Given that molecular formula is
C6H10. First of all let us calculate the degree of unsaturation for
C6H10.
So,
‌DBE=C+1‌‌DBE=6+1‌=2Here, DBE of 2 indicates the presence of one triple bond or two double bonds or one ring and one double bond. So, for alkynes let us see the structure with one triple bond.
Now, the possible number of alkyne monomers of
C6H10 are
Hex-1-yne (Terminal alkyne)
Hex-2-yne (Internal alkyne)
Hex-3-yne (Internal alkyne)
3-methyl pent-1-yne (Terminal alkyne)
4-methylpent-1-yne (Terminal alkyne)
4-methylpent-2-yne (Internal alkyne)
3, 3-Dimethyl but-1-yne (Terminal alkyne)
So, terminal alkynes contain acidic hydrogen while internal alkynes contains no acidic hydrogen. Thus
x and
y are 4,3 respectively.
