitals of high energy levels. While there is an absence of π-back bonding in ${ Me }_{ 3 }{ B }$.
As we know that back bonding is inversely proportional to acidic strength, so ${ Me }_{ 3 }{ B }$ will be more acidic than ${ BF }_{ 3 }$.
Hence, the correct options are C and D.
Note: The possibility to make a mistake is that you may choose option A. But ${BF}_{3 }$$<$${Me }_{ 3 }{ B }$ is more acidic as there is a pi back bonding present in ${BF}_{3}$ molecule which will increase the stability.
Back bonding is an additional pi bond formed between two atoms of which one has an empty orbital and another must-have lone pair.
Complete step-by-step answer:
The order of acidic strength is alkane < alkene < alkyne. and this is because of their hybridization and hence alkanes are the least electronegative whereas, an alkyne is the most electronegative.
${ H-C\equiv HC\rightarrow }{ H-C\equiv C }^{ \_ }$
${ R-CH=CH }_{ 2 }{ \rightarrow R-CH=C^{ \_ }H^{ + } }$
${ H-CH }_{ 2 }{ -CH }_{ 3 }{ \rightarrow H-CH }_{ 2 }{ C }^{ - }{ H }_{ 2 }$
The ${ C }^{ - }$ is sp hybridized, so it can hold negative.
The carbon with a triple bond is more electronegative and hence, can hold a negative charge and is most stable. Hence, an alkyne is more stable than alkene and alkene is more than alkane.
So, the correct order of acidic strength will be
${ CH }_{ 3 }{ \equiv C-H>CH }_{ 2 }{ =CH-H>CH }_{ 3 }{ -CH }_{ 2 }{ -H }$
Now, among boron trihalides, maximum ${ p\Pi -p\Pi }$, back bonding exists in ${ BF }_{ 3 }$. This is because as the size of the halogen atom increases down the group, the overlapping of the vacant ${ 2p }$ orbital of boron cannot take place easily and efficiently with the p-orbitals of high energy levels. While there is an absence of π-back bonding in ${ Me }_{ 3 }{ B }$.
As we know that back bonding is inversely proportional to acidic strength, so ${ Me }_{ 3 }{ B }$ will be more acidic than ${ BF }_{ 3 }$.
Hence, the correct options are C and D.
Note: The possibility to make a mistake is that you may choose option A. But ${BF}_{3 }$$<$${Me }_{ 3 }{ B }$ is more acidic as there is a pi back bonding present in ${BF}_{3}$ molecule which will increase the stability.
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