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Question: How many isomeric alkynes exist for \({C_6}{H_{10}}\) which gives effervescence of heating with \(Na...

How many isomeric alkynes exist for C6H10{C_6}{H_{10}} which gives effervescence of heating with NaNa ?
A. 2
B. 3
C. 4
D. 5

Explanation

Solution

Effervescence is the evolution of gas bubbles in a liquid. When a chemical reaction occurs and if there is an evolution of gas from the solution then the gas will escape from that solution that appears as bubbles and we call it as effervescence.

Complete answer:
Alkene and alkynes are unsaturated hydrocarbons containing C-C double and triple bonds respectively. Like alkenes have the suffix – ene, alkynes use the ending –yne. They have a general empirical formula CnH2n−2{C_n}{H_{2n - 2}}. They are also known as acetylenes.
Terminal and internal alkynes: Internal alkynes, the triple bond between carbons are in between the molecules, while for terminal alkynes, the triple bond will be on the terminal end of the molecule.
Alkynes are more acidic than the alkanes and alkenes. Terminal alkynes only have the hydrogen present which acts as an acidic proton. So we can say terminal alkynes are mildly acidic and can give its proton.
Isomers have the same molecular formula but different arrangements of atoms.
C6H10{C_6}{H_{10}} has the following isomers:
a. Hex-1-yne CH3CH2CH2CH2C≡CHC{H_3}C{H_2}C{H_2}C{H_2}C \equiv CH
b. Hex-2-yne CH3CH2CH2C≡CCH3C{H_3}C{H_2}C{H_2}C \equiv CC{H_3}
c. Hex-3-yne CH3CH2C≡CCH2CH3C{H_3}C{H_2}C \equiv CC{H_2}C{H_3}
d. 4-methylpent-1-yne (CH3)2CHCH2C≡CH{(C{H_3})_2}CHC{H_2}C \equiv CH
e. 3-methylpent-1-yne CH3CH2CH(CH3)C≡CHC{H_3}C{H_2}CH(C{H_3})C \equiv CH
f. 3,3-dimethylbut-1-yne (CH3)3CC≡CH{(C{H_3})_3}CC \equiv CH
g. 4-methylpent-2yne (CH3)2CHC≡CCH3{(C{H_3})_2}CHC \equiv CC{H_3}
Only terminal alkynes have protons. Therefore when they react with sodium metal they act as acid and give its proton and form hydrogen gas. The hydrogen gas will now try to escape from the solution and appear as effervescence.
From the total isomers of C6H10{C_6}{H_{10}} only four have terminal alkyne groups present.

**Therefore the correct option will be C. 4

Note:**
The triple bond in alkyne is very strong. It has one sigma bond and 2 pi bonds. In terms of molecular orbital theory, triple bond arises from the overlap of s and p orbital. So the hybridization of carbons at triple bond is sp. Therefore the triple bond is linear.